Practical Regression And Anova Using R
Practical Regression and Anova using R
Julian J. Faraway
July 2002
1
Copyright c 1999, 2000, 2002 Julian J. Faraway
Permission to reproduce individual copies of this book for personal use is granted. Multiple copies may
be created for nonprofit academic purposes — a nominal charge to cover the expense of reproduction may
be made. Reproduction for profit is prohibited without permission.
Preface
There are many books on regression and analysis of variance. These books expect different levels of pre-
paredness and place different emphases on the material. This book is not introductory. It presumes some
knowledge of basic statistical theory and practice. Students are expected to know the essentials of statistical
inference like estimation, hypothesis testing and confidence intervals. A basic knowledge of data analysis is
presumed. Some linear algebra and calculus is also required.
The emphasis of this text is on the practice of regression and analysis of variance. The objective is to
learn what methods are available and more importantly, when they should be applied. Many examples are
presented to clarify the use of the techniques and to demonstrate what conclusions can be made. There
is relatively less emphasis on mathematical theory, partly because some prior knowledge is assumed and
partly because the issues are better tackled elsewhere. Theory is important because it guides the approach
we take. I take a wider view of statistical theory. It is not just the formal theorems. Qualitative statistical
concepts are just as important in Statistics because these enable us to actually do it rather than just talk about
it. These qualitative principles are harder to learn because they are difficult to state precisely but they guide
the successful experienced Statistician.
Data analysis cannot be learnt without actually doing it. This means using a statistical computing pack-
age. There is a wide choice of such packages. They are designed for different audiences and have different
strengths and weaknesses. I have chosen to use R (ref. Ihaka and Gentleman (1996)). Why do I use R ?
The are several reasons.
1. Versatility. R is a also a programming language, so I am not limited by the procedures that are
preprogrammed by a package. It is relatively easy to program new methods in R .
2. Interactivity. Data analysis is inherently interactive. Some older statistical packages were designed
when computing was more expensive and batch processing of computations was the norm. Despite
improvements in hardware, the old batch processing paradigm lives on in their use. R does one thing
at a time, allowing us to make changes on the basis of what we see during the analysis.
3. R is based on S from which the commercial package S-plus is derived. R itself is open-source
software and may be freely redistributed. Linux, Macintosh, Windows and other UNIX versions
are maintained and can be obtained from the R-project at www.r-project.org. R is mostly
compatible with S-plus meaning that S-plus could easily be used for the examples given in this
book.
4. Popularity. SAS is the most common statistics package in general but R or S is most popular with
researchers in Statistics. A look at common Statistical journals confirms this popularity. R is also
popular for quantitative applications in Finance.
The greatest disadvantage of R is that it is not so easy to learn. Some investment of effort is required
before productivity gains will be realized. This book is not an introduction to R . There is a short introduction
2
3
in the Appendix but readers are referred to the R-project web site at www.r-project.org where you
can find introductory documentation and information about books on R . I have intentionally included in
the text all the commands used to produce the output seen in this book. This means that you can reproduce
these analyses and experiment with changes and variations before fully understanding R . The reader may
choose to start working through this text before learning R and pick it up as you go.
The web site for this book is at www.stat.lsa.umich.edu/˜faraway/book where data de-
scribed in this book appears. Updates will appear there also.
Thanks to the builders of R without whom this book would not have been possible.
Contents
1
Introduction
8
1.1
Before you start . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.1.1
Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.1.2
Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.1.3
Initial Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.2
When to use Regression Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
1.3
History
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
2
Estimation
16
2.1
Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
2.2
Linear Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
2.3
Matrix Representation
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
2.4
Estimating β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.5
Least squares estimation
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
2.6
Examples of calculating ˆβ
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
2.7
Why is ˆβ a good estimate? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.8
Gauss-Markov Theorem
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
2.9
Mean and Variance of ˆβ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.10 Estimating σ2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.11 Goodness of Fit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
2.12 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
3
Inference
26
3.1
Hypothesis tests to compare models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
3.2
Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.2.1
Test of all predictors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.2.2
Testing just one predictor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
3.2.3
Testing a pair of predictors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
3.2.4
Testing a subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
3.3
Concerns about Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
3.4
Confidence Intervals for β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.5
Confidence intervals for predictions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
3.6
Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
3.7
Identifiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
3.8
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
3.9
What can go wrong?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
3.9.1
Source and quality of the data . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
4
CONTENTS
5
3.9.2
Error component . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
3.9.3
Structural Component
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
3.10 Interpreting Parameter Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
4
Errors in Predictors
55
5
Generalized Least Squares
59
5.1
The general case
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
5.2
Weighted Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
5.3
Iteratively Reweighted Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
6
Testing for Lack of Fit
65
6.1
σ2 known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
6.2
σ2 unknown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
7
Diagnostics
72
7.1
Residuals and Leverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72
7.2
Studentized Residuals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
7.3
An outlier test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
7.4
Influential Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
7.5
Residual Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
7.6
Non-Constant Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
7.7
Non-Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
7.8
Assessing Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
7.9
Half-normal plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
7.10 Correlated Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
8
Transformation
95
8.1
Transforming the response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
8.2
Transforming the predictors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
8.2.1
Broken Stick Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
8.2.2
Polynomials
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
8.3
Regression Splines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
8.4
Modern Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
9
Scale Changes, Principal Components and Collinearity
106
9.1
Changes of Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
9.2
Principal Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
9.3
Partial Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
9.4
Collinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
9.5
Ridge Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
10 Variable Selection
124
10.1 Hierarchical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
10.2 Stepwise Procedures
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
10.2.1 Forward Selection
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
10.2.2 Stepwise Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
10.3 Criterion-based procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
CONTENTS
6
10.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
11 Statistical Strategy and Model Uncertainty
134
11.1 Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
11.2 Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
11.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
12 Chicago Insurance Redlining - a complete example
138
13 Robust and Resistant Regression
150
14 Missing Data
156
15 Analysis of Covariance
160
15.1 A two-level example
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
15.2 Coding qualitative predictors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
15.3 A Three-level example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
16 ANOVA
168
16.1 One-Way Anova . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
16.1.1 The model
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
16.1.2 Estimation and testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
16.1.3 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
16.1.4 Diagnostics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
16.1.5 Multiple Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
16.1.6 Contrasts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
16.1.7 Scheff´e’s theorem for multiple comparisons . . . . . . . . . . . . . . . . . . . . . . 177
16.1.8 Testing for homogeneity of variance . . . . . . . . . . . . . . . . . . . . . . . . . . 179
16.2 Two-Way Anova
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
16.2.1 One observation per cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
16.2.2 More than one observation per cell . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
16.2.3 Interpreting the interaction effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
16.2.4 Replication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
16.3 Blocking designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
16.3.1 Randomized Block design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
16.3.2 Relative advantage of RCBD over CRD . . . . . . . . . . . . . . . . . . . . . . . . 190
16.4 Latin Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
16.5 Balanced Incomplete Block design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
16.6 Factorial experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
A Recommended Books
204
A.1 Books on R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
A.2 Books on Regression and Anova . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
B R functions and data
205
CONTENTS
7
C Quick introduction to R
207
C.1 Reading the data in . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
C.2 Numerical Summaries
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
C.3 Graphical Summaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
C.4 Selecting subsets of the data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
C.5 Learning more about R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
Chapter 1
Introduction
1.1
Before you start
Statistics starts with a problem, continues with the collection of data, proceeds with the data analysis and
finishes with conclusions. It is a common mistake of inexperienced Statisticians to plunge into a complex
analysis without paying attention to what the objectives are or even whether the data are appropriate for the
proposed analysis. Look before you leap!
1.1.1
Formulation
The formulation of a problem is often more essential than its solution which may be merely a
matter of mathematical or experimental skill. Albert Einstein
To formulate the problem correctly, you must
1. Understand the physical background. Statisticians often work in collaboration with others and need
to understand something about the subject area. Regard this as an opportunity to learn something new
rather than a chore.
2. Understand the objective. Again, often you will be working with a collaborator who may not be clear
about what the objectives are. Beware of “fishing expeditions” - if you look hard enough, you’ll
almost always find something but that something may just be a coincidence.
3. Make sure you know what the client wants. Sometimes Statisticians perform an analysis far more
complicated than the client really needed. You may find that simple descriptive statistics are all that
are needed.
4. Put the problem into statistical terms. This is a challenging step and where irreparable errors are
sometimes made. Once the problem is translated into the language of Statistics, the solution is often
routine. Difficulties with this step explain why Artificial Intelligence techniques have yet to make
much impact in application to Statistics. Defining the problem is hard to program.
That a statistical method can read in and process the data is not enough. The results may be totally
meaningless.
8
1.1. BEFORE YOU START
9
1.1.2
Data Collection
It’s important to understand how the data was collected.
Are the data observational or experimental? Are the data a sample of convenience or were they
obtained via a designed sample survey. How the data were collected has a crucial impact on what
conclusions can be made.
Is there non-response? The data you don’t see may be just as important as the data you do see.
Are there missing values? This is a common problem that is troublesome and time consuming to deal
with.
How are the data coded? In particular, how are the qualitative variables represented.
What are the units of measurement? Sometimes data is collected or represented with far more digits
than are necessary. Consider rounding if this will help with the interpretation or storage costs.
Beware of data entry errors. This problem is all too common — almost a certainty in any real dataset
of at least moderate size. Perform some data sanity checks.
1.1.3
Initial Data Analysis
This is a critical step that should always be performed. It looks simple but it is vital.
Numerical summaries - means, sds, five-number summaries, correlations.
Graphical summaries
– One variable - Boxplots, histograms etc.
– Two variables - scatterplots.
– Many variables - interactive graphics.
Look for outliers, data-entry errors and skewed or unusual distributions. Are the data distributed as you
expect?
Getting data into a form suitable for analysis by cleaning out mistakes and aberrations is often time
consuming. It often takes more time than the data analysis itself. In this course, all the data will be ready to
analyze but you should realize that in practice this is rarely the case.
Let’s look at an example. The National Institute of Diabetes and Digestive and Kidney Diseases
conducted a study on 768 adult female Pima Indians living near Phoenix. The following variables were
recorded: Number of times pregnant, Plasma glucose concentration a 2 hours in an oral glucose tolerance
test, Diastolic blood pressure (mm Hg), Triceps skin fold thickness (mm), 2-Hour serum insulin (mu U/ml),
Body mass index (weight in kg/(height in m2)), Diabetes pedigree function, Age (years) and a test whether
the patient shows signs of diabetes (coded 0 if negative, 1 if positive). The data may be obtained from UCI
Repository of machine learning databases at http://www.ics.uci.edu/˜mlearn/MLRepository.html.
Of course, before doing anything else, one should find out what the purpose of the study was and more
about how the data was collected. But let’s skip ahead to a look at the data:
1.1. BEFORE YOU START
10
> library(faraway)
> data(pima)
> pima
pregnant glucose diastolic triceps insulin
bmi diabetes age test
1
6
148
72
35
0 33.6
0.627
50
1
2
1
85
66
29
0 26.6
0.351
31
0
3
8
183
64
0
0 23.3
0.672
32
1
... much deleted ...
768
1
93
70
31
0 30.4
0.315
23
0
The library(faraway) makes the data used in this book available while data(pima) calls up
this particular dataset. Simply typing the name of the data frame, pima prints out the data. It’s too long to
show it all here. For a dataset of this size, one can just about visually skim over the data for anything out of
place but it is certainly easier to use more direct methods.
We start with some numerical summaries:
> summary(pima)
pregnant
glucose
diastolic
triceps
insulin
Min.
: 0.00
Min.
:
0
Min.
:
0.0
Min.
: 0.0
Min.
:
0.0
1st Qu.: 1.00
1st Qu.: 99
1st Qu.: 62.0
1st Qu.: 0.0
1st Qu.:
0.0
Median : 3.00
Median :117
Median : 72.0
Median :23.0
Median : 30.5
Mean
: 3.85
Mean
:121
Mean
: 69.1
Mean
:20.5
Mean
: 79.8
3rd Qu.: 6.00
3rd Qu.:140
3rd Qu.: 80.0
3rd Qu.:32.0
3rd Qu.:127.2
Max.
:17.00
Max.
:199
Max.
:122.0
Max.
:99.0
Max.
:846.0
bmi
diabetes
age
test
Min.
: 0.0
Min.
:0.078
Min.
:21.0
Min.
:0.000
1st Qu.:27.3
1st Qu.:0.244
1st Qu.:24.0
1st Qu.:0.000
Median :32.0
Median :0.372
Median :29.0
Median :0.000
Mean
:32.0
Mean
:0.472
Mean
:33.2
Mean
:0.349
3rd Qu.:36.6
3rd Qu.:0.626
3rd Qu.:41.0
3rd Qu.:1.000
Max.
:67.1
Max.
:2.420
Max.
:81.0
Max.
:1.000
The summary() command is a quick way to get the usual univariate summary information. At this stage,
we are looking for anything unusual or unexpected perhaps indicating a data entry error. For this purpose, a
close look at the minimum and maximum values of each variable is worthwhile. Starting with pregnant,
we see a maximum value of 17. This is large but perhaps not impossible. However, we then see that the next
5 variables have minimum values of zero. No blood pressure is not good for the health — something must
be wrong. Let’s look at the sorted values:
> sort(pima$diastolic)
[1]
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
[19]
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
24
[37]
30
30
38
40
44
44
44
44
46
46
48
48
48
48
48
50
50
50
...etc...
We see that the first 36 values are zero. The description that comes with the data says nothing about it but
it seems likely that the zero has been used as a missing value code. For one reason or another, the researchers
did not obtain the blood pressures of 36 patients. In a real investigation, one would likely be able to question
the researchers about what really happened. Nevertheless, this does illustrate the kind of misunderstanding
1.1. BEFORE YOU START
11
that can easily occur. A careless statistician might overlook these presumed missing values and complete an
analysis assuming that these were real observed zeroes. If the error was later discovered, they might then
blame the researchers for using 0 as a missing value code (not a good choice since it is a valid value for
some of the variables) and not mentioning it in their data description. Unfortunately such oversights are
not uncommon particularly with datasets of any size or complexity. The statistician bears some share of
responsibility for spotting these mistakes.
We set all zero values of the five variables to NA which is the missing value code used by R .
> pima$diastolic[pima$diastolic == 0]
<- NA
> pima$glucose[pima$glucose == 0] <- NA
> pima$triceps[pima$triceps == 0]
<- NA
> pima$insulin[pima$insulin == 0] <- NA
> pima$bmi[pima$bmi == 0] <- NA
The variable test is not quantitative but categorical. Such variables are also called factors. However,
because of the numerical coding, this variable has been treated as if it were quantitative. It’s best to designate
such variables as factors so that they are treated appropriately. Sometimes people forget this and compute
stupid statistics such as “average zip code”.
> pima$test <- factor(pima$test)
> summary(pima$test)
0
1
500 268
We now see that 500 cases were negative and 268 positive. Even better is to use descriptive labels:
> levels(pima$test) <- c("negative","positive")
> summary(pima)
pregnant
glucose
diastolic
triceps
insulin
Min.
: 0.00
Min.
: 44
Min.
: 24.0
Min.
:
7.0
Min.
: 14.0
1st Qu.: 1.00
1st Qu.: 99
1st Qu.: 64.0
1st Qu.: 22.0
1st Qu.: 76.2
Median : 3.00
Median :117
Median : 72.0
Median : 29.0
Median :125.0
Mean
: 3.85
Mean
:122
Mean
: 72.4
Mean
: 29.2
Mean
:155.5
3rd Qu.: 6.00
3rd Qu.:141
3rd Qu.: 80.0
3rd Qu.: 36.0
3rd Qu.:190.0
Max.
:17.00
Max.
:199
Max.
:122.0
Max.
: 99.0
Max.
:846.0
NA’s
:
5
NA’s
: 35.0
NA’s
:227.0
NA’s
:374.0
bmi
diabetes
age
test
Min.
:18.2
Min.
:0.078
Min.
:21.0
negative:500
1st Qu.:27.5
1st Qu.:0.244
1st Qu.:24.0
positive:268
Median :32.3
Median :0.372
Median :29.0
Mean
:32.5
Mean
:0.472
Mean
:33.2
3rd Qu.:36.6
3rd Qu.:0.626
3rd Qu.:41.0
Max.
:67.1
Max.
:2.420
Max.
:81.0
NA’s
:11.0
Now that we’ve cleared up the missing values and coded the data appropriately we are ready to do some
plots. Perhaps the most well-known univariate plot is the histogram:
hist(pima$diastolic)
1.1. BEFORE YOU START
12
120
200
0.030
100
150
0.020
80
Density
Frequency
100
60
sort(pima$diastolic)
0.010
50
40
0
0.000
20
40
60
80
100
120
20
40
60
80
100 120
0
200
400
600
pima$diastolic
N = 733 Bandwidth = 2.872
Index
Figure 1.1: First panel shows histogram of the diastolic blood pressures, the second shows a kernel density
estimate of the same while the the third shows an index plot of the sorted values
as shown in the first panel of Figure 1.1. We see a bell-shaped distribution for the diastolic blood pressures
centered around 70. The construction of a histogram requires the specification of the number of bins and
their spacing on the horizontal axis. Some choices can lead to histograms that obscure some features of the
data. R attempts to specify the number and spacing of bins given the size and distribution of the data but
this choice is not foolproof and misleading histograms are possible. For this reason, I prefer to use Kernel
Density Estimates which are essentially a smoothed version of the histogram (see Simonoff (1996) for a
discussion of the relative merits of histograms and kernel estimates).
> plot(density(pima$diastolic,na.rm=TRUE))
The kernel estimate may be seen in the second panel of Figure 1.1. We see that it avoids the distracting
blockiness of the histogram. An alternative is to simply plot the sorted data against its index:
plot(sort(pima$diastolic),pch=".")
The advantage of this is we can see all the data points themselves. We can see the distribution and
possible outliers. We can also see the discreteness in the measurement of blood pressure - values are rounded
to the nearest even number and hence we the “steps” in the plot.
Now a couple of bivariate plots as seen in Figure 1.2:
> plot(diabetes ˜ diastolic,pima)
> plot(diabetes ˜ test,pima)
hist(pima$diastolic)
First, we see the standard scatterplot showing two quantitative variables. Second, we see a side-by-side
boxplot suitable for showing a quantitative and a qualititative variable. Also useful is a scatterplot matrix,
not shown here, produced by
1.2. WHEN TO USE REGRESSION ANALYSIS
13
2.5
2.5
2.0
2.0
1.5
1.5
diabetes
1.0
diabetes
1.0
0.5
0.5
0.0
0.0
40
60
80
100
120
negative
positive
diastolic
test
Figure 1.2: First panel shows scatterplot of the diastolic blood pressures against diabetes function and the
second shows boxplots of diastolic blood pressure broken down by test result
> pairs(pima)
We will be seeing more advanced plots later but the numerical and graphical summaries presented here are
sufficient for a first look at the data.
1.2
When to use Regression Analysis
Regression analysis is used for explaining or modeling the relationship between a single variable Y , called
the response, output or dependent variable, and one or more predictor, input, independent or explanatory
variables, X1
Xp. When p
1, it is called simple regression but when p ¥
1 it is called multiple re-
¤
¢¡¢¡¢¡£
gression or sometimes multivariate regression. When there is more than one Y , then it is called multivariate
multiple regression which we won’t be covering here.
The response must be a continuous variable but the explanatory variables can be continuous, discrete
or categorical although we leave the handling of categorical explanatory variables to later in the course.
Taking the example presented above, a regression of diastolic and bmi on diabetes would be a
multiple regression involving only quantitative variables which we shall be tackling shortly. A regression of
diastolic and bmi on test would involve one predictor which is quantitative which we will consider
in later in the chapter on Analysis of Covariance. A regression of diastolic on just test would involve
just qualitative predictors, a topic called Analysis of Variance or ANOVA although this would just be a simple
two sample situation. A regression of test (the response) on diastolic and bmi (the predictors) would
involve a qualitative response. A logistic regression could be used but this will not be covered in this book.
Regression analyses have several possible objectives including
1. Prediction of future observations.
2. Assessment of the effect of, or relationship between, explanatory variables on the response.
3. A general description of data structure.
1.3. HISTORY
14
Extensions exist to handle multivariate responses, binary responses (logistic regression analysis) and
count responses (poisson regression).
1.3
History
Regression-type problems were first considered in the 18th century concerning navigation using astronomy.
Legendre developed the method of least squares in 1805. Gauss claimed to have developed the method a
few years earlier and showed that the least squares was the optimal solution when the errors are normally
distributed in 1809. The methodology was used almost exclusively in the physical sciences until later in the
19th century. Francis Galton coined the term regression to mediocrity in 1875 in reference to the simple
regression equation in the form
¡
y
¯
y
x
¯
x
r
¢
¤
SD
¡
y
SDx
Galton used this equation to explain the phenomenon that sons of tall fathers tend to be tall but not as tall as
their fathers while sons of short fathers tend to be short but not as short as their fathers. This effect is called
the regression effect.
We can illustrate this effect with some data on scores from a course taught using this book. In Figure 1.3,
we see a plot of midterm against final scores. We scale each variable to have mean 0 and SD 1 so that we are
not distracted by the relative difficulty of each exam and the total number of points possible. Furthermore,
this simplifies the regression equation to
y
rx
¤
> data(stat500)
> stat500 <- data.frame(scale(stat500))
> plot(final ˜ midterm,stat500)
> abline(0,1)
2
1
0
final
−1
−2
−2
−1
0
1
2
midterm
Figure 1.3: Final and midterm scores in standard units. Least squares fit is shown with a dotted line while
y
x is shown as a solid line
¤
1.3. HISTORY
15
We have added the y
x (solid) line to the plot. Now a student scoring, say one standard deviation
¤
above average on the midterm might reasonably expect to do equally well on the final. We compute the
least squares regression fit and plot the regression line (more on the details later). We also compute the
correlations.
> g <- lm(final ˜ midterm,stat500)
> abline(g$coef,lty=5)
> cor(stat500)
midterm
final
hw
total
midterm 1.00000 0.545228 0.272058 0.84446
final
0.54523 1.000000 0.087338 0.77886
hw
0.27206 0.087338 1.000000 0.56443
total
0.84446 0.778863 0.564429 1.00000
We see that the the student scoring 1 SD above average on the midterm is predicted to score somewhat
less above average on the final (see the dotted regression line) - 0.54523 SD’s above average to be exact.
Correspondingly, a student scoring below average on the midterm might expect to do relatively better in the
final although still below average.
If exams managed to measure the ability of students perfectly, then provided that ability remained un-
changed from midterm to final, we would expect to see a perfect correlation. Of course, it’s too much to
expect such a perfect exam and some variation is inevitably present. Furthermore, individual effort is not
constant. Getting a high score on the midterm can partly be attributed to skill but also a certain amount of
luck. One cannot rely on this luck to be maintained in the final. Hence we see the “regression to mediocrity”.
¡
Of course this applies to any x y situation like this — an example is the so-called sophomore jinx
¢
in sports when a rookie star has a so-so second season after a great first year. Although in the father-son
example, it does predict that successive descendants will come closer to the mean, it does not imply the
same of the population in general since random fluctuations will maintain the variation. In many other
applications of regression, the regression effect is not of interest so it is unfortunate that we are now stuck
with this rather misleading name.
Regression methodology developed rapidly with the advent of high-speed computing. Just fitting a
regression model used to require extensive hand calculation. As computing hardware has improved, then
the scope for analysis has widened.
Chapter 2
Estimation
2.1
Example
Let’s start with an example. Suppose that Y is the fuel consumption of a particular model of car in m.p.g.
Suppose that the predictors are
1. X1 — the weight of the car
2. X2 — the horse power
3. X3 — the no. of cylinders.
X3 is discrete but that’s OK. Using country of origin, say, as a predictor would not be possible within the
current development (we will see how to do this later in the course). Typically the data will be available in
the form of an array like this
y1
x11 x12 x13
y2
x21 x22 x23
¡¢¡¢¡
¡¢¡¢¡
yn
xn1 xn2 xn3
where n is the number of observations or cases in the dataset.
2.2
Linear Model
One very general form for the model would be
¡
Y
f X
ε
1 X2 X3
¤
¢
¡
where f is some unknown function and ε is the error in this representation which is additive in this instance.
Since we usually don’t have enough data to try to estimate f directly, we usually have to assume that it has
some more restricted form, perhaps linear as in
Y
β
β
β
β
ε
0
1X1
2X2
3X3
¤
where βi, i
0 1 2 3 are unknown parameters. β0 is called the intercept term. Thus the problem is reduced
¤
to the estimation of four values rather than the complicated infinite dimensional f .
In a linear model the parameters enter linearly — the predictors do not have to be linear. For example
Y
β
β
β
ε
0
1X1
2 log X2
¤
16
2.3. MATRIX REPRESENTATION
17
is linear but
β
Y
β
β
2
ε
0
1X
¤
1
β
is not. Some relationships can be transformed to linearity — for example y
β
ε
0x
can be linearized by
¤
1
taking logs. Linear models seem rather restrictive but because the predictors can transformed and combined
in any way, they are actually very flexible. Truly non-linear models are rarely absolutely necessary and most
often arise from a theory about the relationships between the variables rather than an empirical investigation.
2.3
Matrix Representation
Given the actual data, we may write
y
β
β
β
β
ε
i
0
1x1i
2x2i
3x3i
i
i
1
n
¤
¤
¢¡¢¡¢¡£
but the use of subscripts becomes inconvenient and conceptually obscure. We will find it simpler both
notationally and theoretically to use a matrix/vector representation. The regression equation is written as
y
X β
ε
¤
¡
¡
¡
where y
y
T
ε
ε T
β
β T
1
yn , ε
1
n
, β
0
3
and
¤
¢
¤
¢
¤
¢
¡¢¡¢¡
¡¢¡¢¡
¡¢¡¢¡
¡
£¥¤
1
x
¡
11
x12 x13 ¤
1
x
X
21
x22 x23
¤
¢
¦
¡¢¡¢¡
¡¢¡¢¡
1
xn1 xn2 xn3
The column of ones incorporates the intercept term. A couple of examples of using this notation are the
simple no predictor, mean only model y
µ
ε
¤
£
£
£
y
ε
1
1
1
¢
¦
¢
¦
µ
¢
¦
¤
¡¢¡¢¡
¡¢¡¢¡
¡¢¡¢¡
y
ε
n
1
n
We can assume that Eε
0 since if this were not so, we could simply absorb the non-zero expectation for
¤
the error into the mean µ to get a zero expectation. For the two sample problem with a treatment group
having the response y1
ym with mean µy and control group having response z1
zn with mean µz we
¢¡¢¡¢¡£
¢¡¢¡¢¡
have
¡
£¥¤
¡
£¥¤
y
1
0
¡
1
¤
¡
¤
¡
ε
£¥¤
¡
¤
¡
¤
¡
1
¤
¡
¤
¡
¤
¡
¤
¡¢¡¢¡
¡¢¡¢¡
§
¡
¤
¡
¤
¡
¤
¡¢¡¢¡
¡
ym ¤
¡
1
0
¤
µy
z
¤
¡¢¡¢¡
1
0
1
µz ¨
¢
¦
¡¢¡¢¡
¢
¦
¢
¦
¡¢¡¢¡
¡
¡
ε
z
m n
©
n
0
1
2.4
Estimating β
We have the regression equation y
X β
ε - what estimate of β would best separate the systematic com-
¤
ponent X β from the random component ε. Geometrically speaking, y
IRn while β
IRp where p is the
number of parameters (if we include the intercept then p is the number of predictors plus one).
2.5. LEAST SQUARES ESTIMATION
18
y in n dimensions
Residual in
n−p dimensions
Space spanned by X
Fitted in p dimensions
Figure 2.1: Geometric representation of the estimation β. The data vector Y is projected orthogonally onto
the model space spanned by X . The fit is represented by projection ˆ
y
X ˆβ with the difference between the
¤
fit and the data represented by the residual vector ˆε.
The problem is to find β such that X β is close to Y . The best choice of ˆβ is apparent in the geometrical
representation shown in Figure 2.4.
ˆβ is in some sense the best estimate of β within the model space. The response predicted by the model
is ˆ
y
X ˆβ or Hy where H is an orthogonal projection matrix. The difference between the actual response
¤
and the predicted response is denoted by ˆε — the residuals.
The conceptual purpose of the model is to represent, as accurately as possible, something complex — y
which is n-dimensional — in terms of something much simpler — the model which is p-dimensional. Thus
if our model is successful, the structure in the data should be captured in those p dimensions, leaving just
random variation in the residuals which lie in an n
p dimensional space. We have
Data
Systematic Structure
Random Variation
¤
¡
n dimensions
p dimensions
n
p dimensions
¤
¢
2.5
Least squares estimation
The estimation of β can be considered from a non-geometric point of view. We might define the best estimate
of β as that which minimizes the sum of the squared errors, εT ε. That is to say that the least squares estimate
of β, called ˆβ minimizes
∑ε2 εTε ¡
T ¡
i
y
X β
y
X β
¤
¤
¢
¢
Expanding this out, we get
yT y
2βX T y
βT XTXβ
Differentiating with respect to β and setting to zero, we find that ˆβ satisfies
X T X ˆβ
X T y
¤
These are called the normal equations. We can derive the same result using the geometric approach. Now
provided X T X is invertible
ˆβ
¡
X T X
1XT y
¤
¢
¡
¡
X ˆβ
X X T X
1XT y
¤
¢
Hy
¤
2.6. EXAMPLES OF CALCULATING ˆβ
19
¡
H
X X T X
1XT is called the “hat-matrix” and is the orthogonal projection of y onto the space spanned
¤
¢
by X . H is useful for theoretical manipulations but you usually don’t want to compute it explicitly as it is an
n n matrix.
Predicted values: ˆ
y
Hy
X ˆβ.
¤
¤
¡
Residuals: ˆε
y
X ˆβ
y
ˆ
y
I
H y
¤
¤
¤
¢
¡
Residual sum of squares: ˆεT ˆε
yT ¡ I
H I
H y
yT ¡ I
H y
¤
¢
¢
¤
¢
Later we will show that the least squares estimate is the best possible estimate of β when the errors ε are
uncorrelated and have equal variance - i.e. var ε
σ2I.
¤
2.6
Examples of calculating ˆ
β
1. When y
µ
ε, X
1 and β
µ so X T X
1T 1
n so
¤
¤
¤
¤
¤
ˆβ
1
¡
X T X
1XT y
1T y
¯
y
¤
¢
¤
¤
n
2. Simple linear regression (one predictor)
y
α β
ε
i
xi
i
¤
£
£
£
y
ε
1
1
x1
§
α
1
¢
¦
¢
¦
¢
¦
¤
¡¢¡¢¡
¡¢¡¢¡
β
¡¢¡¢¡
y
¨
ε
n
1
xn
n
We can now apply the formula but a simpler approach is to rewrite the equation as
α¡
¢
£¥¤
¦
y
α β
β¡
ε
i
¯
x
xi
¯
x
i
¤
¢
so now
£
1
x1
¯
x
§
n
0
X
¢
¦
X T X
¡
¤
¤
2
¡¢¡¢¡
0
∑n x
¯
x
1
x
i 1
i
¢
¨
§
n
¯
x
Now work through the rest of the calculation to reconstruct the familiar estimates, i.e.
ˆβ
∑¡ xi ¯x yi
¢
¤
∑¡ x
2
i
¯
x
¢
In higher dimensions, it is usually not possible to find such explicit formulae for the parameter estimates
unless X T X happens to be a simple form.
2.7
Why is ˆ
β a good estimate?
1. It results from an orthogonal projection onto the model space. It makes sense geometrically.
2. If the errors are independent and identically normally distributed, it is the maximum likelihood esti-
mator. Loosely put, the maximum likelihood estimate is the value of β that maximizes the probability
of the data that was observed.
3. The Gauss-Markov theorem states that it is best linear unbiased estimate. (BLUE).
2.8. GAUSS-MARKOV THEOREM
20
2.8
Gauss-Markov Theorem
First we need to understand the concept of an estimable function. A linear combination of the parameters
ψ cT β is estimable if and only if there exists a linear combination aT y such that
¤
EaT y
cT β
β
¤
Estimable functions include predictions of future observations which explains why they are worth consid-
ering. If X is of full rank (which it usually is for observational data), then all linear combinations are
estimable.
Gauss-Markov theorem
Suppose Eε
0 and var ε
σ2I. Suppose also that the structural part of the model, EY
X β is correct.
¤
¤
¤
Let ψ
cT β be an estimable function, then in the class of all unbiased linear estimates of ψ, ˆ
ψ cT ˆβ has
¤
¤
the minimum variance and is unique.
Proof:
We start with a preliminary calculation:
Suppose aT y is some unbiased estimate of cT β so that
EaT y
cT β
β
¤
aT X β
cT β
β
¤
which means that aT X
cT . This implies that c must be in the range space of X T which in turn implies that
¤
c is also in the range space of X T X which means there exists a λ such that
c
X T X λ
¤
cT ˆβ
λT XT X ˆβ λT XT y
¤
¤
Now we can show that the least squares estimator has the minimum variance — pick an arbitrary es-
timable function aT y and compute its variance:
¡
¡
var aT y
var aT y
cT ˆβ
cT ˆβ
¢
¤
¢
¡
var aT y
λT XTy cT ˆβ
¤
¢
¡
¡
¡
var aT y
λT XTy
var cT ˆβ
2cov aT y
λT XT y λT XTy
¤
¢
¢
¢
but
¡
¡
cov aT y
λT XT y λT XTy
aT
λTXT σ2IXλ
¢
¤
¢
¡
aT X
λXTX σ2Iλ
¤
¢
¡
cT
cT σ2Iλ
0
¤
¢
¤
so
¡
¡
¡
var aT y
var aT y
λT XT y
var cT ˆβ
¢
¤
¢
¢
Now since variances cannot be negative, we see that
¡
¡
var aT y
var cT ˆβ
¢
¢¡
¢
In other words cT ˆβ has minimum variance. It now remains to show that it is unique. There will be equality
¡
in above relation if var aT y
λT XT y
0 which would require that aT
λT XT
0 which means that
¢
¤
¤
aT y
λT XTy cT ˆβ so equality occurs only if aT y cT ˆβ so the estimator is unique.
¤
¤
¤
2.9. MEAN AND VARIANCE OF ˆβ
21
Implications
The Gauss-Markov theorem shows that the least squares estimate ˆβ is a good choice, but if the errors
are correlated or have unequal variance, there will be better estimators. Even if the errors behave but are
non-normal then non-linear or biased estimates may work better in some sense. So this theorem does not
tell one to use least squares all the time, it just strongly suggests it unless there is some strong reason to do
otherwise.
Situations where estimators other than ordinary least squares should be considered are
1. When the errors are correlated or have unequal variance, generalized least squares should be used.
2. When the error distribution is long-tailed, then robust estimates might be used. Robust estimates are
typically not linear in y.
3. When the predictors are highly correlated (collinear), then biased estimators such as ridge regression
might be preferable.
2.9
Mean and Variance of ˆ
β
¡
Now ˆβ
X T X
1XT y so
¤
¢
¡
1
Mean E ˆβ
X T X
X T X β
β (unbiased)
¤
¢
¤
¡
1
¡
1
¡
1
var ˆβ
X T X
X T σ2IX X T X
X T X
σ2
¤
¢
¢
¤
¢
¡
Note that since ˆβ is a vector, X T X
1σ2 is a variance-covariance matrix. Sometimes you want the
¢
¡
¡
standard error for a particular component which can be picked out as in se ˆβ
1
i
X T X
ˆ
σ.
¢
¤
¢
ii
2.10
Estimating σ2
Recall that the residual sum of squares was ˆεT ˆε
yT ¡ I
H y. Now after some calculation, one can show
¤
¢
that E ˆεT ˆε
σ2 ¡ n p which shows that
¤
¢
ˆεT ˆε
ˆ
σ2 ¤ n p
is an unbiased estimate of σ2. n
p is the degrees of freedom of the model. Actually a theorem parallel to
the Gauss-Markov theorem shows that it has the minimum variance among all quadratic unbiased estimators
of σ2.
2.11
Goodness of Fit
How well does the model fit the data? One measure is R2, the so-called coefficient of determination or
percentage of variance explained
∑¡ ˆy
2
i
yi
RSS
R2
1
¢
1
¡
¤
∑¡
y
2 ¤
i
¯
y
Total SS corrected for mean
¢
¢
2.11. GOODNESS OF FIT
22
1.0
0.6
y
0.2
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
Figure 2.2: Variation in the response y when x is known is denoted by dotted arrows while variation in y
when x is unknown is shown with the solid arrows
The range is 0
R2
1 - values closer to 1 indicating better fits. For simple linear regression R2
r2 where
¤
r is the correlation between x and y. An equivalent definition is
∑¡ ˆy
2
i
¯
y
R2
¢
¤
∑¡ y
2
i
¯
y
¢
The graphical intuition behind R2 is shown in Figure 2.2. Suppose you want to predict y. If you don’t
know x, then your best prediction is ¯
y but the variability in this prediction is high. If you do know x, then
your prediction will be given by the regression fit. This prediction will be less variable provided there is
some relationship between x and y. R2 is one minus the ratio of the sum of squares for these two predictions.
Thus for perfect predictions the ratio will be zero and R2 will be one.
Warning: R2 as defined here doesn’t make any sense if you do not have an intercept in your model. This
is because the denominator in the definition of R2 has a null model with an intercept in mind when the sum
of squares is calculated. Alternative definitions of R2 are possible when there is no intercept but the same
graphical intuition is not available and the R2’s obtained should not be compared to those for models with
an intercept. Beware of high R2’s reported from models without an intercept.
What is a good value of R2? It depends on the area of application. In the biological and social sciences,
variables tend to be more weakly correlated and there is a lot of noise. We’d expect lower values for R2
in these areas — a value of 0.6 might be considered good. In physics and engineering, where most data
comes from closely controlled experiments, we expect to get much higher R2’s and a value of 0.6 would
be considered low. Of course, I generalize excessively here so some experience with the particular area is
necessary for you to judge your R2’s well.
An alternative measure of fit is ˆ
σ. This quantity is directly related to the standard errors of estimates
of β and predictions. The advantage is that ˆσ is measured in the units of the response and so may be
directly interpreted in the context of the particular dataset. This may also be a disadvantage in that one
2.12. EXAMPLE
23
must understand whether the practical significance of this measure whereas R2, being unitless, is easy to
understand.
2.12
Example
Now let’s look at an example concerning the number of species of tortoise on the various Galapagos Islands.
There are 30 cases (Islands) and 7 variables in the dataset. We start by reading the data into R and examining
it
> data(gala)
> gala
Species Endemics
Area Elevation Nearest Scruz Adjacent
Baltra
58
23
25.09
346
0.6
0.6
1.84
Bartolome
31
21
1.24
109
0.6
26.3
572.33
--- cases deleted ---
Tortuga
16
8
1.24
186
6.8
50.9
17.95
Wolf
21
12
2.85
253
34.1 254.7
2.33
The variables are
Species The number of species of tortoise found on the island
Endemics The number of endemic species
Elevation The highest elevation of the island (m)
Nearest The distance from the nearest island (km)
Scruz The distance from Santa Cruz island (km)
Adjacent The area of the adjacent island (km2)
The data were presented by Johnson and Raven (1973) and also appear in Weisberg (1985). I have filled
in some missing values for simplicity (see Chapter 14 for how this can be done). Fitting a linear model in R
is done using the lm() command. Notice the syntax for specifying the predictors in the model. This is the
so-called Wilkinson-Rogers notation. In this case, since all the variables are in the gala data frame, we must
use the data= argument:
> gfit <- lm(Species ˜ Area + Elevation + Nearest + Scruz + Adjacent,
data=gala)
> summary(gfit)
Call:
lm(formula = Species ˜ Area + Elevation + Nearest + Scruz + Adjacent,
data = gala)
Residuals:
Min
1Q
Median
3Q
Max
-111.68
-34.90
-7.86
33.46
182.58
2.12. EXAMPLE
24
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.06822
19.15420
0.37
0.7154
Area
-0.02394
0.02242
-1.07
0.2963
Elevation
0.31946
0.05366
5.95
3.8e-06
Nearest
0.00914
1.05414
0.01
0.9932
Scruz
-0.24052
0.21540
-1.12
0.2752
Adjacent
-0.07480
0.01770
-4.23
0.0003
Residual standard error: 61 on 24 degrees of freedom
Multiple R-Squared: 0.766,
Adjusted R-squared: 0.717
F-statistic: 15.7 on 5 and 24 degrees of freedom,
p-value: 6.84e-07
We can identify several useful quantities in this output. Other statistical packages tend to produce output
quite similar to this. One useful feature of R is that it is possible to directly calculate quantities of interest.
Of course, it is not necessary here because the lm() function does the job but it is very useful when the
statistic you want is not part of the pre-packaged functions.
First we make the X-matrix
> x <- cbind(1,gala[,-c(1,2)])
and here’s the response y:
> y <- gala$Species
Now let’s construct X T X : t() does transpose and %*% does matrix multiplication:
> t(x) %*% x
Error: %*% requires numeric matrix/vector arguments
Gives a somewhat cryptic error. The problem is that matrix arithmetic can only be done with numeric
values but x here derives from the data frame type. Data frames are allowed to contain character variables
which would disallow matrix arithmetic. We need to force x into the matrix form:
> x <- as.matrix(x)
> t(x) %*% x
Inverses can be taken using the solve() command:
> xtxi <- solve(t(x) %*% x)
> xtxi
¡
A somewhat more direct way to get X T X
1 is as follows:
¢
> gfit <- lm(Species ˜ Area + Elevation + Nearest + Scruz + Adjacent,
data=gala)
> gs <- summary(gfit)
> gs$cov.unscaled
2.12. EXAMPLE
25
The names() command is the way to see the components of an Splus object - you can see that there
are other useful quantities that are directly available:
> names(gs)
> names(gfit)
In particular, the fitted (or predicted) values and residuals are
> gfit$fit
> gfit$res
We can get ˆβ directly:
> xtxi %*% t(x) %*% y
[,1]
[1,]
7.068221
[2,] -0.023938
[3,]
0.319465
[4,]
0.009144
[5,] -0.240524
[6,] -0.074805
or in a computationally efficient and stable manner:
> solve(t(x) %*% x, t(x) %*% y)
[,1]
[1,]
7.068221
[2,] -0.023938
[3,]
0.319465
[4,]
0.009144
[5,] -0.240524
[6,] -0.074805
We can estimate σ using the estimator in the text:
> sqrt(sum(gfit$resˆ2)/(30-6))
[1] 60.975
Compare this to the results above.
We may also obtain the standard errors for the coefficients. Also diag() returns the diagonal of a
matrix):
> sqrt(diag(xtxi))*60.975
[1] 19.154139
0.022422
0.053663
1.054133
0.215402
0.017700
Finally we may compute R2:
> 1-sum(gfit$resˆ2)/sum((y-mean(y))ˆ2)
[1] 0.76585
Chapter 3
Inference
Up till now, we haven’t found it necessary to assume any distributional form for the errors ε. However, if we
want to make any confidence intervals or perform any hypothesis tests, we will need to do this. The usual
assumption is that the errors are normally distributed and in practice this is often, although not always, a
reasonable assumption. We’ll assume that the errors are independent and identically normally distributed
with mean 0 and variance σ2, i.e.
ε
¡
N 0 σ2I
¢
We can handle non-identity variance matrices provided we know the form — see the section on gener-
alized least squares later. Now since y
X β
ε,
¤
¡
y
N X β σ2I
¢
is a compact description of the regression model and from this we find that (using the fact that linear com-
binations of normally distributed values are also normal)
ˆβ
¡
¡
¡
X T X
1XT y
N β X T X
1σ2
¤
¢
¢
¢
3.1
Hypothesis tests to compare models
Given several predictors for a response, we might wonder whether all are needed. Consider a large model,
Ω, and a smaller model, ω, which consists of a subset of the predictors that are in Ω. By the principle of
Occam’s Razor (also known as the law of parsimony), we’d prefer to use ω if the data will support it. So
we’ll take ω to represent the null hypothesis and Ω to represent the alternative. A geometric view of the
problem may be seen in Figure 3.1.
If RSSω
RSSΩ is small, then ω is an adequate model relative to Ω. This suggests that something like
RSSω
RSSΩ
RSSΩ
would be a potentially good test statistic where the denominator is used for scaling purposes.
As it happens the same test statistic arises from the likelihood-ratio testing approach. We give an outline
¡
of the development: If L β σ ¡ y is likelihood function, then the likelihood ratio statistic is
¢
¡
maxβ σ Ω L β σ¡ y¢
¢
£
¡
maxβ σ ω L β σ¡ y¢
¢
£
26
3.1. HYPOTHESIS TESTS TO COMPARE MODELS
27
Y
Residual for small model
Residual for large model
Difference between
two models
Large model space
Small model space
Figure 3.1: Geometric view of the comparison between big model, Ω, and small model, ω. The squared
length of the residual vector for the big model is RSSΩ while that for the small model is RSSω. By Pythago-
ras’ theorem, the squared length of the vector connecting the two fits is RSSω
RSSΩ. A small value for this
indicates that the small model fits almost as well as the large model and thus might be preferred due to its
simplicity.
The test should reject if this ratio is too large. Working through the details, we find that
¡
L ˆβ ˆσ¡ y ∝ ˆσ n
¢
which gives us a test that rejects if
ˆ
σ2ω ¥ a constant
ˆ
σ2Ω
which is equivalent to
RSSω ¥ a constant
RSSΩ
(constants are not the same) or
RSSω
1 ¥
a constant
1
RSSΩ
which is
RSSω
RSSΩ
¥
a constant
RSSΩ
which is the same statistics suggested by the geometric view. It remains for us to discover the null distribu-
tion of this statistic.
Now suppose that the dimension (no. of parameters) of Ω is q and dimension of ω is p. Now by
Cochran’s theorem, if the null (ω) is true then
RSSω
RSSΩ
RSSΩ
σ2χ2
σ2χ2
q
p
q p
n
q
n q
and these two quantities are independent. So we find that
¡
¡
RSSω
RSSΩ
q
p
F
¢
¡
¢
F
¡
q p n q
¤
RSSΩ n
q
¢
¢
3.2. SOME EXAMPLES
28
α
Thus we would reject the null hypothesis if F ¥ F ¡
q p n q The degrees of freedom of a model is (usually) the
¢
number of observations minus the number of parameters so this test statistic can be written
¡
¡
RSSω
RSSΩ
d fω
d fΩ
F
¢
¡
¢
¤
RSSΩ d fΩ
where d fΩ
n
q and d fω
n
p. The same test statistic applies not just when ω is a subset of Ω but
¤
¤
also to a subspace. This test is very widely used in regression and analysis of variance. When it is applied
in different situations, the form of test statistic may be re-expressed in various different ways. The beauty
of this approach is you only need to know the general form. In any particular case, you just need to figure
out which models represents the null and alternative hypotheses, fit them and compute the test statistic. It is
very versatile.
3.2
Some Examples
3.2.1
Test of all predictors
Are any of the predictors useful in predicting the response?
Full model (Ω) : y
X β
ε where X is a full-rank n p matrix.
¤
Reduced model (ω) : y
µ
ε — predict y by the mean.
¤
We could write the null hypothesis in this case as
H
β
0 : β1
p 1
0
¤
¤
¡¢¡¢¡
Now
¡
T ¡
RSSΩ
y
X ˆβ
y
X ˆβ
ˆεT ˆε
RSS
¤
¢
¢
¤
¤
¡
T ¡
RSSω
y
¯
y
y
¯
y
SYY, which is sometimes known as the sum of squares corrected for the
¤
¢
¢
¤
mean.
So in this case
¡
¡
SYY
RSS
p
1
F
¢
¡
¢
¡
¤
RSS n
p
¢
We’d now refer to Fp 1 n p for a critical value or a p-value. Large values of F would indicate rejection
¢
of the null. Traditionally, the information in the above test is presented in an analysis of variance table.
Most computer packages produce a variant on this. See Table 3.1. It is not really necessary to specifically
compute all the elements of the table. As the originator of the table, Fisher said in 1931, it is “nothing but a
convenient way of arranging the arithmetic”. Since he had to do his calculations by hand, the table served
some purpose but it is less useful now.
A failure to reject the null hypothesis is not the end of the game — you must still investigate the pos-
sibility of non-linear transformations of the variables and of outliers which may obscure the relationship.
Even then, you may just have insufficient data to demonstrate a real effect which is why we must be care-
ful to say “fail to reject” the null rather than “accept” the null. It would be a mistake to conclude that no
real relationship exists. This issue arises when a pharmaceutical company wishes to show that a proposed
generic replacement for a brand-named drug is equivalent. It would not be enough in this instance just to
fail to reject the null. A higher standard would be required.
3.2. SOME EXAMPLES
29
Source
Deg. of Freedom
Sum of Squares
Mean Square
F
¡
Regression
p
1
SSreg
SSreg
p
1
F
¢
¡
Residual
n-p
RSS
RSS n
p
¢
Total
n-1
SYY
Table 3.1: Analysis of Variance table
When the null is rejected, this does not imply that the alternative model is the best model. We don’t
know whether all the predictors are required to predict the response or just some of them. Other predictors
might also be added — for example quadratic terms in the existing predictors. Either way, the overall F-test
is just the beginning of an analysis and not the end.
Let’s illustrate this test and others using an old economic dataset on 50 different countries. These data
are averages over 1960-1970 (to remove business cycle or other short-term fluctuations). dpi is per-capita
disposable income in U.S. dollars; ddpi is the percent rate of change in per capita disposable income; sr
is aggregate personal saving divided by disposable income. The percentage population under 15 (pop15)
and over 75 (pop75) are also recorded. The data come from Belsley, Kuh, and Welsch (1980). Take a look
at the data:
> data(savings)
> savings
sr pop15 pop75
dpi
ddpi
Australia
11.43 29.35
2.87 2329.68
2.87
Austria
12.07 23.32
4.41 1507.99
3.93
--- cases deleted ---
Malaysia
4.71 47.20
0.66
242.69
5.08
First consider a model with all the predictors:
> g <- lm(sr ˜ pop15 + pop75 + dpi + ddpi, data=savings)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 28.566087
7.354516
3.88
0.00033
pop15
-0.461193
0.144642
-3.19
0.00260
pop75
-1.691498
1.083599
-1.56
0.12553
dpi
-0.000337
0.000931
-0.36
0.71917
ddpi
0.409695
0.196197
2.09
0.04247
Residual standard error: 3.8 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
We can see directly the result of the test of whether any of the predictors have significance in the model.
In other words, whether β
β
β
β
1
2
3
4
0. Since the p-value is so small, this null hypothesis is rejected.
¤
¤
¤
¤
We can also do it directly using the F-testing formula:
3.2. SOME EXAMPLES
30
> sum((savings$sr-mean(savings$sr))ˆ2)
[1] 983.63
> sum(g$resˆ2)
[1] 650.71
> ((983.63-650.71)/4)/(650.706/45)
[1] 5.7558
> 1-pf(5.7558,4,45)
[1] 0.00079026
Do you know where all the numbers come from? Check that they match the regression summary above.
3.2.2
Testing just one predictor
Can one particular predictor be dropped from the model? The null hypothesis would be H0 : βi
0. Set it
¤
up like this
RSSΩ is the RSS for the model with all the predictors of interest (p parameters).
RSSω is the RSS for the model with all the above predictors except predictor i.
The F-statistic may be computed using the formula from above. An alternative approach is to use a
t-statistic for testing the hypothesis:
¡
t
ˆβ
ˆβ
i
i se
i
¤
¢
and check for significance using a t distribution with n
p degrees of freedom.
However, squaring the t-statistic here, i.e. t 2i gives you the F-statistic, so the two approaches are identical.
For example, to test the null hypothesis that β1
0 i.e. that p15 is not significant in the full model, we
¤
can simply observe that the p-value is 0.0026 from the table and conclude that the null should be rejected.
Let’s do the same test using the general F-testing approach: We’ll need the RSS and df for the full model
— these are 650.71 and 45 respectively.
and then fit the model that represents the null:
> g2 <- lm(sr ˜ pop75 + dpi + ddpi, data=savings)
and compute the RSS and the F-statistic:
> sum(g2$resˆ2)
[1] 797.72
> (797.72-650.71)/(650.71/45)
[1] 10.167
The p-value is then
> 1-pf(10.167,1,45)
[1] 0.0026026
We can relate this to the t-based test and p-value by
> sqrt(10.167)
[1] 3.1886
> 2*(1-pt(3.1886,45))
[1] 0.0026024
3.2. SOME EXAMPLES
31
A somewhat more convenient way to compare two nested models is
> anova(g2,g)
Analysis of Variance Table
Model 1: sr ˜ pop75 + dpi + ddpi
Model 2: sr ˜ pop15 + pop75 + dpi + ddpi
Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
1
46
798
2
45
651
1
147
10.2 0.0026
Understand that this test of pop15 is relative to the other predictors in the model, namely pop75, dpi
and ddpi. If these other predictors were changed, the result of the test may be different. This means that it is
not possible to look at the effect of pop15 in isolation. Simply stating the null hypothesis as H0 : βpop15
0
¤
is insufficient — information about what other predictors are included in the null is necessary. The result of
the test may be different if the predictors change.
3.2.3
Testing a pair of predictors
Suppose we wish to test the significance of variables X j and Xk. We might construct a table as shown just
above and find that both variables have p-values greater than 0.05 thus indicating that individually neither is
significant. Does this mean that both X j and Xk can be eliminated from the model? Not necessarily
Except in special circumstances, dropping one variable from a regression model causes the estimates of
the other parameters to change so that we might find that after dropping X j, that a test of the significance of
Xk shows that it should now be included in the model.
If you really want to check the joint significance of X j and Xk, you should fit a model with and then
without them and use the general F-test discussed above. Remember that even the result of this test may
depend on what other predictors are in the model.
Can you see how to test the hypothesis that both pop75 and ddpi may be excluded from the model?
¡
£
¢
y ˜ x1 + x2 + x3
¨
¨
¨©
¡
¢
£
¡
£
¢
y ˜ x1 + x2
y ˜ x1 + x3
¤
¥
§
¦
y ˜ x1
Figure 3.2: Testing two predictors
The testing choices are depicted in Figure 3.2. Here we are considering two predictors, x2 and x3 in
the presence of x1. Five possible tests may be considered here and the results may not always be appar-
ently consistent. The results of each test need to be considered individually in the context of the particular
example.
3.2. SOME EXAMPLES
32
3.2.4
Testing a subspace
Consider this example. Suppose that y is the miles-per-gallon for a make of car and X j is the weight of the
engine and Xk is the weight of the rest of the car. There would also be some other predictors. We might
wonder whether we need two weight variables — perhaps they can be replaced by the total weight, X j
Xk.
So if the original model was
y
β
β
β
ε
0
jX j
kXk
¤
¡¢¡¢¡
¡¢¡¢¡
then the reduced model is
¡
y
β
β
ε
0
l X j
Xk
¤
¢
¡
¡¢¡¢¡
¡¢¡¢¡
which requires that β
β
j
k for this reduction to be possible. So the null hypothesis is
¤
H
β
0 : β j
k
¤
This defines a linear subspace to which the general F-testing procedure applies. In our example, we might
hypothesize that the effect of young and old people on the savings rate was the same or in other words that
H
β
0 : βpop15
pop75
¤
In this case the null model would take the form
¡
y
β
β
β
β
ε
0
pop15 po p15
pop75
d pi d pi
dd pi dd pi
¤
¢
We can then compare this to the full model as follows:
> g <- lm(sr ˜ .,savings)
> gr <- lm(sr ˜ I(pop15+pop75)+dpi+ddpi,savings)
> anova(gr,g)
Analysis of Variance Table
Model 1: sr ˜ I(pop15 + pop75) + dpi + ddpi
Model 2: sr ˜ pop15 + pop75 + dpi + ddpi
Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
1
46
674
2
45
651
1
23
1.58
0.21
The period in the first model formula is short hand for all the other variables in the data frame. The
function I() ensures that the argument is evaluated rather than interpreted as part of the model formula.
The p-value of 0.21 indicates that the null cannot be rejected here meaning that there is not evidence here
that young and old people need to be treated separately in the context of this particular model.
Suppose we want to test whether one of the coefficients can be set to a particular value. For example,
H0 : βddpi
1
¤
Here the null model would take the form:
y
β
β
β
β
ε
0
pop15 po p15
pop75 po p75
d pi d pi
dd pi
¤
Notice that there is now no coefficient on the ddpi term. Such a fixed term in the regression equation is
called an offset. We fit this model and compare it to the full:
3.3. CONCERNS ABOUT HYPOTHESIS TESTING
33
> gr <- lm(sr ˜ pop15+pop75+dpi+offset(ddpi),savings)
> anova(gr,g)
Analysis of Variance Table
Model 1: sr ˜ pop15 + pop75 + dpi + offset(ddpi)
Model 2: sr ˜ pop15 + pop75 + dpi + ddpi
Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
1
46
782
2
45
651
1
131
9.05 0.0043
We see that the p-value is small and the null hypothesis here is soundly rejected. A simpler way to test such
point hypotheses is to use a t-statistic:
¡
¡
t
ˆβ c se ˆβ
¤
¢
¡
¢
where c is the point hypothesis. So in our example the statistic and corresponding p-value is
> tstat <- (0.409695-1)/0.196197
> tstat
[1] -3.0087
> 2*pt(tstat,45)
[1] 0.0042861
We can see the p-value is the same as before and if we square the t-statistic
> tstatˆ2
[1] 9.0525
we find we get the F-value. This latter approach is preferred in practice since we don’t need to fit two
models but it is important to understand that it is equivalent to the result obtained using the general F-testing
approach.
Can we test a hypothesis such as
H
β
0 : β j k
1
¤
using our general theory?
No. This hypothesis is not linear in the parameters so we can’t use our general method. We’d need to fit
a non-linear model and that lies beyond the scope of this book.
3.3
Concerns about Hypothesis Testing
1. The general theory of hypothesis testing posits a population from which a sample is drawn — this is
our data. We want to say something about the unknown population values β using estimated values
ˆβ that are obtained from the sample data. Furthermore, we require that the data be generated using a
simple random sample of the population. This sample is finite in size, while the population is infinite
in size or at least so large that the sample size is a negligible proportion of the whole. For more
complex sampling designs, other procedures should be applied, but of greater concern is the case
when the data is not a random sample at all. There are two cases:
(a) A sample of convenience is where the data is not collected according to a sampling design.
In some cases, it may be reasonable to proceed as if the data were collected using a random
mechanism. For example, suppose we take the first 400 people from the phonebook whose
3.3. CONCERNS ABOUT HYPOTHESIS TESTING
34
names begin with the letter P. Provided there is no ethnic effect, it may be reasonable to consider
this a random sample from the population defined by the entries in the phonebook. Here we
are assuming the selection mechanism is effectively random with respect to the objectives of the
study. An assessment of exchangeability is required - are the data as good as random? Other
situations are less clear cut and judgment will be required. Such judgments are easy targets for
criticism. Suppose you are studying the behavior of alcoholics and advertise in the media for
study subjects. It seems very likely that such a sample will be biased perhaps in unpredictable
ways. In cases such as this, a sample of convenience is clearly biased in which case conclusions
must be limited to the sample itself. This situation reduces to the next case, where the sample is
the population.
Sometimes, researchers may try to select a “representative” sample by hand. Quite apart from
the obvious difficulties in doing this, the logic behind the statistical inference depends on the
sample being random. This is not to say that such studies are worthless but that it would be
unreasonable to apply anything more than descriptive statistical techniques. Confidence in the
of conclusions from such data is necessarily suspect.
(b) The sample is the complete population in which case one might argue that inference is not
required since the population and sample values are one and the same. For both regression
datasets we have considered so far, the sample is effectively the population or a large and biased
proportion thereof.
In these situations, we can put a different meaning to the hypothesis tests we are making. For
the Galapagos dataset, we might suppose that if the number of species had no relation to the
five geographic variables, then the observed response values would be randomly distributed
between the islands without relation to the predictors. We might then ask what the chance would
be under this assumption that an F-statistic would be observed as large or larger than one we
actually observed. We could compute this exactly by computing the F-statistic for all possible
(30!) permutations of the response variable and see what proportion exceed the observed F-
statistic. This is a permutation test. If the observed proportion is small, then we must reject the
contention that the response is unrelated to the predictors. Curiously, this proportion is estimated
by the p-value calculated in the usual way based on the assumption of normal errors thus saving
us from the massive task of actually computing the regression on all those computations.
Let see how we can apply the permutation test to the savings data. I chose a model with just
pop75 and dpi so as to get a p-value for the F-statistic that is not too small.
> g <- lm(sr ˜ pop75+dpi,data=savings)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.056619
1.290435
5.47
1.7e-06
pop75
1.304965
0.777533
1.68
0.10
dpi
-0.000341
0.001013
-0.34
0.74
Residual standard error: 4.33 on 47 degrees of freedom
Multiple R-Squared: 0.102,
Adjusted R-squared: 0.0642
F-statistic: 2.68 on 2 and 47 degrees of freedom, p-value: 0.0791
We can extract the F-statistic as
> gs <- summary(g)
3.3. CONCERNS ABOUT HYPOTHESIS TESTING
35
> gs$fstat
value
numdf
dendf
2.6796
2.0000 47.0000
The function sample() generates random permutations. We compute the F-statistic for 1000
randomly selected permutations and see what proportion exceed the the F-statistic for the origi-
nal data:
> fstats <- numeric(1000)
> for(i in 1:1000){
+ ge <- lm(sample(sr) ˜ pop75+dpi,data=savings)
+ fstats[i] <- summary(ge)$fstat[1]
+ }
> length(fstats[fstats > 2.6796])/1000
[1] 0.092
So our estimated p-value using the permutation test is 0.092 which is close to the normal theory
based value of 0.0791. We could reduce variability in the estimation of the p-value simply
by computing more random permutations. Since the permutation test does not depend on the
assumption of normality, we might regard it as superior to the normal theory based value.
Thus it is possible to give some meaning to the p-value when the sample is the population or
for samples of convenience although one has to be clear that one’s conclusion apply only the
particular sample.
Tests involving just one predictor also fall within the permutation test framework. We permute
that predictor rather than the response
Another approach that gives meaning to the p-value when the sample is the population involves
the imaginative concept of “alternative worlds” where the sample/population at hand is sup-
posed to have been randomly selected from parallel universes. This argument is definitely more
tenuous.
2. A model is usually only an approximation of underlying reality which makes the meaning of the pa-
rameters debatable at the very least. We will say more on the interpretation of parameter estimates
later but the precision of the statement that β1
0 exactly is at odds with the acknowledged approx-
¤
imate nature of the model. Furthermore, it is highly unlikely that a predictor that one has taken the
trouble to measure and analyze has exactly zero effect on the response. It may be small but it won’t
be zero.
This means that in many cases, we know that the point null hypothesis is false without even looking
at the data. Furthermore, we know that the more data we have, the greater the power of our tests.
Even small differences from zero will be detected with a large sample. Now if we fail to reject the
null hypothesis, we might simply conclude that we didn’t have enough data to get a significant result.
According to this view, the hypothesis test just becomes a test of sample size. For this reason, I prefer
confidence intervals.
3. The inference depends on the correctness of the model we use. We can partially check the assumptions
about the model but there will always be some element of doubt. Sometimes the data may suggest
more than one possible model which may lead to contradictory results.
4. Statistical significance is not equivalent to practical significance. The larger the sample, the smaller
your p-values will be so don’t confuse p-values with a big predictor effect. With large datasets it will
3.4. CONFIDENCE INTERVALS FOR β
36
be very easy to get statistically significant results, but the actual effects may be unimportant. Would
we really care if test scores were 0.1% higher in one state than another? Or that some medication
reduced pain by 2%? Confidence intervals on the parameter estimates are a better way of assessing
the size of an effect. There are useful even when the null hypothesis is not rejected because they tell
us how confident we are that the true effect or value is close to the null.
Even so, hypothesis tests do have some value, not least because they impose a check on unreasonable
conclusions which the data simply does not support.
3.4
Confidence Intervals for β
Confidence intervals provide an alternative way of expressing the uncertainty in our estimates. Even so, they
are closely linked to the tests that we have already constructed. For the confidence intervals and regions that
¡
we will consider here, the following relationship holds. For a 100 1
α % confidence region, any point
¢
that lies within the region represents a null hypothesis that would not be rejected at the 100α% level while
every point outside represents a null hypothesis that would be rejected. So, in a sense, the confidence region
provides a lot more information than a single hypothesis test in that it tells us the outcome of a whole range
of hypotheses about the parameter values. Of course, by selecting the particular level of confidence for the
region, we can only make tests at that level and we cannot determine the p-value for any given test simply
from the region. However, since it is dangerous to read too much into the relative size of p-values (as far as
how much evidence they provide against the null), this loss is not particularly important.
The confidence region tells us about plausible values for the parameters in a way that the hypothesis test
cannot. This makes it more valuable.
As with testing, we must decide whether to form confidence regions for parameters individually or
simultaneously. Simultaneous regions are preferable but for more than two dimensions they are difficult to
display and so there is still some value in computing the one-dimensional confidence intervals.
We start with the simultaneous regions. Some results from multivariate analysis show that
¡
ˆβ β T
¡
X T X ˆβ
β
¢
¢
χ2
σ2
p
and
¡
n
p ˆ
σ2
¢
χ2
σ2
n p
and these two quantities are independent. Hence
¡
ˆβ β T
¡
X T X ˆβ
β
χ2p p
¢
¢
F
¡
p n p
p ˆ
σ2
χ2
¢
n p
n
p
¢
¡
¡
So to form a 100 1
α % confidence region for β, take β such that
¢
α
¡
ˆβ β T
¡
X T X ˆβ
β
p ˆ
σ2F ¡
¢
¢
p n p
¢
These regions are ellipsoidally shaped. Because these ellipsoids live in higher dimensions, they cannot
easily be visualized.
Alternatively, one could consider each parameter individually which leads to confidence intervals which
take the general form of
estimate ¢ critical value s e of estimate
¡
¡
3.4. CONFIDENCE INTERVALS FOR β
37
or specifically in this case:
ˆβ
α 2
¡
¢
1
¡
i
t
ˆ
σ
X T X ¢¡
n p
ii
It’s better to consider the joint confidence intervals when possible, especially when the ˆβ are heavily
correlated.
Consider the full model for the savings data. The . in the model formula stands for “every other variable
in the data frame” which is a useful abbreviation.
> g <- lm(sr ˜ ., savings)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 28.566087
7.354516
3.88
0.00033
pop15
-0.461193
0.144642
-3.19
0.00260
pop75
-1.691498
1.083599
-1.56
0.12553
dpi
-0.000337
0.000931
-0.36
0.71917
ddpi
0.409695
0.196197
2.09
0.04247
Residual standard error: 3.8 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
We can construct individual 95% confidence intervals for the regression parameters of pop75:
> qt(0.975,45)
[1] 2.0141
> c(-1.69-2.01*1.08,-1.69+2.01*1.08)
[1] -3.8608
0.4808
and similarly for growth
> c(0.41-2.01*0.196,0.41+2.01*0.196)
[1] 0.01604 0.80396
Notice that this confidence interval is pretty wide in the sense that the upper limit is about 50 times larger
than the lower limit. This means that we are not really that confident about what the exact effect of growth
on savings really is.
Confidence intervals often have a duality with two-sided hypothesis tests. A 95% confidence interval
contains all the null hypotheses that would not be rejected at the 5% level. Thus the interval for pop75
contains zero which indicates that the null hypothesis H0 : βpop75
0 would not be rejected at the 5% level.
¤
We can see from the output above that the p-value is 12.5% — greater than 5% — confirming this point. In
contrast, we see that the interval for ddpi does not contain zero and so the null hypothesis is rejected for
its regression parameter.
Now we construct the joint 95% confidence region for these parameters. First we load in a ”library” for
drawing confidence ellipses which is not part of base R:
> library(ellipse)
and now the plot:
3.4. CONFIDENCE INTERVALS FOR β
38
> plot(ellipse(g,c(2,3)),type="l",xlim=c(-1,0))
add the origin and the point of the estimates:
> points(0,0)
> points(g$coef[2],g$coef[3],pch=18)
How does the position of the origin relate to a test for removing pop75 and pop15?
Now we mark the one way confidence intervals on the plot for reference:
> abline(v=c(-0.461-2.01*0.145,-0.461+2.01*0.145),lty=2)
> abline(h=c(-1.69-2.01*1.08,-1.69+2.01*1.08),lty=2)
See the plot in Figure 3.3.
1
0
−1
pop75
−2
−3
−4
−1.0
−0.8
−0.6
−0.4
−0.2
0.0
pop15
Figure 3.3: Confidence ellipse and regions for βpop75 and βpop15
Why are these lines not tangential to the ellipse? The reason for this is that the confidence intervals are
calculated individually. If we wanted a 95% chance that both intervals contain their true values, then the
lines would be tangential.
In some circumstances, the origin could lie within both one-way confidence intervals, but lie outside the
ellipse. In this case, both one-at-a-time tests would not reject the null whereas the joint test would. The latter
test would be preferred. It’s also possible for the origin to lie outside the rectangle but inside the ellipse. In
this case, the joint test would not reject the null whereas both one-at-a-time tests would reject. Again we
prefer the joint test result.
Examine the correlation of the two predictors:
> cor(savings$pop15,savings$pop75)
[1] -0.90848
But from the plot, we see that coefficients have a positive correlation. The correlation between predictors
and the correlation between the coefficients of those predictors are often different in sign. Intuitively, this
3.5. CONFIDENCE INTERVALS FOR PREDICTIONS
39
can be explained by realizing that two negatively correlated predictors are attempting to the perform the
same job. The more work one does, the less the other can do and hence the positive correlation in the
coefficients.
3.5
Confidence intervals for predictions
Given a new set of predictors, x
ˆβ
0 what is the predicted response? Easy — just ˆ
y0
xT . However, we need
¤
0
to distinguish between predictions of the future mean response and predictions of future observations. To
make the distinction, suppose we have built a regression model that predicts the selling price of homes in a
given area that is based on predictors like the number of bedrooms, closeness to a major highway etc. There
are two kinds of predictions that can be made for a given x0.
1. Suppose a new house comes on the market with characteristics x
β ε
0. Its selling price will be xT
.
0
Since Eε
0, the predicted price is xT ˆβ but in assessing the variance of this prediction, we must
¤
0
include the variance of ε.
2. Suppose we ask the question — “What would the house with characteristics x0” sell for on average.
This selling price is xT β and is again predicted by xT ˆβ but now only the variance in ˆβ needs to be
0
0
taken into account.
Most times, we will want the first case which is called “prediction of a future value” while the second case,
called “prediction of the mean response” is less common.
¡
¡
Now var xT ˆβ
xT X T X
1x σ2.
0
0
¢
¤
0
¢
A future observation is predicted to be xT ˆβ
ε (where we don’t what the future ε will turn out to be).
0
¡
So a 100 1
α % confidence interval for a single future response is
¢
α
2
¡
ˆ
y ¢
¡
1
0
t
ˆ
σ 1 xT XTX
x0
n p
0
¢
If on the other hand, you want a confidence interval for the average of the responses for given x0 then use
α
2
¡
ˆ
y ¢
¡
1
0
t
ˆ
σ xT XT X
x0
n p
0
¢
We return to the Galapagos data for this example.
> g <- lm(Species ˜ Area+Elevation+Nearest+Scruz+Adjacent,data=gala)
Suppose we want to predict the number of species (of tortoise) on an island with predictors 0.08,93,6.0,12.0,0.34
(same order as in the dataset). Of course it is difficult to see why in practice we would want to do this be-
cause a new island is unlikely to present itself. For a dataset like this interest would center on the structure of
the model and relative importance of the predictors, so we should regard this more as a ”what if?” exercise.
Do it first directly from the formula:
> x0 <- c(1,0.08,93,6.0,12.0,0.34)
> y0 <- sum(x0*g$coef)
> y0
[1] 33.92
3.5. CONFIDENCE INTERVALS FOR PREDICTIONS
40
This is the predicted no. of species which is not a whole number as the response is. We could round up
to 34.
Now if we want a 95% confidence interval for the prediction, we must decide whether we are predicting
the number of species on one new island or the mean response for all islands with same predictors x0.
Possibly, an island might not have been surveyed for the original dataset in which case the former interval
would be the one we want. For this dataset, the latter interval would be more valuable for “what if?” type
calculations.
First we need the t-critical value:
> qt(0.975,24)
[1] 2.0639
¡
You may need to recalculate the X T X
1 matrix:
¢
> x <- cbind(1,gala[,3:7])
> x <- as.matrix(x)
> xtxi <- solve(t(x) %*% x)
The width of the bands for mean response CI is
> bm <- sqrt(x0 %*% xtxi %*% x0) *2.064 * 60.98
> bm
[,1]
[1,] 32.89
and the interval is
> c(y0-bm,y0+bm)
[1]
1.0296 66.8097
Now we compute the prediction interval for the single future response.
> bm <- sqrt(1+x0 %*% xtxi %*% x0) *2.064 * 60.98
> c(y0-bm,y0+bm)
[1]
-96.17
164.01
What physically unreasonable feature do you notice about it? In such instances, impossible values
in the confidence interval can be avoided by transforming the response, say taking logs, (explained in a
later chapter) or by using a probability model more appropriate to the response. The normal distribution is
supported on the whole real line and so negative values are always possible. A better choice for this example
might be the Poisson distribution which is supported on the non-negative integers.
There is a more direct method for computing the CI. The function predict() requires that its second
argument be a data frame with variables named in the same way as the original dataset:
> predict(g,data.frame(Area=0.08,Elevation=93,Nearest=6.0,Scruz=12,
Adjacent=0.34),se=T)
$fit:
33.92
3.6. ORTHOGONALITY
41
$se.fit:
15.934
$df:
[1] 24
$residual.scale:
[1] 60.975
The width of the mean response interval can then be calculated by multiplying the se for the fit by the
appropriate t-critical value:
> 15.934*2.064
[1] 32.888
which matches what we did before. CI’s for the single future response could also be derived.
3.6
Orthogonality
¡
Suppose we can partition X in two, X
X1 X2 such that X T X2
0. So now
¤
¡
1
¤
¢
Y
X β
ε X β
β
ε
1 1
X2 2
¤
¤
and
§
§
X T X
X
X T X
X T X
1
1
X T
1
2
1
1
0
¤
X T X
X
¤
0
X T X
2
1
X T
2
2
2
2
¨
¨
which means
ˆβ
¡
1
¡
1
1
X T
X T
X T
X T
¤
1 X1 ¢¡
1 y
ˆβ2 ¤
2 X2 ¢¡
2 y
Notice that ˆβ1 will be the same regardless of whether X2 is in the model or not (and vice versa). Now if we
wish to test H0 : β1
0, it should be noted that RSSΩ d f
ˆ
σ2
¤
¤
Ω will be different depending on whether X2
is included in the model or not but the difference in F is not liable to be so large as in non-orthogonal cases.
Orthogonality is a desirable property but will only occur when X is chosen by the experimenter (it is a
feature of a good design). In observational data, we do not have direct control over X which is the source of
much of the interpretational difficulties associated with non-experimental data.
Here’s an example of an experiment to determine the effects of column temperature, gas/liquid ratio and
packing height in reducing unpleasant odor of chemical product that was being sold for household use.
Read the data in and display.
> data(odor)
> odor
odor temp gas pack
1
66
-1
-1
0
2
39
1
-1
0
3
43
-1
1
0
4
49
1
1
0
5
58
-1
0
-1
3.6. ORTHOGONALITY
42
6
17
1
0
-1
7
-5
-1
0
1
8
-40
1
0
1
9
65
0
-1
-1
10
7
0
1
-1
11
43
0
-1
1
12
-22
0
1
1
13
-31
0
0
0
14
-35
0
0
0
15
-26
0
0
0
The three predictors have been transformed from their original scale of measurement, for example temp
= (Fahrenheit-80)/40 so the original values of the predictor were 40,80 and 120. I don’t know the scale of
measurement for odor.
Here’s the X-matrix:
> x <- as.matrix(cbind(1,odor[,-1]))
and X T X :
> t(x) %*% x
1 temp gas pack
1 15
0
0
0
temp
0
8
0
0
gas
0
0
8
0
pack
0
0
0
8
The matrix is diagonal. What would happen if temp was measured in the original Fahrenheit scale? The
matrix would still be diagonal but the entry corresponding to temp would change.
Now fit a model:
> g <- lm(odor ˜ temp + gas + pack, data=odor)
> summary(g,cor=T)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
15.2
9.3
1.63
0.13
temp
-12.1
12.7
-0.95
0.36
gas
-17.0
12.7
-1.34
0.21
pack
-21.4
12.7
-1.68
0.12
Residual standard error: 36 on 11 degrees of freedom
Multiple R-Squared: 0.334,
Adjusted R-squared: 0.152
F-statistic: 1.84 on 3 and 11 degrees of freedom,
p-value: 0.199
Correlation of Coefficients:
(Intercept)
temp gas
temp
-1.52e-17
gas
-1.52e-17 4.38e-17
pack
0.00e+00 0.00e+00
0
3.6. ORTHOGONALITY
43
Check out the correlation of the coefficients - why did that happen?. Notice that the standard errors for
the coefficients are equal due to the balanced design. Now drop one of the variables:
> g <- lm(odor ˜ gas + pack, data=odor)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
15.20
9.26
1.64
0.13
gas
-17.00
12.68
-1.34
0.20
pack
-21.37
12.68
-1.69
0.12
Residual standard error: 35.9 on 12 degrees of freedom
Multiple R-Squared: 0.279,
Adjusted R-squared: 0.159
F-statistic: 2.32 on 2 and 12 degrees of freedom,
p-value: 0.141
Which things changed - which stayed the same? The coefficients themselves do not change but the resid-
ual standard error does change slightly which causes small changes in the standard errors of the coefficients,
t-statistics and p-values, but nowhere near enough to change our qualitative conclusions.
That was data from an experiment so it was possible to control the values of the predictors to ensure
orthogonality. Now consider the savings data which is observational:
> g <- lm(sr ˜ pop15 + pop75 + dpi + ddpi, savings)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 28.566087
7.354516
3.88
0.00033
pop15
-0.461193
0.144642
-3.19
0.00260
pop75
-1.691498
1.083599
-1.56
0.12553
dpi
-0.000337
0.000931
-0.36
0.71917
ddpi
0.409695
0.196197
2.09
0.04247
Residual standard error: 3.8 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
Drop pop15 from the model:
> g <- update(g, . ˜ . - pop15)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.487494
1.427662
3.84
0.00037
pop75
0.952857
0.763746
1.25
0.21849
dpi
0.000197
0.001003
0.20
0.84499
ddpi
0.473795
0.213727
2.22
0.03162
3.7. IDENTIFIABILITY
44
Residual standard error: 4.16 on 46 degrees of freedom
Multiple R-Squared: 0.189,
Adjusted R-squared: 0.136
F-statistic: 3.57 on 3 and 46 degrees of freedom,
p-value: 0.0209
What changed? By how much? Pay particular attention to pop75. The effect has now become positive
whereas it was negative. Granted, in neither case is it significant, but it is not uncommon in other datasets
for such sign changes to occur and for them to be significant.
3.7
Identifiability
The least squares estimate is the solution to the normal equations:
X T X ˆβ
X T y
¤
where X is an n p matrix. If X T X is singular and cannot be inverted, then there will be infinitely many
solutions to the normal equations and ˆβ is at least partially unidentifiable.
Unidentifiability will occur when X is not of full rank — when its columns are linearly dependent. With
observational data, unidentifiability is usually caused by some oversight: Here are some examples:
1. A person’s weight is measured both in pounds and kilos and both variables are entered into the model.
2. For each individual we record no. of years of education K-12 and no. of years of post-HS education
and also the total no. of years of education and put all three variables into the model.
3. p ¥
n — more variables than cases. When p
n, we may perhaps estimate all the parameters, but
¤
with no degrees of freedom left to estimate any standard errors or do any testing. Such a model is
called saturated. When p ¥
n, then the model is called supersaturated. Oddly enough, such models
are considered in large scale screening experiments used in product design and manufacture, but there
is no hope of uniquely estimating all the parameters in such a model.
Such problems can be avoided by paying attention. Identifiability is more of an issue in designed experi-
ments. Consider a simple two sample experiment:
Response
Treatment
y1
yn
¢¡¢¡¢¡£
Control
yn 1
ym n
©
©
¢¡¢¡¢¡£
Suppose we try to model the response by an overall mean µ and group effects α1 and α2:
y
α
ε
j
µ
i
j
i
1 2
j
1
m
n
¤
¤
¤
¢¡¢¡¢¡
¡
£¥¤
¡
£¥¤
y
1
1
0
¡
1
¤
¡
¤
¡
ε
£¥¤
¡
¤
¡
¤
¡
1
¤
¡
¤
¡
¤
£
¡
¤
¡¢¡¢¡
¡¢¡¢¡
µ
¡
¤
¡
¤
¡
¤
y
1
1
0
¡¢¡¢¡
¡
n
¤
¡
¤
α
¢
¦
y
1
¤
¡¢¡¢¡
n 1
1
0
1
©
α2
¢
¦
¡¢¡¢¡
¢
¦
¢
¦
¡¢¡¢¡
¡
¡
¡
ε
y
m n
©
m n
1
0
1
©
¡
Now although X has 3 columns, it has only rank 2 — µ α α
1
2
are not identifiable and the normal
¢
equations have infinitely many solutions. We can solve this problem by imposing some constraints, µ
0
¤
or α
α
1
2
0 for example.
¤
3.7. IDENTIFIABILITY
45
Statistics packages handle non-identifiability differently. In the regression case above, some may return
error messages and some may fit models because rounding error may remove the exact identifiability. In
other cases, constraints may be applied but these may be different from what you expect.
Identifiability means that
1. You have insufficient data to estimate the parameters of interest or
2. You have more parameters than are necessary to model the data.
Here’s an example. Suppose we create a new variable for the savings dataset - the percentage of people
between 15 and 75:
> pa <- 100-savings$pop15-savings$pop75
and add that to the model:
> g <- lm(sr ˜ pa + pop15 + pop75 + dpi + ddpi, data=savings)
> summary(g)
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.41e+02
1.03e+02
-1.37
0.177
pa
1.69e+00
1.08e+00
1.56
0.126
pop15
1.23e+00
9.77e-01
1.26
0.215
dpi
-3.37e-04
9.31e-04
-0.36
0.719
ddpi
4.10e-01
1.96e-01
2.09
0.042
We get a message about one undefined coefficient because the rank of the design matrix X is 5 but should
be 6.
Let’s take a look at the X-matrix:
> x <- as.matrix(cbind(1,pa,savings[,-1]))
> dimnames(x) <- list(row.names(savings),c("int","pa","p15","p75",
"dpi","ddpi"))
If we didn’t know which linear combination was causing the trouble, how would we find out? An eigen
decomposition of X T X can help:
> e <- eigen(t(x) %*% x)
> signif(e$values,3)
[1] 1.10e+08 1.10e+05 3.19e+03 3.74e+02 1.37e+01 1.09e-14
> signif(e$vectors,3)
int
0.000506
0.0141 -0.00125
0.000603
0.00989
1.00e+00
pa
0.034300
0.7940
0.59700
0.098100 -0.05630 -1.00e-02
p15
0.014700
0.6040 -0.79500 -0.031000
0.04800 -1.00e-02
p75
0.001610
0.0164
0.07310 -0.006840
0.99700 -1.00e-02
dpi
0.999000 -0.0363 -0.00906 -0.001170 -0.00036 -1.07e-17
ddpi 0.001740
0.0594
0.08310 -0.995000 -0.01390 -4.97e-16
3.8. SUMMARY
46
Only the last eigenvalue is zero, indicating one linear combination is the problem. We can deter-
mine which linear combination from the last eigenvalue (last column of the matrix. From this we see
that 100-pa-p15-p75=0 is the offending combination.
Lack of identifiability is obviously a problem but it is usually easy to identify and work around. More
problematic are cases where we are close to unidentifiability. To demonstrate this, suppose we add a small
random perturbation to the third decimal place of pa by adding a random variate from U
0 005 0 005
¢
¡
¡
where U denotes the uniform distribution:
> pae <- pa +0.001*(runif(50)-0.5)
and now refit the model:
> ge <- lm(sr ˜ pae+pop15+pop75+dpi+ddpi,savings)
> summary(ge)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
1.57e+05
1.81e+05
0.87
0.391
pae
-1.57e+03
1.81e+03
-0.87
0.391
pop15
-1.57e+03
1.81e+03
-0.87
0.391
pop75
-1.57e+03
1.81e+03
-0.87
0.390
dpi
-3.34e-04
9.34e-04
-0.36
0.722
ddpi
4.11e-01
1.97e-01
2.09
0.042
Notice the now all parameters can be estimated but the standard errors are very large because we cannot
estimate them in a stable way. We deliberately caused this problem so we know the cause but in general we
need to be able to identify such situations. We do this in Chapter 9.
3.8
Summary
We have described a linear model y
X β
ε. The parameters β may be estimated using least squares
¤
ˆβ
¡
¡
X T X
1XT y. If we further assume that ε
N 0 σ2I then we can test any linear hypothesis about β,
¤
¢
¢
construct confidence regions for β, make predictions with confidence intervals.
3.9
What can go wrong?
Many things, unfortunately — we try to categorize them below:
3.9.1
Source and quality of the data
How the data was collected directly effects what conclusions we can draw.
1. We may have a biased sample, such as a sample of convenience, from the population of interest. This
makes it very difficult to extrapolate from what we see in the sample to general statements about
the population. As we have seen, in some cases the sample is the population, in which case any
generalization of the conclusions is problematic.
2. Important predictors may not have been observed. This means that our predictions may be poor or we
may misinterpret the relationship between the predictors and the response.
3.9. WHAT CAN GO WRONG?
47
3. Observational data make causal conclusions problematic — lack of orthogonality makes disentangling
effects difficult. Missing predictors add to this problem.
4. The range and qualitative nature of the data may limit effective predictions. It is unsafe to extrapolate
too much. Carcinogen trials may apply large doses to mice. What do the results say about small
doses applied to humans? Much of the evidence for harm from substances such as asbestos and radon
comes from people exposed to much larger amounts than that encountered in a normal life. It’s clear
that workers in older asbestos manufacturing plants and uranium miners suffered from their respective
exposures to these substances, but what does that say about the danger to you or I?
3.9.2
Error component
¡
We hope that ε
N 0 σ2I but
¢
1. Errors may be heterogeneous (unequal variance).
2. Errors may be correlated.
3. Errors may not be normally distributed.
The last defect is less serious than the first two because even if the errors are not normal, the ˆβ’s will
tend to normality due to the power of the central limit theorem. With larger datasets, normality of the data
is not much of a problem.
3.9.3
Structural Component
The structural part of linear model, Ey
X β may be incorrect. The model we use may come from different
¤
sources:
1. Physical theory may suggest a model, for example Hooke’s law says that the extension of a spring
is proportional to the weight attached. Models like these usually arise in the physical sciences and
engineering.
2. Experience with past data. Similar data used in the past was modeled in a particular way. It’s natural
to see if the same model will work the current data. Models like these usually arise in the social
sciences.
3. No prior idea - the model comes from an exploration of the data itself.
Confidence in the conclusions from a model declines as we progress through these. Models that derive
directly from physical theory are relatively uncommon so that usually the linear model can only be regarded
as an approximation to a reality which is very complex.
Most statistical theory rests on the assumption that the model is correct. In practice, the best one can
hope for is that the model is a fair representation of reality. A model can be no more than a good portrait.
All models are wrong but some are useful. George Box
is only a slight exaggeration. Einstein said
So far as theories of mathematics are about reality; they are not certain; so far as they are certain,
they are not about reality.
3.10. INTERPRETING PARAMETER ESTIMATES
48
3.10
Interpreting Parameter Estimates
Suppose we fit a model to obtain the regression equation:
ˆ
y
ˆβ
ˆβ
ˆβ
0
1x1
pxp
¤
¡¢¡¢¡
What does ˆβ1 mean? In some case, a β might represent a real physical constant, but often the statistical
model is just a convenience for representing a complex reality and so the real meaning of a particular β is
not obvious.
Let’s start with a naive interpretation: “A unit change in x1 will produce a change of ˆβ1 in the response”.
For a properly designed experiment, this interpretation is reasonable provided one pays attention to
concerns such as extrapolation and appropriateness of the model selected. The effects of other variables
that are included in the experiment can separated out if an orthogonal design is used. For variables not
included in the experiment by choice, we may eliminate their effect by holding them constant. If variables
that impact the response are not included because they are not known, we use randomization to control their
effect. The treatments (predictor values) are assigned to the experimental units or subjects at random. This
ensures that these unknown variables will not be correlated in expectation with the predictors we do examine
and allows us to come to causal conclusions. These unknown predictors do not, on the average, affect
the parameter estimates of interest, but they do contribute to the residual standard error so it’s sometimes
better to incorporate them in the experimental design if the become known, as this allows for more precise
inference.
In a few tightly controlled experiments, it is possible to claim that measurement error is the only kind
of error but usually some of the “error” actually comes from the effects of unmeasured variables. We can
decompose the usual model as follows:
y
X β
ε
¤
X β
Zγ
δ
¤
where Z are unincluded predictors and δ is measurement error in the response. We can assume that Eε
0
¤
without any loss of generality, because if Eε
c, we could simply redefine β0 as β0
c and the error would
¤
again have expectation zero. This is another reason why it is generally unwise to remove the intercept
term from the model since it acts as a sink for the mean effect of unincluded variables. So we see that ε
incorporates both measurement error and the effect of other variables. In a designed experiment, provided
¡
the assignment of the experimental units is random, we have cor X Z
0 so that the estimate of β is
¢
¤
unaffected in expectation by the presence of Z.
For observational data, no randomization can be used in assigning treatments to the units and orthogo-
nality won’t just happen. There are serious objections to any causal conclusions. An inference of causality
is often desired but this is usually too much to expect from observational data. An unmeasured and possible
unsuspected “lurking” variable Z may be the real cause of an observed relationship between y and X . See
Figure 3.4. For example, we will observe a positive correlation among the shoe sizes and reading abilities
of elementary school students but this relationship is driven by a lurking variable — the age of the child.
¡
So in observational studies, because we have no control over the assignment of units, we have cor X Z ¢¡ ¤
0 and the observed or worse, unobserved, presence of Z causes us great difficulty. In Figure 3.5, we see the
effect of possible confounding variables demonstrated.
In observational studies, it is important to adjust for the effects of possible confounding variables such
as the Z shown in Figure 3.5. If such variables can be identified, then at least their effect can be interpreted.
Unfortunately, one can never be sure that the all relevant Z have been identified.
3.10. INTERPRETING PARAMETER ESTIMATES
49
Z
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¢
¥
¦
¢¤£
¡
X
Y
Figure 3.4: Is the relationship between x and y, really caused by z?
What if all relevant variables have been measured? In other words, suppose there are no unidentified
lurking variables. Even then the naive interpretation does not work. Consider
y
ˆβ
ˆβ
ˆβ
0
1x1
2x2
¤
but suppose we change x2
x1
x2 then
§
¡
¡
y
ˆβ
ˆβ
ˆβ
ˆβ
0
1
2 x1
2 x1
x2
¤
¢
¢
The coefficient for x1 has changed. Interpretation cannot be done separately for each variable. This is a
practical problem because it is not unusual for the predictor of interest, x1 in this example, to be mixed up
in some way with other variables like x2.
Let’s try a new interpretation:
“ ˆβ1 is the effect of x1 when all the other (specified) predictors are held constant”.
This too has problems. Often in practice, individual variables cannot be changed without changing
others too. For example, in economics we can’t expect to change tax rates without other things changing
too. Furthermore, this interpretation requires the specification of the other variables - changing which other
variables are included will change the interpretation. Unfortunately, there is no simple solution.
Just to amplify this consider the effect of pop75 on the savings rate in the savings dataset. I’ll fit four
different models, all including pop75 but varying the inclusion of other variables.
> g <- lm(sr ˜ pop15 + pop75 + dpi + ddpi, data=savings)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 28.566087
7.354516
3.88
0.00033
pop15
-0.461193
0.144642
-3.19
0.00260
pop75
-1.691498
1.083599
-1.56
0.12553
dpi
-0.000337
0.000931
-0.36
0.71917
ddpi
0.409695
0.196197
2.09
0.04247
Residual standard error: 3.8 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
3.10. INTERPRETING PARAMETER ESTIMATES
50
Y
Y
Y
A
B
C
Z=1
Z=1
Z=1
Z=0
Z=0
Z=0
X
X
X
Figure 3.5: Possible confounding effects illustrated. Imagine the data is observed within the ellipses. If the
effect of Z is ignored, a strong positive correlation between X and Y is observed in all three cases. In panel
A, we see that when we allow for the effect of Z by observing the relationship between X and Y separately
within each level of Z, that the relationship remains a positive correlation. In panel B, after allowing for Z,
there is no correlation between X and Y, while in panel C, after allowing for Z, the relationship becomes a
negative correlation.
It is perhaps surprising that pop75 is not significant in this model. However, pop75 is negatively correlated
with pop15 since countries with proportionately more younger people are likely to relatively fewer older
ones and vice versa. These two variables are both measuring the nature of the age distribution in a country.
When two variables that represent roughly the same thing are included in a regression equation, it is not
unusual for one (or even both) of them to appear insignificant even though prior knowledge about the effects
of these variables might lead one to expect them to be important.
> g2 <- lm(sr ˜ pop75 + dpi + ddpi, data=savings)
> summary(g2)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.487494
1.427662
3.84
0.00037
pop75
0.952857
0.763746
1.25
0.21849
dpi
0.000197
0.001003
0.20
0.84499
ddpi
0.473795
0.213727
2.22
0.03162
Residual standard error: 4.16 on 46 degrees of freedom
Multiple R-Squared: 0.189,
Adjusted R-squared: 0.136
F-statistic: 3.57 on 3 and 46 degrees of freedom,
p-value: 0.0209
We note that the income variable dpi and pop75 are both not significant in this model and yet one might
expect both of them to have something to do with savings rates. Higher values of these variables are both
associated with wealthier countries. Let’s see what happens when we drop dpi from the model:
> g3 <- lm(sr ˜ pop75 + ddpi, data=savings)
> summary(g3)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
5.470
1.410
3.88
0.00033
3.10. INTERPRETING PARAMETER ESTIMATES
51
pop75
1.073
0.456
2.35
0.02299
ddpi
0.464
0.205
2.26
0.02856
Residual standard error: 4.12 on 47 degrees of freedom
Multiple R-Squared: 0.188,
Adjusted R-squared: 0.154
F-statistic: 5.45 on 2 and 47 degrees of freedom,
p-value: 0.00742
Now pop75 is statistically significant with a positive coefficient. We try dropping ddpi:
> g4 <- lm(sr ˜ pop75, data=savings)
> summary(g4)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.152
1.248
5.73
6.4e-07
pop75
1.099
0.475
2.31
0.025
Residual standard error: 4.29 on 48 degrees of freedom
Multiple R-Squared:
0.1,
Adjusted R-squared: 0.0814
F-statistic: 5.34 on 1 and 48 degrees of freedom,
p-value: 0.0251
The coefficient and p-value do not change much here due to the low correlation between pop75 and ddpi.
Compare the coefficients and p-values for pop75 throughout. Notice how the sign and significance
change in Table3.2.
No. of Preds
Sign
Significant?
4
-
no
3
+
no
2
+
yes
1
+
yes
Table 3.2: Sign and Significance of ˆβpop75
We see that the significance and the direction of the effect of pop75 change according to what other
variables are also included in the model. We see that no simple conclusion about the effect of pop75 is
possible. We must find interpretations for a variety of models. We certainly won’t be able to make any
causal conclusions.
In observational studies, there are steps one can take to make a stronger case for causality:
1. Try to include all relevant variables
2. Use non-statistical knowledge of the physical nature of the relationship.
3. Try a variety of models - see if a similar effect is observed. Is ˆβ1 similar, no matter what the model?
4. Multiple studies under different conditions can help confirm a relationship. The connection between
smoking and lung cancer was suspected since the early 50’s but other explanations for the effect were
proposed. It was many years before other plausible explanations were eliminated.
The news media often jump on the results of a single study but one should be suspicious of these one
off results. Publication bias is a problem. Many scientific journal will not publish the results of a
3.10. INTERPRETING PARAMETER ESTIMATES
52
study whose conclusions do not reject the null hypothesis. If different researchers keep studying the
same relationship, sooner or later one of them will come up with a significant effect even if one really
doesn’t exist. It’s not easy to find out about all the studies with negative results so it is easy to make
the wrong conclusions.
Another source of bias is that researchers have a vested interest in obtaining a positive result. There
is often more than one way to analyze the data and the researchers may be tempted to pick the one
that gives them the results they want. This is not overtly dishonest but it does lead to a bias towards
positive results.
It’s difficult to assess the evidence in these situations and one can never be certain. The history of the
study of the link between smoking and lung cancer shows that it takes a great deal of effort to progress
beyond the observation of an association to strong evidence of causation. One can never be 100% sure.
An alternative approach is recognize that the parameters and their estimates are fictional quantities in
most regression situations. The “true” values may never be known (if they even exist in the first place).
Instead concentrate on predicting future values - these may actually be observed and success can then be
measured in terms of how good the predictions were.
Consider a prediction made using each of the four models above:
> x0 <- data.frame(pop15=32,pop75=3,dpi=700,ddpi=3)
> predict(g,x0)
[1] 9.7267
> predict(g2,x0)
[1] 9.9055
> predict(g3,x0)
[1] 10.078
> predict(g4,x0)
[1] 10.448
Prediction is more stable than parameter estimation. This enables a rather cautious interpretation of ˆβ1.
Suppose the predicted value of y is ˆ
y for given x1 and for other given predictor values. Now suppose we
observe x1
1 and the same other given predictor values then the predicted response is increased by ˆβ1.
Notice that I have been careful to not to say that we have taken a specific individual and increased their x1
by 1, rather we have observed a new individual with predictor x1
1. To put it another way, people with
yellow fingers tend to be smokers but making someone’s fingers yellow won’t make them more likely to
smoke.
Prediction is conceptually simpler since interpretation is not an issue but you do need to worry about
extrapolation.
1. Quantitative extrapolation: Is the new x0 within the range of validity of the model. Is it close to
the range of the original data? If not, the prediction may be unrealistic. Confidence intervals for
predictions get wider as we move away from the data. We can compute these bands for our last
model:
> grid <- seq(0,10,0.1)
> p <- predict(g4,data.frame(pop75=grid),se=T)
> cv <- qt(0.975,48)
> matplot(grid,cbind(p$fit,p$fit-cv*p$se,p$fit+cv*p$se),lty=c(1,2,2),
type="l",xlab="pop75",ylab="Saving")
> rug(savings$pop75)
3.10. INTERPRETING PARAMETER ESTIMATES
53
We see that the confidence bands in Figure 3.6 become wider as we move away from the range of the
data. However, this widening does not reflect the possibility that the structure of the model itself may
change as we move into new territory. The uncertainty in the parametric estimates is allowed for but
not uncertainty about the model itself. In Figure 3.7, we see that a model may fit well in the range of
the data, but outside of that range, the predictions may be very bad.
25
20
15
Saving
10
5
0
2
4
6
8
10
pop75
Figure 3.6: Predicted pop75 over a range of values with 95% pointwise confidence bands for the mean
response shown as dotted lines. A “rug” shows the location of the observed values of pop75
2. Qualitative extrapolation: Is the new x0 drawn from the same population from which the original
sample was drawn. If the model was built in the past and is to be used for future predictions, we must
make a difficult judgment as to whether conditions have remained constant enough for this to work.
Let’s end with a quote from the 4th century. Prediction is a tricky business — perhaps the only thing
worse than a prediction is no prediction at all.
The good Christian should beware of mathematicians and all those who make empty prophecies.
The danger already exists that mathematicians have made a covenant with the devil to darken
the spirit and confine man in the bonds of Hell. - St. Augustine
3.10. INTERPRETING PARAMETER ESTIMATES
54
2.0
Range of the data
1.0
y
0.0
−1.0
0.0
0.5
1.0
x
Figure 3.7: Dangers of extrapolation: The model is shown in solid, the real relationship by the dotted line.
The data all lie in the predictor range [0,1]
Chapter 4
Errors in Predictors
The regression model Y
X β
ε allows for Y being measured with error by having the ε term, but what if
¤
the X is measured with error? In other words, what if the X we see is not the X used to generate Y ?
¡
Consider the simple regression xi yi for i
1
n.
¢
¤
¢¡¢¡¢¡
y
η
ε
i
i
i
¤
x
ξ
δ
i
i
i
¤
where the errors ε and δ are independent. Suppose the true underlying relationship is
η
β
β ξ
i
0
1 i
¤
¡
but we only see xi yi . Putting it together, we get
¢
¡
y
β
β
ε β δ
i
0
1xi
i
1 i
¤
¢
Suppose we use least squares to estimate β
σ2
0 and β1. Let’s assume E εi
Eδi
0 and that var εi
,
¤
¤
¤
var δ
σ2
i ¤
δ. Let
σ2
¯2
¡
ξ
∑¡ ξ ξ
ξ δ
i
n
σξδ cov
¤
¢
¤
¢
where ξ are the true values of X and not random variables but we could (theoretically since they are not
observed) compute statistics using their values. Now ˆβ
2
1
∑¡ xi ¯x yi ∑¡ xi ¯x and after some calculation
¤
¢
¢
we find that
¡
σ2ξ σξδ
¢
E ˆβ
β
1
1 ¡
σ2ξ σ2 2σξδ
δ
¢
If there is no relation between ξ and δ, this simplifies to
σ2ξ
1
E ˆβ
β
β
1
1 ¡
1
¤
σ2ξ σ2
1
σ2 σ2
δ¢
δ
ξ
So in general ˆβ1 will be biased (regardless of the sample size and typically towards zero). If σ2δ is small
relative to σ2ξ then the problem can be ignored. In other words, if the variability in the errors of observation
of X are small relative to the range of X then we need not be concerned. If not, it’s a serious problem and
other methods such as fitting using orthogonal rather than vertical distance in the least squares fit should be
considered.
55
CHAPTER 4. ERRORS IN PREDICTORS
56
For prediction, measurement error in the x’s is not such a problem since the same error will apply to the
new x0 and the model used will be the right one.
For multiple predictors, the usual effect of measurement errors is to bias the ˆβ in the direction of zero.
One should not confuse the errors in predictors with treating X as a random variable. For observational
data, X could be regarded as a random variable, but the regression inference proceeds conditional on a fixed
value for X . We make the assumption that the Y is generated conditional on the fixed value of X . Contrast
this with the errors in predictors case where the X we see is not the X that was used to generate the Y .
For real data, the true values of the parameters are usually never known, so it’s hard to know how well
the estimation is working. Here we generate some artificial data from a known model so we know the true
values of the parameters and we can tell how well we do: runif() generates uniform random numbers
and rnorm() generates standard normal random numbers.
Because you will get different random numbers, your results will not exactly match mine if you try to
duplicate this.
> x <- 10*runif(50)
> y <- x+rnorm(50)
> gx <- lm(y ˜ x)
> summary(gx)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-0.1236
0.2765
-0.45
0.66
x
0.9748
0.0485
20.09
<2e-16
Residual standard error: 1.02 on 48 degrees of freedom
Multiple R-Squared: 0.894,
Adjusted R-squared: 0.891
F-statistic:
403 on 1 and 48 degrees of freedom,
p-value:
0
True values of the regression coeffs are 0 and 1 respectively. What happens when we add some noise to
the predictor?
> z <- x + rnorm(50)
> gz <- lm(y ˜ z)
> summary(gz)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
0.3884
0.3248
1.2
0.24
z
0.8777
0.0562
15.6
<2e-16
Residual standard error: 1.27 on 48 degrees of freedom
Multiple R-Squared: 0.835,
Adjusted R-squared: 0.832
F-statistic:
244 on 1 and 48 degrees of freedom,
p-value:
0
Compare the results - notice how the slope has decreased. Now add even more noise:
> z2 <- x+5*rnorm(50)
> gz2 <- lm(y ˜ z2)
> summary(gz2)
Coefficients:
CHAPTER 4. ERRORS IN PREDICTORS
57
Estimate Std. Error t value Pr(>|t|)
(Intercept)
2.736
0.574
4.77
1.8e-05
z2
0.435
0.101
4.32
7.7e-05
Residual standard error: 2.66 on 48 degrees of freedom
Multiple R-Squared: 0.28,
Adjusted R-squared: 0.265
F-statistic: 18.7 on 1 and 48 degrees of freedom,
p-value: 7.72e-05
Compare again — the slope is now very much smaller. We can plot all this information in Figure 4.1.
> matplot(cbind(x,z,z2),y,xlab="x",ylab="y")
> abline(gx,lty=1)
> abline(gz,lty=2)
> abline(gz2,lty=5)
10
3
1
1
2
3
3
1
2
2
2
3 11
1
23
8
2
3 3 12 31
122
3
3
2
3
1
2
1 1
3 2
3
2
2
1 1
2 3 1
3
2 3
3
3
3
11
6
2 1 2 2 2 1
3 1
12 3 12
3
3
1
2
12
2
1
3 21
2
3
3
2 13
y
1
2
3
4
3
3
1 2
3 1
122 1
3 1
1
2
2
3
3
2
3
2 1
1
3
3 22 13 1
1 2 22
33
231
1 22
3
3
1
0
2
3
3
2 1 3
11
1 223
3
2
1
3
−5
0
5
10
x
Figure 4.1: Original x shown with “1”, with small error as “2” and with large error as “3”. The regression
lines for the no measurement error, small error and large error are shown as solid, dotted and dashed lines
respectively.
This was just one realization - to get an idea of average behavior we need to repeat the experiment (I’ll
do it 1000 times here). The slopes from the 1000 experiments are saved in the vector bc:
> bc <- numeric(1000)
> for(i in 1:1000){
+ y <- x + rnorm(50)
+ z <- x + 5*rnorm(50)
+ g <- lm(y ˜ z)
+ bc[i] <- g$coef[2]
+ }
Now look at the distribution of bc.
CHAPTER 4. ERRORS IN PREDICTORS
58
> summary(bc)
Min. 1st Qu.
Median
Mean 3rd Qu.
Max.
-0.0106
0.2220
0.2580
0.2580
0.2950
0.4900
Given that the variance of a standard uniform random variable is 1/12, σ2δ
25 and σ2
100 12, we’d
¤
ξ ¤
expect the mean to be 0.25. Remember that there is some simulation variation and the expression for the
bias is only approximation, so we don’t expect them to match exactly.
Chapter 5
Generalized Least Squares
5.1
The general case
Until now we have assumed that var ε
σ2I but it can happen that the errors have non-constant variance or
¤
are correlated. Suppose instead that var ε
σ2Σ where σ2 is unknown but Σ is known — in other words we
¤
know the correlation and relative variance between the errors but we don’t know the absolute scale.
Generalized least squares minimizes
¡
y
X β T Σ 1 ¡ y
X β
¢
¢
which is solved by
ˆβ
¡
X T Σ 1X
1XT Σ 1y
¤
¢
Since we can write Σ
SST , where S is a triangular matrix using the Choleski Decomposition, we have
¤
¡
¡
y
X β T S T S 1 ¡ y
X β
S 1y
S 1X β T ¡ S 1y
S 1X β
¢
¢
¤
¢
¢
So GLS is like regressing S 1X on S 1y. Furthermore
y
X β
ε
¤
S 1y
S 1X β
S 1ε
¤
y
X β
ε
¤
So we have a new regression equation y
X β
ε where if we examine the variance of the new errors, ε
¤
we find
¡
var ε
var S 1ε
S 1 ¡ var ε S T
S 1σ2SST S T
σ2I
¤
¢
¤
¢
¤
¤
So the new variables y and X are related by a regression equation which has uncorrelated errors with
equal variance. Of course, the practical problem is that Σ may not be known.
We find that
¡
var ˆβ
X T Σ 1X
1σ2
¤
¢
¡
To illustrate this we’ll use a built-in R dataset called Longley’s regression data where the response is
number of people employed, yearly from 1947 to 1962 and the predictors are GNP implicit price deflator
(1954=100), GNP, unemployed, armed forces, noninstitutionalized population 14 years of age and over, and
year. The data originally appeared in Longley (1967)
Fit a linear model.
59
5.1. THE GENERAL CASE
60
> data(longley)
> g <- lm(Employed ˜ GNP + Population, data=longley)
> summary(g,cor=T)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
88.9388
13.7850
6.45
2.2e-05
GNP
0.0632
0.0106
5.93
5.0e-05
Population
-0.4097
0.1521
-2.69
0.018
Residual standard error: 0.546 on 13 degrees of freedom
Multiple R-Squared: 0.979,
Adjusted R-squared: 0.976
F-statistic:
304 on 2 and 13 degrees of freedom,
p-value: 1.22e-11
Correlation of Coefficients:
(Intercept)
GNP
GNP
0.985
Population
-0.999 -0.991
Compare the correlation between the variables gnp, pop and their corresponding coefficients. What do
you notice?
In data collected over time such as this, successive errors could be correlated. Assuming that the errors
take a simple autoregressive form:
ε
ρε
δ
i 1
i
i
¤
©
¡
where δ
τ2
i
N 0
. We can estimate this correlation ρ by
¢
> cor(g$res[-1],g$res[-16])
[1] 0.31041
Under this assumption Σ
ρ i j
i j
. For simplicity, lets assume we know that ρ
0 31041. We now construct
¤
¤
¡
the Σ matrix and compute the GLS estimate of β along with its standard errors.
> x <- model.matrix(g)
> Sigma <- diag(16)
> Sigma <- 0.31041ˆabs(row(Sigma)-col(Sigma))
> Sigi <- solve(Sigma)
> xtxi <- solve(t(x) %*% Sigi %*% x)
> beta <- xtxi %*% t(x) %*% Sigi %*% longley$Empl
> beta
[,1]
[1,] 94.89889
[2,]
0.06739
[3,] -0.47427
> res <- longley$Empl - x %*% beta
> sig <- sqrt(sum(resˆ2)/g$df)
> sqrt(diag(xtxi))*sig
[1] 14.157603
0.010867
0.155726
Compare with the model output above where the errors are assumed to be uncorrelated.
Another way to get the same result is to regress S 1y on S 1x as we demonstrate here:
5.1. THE GENERAL CASE
61
> sm <- chol(Sigma)
> smi <- solve(t(sm))
> sx <- smi %*% x
> sy <- smi %*% longley$Empl
> lm(sy ˜ sx-1)$coef
sx(Intercept)
sxGNP
sxPopulation
94.89889
0.06739
-0.47427
In practice, we would not know that the ρ
0 31 and we’d need to estimate it from the data. Our initial
¤
¡
estimate is 0.31 but once we fit our GLS model we’d need to re-estimate it as
> cor(res[-1],res[-16])
[1] 0.35642
and then recompute the model again with ρ
0 35642 . This process would be iterated until conver-
¤
¡
gence.
The nlme library contains a GLS fitting function. We can use it to fit this model:
> library(nlme)
> g <- gls(Employed ˜ GNP + Population,
correlation=corAR1(form= ˜Year), data=longley)
> summary(g)
Correlation Structure: AR(1)
Formula: ˜Year
Parameter estimate(s):
Phi
0.64417
Coefficients:
Value Std.Error t-value p-value
(Intercept) 101.858
14.1989
7.1736
<.0001
GNP
0.072
0.0106
6.7955
<.0001
Population
-0.549
0.1541 -3.5588
0.0035
Residual standard error: 0.68921
Degrees of freedom: 16 total; 13 residual
We see that the estimated value of ρ is 0.64. However, if we check the confidence intervals for this:
> intervals(g)
Approximate 95% confidence intervals
Coefficients:
lower
est.
upper
(Intercept) 71.183204 101.858133 132.533061
GNP
0.049159
0.072071
0.094983
Population
-0.881491
-0.548513
-0.215536
5.2. WEIGHTED LEAST SQUARES
62
Correlation structure:
lower
est.
upper
Phi -0.44335 0.64417 0.96451
Residual standard error:
lower
est.
upper
0.24772 0.68921 1.91748
we see that it is not significantly different from zero.
5.2
Weighted Least Squares
Sometimes the errors are uncorrelated, but have unequal variance where the form of the inequality is known.
Weighted least squares (WLS) can be used in this situation. When Σ is diagonal, the errors are uncorrelated
¡
but do not necessarily have equal variance. We can write Σ
diag 1 w1
1 wn , where the wi are the
¤
¢
¢¡¢¡¢¡
¡
¡
weights so S
diag
1 w1
1 wn . So we can regress
wixi on
wiyi (although the column of ones
¤
¢
¢
¢
¢¡¢¡¢¡
in the X-matrix needs to be replaced with
wi. Cases with low variability should get a high weight, high
¢
variability a low weight. Some examples
¡
1. Errors proportional to a predictor: var ε ∝
1
i
xi suggests wi
x
¢
¤
i
2. Y
σ2
i are the averages of ni observations then var yi
var εi
ni suggests wi
ni.
¤
¤
¤
Here’s an example from an experiment to study the interaction of certain kinds of elementary particles
on collision with proton targets. The experiment was designed to test certain theories about the nature of
the strong interaction. The cross-section(crossx) variable is believed to be linearly related to the inverse
of the energy(energy - has already been inverted). At each level of the momentum, a very large num-
ber of observations were taken so that it was possible to accurately estimate the standard deviation of the
response(sd).
Read in and check the data:
> data(strongx)
> strongx
momentum energy
crossx sd
1
4
0.345
367 17
2
6
0.287
311
9
3
8
0.251
295
9
4
10
0.225
268
7
5
12
0.207
253
7
6
15
0.186
239
6
7
20
0.161
220
6
8
30
0.132
213
6
9
75
0.084
193
5
10
150
0.060
192
5
Define the weights and fit the model:
> g <- lm(crossx ˜ energy, strongx, weights=sdˆ-2)
> summary(g)
5.2. WEIGHTED LEAST SQUARES
63
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
148.47
8.08
18.4
7.9e-08
energy
530.84
47.55
11.2
3.7e-06
Residual standard error: 1.66 on 8 degrees of freedom
Multiple R-Squared: 0.94,
Adjusted R-squared: 0.932
F-statistic:
125 on 1 and 8 degrees of freedom,
p-value: 3.71e-06
Try fitting the regression without weights and see what the difference is.
> gu <- lm(crossx ˜ energy, strongx)
> summary(gu)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
135.0
10.1
13.4
9.2e-07
energy
619.7
47.7
13.0
1.2e-06
Residual standard error: 12.7 on 8 degrees of freedom
Multiple R-Squared: 0.955,
Adjusted R-squared: 0.949
F-statistic:
169 on 1 and 8 degrees of freedom,
p-value: 1.16e-06
The two fits can be compared
> plot(crossx ˜ energy, data=strongx)
> abline(g)
> abline(gu,lty=2)
and are shown in Figure 5.1.
350
300
250
Cross−section
200
0.05
0.15
0.25
0.35
Energy
Figure 5.1: Weighted least square fit shown in solid. Unweighted is dashed.
5.3. ITERATIVELY REWEIGHTED LEAST SQUARES
64
The unweighted fit appears to fit the data better overall but remember that for lower values of energy,
the variance in the response is less and so the weighted fit tries to catch these points better than the others.
5.3
Iteratively Reweighted Least Squares
In cases, where the form of the variance of ε is not completely known, we may model Σ using a small
number of parameters. For example,
var ε
γ
γ
i
0
1x1
¤
might seem reasonable in a given situation. The IRWLS fitting Algorithm is
1. Start with wi
1
¤
2. Use least squares to estimate β.
3. Use the residuals to estimate γ, perhaps by regressing ˆε2 on x.
4. Recompute the weights and goto 2.
Continue until convergence. There are some concerns about this — how is subsequent inference about
β affected? Also how many degrees of freedom do we have? More details may be found in Carroll and
Ruppert (1988).
An alternative approach is to model the variance and jointly estimate the regression and weighting
parameters using likelihood based method. This can be implemented in R using the gls() function in
the nlme library.
Chapter 6
Testing for Lack of Fit
How can we tell if a model fits the data? If the model is correct then ˆ
σ2 should be an unbiased estimate of
σ2. If we have a model which is not complex enough to fit the data or simply takes the wrong form, then ˆσ2
will overestimate σ2. An example can be seen in Figure 6.1. Alternatively, if our model is too complex and
overfits the data, then ˆ
σ2 will be an underestimate.
2.05
1.95
y
1.85
1.75
0.0
0.2
0.4
0.6
0.8
1.0
x
Figure 6.1: True quadratic fit shown with the solid line and incorrect linear fit shown with the dotted line.
Estimate of σ2 will be unbiased for the quadratic model but far too large for the linear model
This suggests a possible testing procedure — we should compare ˆ
σ2 to σ2. There are two cases — one
where σ2 is known and one where it is not.
65
6.1. σ2 KNOWN
66
6.1
σ2 known
σ2 known may be known from past experience, knowledge of the measurement error inherent in an instru-
ment or by definition. Recall (from Section 3.4) that
ˆ
σ2
χ2n p
σ
2
¡
n
p
¢
which leads to the test: Conclude there is a lack of fit if
¡
n
p ˆ
σ2
1 α
¢
¡
¥
χ2
σ
2
n p
If a lack of fit is found, then a new model is needed.
Continuing with the same data as in the weighted least squares example we test to see if a linear model is
adequate. In this example, we know the variance almost exactly because each response value is the average
of a large number of observations. Because of the way the weights are defined, wi
1 var yi, the known
¤
variance is implicitly equal to one. There is nothing special about one - we could define wi
99 var yi and
¤
the variance would be implicitly 99. However, we would get essentially the same result as the following
analysis.
> data(strongx)
> g <- lm(crossx ˜ energy, weights=sdˆ-2, strongx)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
148.47
8.08
18.4
7.9e-08
energy
530.84
47.55
11.2
3.7e-06
Residual standard error: 1.66 on 8 degrees of freedom
Multiple R-Squared: 0.94,
Adjusted R-squared: 0.932
F-statistic:
125 on 1 and 8 degrees of freedom,
p-value: 3.71e-06
Examine the R2 - do you think the model is a good fit?
Now plot the data and the fitted regression line (shown as a solid line on Figure 6.2.
> plot(strongx$energy,strongx$crossx,xlab="Energy",ylab="Crossection")
> abline(g$coef)
Compute the test statistic and the p-value:
> 1.66ˆ2*8
[1] 22.045
> 1-pchisq(22.045,8)
[1] 0.0048332
We conclude that there is a lack of fit. Just because R2 is large does not mean that you can not do better.
Add a quadratic term to the model and test again:
6.2. σ2 UNKNOWN
67
350
300
Crossection
250
200
0.05
0.10
0.15
0.20
0.25
0.30
0.35
Energy
Figure 6.2: Linear and quadratic fits to the physics data
> g2 <- lm(crossx ˜ energy + I(energyˆ2), weights=sdˆ-2, strongx)
> summary(g2)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
183.830
6.459
28.46
1.7e-08
energy
0.971
85.369
0.01
0.99124
I(energyˆ2) 1597.505
250.587
6.38
0.00038
Residual standard error: 0.679 on 7 degrees of freedom
Multiple R-Squared: 0.991,
Adjusted R-squared: 0.989
F-statistic:
391 on 2 and 7 degrees of freedom,
p-value: 6.55e-08
> 0.679ˆ2*7
[1] 3.2273
> 1-pchisq(3.32273,7)
[1] 0.85363
This time we cannot detect a lack of fit. Plot the fit:
> x <- seq(0.05,0.35,by=0.01)
> lines(x,g2$coef[1]+g2$coef[2]*x+g2$coef[3]*xˆ2,lty=2)
The curve is shown as a dotted line on the plot (thanks to lty=2). This seems clearly more appropriate
than the linear model.
6.2
σ2 unknown
The ˆ
σ2that is based in the chosen regression model needs to be compared to some model-free estimate
of σ2. We can do this if we have repeated y for one or more fixed x. These replicates do need to be
6.2. σ2 UNKNOWN
68
truly independent. They cannot just be repeated measurements on the same subject or unit. Such repeated
measures would only reveal the within subject variability or the measurement error. We need to know the
between subject variability — this reflects the σ2 described in the model.
The “pure error” estimate of σ2 is given by SSpe d fpe where
¡
SS
2
pe
∑ ∑ yi ¯y
¤
¢
distinct x givenx
¡
Degrees of freedom d f pe
∑
#replicates
1
¤
distinctx
¢
If you fit a model that assigns one parameter to each group of observations with fixed x then the ˆ
σ2 from
this model will be the pure error ˆ
σ2. This model is just the one-way anova model if you are familiar with
that. Comparing this model to the regression model amounts to the lack of fit test. This is usually the most
convenient way to compute the test but if you like we can then partition the RSS into that due to lack of fit
and that due to the pure error as in Table 6.1.
df
SS
MS
F
Residual
n-p
RSS
RSS SS
Lack of Fit
n
p
d f
pe
pe
RSS
SSpe
Ratio of MS
n p d f pe
Pure Error
d fpe
SSpe
SSpe d fpe
Table 6.1: ANOVA for lack of fit
Compute the F-statistic and compare to Fn p d f
and reject if the statistic is too large.
pe d f pe
¢
Another way of looking at this is a comparison between the model of interest and a saturated model that
assigns a parameter to each unique combination of the predictors. Because the model of interest represents
a special case of the saturated model where the saturated parameters satisfy the constraints of the model of
interest, we can use the standard F-testing methodology.
The data for this example consist of thirteen specimens of 90/10 Cu-Ni alloys with varying iron content
in percent. The specimens were submerged in sea water for 60 days and the weight loss due to corrosion
was recorded in units of milligrams per square decimeter per day. The data come from Draper and Smith
(1998).
We load in and print the data
> data(corrosion)
> corrosion
Fe
loss
1
0.01 127.6
2
0.48 124.0
3
0.71 110.8
4
0.95 103.9
5
1.19 101.5
6
0.01 130.1
7
0.48 122.0
8
1.44
92.3
9
0.71 113.1
10 1.96
83.7
11 0.01 128.0
12 1.44
91.4
13 1.96
86.2
6.2. σ2 UNKNOWN
69
We fit a straight line model:
> g <- lm(loss ˜ Fe, data=corrosion)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
129.79
1.40
92.5
< 2e-16
Fe
-24.02
1.28
-18.8
1.1e-09
Residual standard error: 3.06 on 11 degrees of freedom
Multiple R-Squared: 0.97,
Adjusted R-squared: 0.967
F-statistic:
352 on 1 and 11 degrees of freedom,
p-value: 1.06e-09
Check the fit graphically — see Figure 6.3.
> plot(corrosion$Fe,corrosion$loss,xlab="Iron content",ylab="Weight loss")
> abline(g$coef)
130
120
110
Weight loss
100
90
0.0
0.5
1.0
1.5
2.0
Iron content
Figure 6.3: Linear fit to the Cu-Ni corrosion data. Group means denoted by black diamonds
We have an R2 of 97% and an apparently good fit to the data. We now fit a model that reserves a
parameter for each group of data with the same value of x. This is accomplished by declaring the predictor
to be a factor. We will describe this in more detail in a later chapter
> ga <- lm(loss ˜ factor(Fe), data=corrosion)
The fitted values are the means in each group - put these on the plot:
> points(corrosion$Fe,ga$fit,pch=18)
6.2. σ2 UNKNOWN
70
We can now compare the two models in the usual way:
> anova(g,ga)
Analysis of Variance Table
Model 1: loss ˜ Fe
Model 2: loss ˜ factor(Fe)
Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
1
11
102.9
2
6
11.8
5
91.1
9.28 0.0086
The low p-value indicates that we must conclude that there is a lack of fit. The reason is that the pure error
¡
sd
11 8 6
1 4 is substantially less than the regression standard error of 3.06. We might investigate
¢
¤
¡
¡
models other than a straight line although no obvious alternative is suggested by the plot. Before considering
other models, I would first find out whether the replicates are genuine — perhaps the low pure error SD can
be explained by some correlation in the measurements. Another possible explanation is unmeasured third
variable is causing the lack of fit.
When there are replicates, it is impossible to get a perfect fit. Even when there is parameter assigned
to each group of x-values, the residual sum of squares will not be zero. For the factor model above, the R2
is 99.7%. So even this saturated model does not attain a 100% value for R2. For these data, it’s a small
difference but in other cases, the difference can be substantial. In these cases, one should realize that the
maximum R2 that may be attained might be substantially less than 100% and so perceptions about what a
good value for R2 should be downgraded appropriately.
These methods are good for detecting lack of fit, but if the null hypothesis is accepted, we cannot
conclude that we have the true model. After all, it may be that we just did not have enough data to detect the
inadequacies of the model. All we can say is that the model is not contradicted by the data.
When there are no replicates, it may be possible to group the responses for similar x but this is not
straightforward. It is also possible to detect lack of fit by less formal, graphical methods.
A more general question is how good a fit do you really want? By increasing the complexity of the
model, it is possible to fit the data more closely. By using as many parameters as data points, we can fit
the data exactly. Very little is achieved by doing this since we learn nothing beyond the data itself and any
predictions made using such a model will tend to have very high variance. The question of how complex a
model to fit is difficult and fundamental. For example, we can fit the mean responses for the example above
exactly using a sixth order polynomial:
> gp <- lm(loss ˜ Fe+I(Feˆ2)+I(Feˆ3)+I(Feˆ4)+I(Feˆ5)+I(Feˆ6),corrosion)
Now look at this fit:
> plot(loss ˜ Fe, data=corrosion,ylim=c(60,130))
> points(corrosion$Fe,ga$fit,pch=18)
> grid <- seq(0,2,len=50)
> lines(grid,predict(gp,data.frame(Fe=grid)))
as shown in Figure 6.4. The fit of this model is excellent — for example:
> summary(gp)$r.squared
[1] 0.99653
but it is clearly riduculous. There is no plausible reason corrosion loss should suddenly drop at 1.7 and
thereafter increase rapidly. This is a consequence of overfitting the data. This illustrates the need not to
become too focused on measures of fit like R2.
6.2. σ2 UNKNOWN
71
120
100
loss
80
60
0.0
0.5
1.0
1.5
2.0
Fe
Figure 6.4: Polynomial fit to the corrosion data
Chapter 7
Diagnostics
Regression model building is often an iterative and interactive process. The first model we try may prove
to be inadequate. Regression diagnostics are used to detect problems with the model and suggest improve-
ments. This is a hands-on process.
7.1
Residuals and Leverage
¡
We start with some basic diagnostic quantities - the residuals and the leverages. Recall that ˆ
y
X X T X
1XT y
¤
¢
¤
Hy where H is the hat-matrix. Now
¡
ˆε
y
ˆ
y
I
H y
¤
¤
¢
¡
¡
I
H X β
I
H ε
¤
¢
¢
¡
I
H ε
¤
¢
¡
¡
So var ˆε
var I
H ε
I
H σ2 assuming that var ε
σ2I. We see that although the errors may
¤
¢
¤
¢
¤
have equal variance and be uncorrelated the residuals do not.
h
σ2 ¡
i
Hii are called leverages and are useful diagnostics. We see that var ˆεi
1
hi so that a large
¤
¤
¢
leverage for hi will make var ˆεi small — in other words the fit will be “forced” to be close to yi. The hi
depends only on X — knowledge of y is required for a full interpretation. Some facts:
∑
hi
p
hi
1 n
i
¤
¡
i
An average value for hi is p n and a “rule of thumb” is that leverages of more than 2p n should be looked
at more closely. Large values of hi are due to extreme values in X. hi corresponds to a Mahalanobis distance
¡
defined by X which is x
¯
xT ˆ
Σ 1 ¡ x ¯x whereˆΣ is the estimated covariance of X.
¢
¢
¡
Also notice that var ˆ
y
var Hy
Hσ2 so v ˆar ˆyi
hi ˆσ2
¤
¢
¤
¤
We’ll use the savings dataset as an example here. First fit the model and make an index plot of the
residuals:
> data(savings)
> g <- lm(sr ˜ pop15 + pop75 + dpi + ddpi, savings)
> plot(g$res,ylab="Residuals",main="Index plot of residuals")
The plot is shown in the first panel of Figure 7.1
We can find which countries correspond to the largest and smallest residuals:
72
7.1. RESIDUALS AND LEVERAGE
73
Index plot of Residuals
Index plot of Leverages
10
Zambia
0.5
Libya
5
United States
0
0.3
Residuals
Leverages
−5
0.1
Chile
0
10
20
30
40
50
0
10
20
30
40
50
Index
Index
Studentized Residuals
Jacknife Residuals
3
Zambia
2
2
1
1
0
0
−1
−1
Jacknife Residuals
Studentized Residuals
−2
−2
0
10
20
30
40
50
0
10
20
30
40
50
Index
Index
Figure 7.1: Residuals and leverages for the savings data
7.2. STUDENTIZED RESIDUALS
74
> sort(g$res)[c(1,50)]
Chile
Zambia
-8.2422
9.7509
or by using the identify() function. We first make up a character vector of the country names using
row.names() which gets the row names from the data frame.
> countries <- row.names(savings)
> identify(1:50,g$res,countries)
Click on the left mouse button next to the points you are interested in to identify them. When you are
done, click on the middle (if not available, the right) mouse button. I have identified Chile and Zambia on
the plot.
Now look at the leverage: We first extract the X-matrix here using model.matrix() and then com-
pute and plot the leverages (also called ”hat” values)
> x <- model.matrix(g)
> lev <- hat(x)
> plot(lev,ylab="Leverages",main="Index plot of Leverages")
> abline(h=2*5/50)
> sum(lev)
[1] 5
Notice that the sum of the leverages is equal to p which is 5 for this data. Which countries have large
leverage? We have marked a horizontal line at 2p n to indicate our “rule of thumb”. We can see which
countries exceed this rather arbitrary cut-off:
> names(lev) <- countries
> lev[lev > 0.2]
Ireland
Japan United States
Libya
0.21224
0.22331
0.33369
0.53146
The command names() assigns the country names to the elements of the vector lev making it easier
to identify them. Alternatively, we can do it interactively like this
identify(1:50,lev,countries)
I have identified Libya and the United States as the points with the highest leverage.
7.2
Studentized Residuals
As we have seen var ˆε
σ2 ¡
i
1
hi this suggests the use of
¤
¢
ˆεi
ri ¤
ˆ
σ 1 h
¢
i
which are called (internally) studentized residuals. If the model assumptions are correct var ri
1 and
¤
¡
corr ri r j tends to be small. Studentized residuals are sometimes preferred in residual plots as they have
¢
been standardized to have equal variance.
7.3. AN OUTLIER TEST
75
Note that studentization can only correct for the natural non-constant variance in residuals when the
errors have constant variance. If there is some underlying heteroscedascity in the errors, studentization
cannot correct for it.
We now get the studentized residuals for the savings data:
> gs <- summary(g)
> gs$sig
[1] 3.8027
> stud <- g$res/(gs$sig*sqrt(1-lev))
> plot(stud,ylab="Studentized Residuals",main="Studentized Residuals")
Notice the range on the axis. Which residuals are large? In this case, there is not much difference
between the studentized and raw residuals apart from the scale. Only when there is unusually large leverage
will the differences be noticeable.
7.3
An outlier test
An outlier is a point that does not fit the current model. We need to be aware of such exceptions. An outlier
test is useful because it enables us to distinguish between truly unusual points and residuals which are large
but not exceptional.
Outliers may effect the fit — see Figure 7.2. The two additional points marked points both have high
leverage because they are far from the rest of the data.
is not an outlier. does not have a large residual if
it is included in the fit. Only when we compute the fit without that point do we get a large residual.
10
8
6
4
y
2
0
−2
−2
0
2
4
6
8
10
x
Figure 7.2: Outliers can conceal themselves. The solid line is the fit including the
point but not the
point. The dotted line is the fit without either additional point and the dashed line is the fit with the point
but not the
point.
¡
We exclude point i and recompute the estimates to get ˆβ i and ˆσ2 where i denotes that the ith case has
i
¢
¡
¡
been excluded. Hence
ˆ
y
ˆβ
i
xT
¤
i
i
¡
¡
7.3. AN OUTLIER TEST
76
If ˆ
y i
yi is large then case i is an outlier. Just looking at ˆεi misses those nasty points which pull the
¡
regression line so close to them that they conceal their true status. How large is large?
¡
¡
var ˆ
y
σ2 ¡
i
yi
1
xT X T X
xi
¢
¤
i
i
i ¢
¢
¡
¡
¡
and so
¡
¡
¡
vˆar ˆ
y i
yi
ˆ
σ2 1 xT XT X xi
¢
¤
i
i
i
i ¢
¢
¡
¡
¡
¡
Define the jackknife (or externally studentized or crossvalidated) residuals as
yi
ˆ
y
i
t
¡
i
¡
¡
¤
ˆ
σ
1 2
i
1
xT X T X
xi
i
i
i ¢
¢
¡
¡
¡
¡
which are distributed t
σ2
n p 1 if the model is correct and ε
N 0
I . Fortunately there is an easy way to
¢
compute ti:
§
ˆε
1 2
i
n
p
1
t
i
ri
¤
¤
ˆ
σ i 1 h
n
p
r2
¢
i
i ¨
¡
which avoids doing n regressions.
Since ti
tn p 1 and we can calculate a p-value to test whether case i is an outlier. However, we are
likely to want to test all cases so we must adjust the level of the test accordingly. Even though it might
seems that we only test one or two large ti’s, by identifying them as large we are implicitly testing all cases.
Suppose we want a level α test. Now P(all tests accept) = 1 - P(At least one rejects)
1
∑
¡
i P(Test i rejects)
1
nα. So this suggests that if an overall level α test is required then a level α n should be used in each of
¤
the tests. This method is called the Bonferroni correction and is used in contexts other than outliers as well.
It’s biggest drawback is that it is conservative — it finds fewer outliers than the nominal level of confidence
would dictate. The larger that n is, the more conservative it gets.
Now get the jacknife residuals for the savings data:
> jack <- rstudent(g)
> plot(jack,ylab="Jacknife Residuals",main="Jacknife Residuals")
> jack[abs(jack)==max(abs(jack))]
Zambia
2.8536
The largest residual of 2.85 is pretty big for a standard normal scale but is it an outlier? Compute the
Bonferroni critical value:
> qt(.05/(50*2),44)
[1] -3.5258
What do you conclude?
Notes
1. Two or more outliers next to each other can hide each other.
2. An outlier in one model may not be an outlier in another when the variables have been changed or
transformed. You will usually need to reinvestigate the question of outliers when you change the
model.
7.3. AN OUTLIER TEST
77
3. The error distribution may not be normal and so larger residuals may be expected. For example,
day-to-day changes in stock indices seem mostly normal but large changes occur not infrequently.
4. Individual outliers are usually much less of a problem in larger datasets. A single point won’t have the
leverage to affect the fit very much. It’s still worth identifying outliers if these type of points are worth
knowing about in the particular application. For large datasets, we need only worry about clusters of
outliers. Such clusters are less likely to occur by chance and more likely to represent actual structure.
Finding these cluster is not always easy.
What should be done about outliers?
1. Check for a data entry error first. These are relatively common. Unfortunately, the original source of
the data may have been lost.
2. Examine the physical context - why did it happen? Sometimes, the discovery of an outlier may be of
singular interest. Some scientific discoveries spring from noticing unexpected aberrations. Another
example of the importance of outliers is in the statistical analysis of credit card transactions. Outliers
in this case may represent fraudulent use.
3. Exclude the point from the analysis but try reincluding it later if the model is changed. The exclusion
of one or more points may make the difference between getting a statistical significant result or having
some unpublishable research. This can lead to difficult decision about what exclusions are reasonable.
To avoid any suggestion of dishonesty, always report the existence of outliers even if you do not
include them in your final model.
It’s dangerous to exclude outliers in an automatic manner. NASA launched the Nimbus 7 satellite to
record atmospheric information. After several years of operation in 1985, the British Antartic Survey
observed a large decrease in atmospheric ozone over the Antartic. On further examination of the
NASA data, it was found that the data processing program automatically discarded observations that
were extremely low and assumed to be mistakes. Thus the discovery of the Antartic ozone hole was
delayed several years. Perhaps, if this had been known earlier, the CFC phaseout would have been
agreed earlier and the damage could have been limited.
Here is an example of a dataset with multiple outliers. Data are available on the log of the surface
temperature and the log of the light intensity of 47 stars in the star cluster CYG OB1, which is in the
direction of Cygnus.
Read in and plot the data:
> data(star)
> plot(star$temp,star$light,xlab="log(Temperature)",
ylab="log(Light Intensity)")
What do you think relationship is between temperature and light intensity? Now fit a linear regression
and add the fitted line to the plot
> ga <- lm(light ˜ temp, data=star)
> abline(ga)
The plot is shown in Figure 7.3 with the regression line in solid type.
Is this what you expected? Are there any outliers in the data? The outlier test does not reveal any.
7.4. INFLUENTIAL OBSERVATIONS
78
6.0
5.5
5.0
4.5
log(Light Intensity)
4.0
3.6
3.8
4.0
4.2
4.4
4.6
log(Temperature)
Figure 7.3: Regression line including four leftmost points is solid and excluding these points is dotted
> range(rstudent(ga))
[1] -2.0494
1.9058
We need not bother to actually compute the critical value since these values are clearly not large enough.
The four stars on the upper left of the plot are giants. See what happens if these are excluded
> ga <- lm(light ˜ temp, data=star, subset=(temp>3.6))
> abline(ga$coef,lty=2)
This illustrates the problem of multiple outliers. We can visualize the problems here, but for higher
dimensional data this is much more difficult.
7.4
Influential Observations
An influential point is one whose removal from the dataset would cause a large change in the fit. An
influential point may or may not be an outlier and may or may not have large leverage but it will tend to
have at least one of those two properties. In Figure 7.2, the
point is not an influential point but the point
is.
¡
Here are some measures of influence, where the subscripted i indicates the fit without case i.
¢
1. Change in the coefficients ˆβ
ˆβ
i¡
2. Change in the fit X T ¡ ˆβ
ˆβ
ˆ
y
ˆ
y
i ¢ ¤
i
¡
¡
These are hard to judge in the sense that the scale varies between datasets. A popular alternative are the
Cook Statistics:
¡
ˆβ ˆβ T ¡
¡
X T X
ˆβ ˆβ
i ¢
¢
i ¢
D
¡
¡
i
¤
p ˆ
σ2
7.4. INFLUENTIAL OBSERVATIONS
79
¡
ˆ
y
ˆ
y
T ¡ ˆy
ˆ
y
i ¢
i ¢
¡
¡
¤
p ˆ
σ2
1
hi
r2
¤
p i 1
hi
The first term, r2, is the residual effect and the second is the leverage. The combination of the two leads to
i
influence. An index plot of Di can be used to identify influential points.
Continuing with our study of the savings data:
> cook <- cooks.distance(g)
> plot(cook,ylab="Cooks distances")
> identify(1:50,cook,countries)
The Cook statistics may be seen in Figure 7.4. I have identified the largest three values.
Ireland
Libya
0.20
0.2
Japan
−0.2
0.10
Zambia
Libya
Cooks distances
Change in pop75 coef
−0.6
Japan
0.00
0
10 20 30 40 50
−0.08
−0.02
0.04
Index
Change in pop15 coef
Figure 7.4: Cook Statistics and ˆβ
ˆβ ’s for the savings data
i¡
Which ones are large? We now exclude the largest one and see how the fit changes:
> gl <- lm(sr ˜ pop15+pop75+dpi+ddpi,savings,subset=(cook < max(cook)))
> summary(gl)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 24.524046
8.224026
2.98
0.0047
pop15
-0.391440
0.157909
-2.48
0.0171
pop75
-1.280867
1.145182
-1.12
0.2694
dpi
-0.000319
0.000929
-0.34
0.7331
ddpi
0.610279
0.268778
2.27
0.0281
Residual standard error: 3.79 on 44 degrees of freedom
Multiple R-Squared: 0.355,
Adjusted R-squared: 0.297
F-statistic: 6.07 on 4 and 44 degrees of freedom,
p-value: 0.000562
7.5. RESIDUAL PLOTS
80
Compared to the full data fit:
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 28.566087
7.354516
3.88
0.00033
pop15
-0.461193
0.144642
-3.19
0.00260
pop75
-1.691498
1.083599
-1.56
0.12553
dpi
-0.000337
0.000931
-0.36
0.71917
ddpi
0.409695
0.196197
2.09
0.04247
Residual standard error: 3.8 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
What changed? The coefficient for ddpi changed by about 50%. We don’t like our estimates to be so
sensitive to the presence of just one country. It would be rather tedious to do this for each country but there’s
a quicker way:
> ginf <- lm.influence(g)
> plot(ginf$coef[,2],ginf$coef[,3],xlab="Change in pop15 coef",
ylab="Change in pop75 coef")
> identify(ginf$coef[,2],ginf$coef[,3],countries)
We just plotted the change in the second and third parameter estimates when a case is left out as seen in
the second panel of Figure 7.4. Try this for the other estimates - which countries stick out? Consider Japan:
> gj <- lm(sr ˜ pop15+pop75+dpi+ddpi,savings,
subset=(countries != "Japan"))
> summary(gj)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 23.940171
7.783997
3.08
0.0036
pop15
-0.367901
0.153630
-2.39
0.0210
pop75
-0.973674
1.155450
-0.84
0.4040
dpi
-0.000471
0.000919
-0.51
0.6112
ddpi
0.334749
0.198446
1.69
0.0987
Residual standard error: 3.74 on 44 degrees of freedom
Multiple R-Squared: 0.277,
Adjusted R-squared: 0.211
F-statistic: 4.21 on 4 and 44 degrees of freedom,
p-value: 0.00565
Compare to the full data fit - what qualitative changes do you observe? Notice that the ddpi term is no
longer significant and that the the R2 value has decreased a lot.
7.5
Residual Plots
Outliers and influential points indicate cases that are in some way individually unusual but we also need
to check the assumptions of the model. Plot ˆε against ˆy. This is the most important diagnostic plot that
7.5. RESIDUAL PLOTS
81
you can make. If all is well, you should see constant variance in the vertical ( ˆε) direction and the scatter
should be symmetric vertically about 0. Things to look for are heteroscedascity (non-constant variance) and
nonlinearity (which indicates some change in the model is necessary). In Figure 7.5, these three cases are
illustrated.
No problem
Heteroscedascity
Nonlinear
0.3
1.0
1
0.2
0.5
0.1
0
0.0
0.0
Residual
Residual
Residual
−0.5
−0.1
−1
−1.0
−0.2
−2
−0.3
−1.5
0.0
0.2
0.4
0.6
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
−0.8
−0.6
−0.4
−0.2
0.0
0.2
Fitted
Fitted
Fitted
Figure 7.5: Residuals vs Fitted plots - the first suggests no change to the current model while the second
shows non-constant variance and the third indicates some nonlinearity which should prompt some change
in the structural form of the model
You should also plot ˆε against xi (for predictors that are both in and out of the model). Look for the same
things except in the case of plots against predictors not in the model, look for any relationship which might
indicate that this predictor should be included.
We illustrate this using the savings dataset as an example again:
> g <- lm(sr ˜ pop15+pop75+dpi+ddpi,savings)
First the residuals vs. fitted plot and the abs(residuals) vs. fitted plot.
> plot(g$fit,g$res,xlab="Fitted",ylab="Residuals")
> abline(h=0)
> plot(g$fit,abs(g$res),xlab="Fitted",ylab="|Residuals|")
The plots may be seen in the first two panels of Figure 7.5. What do you see? The latter plot is
designed to check for non-constant variance only. It folds over the bottom half of the first plot to increase
the resolution for detecting non-constant variance. The first plot is still needed because non-linearity must
be checked.
A quick way to check non-constant variance is this regression:
> summary(lm(abs(g$res) ˜ g$fit))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
4.840
1.186
4.08
0.00017
g$fit
-0.203
0.119
-1.72
0.09250
7.5. RESIDUAL PLOTS
82
10
10
8
5
6
0
4
Residuals
|Residuals|
−5
2
0
6
8
10
12
14
16
6
8
10
12
14
16
Fitted
Fitted
10
10
5
5
0
0
Residuals
Residuals
−5
−5
25
30
35
40
45
1
2
3
4
Pop’n under 15
Pop’n over 75
Figure 7.6: Residual plots for the savings data
Residual standard error: 2.16 on 48 degrees of freedom
Multiple R-Squared: 0.0578,
Adjusted R-squared: 0.0382
F-statistic: 2.95 on 1 and 48 degrees of freedom,
p-value: 0.0925
This test is not quite right as some weighting should be used and the degrees of freedom should be
adjusted but there doesn’t seem to be a clear problem with non-constant variance.
It’s often had to judge residual plots without prior experience so let’s show how to generate some of the
artificial variety. The following four for() loops show
1. Constant Variance
2. Strong non-constant variance
3. Mild non-constant variance
4. Non-linearity
7.6. NON-CONSTANT VARIANCE
83
> par(mfrow=c(3,3))
> for(i in 1:9) plot(1:50,rnorm(50))
> for(i in 1:9) plot(1:50,(1:50)*rnorm(50))
> for(i in 1:9) plot(1:50,sqrt((1:50))*rnorm(50))
> for(i in 1:9) plot(1:50,cos((1:50)*pi/25)+rnorm(50))
> par(mfrow=c(1,1))
In this case we know the truth - do you think you would be able to come to the right conclusions for real
data? Repeat to get an idea of the usual amount of variation. I recommend the artificial generation of plots
as a way to “calibrate” diagnostic plots. It’s often hard to judge whether an apparent feature is real or just
random variation. Repeated generation of plots under a model where there is no violation of the assumption
that the diagnostic plot is designed to check is helpful in making this judgement.
Now look at some residuals against predictor plots:
> plot(savings$pop15,g$res,xlab="Population under 15",ylab="Residuals")
> plot(savings$pop75,g$res,xlab="Population over 75",ylab="Residuals")
The plots may be seen in the second two panels of Figure 7.5. Can you see the two groups? Let’s
compare and test the variances. Given two independent samples from normal distributions, we can test for
equal variance using the test statistic of the ratio of the two variance. The null distribution is a F with degrees
of freedom given by the two samples.
> var(g$res[savings$pop15 > 35])/var(g$res[savings$pop15 <35])
[1] 2.7851
> table(savings$pop15 > 35)
FALSE
TRUE
27
23
> 1-pf(2.7851,22,26)
[1] 0.0067875
A significant difference is seen
7.6
Non-Constant Variance
There are two approaches to dealing with non-constant variance. Weighted least squares is appropriate
when the form of the non-constant variance is either known exactly or there is some known parametric
¡
¡
¡
form. Alternatively, one can transform y to h y where h
is chosen so that var h y is constant. To see how
¢
¢
¢
¡
to choose h
consider this
¢
¡
¡
¡
¡
h y
h Ey
y
Ey h Ey
¢
¤
¢
¢
¢
¡¢¡¢¡
¡
¡
var h y
h Ey 2var y
¢
¤
¢
¡¢¡¢¡
¡
We ignore the higher order terms. For var h y to be constant we need
¢
¡
h Ey ∝ ¡ var y 1 2
¢
¢
which suggests
dy
dy
¡
h y¢ ¤
¤
¤
var y
¡£¢
y
¢
7.6. NON-CONSTANT VARIANCE
84
¡
¡
For example if var y
var ε ∝ ¡ Ey 2 then h y
log y is suggested while if var ε ∝ ¡ Ey then h y
y.
¤
¢
¢
¤
¢
¢
¤
¢
Graphically one tends to see SDy rather then var y. Sometimes yi
0 for some i in which case the
¡
transformations may choke. You can try log y
δ for some small δ but this makes interpretation difficult.
¢
Consider the residual-fitted plot for the Galapagos data:
> gg <- lm(Species ˜ Area + Elevation + Scruz + Nearest + Adjacent, gala)
> plot(gg$fit,gg$res,xlab="Fitted",ylab="Residuals",
main="Untransformed Response")
We can see non-constant variance in the first plot of Figure 7.7.
Untransformed Response
Square root Response
150
4
2
50
0
0
Residuals
Residuals
−2
−4
−100
0
100
300
5
10
15
20
Fitted
Fitted
Figure 7.7: Residual-Fitted plots for the Galapagos data before and after transformation
We guess that a square root transformation will give us constant variance:
> gs <- lm(sqrt(Species) ˜ Area+ Elevation+ Scruz+ Nearest+ Adjacent, gala)
> plot(gs$fit,gs$res,xlab="Fitted",ylab="Residuals",
main="Square root Response")
We see in the second plot of Figure 7.7 that the variance is now constant. Our guess at a variance
stabilizing transformation worked out here, but had it not, we could always have tried something else. The
square root transformation is often appropriate for count response data. The poisson distribution is a good
model for counts and that distribution has the property that the mean is equal to the variance thus suggesting
the square root transformation. It might be even better to go with a poisson regression rather than the
normal-based regression we are using here.
¡
There are more formal tests for non-constant variance — for example one could start by regressing ˆε ¡
on y or xi but there is a problem in specifying the alternative hypothesis for such a test. A formal test may be
good at detecting a particular kind of non-constant variance but have no power to detect another. Residual
plots are more versatile because unanticipated problems may be spotted.
7.7. NON-LINEARITY
85
A formal diagnostic test may have reassuring aura of exactitude about it, but one needs to understand
that any such test may be powerless to detect problems of an unsuspected nature. Graphical techniques are
usually more effective at revealing structure that you may not have suspected. Of course, sometimes the
interpretation of the plot may be ambiguous but at least one can be sure that nothing is seriously wrong with
the assumptions. For this reason, I usually prefer a graphical approach to diagnostics.
7.7
Non-Linearity
Lack of fit tests can be used when there is replication which doesn’t happen too often, but even if you do
have it, the tests don’t tell you how to improve the model. How do we check if the systematic part (Ey
X β)
¤
of the model is correct? We can look at
1. Plots of ˆε against ˆy and xi
2. Plots of y against each xi.
but what about the effect of other x on the y vs. xi plot?
Partial Regression or Added Variable plots can help isolate the effect of xi on y.
1. Regress y on all x except xi, get residuals ˆδ. This represents y with the other X-effect taken out.
2. Regress xi on all x except xi, get residuals ˆγ This represents xi with the other X-effect taken out.
3. Plot ˆδ against ˆγ
The slope of a line fitted to the plot is ˆβi which adds some insight into the meaning of regression
coefficients. Look for non-linearity and outliers and/or influential points.
Partial Residual plots are a competitor to added variable plots. These plot ˆε
ˆβixi against xi. To see
where this comes from, look at the response with the predicted effect of the other X removed:
y
∑x ˆβ
ˆβ
ˆβ
j
j
ˆ
y
ˆε
∑xj j xi i ˆε
¤
¤
j i
j i
§
§
Again the slope on the plot will be ˆβi and the interpretation is the same. Partial residual plots are reckoned
to be better for non-linearity detection while added variable plots are better for outlier/influential detection.
We illustrate using the savings dataset as an example again: First we construct a partial regression (added
variable) plot for pop15:
> d <- lm(sr ˜ pop75 + dpi + ddpi,savings)$res
> m <- lm(pop15 ˜ pop75 + dpi + ddpi,savings)$res
> plot(m,d,xlab="pop15 residuals",ylab="Saving residuals",
main="Partial Regression")
Compare the slope on the plot to the original regression and show the line on the plot (see Figure 7.7).
> lm(d ˜ m)$coef
(Intercept)
m
5.4259e-17 -4.6119e-01
> g$coef
(Intercept)
pop15
pop75
dpi
ddpi
28.5660865
-0.4611931
-1.6914977
-0.0003369
0.4096949
> abline(0,g$coef[’pop15’])
7.7. NON-LINEARITY
86
Partial Regression
Partial Residuals
10
5
−10
−15
0
−20
savings residuals
Savings(adjusted)
−5
−25
−10
−5
0
5
25
35
45
pop15 residuals
pop15
Figure 7.8: Partial regression and residual plots for the savings data
Notice how the slope in the plot and the slope for pop15 in the regression fit are the same.
The partial regression plot also provides some intuition about the meaning of regression coefficients. We
are looking at the marginal relationship between the response and the predictor after the effect of the other
predictors has been removed. Multiple regression is difficult because we cannot visualize the full relation-
ship because of the high dimensionality. The partial regression plot allows us to focus on the relationship
between one predictor and the response, much as in simple regression.
A partial residual plot is easier to do:
> plot(savings$pop15,g$res+g$coef[’pop15’]*savings$pop15,xlab="pop’n under 15",
ylab="Saving(adjusted)",main="Partial Residual")
> abline(0,g$coef[’pop15’])
Or more directly:
> prplot(g,1)
Might there be a different relationship in the two groups?
> g1 <- lm(sr ˜ pop15+pop75+dpi+ddpi,savings,subset=(pop15 > 35))
> g2 <- lm(sr ˜ pop15+pop75+dpi+ddpi,savings,subset=(pop15 < 35))
> summary(g1)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.433969
21.155028
-0.12
0.91
pop15
0.273854
0.439191
0.62
0.54
7.7. NON-LINEARITY
87
pop75
-3.548477
3.033281
-1.17
0.26
dpi
0.000421
0.005000
0.08
0.93
ddpi
0.395474
0.290101
1.36
0.19
Residual standard error: 4.45 on 18 degrees of freedom
Multiple R-Squared: 0.156,
Adjusted R-squared: -0.0319
F-statistic: 0.83 on 4 and 18 degrees of freedom,
p-value: 0.523
> summary(g2)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 23.961795
8.083750
2.96
0.0072
pop15
-0.385898
0.195369
-1.98
0.0609
pop75
-1.327742
0.926063
-1.43
0.1657
dpi
-0.000459
0.000724
-0.63
0.5326
ddpi
0.884394
0.295341
2.99
0.0067
Residual standard error: 2.77 on 22 degrees of freedom
Multiple R-Squared: 0.507,
Adjusted R-squared: 0.418
F-statistic: 5.66 on 4 and 22 degrees of freedom,
p-value: 0.00273
Can you see the difference? The graphical analysis has shown a relationship in the data that a purely
numerical analysis might easily have missed.
Higher dimensional plots can also be useful for detecting structure that cannot be seen in two dimensions.
These are interactive in nature so you need to try them to see how they work. Two ideas are
1. Spinning - 3D plots where color, point size and rotation are used to give illusion of a third dimension.
2. Brushing - Two or more plots are linked so that point which are brushed in one plot are highlighted in
another.
These tools look good but it’s not clear whether they actually are useful in practice. Certainly there are
communication difficulties as these plots cannot be easily printed. Many statistical packages allow for this
kind of investigation. XGobi is a useful free UNIX-based tool for exploring higher dimensional data that
has now been made extended to Windows also as Ggobi. See www.ggobi.org
> library(xgobi)
> xgobi(savings)
or
> library(Rggobi)
> ggobi(savings)
Most of the functionality can be discovered by experimentation and the online help.
7.8. ASSESSING NORMALITY
88
7.8
Assessing Normality
The test and confidence intervals we use are based on the assumption of normal errors. The residuals can be
assessed for normality using a Q-Q plot. The steps are:
1. Sort the residuals: ˆε
ˆε
1
n
¡¢¡¢¡
¡
¡
2. Compute u
Φ 1 ¡ i
i ¤
n 1 ¢
©
3. Plot ˆε against ui. If the residuals are normally distributed an approximately straight-line relationship
i¡
will be observed.
Let’s try it out on the same old data:
> qqnorm(g$res,ylab="Raw Residuals")
> qqline(g$res)
See the first plot of Figure 7.8 - qqline() adds a line joining the first and third quartiles - it’s useful
as a guide. We can plot the (externally) studentized residuals:
> qqnorm(rstudent(g),ylab="Studentized residuals")
> abline(0,1)
See the second plot of the figure. Because these residuals have been normalized, they should lie along a
45 degree line.
Histograms and boxplots are not as sensitive for checking normality:
> hist(g$res,10)
> boxplot(g$res,main="Boxplot of savings residuals")
We can get an idea of the variation to be expected in QQ-plots in the following experiment. I generate
data from different distributions:
1. Normal
2. Lognormal - an example of a skewed distribution
3. Cauchy - an example of a long-tailed (platykurtic) distribution
4. Uniform - an example of a short-tailed (leptokurtic) distribution
Here’s how to generate 9 replicates at a time from each of these test cases:
> oldpar <- par()
> par(mfrow=c(3,3))
> for(i in 1:9) qqnorm(rnorm(50))
> for(i in 1:9) qqnorm(exp(rnorm(50)))
> for(i in 1:9) qqnorm(rcauchy(50))
> for(i in 1:9) qqnorm(runif(50))
> par(oldpar)
7.8. ASSESSING NORMALITY
89
Normal Q−Q Plot
Normal Q−Q Plot
10
2
5
1
0
0
Raw Residuals
−1
−5
Studentized residuals
−2
−2
−1
0
1
2
−2
−1
0
1
2
Theoretical Quantiles
Theoretical Quantiles
Histogram of Residuals
Boxplot of savings residuals
10
10
8
5
6
0
4
Frequency
2
−5
0
−10
−5
0
5
10
Residuals
Figure 7.9: Normality checks for the savings data
We save the original settings for the graphics layout in oldpar and restore it after we are done. This is a
useful trick when you want to experiment with changing these settings.
In Figure 7.8, you can see examples of all four cases:
It’s not always easy to diagnose the problem in QQ plots.
The consequences of non-normality are
1. that the least squares estimates may not be optimal - they will still be BLUE but other robust estimators
may be more effective.
2. that the tests and confidence intervals are invalid. However, it has been shown that only really long-
tailed distributions cause a problem. Mild non-normality can safely be ignored and the larger the
sample size the less troublesome the non-normality.
What to do?
7.8. ASSESSING NORMALITY
90
Normal Q−Q Plot
Normal Q−Q Plot
6
2
5
1
4
3
0
2
Normal Residuals
−1
1
Lognormal Residuals
0
−2
−1
0
1
2
−2
−1
0
1
2
Theoretical Quantiles
Theoretical Quantiles
Normal Q−Q Plot
Normal Q−Q Plot
250
0.8
150
0.4
50
Cauchy Residuals
Uniform Residuals
0
0.0
−2
−1
0
1
2
−2
−1
0
1
2
Theoretical Quantiles
Theoretical Quantiles
Figure 7.10: QQ plots of simulated data
1. A transformation of the response may solve the problem - this is often true for skewed errors.
2. Other changes in the model may help.
3. Accept non-normality and base the inference on the assumption of another distribution or use resam-
pling methods such as the bootstrap or permutation tests. You don’t want to do this unless absolutely
necessary. Alternatively use robust methods which give less weight to outlying points. This is appro-
priate for long tailed distributions.
4. For short-tailed distributions, the consequences of non-normality are not serious and can reasonably
be ignored.
There are formal tests for normality such as the Kolmogorov-Smirnov test but these are not as flexible
as the Q-Q plot. The p-value is not very helpful as an indicator of what action to take. After all, with a
large dataset, even mild deviations from non-normality may be detected, but there would be little reason to
7.9. HALF-NORMAL PLOTS
91
abandon least squares because the effects of non-normality are mitigated by large sample sizes. For smaller
sample sizes, formal tests lack power.
7.9
Half-normal plots
¡
Half-normal plots are designed for the assessment of positive data. They could be used for ˆε ¡ but are more
typically useful for diagnostic quantities like the leverages or the Cook Statistics. The idea is to plot the data
against the positive normal quantiles
The steps are:
1. Sort the data: x
x
1
n
¡¢¡¢¡
¡
¡
2. Compute u
Φ 1 ¡ n i
©
i ¤
2n 1 ¢
©
3. Plot x
against ui.
i¡
We are usually not looking for a straight line relationship since we do not necessarily expect a positive
normal distribution for quantities like the leverages. (If the X is multivariate normal, the leverages will have
a χ2p distribution but there is usually no good reason to assume multivariate normality for the X.) We are
looking for outliers which will be apparent as points that diverge substantially from the rest of the data.
We demonstrate the half-normal plot on the leverages and Cook statistics for the savings data:
> halfnorm(lm.influence(g)$hat,labs=countries,ylab="Leverages")
> halfnorm(cooks.distance(g),labs=countries,ylab="Cook Statistics")
The plots are chosen in Figure 7.11 — I have plotted the country name instead of just a dot for the largest
two cases respectively to aid identification. The halfnorm() function comes from the book library.
Libya
Libya
0.5
0.4
0.20
United States
0.3
Japan
Leverages
0.2
0.10
Cook Statistics
0.1
0.00
0.0
0.5
1.0
1.5
2.0
0.0
0.5
1.0
1.5
2.0
Half−normal quantiles
Half−normal quantiles
Figure 7.11: Half-normal plots for the leverages and Cook statistics
Libya shows up clearly as unusual in both plots
7.10. CORRELATED ERRORS
92
7.10
Correlated Errors
We assume that the errors are uncorrelated but for temporally or spatially related data this may well be
untrue. For this type of data, it is wise to check the uncorrelated assumption.
1. Plot ˆε against time.
2. Use formal tests like the Durbin-Watson or the run test.
If you do have correlated errors, you can use GLS. This does require that you know Σ or more usually
that you can estimate it. In the latter case, an iterative fitting procedure will be necessary as in IRWLS. Such
problems are common in Econometrics.
For the example, we use some taken from an environmental study that measured the four variables
ozone, solar radiation, temperature, and wind speed for 153 consecutive days in New York.
> data(airquality)
> airquality
Ozone Solar.R Wind Temp Month Day
1
41
190
7.4
67
5
1
2
36
118
8.0
72
5
2
3
12
149 12.6
74
5
3
4
18
313 11.5
62
5
4
5
NA
NA 14.3
56
5
5
etc..
We notice that there are some missing values. Take a look at the data: (plot not shown)
> pairs(airquality,panel=panel.smooth)
We fit a standard linear model and check the residual-fitted plot in Figure 7.10.
> g <- lm(Ozone ˜ Solar.R + Wind + Temp,airquality)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -64.3421
23.0547
-2.79
0.0062
Solar.R
0.0598
0.0232
2.58
0.0112
Wind
-3.3336
0.6544
-5.09
1.5e-06
Temp
1.6521
0.2535
6.52
2.4e-09
Residual standard error: 21.2 on 107 degrees of freedom
Multiple R-Squared: 0.606,
Adjusted R-squared: 0.595
F-statistic: 54.8 on 3 and 107 degrees of freedom,
p-value:
0
> plot(g$fit,g$res,xlab="Fitted",ylab="Residuals",
main="Untransformed Response")
Notice how there are only 107 degrees corresponding to the 111 complete observations. The default
behavior in R when performing a regression with missing values is to exclude any case that contains a
missing value. We see some non-constant variance and nonlinearity and so we try transforming the response:
7.10. CORRELATED ERRORS
93
> gl <- lm(log(Ozone) ˜ Solar.R + Wind + Temp,airquality)
> plot(gl$fit,gl$res,xlab="Fitted",ylab="Residuals",main="Logged Response")
Suppose we are now otherwise satisfied with this model and want to check for serial correlation. The
missing values in the data were not used in the construction of the model but this also breaks up the sequential
pattern in the data. I get round this by reintroducing missing values into the residuals corresponding to the
omitted cases.
> res <- rep(NA,153)
> res[as.numeric(row.names(na.omit(airquality)))] <- gl$res
First make an index plot of the residuals — see Figure 7.10.
> plot(res,ylab="Residuals",main="Index plot of residuals")
Is there any evidence of serial correlation? Now plot successive residuals:
> plot(res[-153],res[-1],xlab=expression(hat(epsilon)[i]),
ylab=expression(hat(epsilon)[i+1]))
Do you see any problem? Let’s check
> summary(lm(res[-1] ˜ -1+res[-153]))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
res[-153]
0.110
0.105
1.05
0.3
Residual standard error: 0.508 on 91 degrees of freedom
Multiple R-Squared: 0.0119,
Adjusted R-squared: 0.00107
F-statistic:
1.1 on 1 and 91 degrees of freedom,
p-value: 0.297
We omitted the intercept term because the residuals have mean zero. We see that there is no significant
correlation.
You can plot more than just successive pairs if you suspect a more complex dependence. For spatial
data, more complex checks are required.
7.10. CORRELATED ERRORS
94
Untransformed Response
Logged Response
80
1.0
40
0.0
Residuals
0
Residuals
−1.0
−40
−2.0
−20
0
20
40
60
80
1.5
2.5
3.5
4.5
Fitted
Fitted
Index plot of residuals
1.0
1.0
0.0
0.0
^ ε i+1
Residuals
−1.0
−1.0
−2.0
−2.0
0
50
100
150
−2.0
−1.0
0.0
1.0
Index
ε
^i
Figure 7.12: Checking for correlated errors - Index plot and scatterplot of successive residuals
Chapter 8
Transformation
Transformations of the response and predictors can improve the fit and correct violations of model assump-
tions such as constant error variance. We may also consider adding additional predictors that are functions
of the existing predictors like quadratic or crossproduct terms.
8.1
Transforming the response
Let’s start with some general considerations about transforming the response.
Suppose that you are contemplating a logged response in a simple regression situation:
log y
β
β
ε
0
1x
¤
In the original scale of the response, this model becomes
¡
¡
y
exp β
β
ε
0
1x
exp
¤
¢
¡
¢
In this model, the errors enter multiplicatively and not additively as they usually do. So the use of
standard regression methods for the logged response model requires that we believe that the errors
enter multiplicatively in the original scale. Notice that if we believe the proper model for y to be
¡
y
exp β
β
ε
0
1x
¤
¢
then we cannot linearize this model and non-linear regression methods would need to be applied.
As a practical matter, we often do not know how the errors enter the model, additively, multiplicatively
or otherwise. The usual approach is to try different transforms and then check the residuals to see
whether they satisfy the conditions required for linear regression. Unless you have good information
that the error enters in some particular way, this is the simplest and most appropriate way to go.
Although you may transform the response, you will probably need to express predictions in the orig-
inal scale. This is simply a matter of back-transforming. For example, in the logged model above,
¡
your prediction would be exp ˆ
y0 . If your prediction confidence interval in the logged scale was l u ,
¢
¢
then you would use exp l exp u . This interval will not be symmetric but this may be desirable —
¢
remember what happened with the prediction confidence intervals for Galapagos data.
95
8.1. TRANSFORMING THE RESPONSE
96
Regression coefficients will need to be interpreted with respect to the transformed scale. There is
no straightforward way of backtransforming them to values that can interpreted in the original scale.
You cannot directly compare regression coefficients for models where the response transformation
is different. Difficulties of this type may dissuade one from transforming the response even if this
requires the use of another type of model such as a generalized linear model.
When you use a log transformation on the response, the regression coefficients have a particular inter-
pretation:
log ˆ
y
ˆβ
ˆβ
ˆβ
0
1x1
pxp
¤
¡¢¡¢¡
ˆ
ˆ
ˆ
ˆ
y
eβ0 eβ1x1
eβpxp
¤
¡¢¡¢¡
ˆ
An increase of one in x
β1
1 would multiply the predicted response (in the original scale) by e
. Thus when
a log scale is used the regression coefficients can be interpreted in a multiplicative rather than the usual
additive manner.
The Box-Cox method is a popular way to determine a tranformation on the response. It is designed
for strictly positive responses and chooses the transformation to find the best fit to the data. The method
¡
transforms the response y
tλ y where the family of transformations indexed by λ is
¢
§
yλ 1
¡
0
t
λ y
λ
λ ¤
¢
¤
log y
λ 0
¤
¡
For fixed y ¥
0, tλ y is continuous in λ. Choose λ using maximum likelihood. The profile log-likelihood
¢
assuming normality of the errors is
n
¡
¡
¡
L λ
log RSSλ n
λ 1 ∑logyi
¢
¤
¢
¢
2
¡
where RSSλ is the RSS when tλ y is the response. You can compute ˆλ exactly to maximize this but usually
¢
¡
L λ is just maximized over a grid of values such as
2
1
1 2 0 1 2 1 2 . This ensures that the
¢
¡
¢
chosen ˆλ is more easily interpreted. For example, if ˆλ
0 46, it would be hard to explain what this new
¤
¡
response means, but
y would be easier.
¢
Transforming the response can make the model harder to interpret so we don’t want to do it unless it’s
¡
really necessary. One way to check this is to form a confidence interval for λ. A 100 1
α % confidence
¢
interval for λ is
λ
1
1 α
¡
¡
:
L λ
¡
¥
L ˆλ
χ2
¡
¢
¢
¢
2 1
This interval can be derived by inverting the likelihood ratio test of the hypothesis that H
λ
0 : λ
0 which
¤
¡
¡
¡
uses the statistic 2 L ˆλ
L λ0
having approximate null distribution χ2.
¢
¢
¢
1
Does the response in the savings data need transformation? You’ll need a function from the MASS
library:
> library(MASS)
Try it out on the savings dataset and plot the results.
> data(savings)
> g <- lm(sr ˜ pop15+pop75+dpi+ddpi,savings)
> boxcox(g,plotit=T)
> boxcox(g,plotit=T,lambda=seq(0.5,1.5,by=0.1))
8.1. TRANSFORMING THE RESPONSE
97
−162.0
−200
−163.0
95%
−250
log−Likelihood
log−Likelihood
−164.0
−300
−165.0
−2
−1
0
1
2
0.6
1.0
1.4
lambda
lambda
Figure 8.1: Log-likelihood plots for the Box-Cox transformation of the savings data
The first plot shown in Figure 8.1 is too broad. I narrowed the range of λ in the second plot so that we
can read off the confidence interval more easily.
The confidence interval for λ runs from about 0.6 to about 1.4. We can see that there is no good reason
to transform.
Now consider the Gal´apagos Islands dataset analyzed earlier:
> data(gala)
> g <- lm(Species ˜ Area + Elevation + Nearest + Scruz + Adjacent,gala)
> boxcox(g,plotit=T)
> boxcox(g,lambda=seq(0.0,1.0,by=0.05),plotit=T)
The plots are shown in Figure 8.2. We see that perhaps a cube-root transformation might be best here. A
square root is also a possibility as this falls just within the confidence intervals. Certainly there is a strong
need to transform.
Notes
1. The Box-Cox method gets upset by outliers - if you find ˆλ
5 then this is probably the reason —
¤
there can be little justification for actually making such an extreme transformation.
2. What if some yi
0? Sometimes adding a constant to all y can work provided that constant is small.
3. If maxi yi mini yi is small then the Box-Cox won’t do anything because power transforms are well
approximated by linear transformations over short intervals.
4. Should the estimation of λ count as an extra parameter to be taken account of in the degrees of
freedom? This is a difficult question since λ is not a linear parameter and its estimation is not part of
the least squares fit.
8.2. TRANSFORMING THE PREDICTORS
98
95%
−150 95%
−160
−180
−155
−200
−160
log−Likelihood
−220
log−Likelihood
−165
−240
−260
−170
−2
−1
0
1
2
0.0
0.2
0.4
0.6
0.8
1.0
lambda
lambda
Figure 8.2: Log-likelihood plots for the Box-Cox transformation of the Gal´apagos data
The Box-Cox method is not the only way of transforming the predictors. For responses, that are pro-
¡
¡
portions (or percentages), the logit transformation, log y 1
y
is often used, while for responses that are
¢
¢
¡
¡
¡
correlations, Fisher’s z transform, y
0 5 log 1
y
1
y
is worth considering.
¤
¢
¡
¢
¢
¡
8.2
Transforming the predictors
You can take a Box-Cox style approach for each of the predictors, choosing the transformation to minimize
the RSS. However, this takes time and furthermore the correct transformation for each predictor may depend
on getting the others right too. Partial residuals are a good way of finding suggestions for transforming the
predictors
8.2.1
Broken Stick Regression
Sometimes we have reason to believe that different linear regression models apply in different regions of the
data. For example, in the analysis of the savings data, we observed that there were two groups in the data
and we might want to fit a different model to the two parts. Suppose we focus attention on just the pop15
predictor for ease of presentation. We fit the two regression models depending on whether pop15 is greater
or less than 35%. The two fits are shown in Figure 8.3.
> g1 <- lm(sr ˜ pop15, savings, subset=(pop15 < 35))
> g2 <- lm(sr ˜ pop15, savings, subset=(pop15 > 35))
> plot(savings$pop15,savings$sr,xlab="Pop’n under 15",ylab="Savings Rate")
8.2. TRANSFORMING THE PREDICTORS
99
> abline(v=35,lty=5)
> segments(20,g1$coef[1]+g1$coef[2]*20,35,g1$coef[1]+g1$coef[2]*35)
> segments(48,g2$coef[1]+g2$coef[2]*48,35,g2$coef[1]+g2$coef[2]*35)
20
15
10
Savings Rate
5
0
25
30
35
40
45
Pop’n under 15
Figure 8.3: Subset regression fit is shown with the solid line while the broken stick regression is shown with
the dotted line
A possible objection to this subsetted regression fit is that the two parts of the fit do not meet at the join.
If we believe the fit should be continuous as the predictor varies, then this is unsatisfactory. One solution to
this problem is the broken stick regression fit. Define two basis functions:
c
x
x
c
¡
B
¡£¢
l x¢ ¤
0
¤¦¥¨§©¨¡£©
and
x
c
x ¥
c
¡
B
¡¢
r x¢ ¤
0
¤¨¥¦§©¦¡ ©
where c marks the division between the two groups. Bl and Br form a first-order spline basis with a knotpoint
at c. Sometimes Bl and Br are called hockey-stick functions because of their shape. We can now fit a model
of the form
¡
¡
y
β
β
β
ε
0
1Bl x
2Br x
¤
¢
¢
¡
using standard regression methods. The two linear parts are guaranteed to meet at c. Notice that this model
uses only three parameters in contrast to the four total parameters used in the subsetted regression illustrated
above. A parameter has been saved by insisting on the continuity of the fit at c.
We define the two hockey stick functions, compute and display the fit:
> lhs <- function(x) ifelse(x < 35,35-x,0)
> rhs <- function(x) ifelse(x < 35,0,x-35)
> gb <- lm(sr ˜ lhs(pop15) + rhs(pop15), savings)
8.2. TRANSFORMING THE PREDICTORS
100
> x <- seq(20,48,by=1)
> py <- gb$coef[1]+gb$coef[2]*lhs(x)+gb$coef[3]*rhs(x)
> lines(x,py,lty=2)
The two (dotted) lines now meet at 35 as shown in Figure 8.3. The intercept of this model is the value
of the response at the join.
We might question which fit is preferable in this particular instance. For the high pop15 countries, we
see that the imposition of continuity causes a change in sign for the slope of the fit. We might argue that
the two groups of countries are so different and that there are so few countries in the middle region, that we
might not want to impose continuity at all.
We can have more than one knotpoint simply by defining more pairs of basis functions with different
knotpoints. Broken stick regression is sometimes called segmented regression. Allowing the knotpoints to
be parameters is worth considering but this will result in a nonlinear model.
8.2.2
Polynomials
Another way of generalizing the X β part of the model is to add polynomial terms. In the one-predictor case,
we have
y
β
β
β
ε
0
1x
d xd
¤
¡¢¡¢¡
which allows for a more flexible relationship although we usually don’t believe it exactly represents any
underlying reality.
There are two ways to choose d:
1. Keep adding terms until the added term is not statistically significant.
2. Start with a large d — eliminate not statistically significant terms starting with the highest order term.
Warning: Do not eliminate lower order terms from the model even if they are not statistically significant.
An additive change in scale would change the t-statistic of all but the highest order term. We would not want
the conclusions of our study to be so brittle to such changes in the scale which ought to be inconsequential.
Let’s see if we can use polynomial regression on the ddpi variable in the savings data. First fit a linear
model:
> summary(lm(sr ˜ ddpi,savings))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.883
1.011
7.80
4.5e-10
ddpi
0.476
0.215
2.22
0.031
Residual standard error: 4.31 on 48 degrees of freedom
Multiple R-Squared: 0.0929,
Adjusted R-squared: 0.074
F-statistic: 4.92 on 1 and 48 degrees of freedom,
p-value: 0.0314
p-value of ddpi is significant so move on to a quadratic term:
> summary(lm(sr ˜ ddpi+I(ddpiˆ2),savings))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
5.1304
1.4347
3.58
0.00082
8.2. TRANSFORMING THE PREDICTORS
101
ddpi
1.7575
0.5377
3.27
0.00203
I(ddpiˆ2)
-0.0930
0.0361
-2.57
0.01326
Residual standard error: 4.08 on 47 degrees of freedom
Multiple R-Squared: 0.205,
Adjusted R-squared: 0.171
F-statistic: 6.06 on 2 and 47 degrees of freedom,
p-value: 0.00456
Again the p-value of ddpi2 is significant so move on to a cubic term:
> summary(lm(sr ˜ ddpi+I(ddpiˆ2)+I(ddpiˆ3),savings))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
5.145360
2.198606
2.34
0.024
ddpi
1.746017
1.380455
1.26
0.212
I(ddpiˆ2)
-0.090967
0.225598
-0.40
0.689
I(ddpiˆ3)
-0.000085
0.009374
-0.01
0.993
Residual standard error: 4.12 on 46 degrees of freedom
Multiple R-Squared: 0.205,
Adjusted R-squared: 0.153
F-statistic: 3.95 on 3 and 46 degrees of freedom,
p-value: 0.0137
p-value of ddpi3 is not significant so stick with the quadratic. What do you notice about the other
p-values? Why do we find a quadratic model when the previous analysis on transforming predictors found
that the ddpi variable did not need transformation? Check that starting from a large model (including the
fourth power) and working downwards gives the same result.
To illustrate the point about the significance of lower order terms, suppose we transform ddpi by
subtracting 10 and refit the quadratic model:
> savings <- data.frame(savings,mddpi=savings$ddpi-10)
> summary(lm(sr ˜ mddpi+I(mddpiˆ2),savings))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
13.4070
1.4240
9.41
2.2e-12
mddpi
-0.1022
0.3027
-0.34
0.737
I(mddpiˆ2)
-0.0930
0.0361
-2.57
0.013
Residual standard error: 4.08 on 47 degrees of freedom
Multiple R-Squared: 0.205,
Adjusted R-squared: 0.171
F-statistic: 6.06 on 2 and 47 degrees of freedom,
p-value: 0.00456
We see that the quadratic term remains unchanged but the linear term is now insignificant. Since there is
often no necessary importance to zero on a scale of measurement, there is no good reason to remove the
linear term in this model but not in the previous version. No advantage would be gained.
You have to refit the model each time a term is removed and for large d there can be problem with
numerical stability. Orthogonal polynomials get round this problem by defining
z1
a1
b1x
¤
z2
a2
b2x
c2x2
¤
z3
a3
b3x
c3x2
d3x3
¤
8.3. REGRESSION SPLINES
102
etc. where the coefficients a b c
are chosen so that zT z
0 when i
j. The z are called orthogonal
i
j ¤
¤
¢¡¢¡¢¡
polynomials. The value of orthogonal polynomials has declined with advances in computing speeds al-
though they are still worth knowing about because of their numerical stability and ease of use. The poly()
function constructs Orthogonal polynomials.
> g <- lm(sr ˜ poly(ddpi,4),savings)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
9.6710
0.5846
16.54
<2e-16
poly(ddpi, 4)1
9.5590
4.1338
2.31
0.025
poly(ddpi, 4)2 -10.4999
4.1338
-2.54
0.015
poly(ddpi, 4)3
-0.0374
4.1338
-0.01
0.993
poly(ddpi, 4)4
3.6120
4.1338
0.87
0.387
Residual standard error: 4.13 on 45 degrees of freedom
Multiple R-Squared: 0.218,
Adjusted R-squared: 0.149
F-statistic: 3.14 on 4 and 45 degrees of freedom,
p-value: 0.0232
Can you see how we come to the same conclusion as above with just this summary? We can verify the
orthogonality of the design matrix when using orthogonal polynomials:
> x <- model.matrix(g)
> dimnames(x) <- list(NULL,c("Int","power1","power2","power3","power4"))
> round(t(x) %*% x,3)
Int power1 power2 power3 power4
Int
50
0
0
0
0
power1
0
1
0
0
0
power2
0
0
1
0
0
power3
0
0
0
1
0
power4
0
0
0
0
1
You can have more than two predictors as can be seen in this response surface model:
y
β
β
β
β
β
β
0
1x1
2x2
11x2
22x2
12x1x2
¤
1
2
8.3
Regression Splines
Polynomials have the advantage of smoothness but the disadvantage that each data point affects the fit
globally. This is because the power functions used for the polynomials take non-zero values across the
whole range of the predictor. In contrast, the broken stick regression method localizes the influence of each
data point to its particular segment which is good but we do not have the same smoothness as with the
polynomials. There is a way we can combine the beneficial aspects of both these methods — smoothness
and local influence — by using B-spline basis functions.
We may define a cubic B-spline basis on the interval a b by the following requirements on the interior
¢
basis functions with knot-points at t1
tk.
¢¡¢¡¢¡£
1. A given basis function is non-zero on interval defined by four successive knots and zero elsewhere.
This property ensures the local influence property.
8.3. REGRESSION SPLINES
103
2. The basis function is a cubic polynomial for each sub-interval between successive knots
3. The basis function is continuous and continuous in its first and second derivatives at each knot point.
This property ensures the smoothness of the fit.
4. The basis function integrates to one over its support
The basis functions at the ends of the interval are defined a little differently to ensure continuity in
derivatives at the edge of the interval. A full definition of B-splines and more details about their properties
may be found in “A practical guide to splines” by Carl De Boor.
Let’s see how the competing methods do on a constructed example. Suppose we know the true model is
¡
¡
y
sin3 ¡ 2πx3
ε
ε N 0 0 1 2
¤
¢
¢
¢
¡
The advantage of using simulated data is that we can see how close our methods come to the truth. We
generate the data and display it in Figure 8.3.
> funky <- function(x) sin(2*pi*xˆ3)ˆ3
> x <- seq(0,1,by=0.01)
> y <- funky(x) + 0.1*rnorm(101)
> matplot(x,cbind(y,funky(x)),type="pl",ylab="y",pch=18,lty=1,
main="True Model")
We see how an orthogonal polynomial bases of orders 4 and 12 do in fitting this data:
> g4 <- lm(y ˜ poly(x,4))
> g12 <- lm(y ˜ poly(x,12))
> matplot(x,cbind(y,g4$fit,g12$fit),type="pll",ylab="y",pch=18,
lty=c(1,2),main="Orthogonal Polynomials")
The two fits are shown in the second panel of Figure 8.3. We see that order 4 is a clear underfit. Order
12 is much better although the fit is too wiggly in the first section and misses the point of inflection.
We now create the B-spline basis. You need to have three additional knots at the start and end to get the
right basis. I have chosen to the knot locations to put more in regions of greater curvature. I have used 12
basis functions for comparability to the orthogonal polynomial fit.
> library(splines)
> knots <- c(0,0,0,0,0.2,0.4,0.5,0.6,0.7,0.8,0.85,0.9,1,1,1,1)
> bx <- splineDesign(knots,x)
> gs <- lm(y ˜ bx)
> matplot(x,bx,type="l",main="B-spline basis functions")
> matplot(x,cbind(y,gs$fit),type="pl",ylab="y",pch=18,lty=1,
main="Spline fit")
The basis functions themselves are shown in the third panel of Figure 8.3 while the fit itself appears in
the fourth panel. We see that the fit comes very close to the truth.
Regression splines are useful for fitting functions with some flexibility provided we have enough data.
We can form basis functions for all the predictors in our model but we need to be careful not to use up too
many degrees of freedom.
8.4. MODERN METHODS
104
True Model
Orthogonal Polynomials
1.0
1.0
0.5
0.5
y
y
0.0
0.0
−1.0
−1.0
0.0
0.2
0.4
0.6
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
x
B−spline basis functions
Spline fit
1.0
0.8
0.5
bx
y
0.0
0.4
0.0
−1.0
0.0
0.2
0.4
0.6
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
x
Figure 8.4: Orthogonal Splines compared to B-splines
8.4
Modern Methods
The methods described above are somewhat awkward to apply exhaustively and even then they may miss
important structure because of the problem of trying to find good transformations on several variables si-
multaneously. One recent approach is the additive model:
¡
¡
¡
y
β
ε
0
f1 x1
f2 x2
fp xp
¤
¢
¡
¢
¢
¡¢¡¢¡
where nonparametric regression techniques are used to estimated the fi’s. Alternatively, you could imple-
ment this using the regression spline bases for each predictor variable. Other techniques are ACE, AVAS,
Regression Trees, MARS and neural networks.
It is important to realize the strengths and weaknesses of regression analysis. For larger data sets with
relatively little noise, more recently developed complex models will be able to fit the data better while
keeping the number of parameters under control. For smaller data sets or where the noise level is high (as
8.4. MODERN METHODS
105
is typically found in the social sciences), more complex models are not justified and standard regression is
most effective. One relative advantage of regression is that the models are easier to interpret in contrast to
techniques like neural networks which are usually only good for predictive purposes.
Chapter 9
Scale Changes, Principal Components and
Collinearity
9.1
Changes of Scale
Suppose we re-express x
a
i as xi ©
. We might want to do this because
b
1. Predictors of similar magnitude are easier to compare. ˆβ
3 51 is easier to parse than ˆβ
0 000000351.
¤
¤
¡
¡
2. A change of units might aid interpretability.
3. Numerical stability is enhanced when all the predictors are on a similar scale.
Rescaling xi leaves the t and F tests and ˆσ2 and R2 unchanged and ˆβi
b ˆβi.
§
Rescaling y in the same way leaves the t and F tests and R2 unchanged but ˆ
σ and ˆβ will rescaled by b.
To demonstrate this, we use same old model:
> g <- lm(sr ˜ pop15+pop75+dpi+ddpi,savings)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 28.566087
7.354516
3.88
0.00033
pop15
-0.461193
0.144642
-3.19
0.00260
pop75
-1.691498
1.083599
-1.56
0.12553
dpi
-0.000337
0.000931
-0.36
0.71917
ddpi
0.409695
0.196197
2.09
0.04247
Residual standard error: 3.8 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
The coefficient for income is rather small - let’s measure income in thousands of dollars instead and
refit:
> g <- lm(sr ˜ pop15+pop75+I(dpi/1000)+ddpi,savings)
> summary(g)
Coefficients:
106
9.2. PRINCIPAL COMPONENTS
107
Estimate Std. Error t value Pr(>|t|)
(Intercept)
28.566
7.355
3.88
0.00033
pop15
-0.461
0.145
-3.19
0.00260
pop75
-1.691
1.084
-1.56
0.12553
I(dpi/1000)
-0.337
0.931
-0.36
0.71917
ddpi
0.410
0.196
2.09
0.04247
Residual standard error: 3.8 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
What changed and what stayed the same?
One rather thorough approach to scaling is to convert all the variables to standard units (mean 0 and
variance 1) using the scale() command:
> scsav <- data.frame(scale(savings))
> g <- lm(sr ˜ ., scsav)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
4.0e-16
0.1200 3.3e-15
1.0000
pop15
-0.9420
0.2954
-3.19
0.0026
pop75
-0.4873
0.3122
-1.56
0.1255
dpi
-0.0745
0.2059
-0.36
0.7192
ddpi
0.2624
0.1257
2.09
0.0425
Residual standard error: 0.849 on 45 degrees of freedom
Multiple R-Squared: 0.338,
Adjusted R-squared: 0.28
F-statistic: 5.76 on 4 and 45 degrees of freedom,
p-value: 0.00079
As may be seen, the intercept is zero. This is because the regression plane always runs through the
point of the averages which because of the centering is now at the origin. Such scaling has the advantage
of putting all the predictors and the response on a comparable scale, which makes comparisons simpler. It
also allows the coefficients to be viewed as kind of partial correlation — the values will always be between
-1 and 1. It also avoids some numerical problems that can arise when variables are of very different scales.
The downside of this scaling is that the regression coefficients now represent the effect of a one standard
unit increase in the predictor on the response in standard units — this might not always be easy to interpret.
9.2
Principal Components
Recall that if the X matrix is orthogonal then testing and interpretation are greatly simplified. One purpose
for principal components is to transform the X to orthogonality. For example, consider the case with two
predictors depicted in Figure 9.1.
The original predictors, x1 and x2, are clearly correlated and so the X-matrix will not be orthogonal.
This will complicate the interpretation of the effects of x1 and x2 on the response. Suppose we rotate the
coordinate axes so that in the new system, the predictors are orthogonal. Furthermore, suppose we make
the rotation so that the first axis lies in the direction of the greatest variation in the data, the second in the
9.2. PRINCIPAL COMPONENTS
108
2
1
z2
z1
x 2
0
−1
−2
−2
−1
0
1
2
x1
Figure 9.1: Original predictors are x1 and x2, principal components are z1 and z2
second greatest direction of variation in those dimensions remaining and so on. These rotated directions, z 1
and z2 in our two predictor example, are simply linear combinations of the original predictors. This is the
geometrical description of principal components. We now indicate how these directions may be calculated.
We wish to find a rotation p p matrix U such that
Z
XU
¤
¡
and ZT Z
diag λ
λ
λ
λ
1
p
and λ1
2
p
0. Zero eigenvalues indicate non-identifiability. Since
¤
¢
¡
¡
¡
¡
¢¡¢¡¢¡£
¡¢¡¢¡
ZT Z
U T X T XU
¤
the eigenvalues of X T X are λ
λ
1
p and the eigenvectors of X T X are the columns of U . The columns of Z
¢¡¢¡¢¡£
are called the principal components and these are orthogonal to each other. λi is the variance of Zi.
Another way of looking at it is to try to find the linear combinations of X which have the maximum
¡
variation. We find the u1 such that var uT X is maximized subject to uT u
1. Now find u
1
1
2 such that
¢
1
¤
¡
var uT X is maximized subject to uT u
0 and uT u
1. We keep finding directions of greatest variation
2
2
2
¢
1
¤
2
¤
orthogonal to those directions we have already found.
Ideally, only a few eigenvalues will be large so that almost all the variation in X will be representable by
those first few principal components.
Principal components can be effective in some situations but
1. The principal components are linear combinations of the predictors — little is gained if these are
not interpretable. Generally the predictors have to be measurements of comparable quantities for
interpretation to be possible
2. Principal components does not use the response. It’s possible that a lesser principal component is
actually very important in predicting the response.
3. There are variations which involve not using the intercept in X T X or using the correlation matrix of
the predictors instead of X T X .
9.2. PRINCIPAL COMPONENTS
109
We use the Longley data for this example: First we compute the eigendecomposition for X T X :
> data(longley)
> x <- as.matrix(longley[,-7])
> e <- eigen(t(x) %*% x)
Look at the eigenvalues:
> e$values
[1] 6.6653e+07 2.0907e+05 1.0536e+05 1.8040e+04 2.4557e+01 2.0151e+00
Look at the relative size - the first is big. Consider the first eigenvector (column) below:
> dimnames(e$vectors)[[2]] <- paste("EV",1:6)
> round(e$vec,3)
EV 1
EV 2
EV 3
EV 4
EV 5
EV 6
GNP def 0.050 -0.070 -0.034
0.043 -0.957 -0.273
GNP
0.191 -0.725 -0.343
0.554
0.075
0.087
Unem
0.157 -0.622
0.564 -0.521
0.007
0.011
Armed
0.128 -0.104 -0.746 -0.645
0.012
0.000
Popn
0.058 -0.038 -0.011
0.036
0.281 -0.956
Year
0.957
0.266
0.078
0.057
0.015
0.053
The first eigenvector is dominated by year. Now examining the X-matrix. What are the scales of the
variables?
> x
GNP deflator
GNP Unemployed Armed Forces Population Year
1947
83.0 234.289
235.6
159.0
107.608 1947
1948
88.5 259.426
232.5
145.6
108.632 1948
1949
88.2 258.054
368.2
161.6
109.773 1949
1950
89.5 284.599
335.1
165.0
110.929 1950
1951
96.2 328.975
209.9
309.9
112.075 1951
1952
98.1 346.999
193.2
359.4
113.270 1952
1953
99.0 365.385
187.0
354.7
115.094 1953
1954
100.0 363.112
357.8
335.0
116.219 1954
1955
101.2 397.469
290.4
304.8
117.388 1955
1956
104.6 419.180
282.2
285.7
118.734 1956
1957
108.4 442.769
293.6
279.8
120.445 1957
1958
110.8 444.546
468.1
263.7
121.950 1958
1959
112.6 482.704
381.3
255.2
123.366 1959
1960
114.2 502.601
393.1
251.4
125.368 1960
1961
115.7 518.173
480.6
257.2
127.852 1961
1962
116.9 554.894
400.7
282.7
130.081 1962
We see that the variables have different scales. It might make more sense to standardize the predictors before
trying principal components. This is equivalent to doing principal components on the correlation matrix:
9.2. PRINCIPAL COMPONENTS
110
> e <- eigen(cor(x))
> e$values
[1] 4.60337710 1.17534050 0.20342537 0.01492826 0.00255207 0.00037671
> dimnames(e$vectors) <- list(c("GNP def","GNP","Unem","Armed","Popn",
"Year"),paste("EV",1:6))
> round(e$vec,3)
EV 1
EV 2
EV 3
EV 4
EV 5
EV 6
GNP def 0.462
0.058 -0.149
0.793
0.338
0.135
GNP
0.462
0.053 -0.278 -0.122 -0.150 -0.818
Unem
0.321 -0.596
0.728
0.008
0.009 -0.107
Armed
0.202
0.798
0.562 -0.077
0.024 -0.018
Popn
0.462 -0.046 -0.196 -0.590
0.549
0.312
Year
0.465
0.001 -0.128 -0.052 -0.750
0.450
One commonly used method of judging how many principal components are worth considering is the
scree plot — see Figure 9.2, which is produced by
> plot(e$values,type="l",xlab="EV no.")
4
3
2
1
0
1
2
3
4
5
6
EV no.
Figure 9.2: Scree plot for the principal components on the correlation of longley predictors
Often, these plots have a noticeable “elbow” — the point at which further eigenvalues are negligible in
size compared to the earlier ones. Here the elbow is at 3 telling us that we need only consider 2 principal
components.
One advantage of principal components is that it transforms the predictors to an orthogonal basis. To
figure out the orthogonalized predictors for this data based on the eigendecomposition for the correlation
matrix we must first standardize the data: The functions scale() does this:
> nx <- scale(x)
We can now create the orthogonalized predictors — the Z
XU operation in our description above.
¤
9.2. PRINCIPAL COMPONENTS
111
> enx <- nx %*% e$vec
and fit:
> g <- lm(longley$Emp ˜ enx)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
65.3170
0.0762
857.03
< 2e-16
enxEV 1
1.5651
0.0367
42.66
1.1e-11
enxEV 2
0.3918
0.0726
5.40
0.00043
enxEV 3
-1.8604
0.1745
-10.66
2.1e-06
enxEV 4
0.3573
0.6442
0.55
0.59267
enxEV 5
-6.1698
1.5581
-3.96
0.00331
enxEV 6
6.9634
4.0555
1.72
0.12011
Residual standard error: 0.305 on 9 degrees of freedom
Multiple R-Squared: 0.995,
Adjusted R-squared: 0.992
F-statistic:
330 on 6 and 9 degrees of freedom,
p-value: 4.98e-10
Notice that the p-values of the 4th and 6th eigenvectors are not significant while the the 5th is. Because
the directions of the eigenvectors are set successively in the greatest remaining direction of variation in the
predictors, it is natural that they be ordered in significance in predicting the response. However, there is no
guarantee of this — we see here that the 5th eigenvectors is significant while the fourth is not even though
there is about six times greater variation in the fourth direction than the fifth. In this example, it hardly
matters since most of the variation is explained by the earlier values, but look out for this effect in other
dataset in the first few eigenvalues.
¡
We can take a look at the X T X
1 matrix:
¢
> round(summary(g)$cov.unscaled,2)
(Intercept) enxEV 1 enxEV 2 enxEV 3 enxEV 4 enxEV 5 enxEV 6
(Intercept)
0.06
0.00
0.00
0.00
0.00
0.00
0.00
enxEV 1
0.00
0.01
0.00
0.00
0.00
0.00
0.00
enxEV 2
0.00
0.00
0.06
0.00
0.00
0.00
0.00
enxEV 3
0.00
0.00
0.00
0.33
0.00
0.00
0.00
enxEV 4
0.00
0.00
0.00
0.00
4.47
0.00
0.00
enxEV 5
0.00
0.00
0.00
0.00
0.00
26.12
0.00
enxEV 6
0.00
0.00
0.00
0.00
0.00
0.00
176.97
Principal components are really only useful if we can interpret the meaning of the new linear combi-
nations. Look back at the first eigenvector - this is roughly a linear combination of all the (standardized
variables). Now plot each of the variables as they change over time — see Figure 9.2.
> par(mfrow=c(3,2))
> for(i in 1:6) plot(longley[,6],longley[,i],xlab="Year",
ylab=names(longley)[i])
What do you notice? This suggests we identify the first principal component with a time trend effect.
The second principal component is roughly a contrast between numbers unemployed and in the armed
forces. Let’s try fitting a regression with those two components:
9.2.
PRINCIP
Population
Unemployed
GNP.deflator
110
115
120
125
130
200
250
300
350
400
450
85
90
95
100
105
110
115
AL
1950
1950
1950
COMPONENTS
Year
Year
Year
1955
1955
1955
Figure
1960
1960
1960
9.3:
The
Longle
Year
Armed.Forces
GNP
1950
1955
1960
150
200
250
300
350
250
300
350
400
450
500
550
y
data
1950
1950
1950
Year
Year
Year
1955
1955
1955
1960
1960
1960
112
9.3. PARTIAL LEAST SQUARES
113
> summary(lm(Employed ˜ Year + I(Unemployed-Armed.Forces),longley))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-1.39e+03
7.89e+01
-17.6
1.8e-10
Year
7.45e-01
4.04e-02
18.5
1.0e-10
I(Unemployed - Armed.Forces) -4.12e-03
1.53e-03
-2.7
0.018
Residual standard error: 0.718 on 13 degrees of freedom
Multiple R-Squared: 0.964,
Adjusted R-squared: 0.958
F-statistic:
173 on 2 and 13 degrees of freedom,
p-value: 4.29e-10
This approaches the fit of the full model and is easily interpretable. We could do more work on the other
principal components.
This is illustrates a typical use of principal components for regression. Some intuition is required to form
new variables as combinations of older ones. If it works, a simplified and interpretable model is obtained,
but it doesn’t always work out that way.
9.3
Partial Least Squares
Partial Least Squares is a method for relating a set of input variables X1
Xm and outputs Y1
Yl. PLS
¢¡¢¡¢¡
¢¡¢¡¢¡
has some relationship to principal component regression (PCR). PCR regresses the response on the principal
components of X while PLS finds the best orthogonal linear combinations of X for predicting Y.
We will consider only univariate PLS — that is to say l
1 so that Y is scalar. This is the typical multiple
¤
regression setting. We will attempt to find models of the form
ˆ
y
β
β
1T1
pTp
¤
¡¢¡¢¡
where Tk is a linear combination of the X’s. See Figure 9.4
X
1
T 1
X
Y
2
T 2
X
p
Figure 9.4: Schematic representation of Partial Least Squares
9.3. PARTIAL LEAST SQUARES
114
We’ll assume that all variables have been centered to have mean 0 — this means that our intercept terms
will always be zero. Here is the algorithm for determining the T ’s.
1. Regress Y on each Xi in turn to get b1i.
2. Form
m
T1
∑w1ib1iX1i
¤
i 1
§
where the weights w1i sum to one.
3. Regress Y on T1 and each Xi on T1. The residuals from these regressions have the effect of T1 removed.
Replace Y and each Xi by the residuals of each corresponding regression.
4. Go back to step one updating the index.
There are various choices for the weighting scheme:
1. Set wi j
1 m giving each predictor equal weight.
¤
2. Set w ∝
i j
varX j. This is the most common choice. The variances of the bi j are then inversely propor-
tional to varX j which does make some sense.
The algorithm ensures that the components Ti are uncorrelated just as the principal components are
uncorrelated. This means that ˆβi will not change as more components are added to or subtracted from the
model.
For this example, we again use the Longley data. We will not need intercept terms in any of the regres-
sions because of the centering.
> x <- as.matrix(longley[,-7])
> y <- longley$Emp
> cx <- sweep(x,2,apply(x,2,mean))
> cy <- y -mean(y)
Now do the PCR using a more direct method than we used above:
> library(mva)
> ex <- princomp(cx)
> g <- lm(cy ˜ ex$scores -1)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
ex$scoresComp.1
0.025095
0.000603
41.65
1.5e-12
ex$scoresComp.2
0.014038
0.000888
15.82
2.1e-08
ex$scoresComp.3
0.029991
0.002152
13.94
7.1e-08
ex$scoresComp.4 -0.061765
0.058332
-1.06
0.3146
ex$scoresComp.5 -0.489877
0.200645
-2.44
0.0348
ex$scoresComp.6
1.762076
0.443363
3.97
0.0026
Residual standard error: 0.289 on 10 degrees of freedom
Multiple R-Squared: 0.995,
Adjusted R-squared: 0.993
F-statistic:
367 on 6 and 10 degrees of freedom,
p-value: 3.94e-11
9.3. PARTIAL LEAST SQUARES
115
Are the principal component scores ordered in terms of their importance in predicting the response?
Now for later comparison, we have the regression on just first PC.
> g <- lm(cy ˜ ex$scores[,1] -1)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
ex$scores[, 1]
0.0251
0.0034
7.38
2.3e-06
Residual standard error: 1.63 on 15 degrees of freedom
Multiple R-Squared: 0.784,
Adjusted R-squared: 0.77
F-statistic: 54.5 on 1 and 15 degrees of freedom,
p-value: 2.28e-06
Now we compute the first component of the partial least squares regression:
> b1 <- numeric(6)
> for(i in 1:6){
+ b1[i] <- crossprod(cx[,i],cy)/crossprod(cx[,i],cx[,i])
+ }
> b1
[1] 0.315966 0.034752 0.018885 0.023078 0.484878 0.716512
> ncx <- sweep(cx,2,b1,"*")
> t1 <- apply(ncx,1,mean)
Here we have a chosen an equal weighting for the variables. Now see how well this component predicts
the response:
> gpls1 <- lm(cy ˜ t1 -1)
> summary(gpls1)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
t1
1.3108
0.0959
13.7
7.1e-10
Residual standard error: 0.957 on 15 degrees of freedom
Multiple R-Squared: 0.926,
Adjusted R-squared: 0.921
F-statistic:
187 on 1 and 15 degrees of freedom,
p-value: 7.11e-10
Compare this to the result above for one principal component.
An alternative weighting scheme assigns the weights proportional to the variances of the variables:
> varx <- apply(cx,2,var)
> vncx <- sweep(ncx,2,varx,"*")
> t1a <- apply(vncx,1,sum)/sum(varx)
> gpls1a <- lm(cy ˜ t1a -1)
> summary(gpls1a)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
t1a
1.605
0.154
10.4
2.8e-08
9.3. PARTIAL LEAST SQUARES
116
Residual standard error: 1.22 on 15 degrees of freedom
Multiple R-Squared: 0.879,
Adjusted R-squared: 0.871
F-statistic:
109 on 1 and 15 degrees of freedom,
p-value: 2.81e-08
Compare this to the other output. Now we compute the second component of the PLS. We need to
regress out the effect of the first component and then use the same computational method as above.
> cx2 <- matrix(0,16,6)
> for(i in 1:6){
+ cx2[,i] <- lm(cx[,i] ˜ t1-1)$res
+ }
> cy2 <- lm(cy ˜ t1 -1)$res
> b2 <- numeric(6)
> for(i in 1:6){
+ b2[i] <- crossprod(cx2[,i],cy2)/crossprod(cx2[,i],cx2[,i])
+ }
> ncx2 <- sweep(cx2,2,b2,"*")
> t2 <- apply(ncx2,1,mean)
Notice the correlation of the components:
> cor(t1,t2)
[1] 9.0843e-19
Now add t2 to the regression:
> gpls2 <- lm(cy ˜ t1+t2 -1)
> summary(gpls2)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
t1
1.3108
0.0749
17.49
6.5e-11
t2
10.9309
3.3658
3.25
0.0058
Residual standard error: 0.748 on 14 degrees of freedom
Multiple R-Squared: 0.958,
Adjusted R-squared: 0.952
F-statistic:
158 on 2 and 14 degrees of freedom,
p-value: 2.44e-10
Compare the coefficient of t1 with that above. Now compare this fit to the two component PCR.
> g <- lm(cy ˜ ex$scores[,1:2] -1)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
ex$scores[, 1:2]Comp.1
0.02510
0.00243
10.34
6.2e-08
ex$scores[, 1:2]Comp.2
0.01404
0.00358
3.93
0.0015
Residual standard error: 1.16 on 14 degrees of freedom
Multiple R-Squared: 0.897,
Adjusted R-squared: 0.883
F-statistic: 61.2 on 2 and 14 degrees of freedom,
p-value: 1.2e-07
9.4. COLLINEARITY
117
Which one is superior in explaining y?
Notes:
The tricky part is choosing how many components are required. Crossvalidation is a possible way of
selecting the number of components.
There are other faster versions of the algorithm described above but these generally provide less
insight into the method.
PLS has been criticized as an algorithm that solves no well-defined modeling problem.
PLS has the biggest advantage over ordinary least squares and PCR when there are large numbers of
variables relative to the number of case. It does not even require that n
p.
¡
PCR and PLS compared
PCR attempts to find linear combinations of the predictors that explain most of the variation in these
predictors using just a few components. The purpose is dimension reduction. Because the principal compo-
nents can be linear combinations of all the predictors, the number of variables used is not always reduced.
Because the principal components are selected using only the X-matrix and not the response, there is no
definite guarantee that the PCR will predict the response particularly well although this often happens. If
it happens that we can interpret the principal components in a meaningful way, we may achieve a much
simpler explanation of the response. Thus PCR is geared more towards explanation than prediction.
In contrast, PLS finds linear combinations of the predictors that best explain the response. It is most
effective when ther are large numbers of variables to be considered. If successful, the variablity of prediction
is substantially reduced. On the other hand, PLS is virtually useless for explanation purposes.
9.4
Collinearity
If X T X is singular, i.e. some predictors are linear combinations of others, we have (exact) collinearity and
there is no unique least squares estimate of β. If X T X is close to singular, we have (approximate) collinearity
or multicollinearity (some just call it collinearity). This causes serious problems with the estimation of β
and associated quantities as well as the interpretation. Collinearity can be detected in several ways:
1. Examination of the correlation matrix of the predictors will reveal large pairwise collinearities.
2. A regression of xi on all other predictors gives R2. Repeat for all predictors. R2 close to one indicates
i
i
a problem — the offending linear combination may be found.
3. Examine the eigenvalues of X T X - small eigenvalues indicate a problem. The condition number is
defined as
λ
κ
1
¤
λp
where κ
30 is considered large. κ is called the condition number. Other condition numbers,
λ λ
1
i
¡
are also worth considering because they indicate whether more than just one independent linear com-
bination is to blame.
Collinearity makes some of the parameters hard to estimate. Define
¡
S
2
x
∑ x ¯x
j x j
i j
j
¤
¢
i
9.4. COLLINEARITY
118
then
1
1
var ˆβ
σ2
j ¤
1
R2
Sx
j
j x j
¡
We can see that if x j doesn’t vary much then the variance of ˆβ j will be large. As an aside, the variance of
the first principal component is maximized and so the variance of the corresponding regression coefficient
will tend to be small. Another consequence of this equation is that it tells us what designs will minimize
the variance of the regression coefficients if we have the ability to place the X . Orthogonality means that
R2j
0 which minimizes the variance. Also we can maximize Sx
by spreading X as much as possible. The
¤
j x j
maximum is attained by placing half the points at the minimum practical value and half at the maximum.
Unfortunately, this design assumes the linearity of the effect and would make it impossible to check for any
curvature. So, in practice, most would put some design points in the middle of the range to allow checking
of the fit.
If R2 is close to one then the variance inflation factor
1
will be large. For orthogonal designs and
j
1 R2j
principal components, R2
0, so in these case, we see that the regression coefficient estimation suffers no
j ¤
additional penalty in terms of precision.
Collinearity leads to
1. imprecise estimates of β — the signs of the coefficients may be misleading.
2. t-tests which fail to reveal significant factors
3. missing importance of predictors
The Longley dataset is a good example of collinearity:
> g <- lm(Employed ˜ ., longley)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-3.48e+03
8.90e+02
-3.91
0.00356
GNP.deflator
1.51e-02
8.49e-02
0.18
0.86314
GNP
-3.58e-02
3.35e-02
-1.07
0.31268
Unemployed
-2.02e-02
4.88e-03
-4.14
0.00254
Armed.Forces -1.03e-02
2.14e-03
-4.82
0.00094
Population
-5.11e-02
2.26e-01
-0.23
0.82621
Year
1.83e+00
4.55e-01
4.02
0.00304
Residual standard error: 0.305 on 9 degrees of freedom
Multiple R-Squared: 0.995,
Adjusted R-squared: 0.992
F-statistic:
330 on 6 and 9 degrees of freedom,
p-value: 4.98e-10
Recall that the response is number employed. Three of the predictors have large p-values but all are
variables that might be expected to affect the response. Why aren’t they significant? Check the correlation
matrix first (rounding to 3 digits for convenience)
> round(cor(longley[,-7]),3)
GNP deflator
GNP Unemployed Armed Forces Population
Year
GNP deflator
1.000 0.992
0.621
0.465
0.979 0.991
9.4. COLLINEARITY
119
GNP
0.992 1.000
0.604
0.446
0.991 0.995
Unemployed
0.621 0.604
1.000
-0.177
0.687 0.668
Armed Forces
0.465 0.446
-0.177
1.000
0.364 0.417
Population
0.979 0.991
0.687
0.364
1.000 0.994
Year
0.991 0.995
0.668
0.417
0.994 1.000
There are several large pairwise correlations. Now we check the eigendecomposition:
> x <- as.matrix(longley[,-7])
> e <- eigen(t(x) %*% x)
> e$val
[1] 6.6653e+07 2.0907e+05 1.0536e+05 1.8040e+04 2.4557e+01 2.0151e+00
> sqrt(e$val[1]/e$val)
[1]
1.000
17.855
25.153
60.785 1647.478 5751.216
There is a wide range in the eigenvalues and several condition numbers are large. This means that
problems are being caused by more than just one linear combination. Now check out the variance inflation
factors. For the first variable this is
> summary(lm(x[,1] ˜ x[,-1]))$r.squared
[1] 0.99262
> 1/(1-0.99262)
[1] 135.5
which is large - the VIF for orthogonal predictors is 1. Now we compute all the VIF’s in one go:
> vif(x)
[1]
135.5324 1788.5135
33.6189
3.5889
399.1510
758.9806
¡
There’s definitely a lot of variance inflation! For example, we can interpret
1788
42 as telling us
¢
that the standard error for GNP is 42 times larger than it would have been without collinearity. How can
we get rid of this problem? One way is to throw out some of the variables. Examine the full correlation
matrix above. Notice that variables 3 and 4 do not have extremely large pairwise correlations with the other
variables so we should keep them and focus on the others for candidates for removal:
> cor(x[,-c(3,4)])
GNP.deflator
GNP Population
Year
GNP.deflator
1.00000 0.99159
0.97916 0.99115
GNP
0.99159 1.00000
0.99109 0.99527
Population
0.97916 0.99109
1.00000 0.99395
Year
0.99115 0.99527
0.99395 1.00000
These four variables are strongly correlated with each other - any one of them could do the job of
representing the other. We pick year arbitrarily:
> summary(lm(Employed ˜ Armed.Forces + Unemployed + Year,longley))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-1.80e+03
6.86e+01
-26.18
5.9e-12
9.5. RIDGE REGRESSION
120
Armed.Forces -7.72e-03
1.84e-03
-4.20
0.0012
Unemployed
-1.47e-02
1.67e-03
-8.79
1.4e-06
Year
9.56e-01
3.55e-02
26.92
4.2e-12
Residual standard error: 0.332 on 12 degrees of freedom
Multiple R-Squared: 0.993,
Adjusted R-squared: 0.991
F-statistic:
555 on 3 and 12 degrees of freedom,
p-value: 3.92e-13
Comparing this with the original fit, we see that the fit is very similar but only three rather than six
predictors are used.
One final point - extreme collinearity can cause problems in computing the estimates - look what happens
when we use the direct formula for ˆβ.
> x <- as.matrix(cbind(1,longley[,-7]))
> solve(t(x) %*% x) %*% t(x) %*% longley[,7]
Error: singular matrix ‘a’ in solve
R , like most statistical packages, uses a more numerically stable method for computing the estimates in
lm(). Something more like this:
> solve(t(x) %*% x , t(x) %*% longley$Emp, tol = 1e-12)
[,1]
[1,] -3.4822e+03
[2,]
1.5061e-02
[3,] -3.5818e-02
[4,] -2.0202e-02
[5,] -1.0332e-02
[6,] -5.1110e-02
[7,]
1.8291e+00
Collinearity can be interpreted geometrically. Imagine a table — as two diagonally opposite legs are
moved closer together, the table becomes increasing unstable.
The effect of collinearity on prediction depends on where the prediction is to be made. The greater the
distance from the observed data, the more unstable the prediction. Distance needs to be considered in a
Mahalanobis sense rather than Euclidean.
One cure for collinearity is amputation — too many variables are trying to do the same job of explaining
the response. When several variables which are highly correlated are each associated with the response, we
have to take care that we don’t conclude that the variables we drop have nothing to do with the response.
9.5
Ridge Regression
Ridge regression makes the assumption that the regression coefficients (after normalization) are not likely
to be very large. It is appropriate for use when the design matrix is collinear and the usual least squares
estimates of β appear to be unstable.
Suppose that the predictors have been centered by their means and scaled by their standard deviations
and that the response has been centered. The ridge regression estimates of β are then given by
ˆβ
¡
X T X
λI 1XT y
¤
¢
9.5. RIDGE REGRESSION
121
The ridge constant λ is usually selected from the range 0 1 .
¢
The use of ridge regression can be motivated in two ways. Suppose we take a Bayesian point of view and
put a prior (multivariate normal) distribution on β that expresses the belief that smaller values of β are more
likely than larger ones. Large values of λ correspond to a belief that the β are really quite small whereas
smaller values of λ correspond to a more relaxed belief about β. This is illustrated in Figure 9.5.
4
3
x 2
2
1
0
0
1
2
3
4
5
6
x1
Figure 9.5: Ridge regression illustrated. The least squares estimate is at the center of the ellipse while the
ridge regression is the point on the ellipse closest to the origin. The ellipse is a contour of equal density of
the posterior probability, which in this case will be comparable to a confidence ellipse. λ controls the size
of the ellipse - the larger λ is, the larger the ellipse will be
Another way of looking at is to suppose we place to some upper bound on βT β and then compute the
least squares estimate of β subject to this restriction. Use of Lagrange multipliers leads to ridge regression.
The choice of λ corresponds to the choice of upper bound in this formulation.
λ may be chosen by automatic methods but it is probably safest to plot the values of ˆβ as a function of
λ. You should pick the smallest value of λ that produces stable estimates of β.
We demonstrate the method on the Longley data. λ
0 corresponds to least squares while we see that
¤
as λ
∞, ˆβ
0.
§
§
> library(MASS)
> gr <- lm.ridge(Employed ˜ .,longley,lambda = seq(0,0.1,0.001))
> matplot(gr$lambda,t(gr$coef),type="l",xlab=expression(lambda),
ylab=expression(hat(beta)))
> abline(h=0,lwd=2)
9.5. RIDGE REGRESSION
122
8
6
4
^ β
2
0
−2
0.00
0.02
0.04
0.06
0.08
0.10
λ
Figure 9.6: Ridge trace plot for the Longley data. The vertical line is the Hoerl-Kennard choice of λ. The
topmost curve represent the coefficient for year. The dashed line that starts well below zero but ends above
is for GNP.
The ridge trace plot is shown in Figure 9.5.
Various automatic selections for λ are available
> select(gr)
modified HKB estimator is 0.0042754
modified L-W estimator is 0.032295
smallest value of GCV
at 0.003
> abline(v=0.00428)
The Hoerl-Kennard (the originators of ridge regression) choice of λ has been shown on the plot but I
would prefer a larger value of 0.03. For this choice of λ, the ˆβ’s are
> gr$coef[,gr$lam == 0.03]
GNP.deflator
GNP
Unemployed Armed.Forces
Population
Year
0.22005
0.76936
-1.18941
-0.52234
-0.68618
4.00643
in contrast to the least squares estimates of
> gr$coef[,1]
GNP.deflator
GNP
Unemployed Armed.Forces
Population
Year
0.15738
-3.44719
-1.82789
-0.69621
-0.34420
8.43197
Note that these values are based on centered and scaled predictors which explains the difference from
previous fits. Consider the change in the coefficient for GNP. For the least squares fit, the effect of GNP is
negative on the response - number of people employed. This is counter-intuitive since we’d expect the effect
to be positive. The ridge estimate is positive which is more in line with what we’d expect.
9.5. RIDGE REGRESSION
123
Ridge regression estimates of coefficients are biased. Bias is undesirable but there are other considera-
tions. The mean squared error can be decomposed in the following way:
¡
¡
¡
¡
E ˆβ
β 2
E ˆβ
β 2 E ˆβ E ˆβ 2
¢
¤
¢
¢
¢
Thus the mean-squared error of an estimate can be represented as the square of the bias plus the variance.
Sometimes a large reduction in the variance may obtained at the price of an increase in the bias. If the MSE
is reduced as a consequence then we may be willing to accept some bias. This is the trade-off that Ridge
Regression makes - a reduction in variance at the price of an increase in bias. This is a common dilemma.
Chapter 10
Variable Selection
Variable selection is intended to select the “best” subset of predictors. But why bother?
1. We want to explain the data in the simplest way — redundant predictors should be removed. The
principle of Occam’s Razor states that among several plausible explanations for a phenomenon, the
simplest is best. Applied to regression analysis, this implies that the smallest model that fits the data
is best.
2. Unnecessary predictors will add noise to the estimation of other quantities that we are interested in.
Degrees of freedom will be wasted.
3. Collinearity is caused by having too many variables trying to do the same job.
4. Cost: if the model is to be used for prediction, we can save time and/or money by not measuring
redundant predictors.
Prior to variable selection:
1. Identify outliers and influential points - maybe exclude them at least temporarily.
2. Add in any transformations of the variables that seem appropriate.
10.1
Hierarchical Models
Some models have a natural hierarchy. For example, in polynomial models, x2 is a higher order term than x.
When selecting variables, it is important to respect the hierarchy. Lower order terms should not be removed
from the model before higher order terms in the same variable. There two common situations where this
situation arises:
Polynomials models. Consider the model
y
β
β
β
ε
0
1x
2x2
¤
Suppose we fit this model and find that the regression summary shows that the term in x is not signif-
icant but the term in x2 is. If we then removed the x term, our reduced model would then become
y
β
β
ε
0
2x2
¤
124
10.2. STEPWISE PROCEDURES
125
but suppose we then made a scale change x
x
a, then the model would become
§
y
β
β
β
ε
0
2a2
2β2ax
2x2
¤
¡
The first order x term has now reappeared. Scale changes should not make any important change to
the model but in this case an additional term has been added. This is not good. This illustrates why
we should not remove lower order terms in the presence of higher order terms. We would not want
interpretation to depend on the choice of scale. Removal of the first order term here corresponds to
the hypothesis that the predicted response is symmetric about and has an optimum at x
0. Often this
¤
hypothesis is not meaningful and should not be considered. Only when this hypothesis makes sense
in the context of the particular problem could we justify the removal of the lower order term.
Models with interactions. Consider the second order response surface model:
y
β
β
β
β
β
β
0
1x1
2x2
11x2
22x2
12x1x2
¤
1
2
We would not normally consider removing the x1x2 interaction term without simultaneously consid-
ering the removal of the x2 and x2 terms. A joint removal would correspond to the clearly meaningful
1
2
comparison of a quadratic surface and linear one. Just removing the x1x2 term would correspond to
a surface that is aligned with the coordinate axes. This is hard to interpret and should not be con-
sidered unless some particular meaning can be attached. Any rotation of the predictor space would
reintroduce the interaction term and, as with the polynomials, we would not ordinarily want our model
interpretation to depend on the particular basis for the predictors.
10.2
Stepwise Procedures
Backward Elimination
This is the simplest of all variable selection procedures and can be easily implemented without special
software. In situations where there is a complex hierarchy, backward elimination can be run manually while
taking account of what variables are eligible for removal.
1. Start with all the predictors in the model
2. Remove the predictor with highest p-value greater than αcrit
3. Refit the model and goto 2
4. Stop when all p-values are less than αcrit .
The αcrit is sometimes called the “p-to-remove” and does not have to be 5%. If prediction performance
is the goal, then a 15-20% cut-off may work best, although methods designed more directly for optimal
prediction should be preferred.
10.2.1
Forward Selection
This just reverses the backward method.
1. Start with no variables in the model.
2. For all predictors not in the model, check their p-value if they are added to the model. Choose the one
with lowest p-value less than αcrit .
3. Continue until no new predictors can be added.
10.2. STEPWISE PROCEDURES
126
10.2.2
Stepwise Regression
This is a combination of backward elimination and forward selection. This addresses the situation where
variables are added or removed early in the process and we want to change our mind about them later. At
each stage a variable may be added or removed and there are several variations on exactly how this is done.
Stepwise procedures are relatively cheap computationally but they do have some drawbacks.
1. Because of the “one-at-a-time” nature of adding/dropping variables, it’s possible to miss the “optimal”
model.
2. The p-values used should not be treated too literally. There is so much multiple testing occurring that
the validity is dubious. The removal of less significant predictors tends to increase the significance of
the remaining predictors. This effect leads one to overstate the importance of the remaining predictors.
3. The procedures are not directly linked to final objectives of prediction or explanation and so may not
really help solve the problem of interest. With any variable selection method, it is important to keep
in mind that model selection cannot be divorced from the underlying purpose of the investigation.
Variable selection tends to amplify the statistical significance of the variables that stay in the model.
Variables that are dropped can still be correlated with the response. It would be wrong to say these
variables are unrelated to the response, it’s just that they provide no additional explanatory effect
beyond those variables already included in the model.
4. Stepwise variable selection tends to pick models that are smaller than desirable for prediction pur-
poses. To give a simple example, consider the simple regression with just one predictor variable.
Suppose that the slope for this predictor is not quite statistically significant. We might not have
enough evidence to say that it is related to y but it still might be better to use it for predictive purposes.
We illustrate the variable selection methods on some data on the 50 states - the variables are popula-
tion estimate as of July 1, 1975; per capita income (1974); illiteracy (1970, percent of population); life
expectancy in years (1969-71); murder and non-negligent manslaughter rate per 100,000 population (1976);
percent high-school graduates (1970); mean number of days with min temperature
32 degrees (1931-
1960) in capital or large city; and land area in square miles. The data was collected from US Bureau of
the Census. We will take life expectancy as the response and the remaining variables as predictors - a fix is
necessary to remove spaces in some of the variable names.
> data(state)
> statedata <- data.frame(state.x77,row.names=state.abb,check.names=T)
> g <- lm(Life.Exp ˜ ., data=statedata)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.09e+01
1.75e+00
40.59
< 2e-16
Population
5.18e-05
2.92e-05
1.77
0.083
Income
-2.18e-05
2.44e-04
-0.09
0.929
Illiteracy
3.38e-02
3.66e-01
0.09
0.927
Murder
-3.01e-01
4.66e-02
-6.46
8.7e-08
HS.Grad
4.89e-02
2.33e-02
2.10
0.042
Frost
-5.74e-03
3.14e-03
-1.82
0.075
Area
-7.38e-08
1.67e-06
-0.04
0.965
10.2. STEPWISE PROCEDURES
127
Residual standard error: 0.745 on 42 degrees of freedom
Multiple R-Squared: 0.736,
Adjusted R-squared: 0.692
F-statistic: 16.7 on 7 and 42 degrees of freedom,
p-value: 2.53e-10
Which predictors should be included - can you tell from the p-values? Looking at the coefficients, can
you see what operation would be helpful? Does the murder rate decrease life expectancy - that’s obvious a
priori, but how should these results be interpreted?
We illustrate the backward method - at each stage we remove the predictor with the largest p-value over
0.05:
> g <- update(g, . ˜ . - Area)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.10e+01
1.39e+00
51.17
< 2e-16
Population
5.19e-05
2.88e-05
1.80
0.079
Income
-2.44e-05
2.34e-04
-0.10
0.917
Illiteracy
2.85e-02
3.42e-01
0.08
0.934
Murder
-3.02e-01
4.33e-02
-6.96
1.5e-08
HS.Grad
4.85e-02
2.07e-02
2.35
0.024
Frost
-5.78e-03
2.97e-03
-1.94
0.058
> g <- update(g, . ˜ . - Illiteracy)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.11e+01
1.03e+00
69.07
< 2e-16
Population
5.11e-05
2.71e-05
1.89
0.066
Income
-2.48e-05
2.32e-04
-0.11
0.915
Murder
-3.00e-01
3.70e-02
-8.10
2.9e-10
HS.Grad
4.78e-02
1.86e-02
2.57
0.014
Frost
-5.91e-03
2.47e-03
-2.39
0.021
> g <- update(g, . ˜ . - Income)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
7.10e+01
9.53e-01
74.54
< 2e-16
Population
5.01e-05
2.51e-05
2.00
0.0520
Murder
-3.00e-01
3.66e-02
-8.20
1.8e-10
HS.Grad
4.66e-02
1.48e-02
3.14
0.0030
Frost
-5.94e-03
2.42e-03
-2.46
0.0180
> g <- update(g, . ˜ . - Population)
> summary(g)
10.3. CRITERION-BASED PROCEDURES
128
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 71.03638
0.98326
72.25
<2e-16
Murder
-0.28307
0.03673
-7.71
8e-10
HS.Grad
0.04995
0.01520
3.29
0.0020
Frost
-0.00691
0.00245
-2.82
0.0070
Residual standard error: 0.743 on 46 degrees of freedom
Multiple R-Squared: 0.713,
Adjusted R-squared: 0.694
F-statistic:
38 on 3 and 46 degrees of freedom,
p-value: 1.63e-12
The final removal of the Population variable is a close call. We may want to consider including this
variable if interpretation is aided. Notice that the R2 for the full model of 0.736 is reduced only slightly to
0.713 in the final model. Thus the removal of four predictors causes only a minor reduction in fit.
10.3
Criterion-based procedures
If there are p potential predictors, then there are 2 p possible models. We fit all these models and choose the
best one according to some criterion. Clever algorithms such as the “branch-and-bound” method can avoid
actually fitting all the models — only likely candidates are evaluated. Some criteria are
1. The Akaike Information Criterion (AIC) and the Bayes Information Criterion (BIC) are some other
commonly used criteria. In general,
AIC
2
2p
¤
¦¤¢¡
£
¡
¥¤©¦
¡
§¤¤¨§
while
BIC
2
p log n
¤
¤¨¡
©
¡
¤
©
¡
§¤¤¢§
¡
For linear regression models, the -2log-likelihood (known as the deviance is n log RSS n . We want to
¢
minimize AIC or BIC. Larger models will fit better and so have smaller RSS but use more parameters.
Thus the best choice of model will balance fit with model size. BIC penalizes larger models more
heavily and so will tend to prefer smaller models in comparison to AIC. AIC and BIC can be used as
selection criteria for other types of model too.
We can apply the AIC (and optionally the BIC) to the state data. The function does not evaluate
the AIC for all possible models but uses a search method that compares models sequentially. Thus
it bears some comparison to the stepwise method described above but with the advantage that no
dubious p-values are used.
> g <- lm(Life.Exp ˜ ., data=statedata)
> step(g)
Start:
AIC= -22.18
Life.Exp ˜ Population + Income + Illiteracy + Murder + HS.Grad +
Frost + Area
Df Sum of Sq
RSS
AIC
- Area
1
0.0011
23.3 -24.2
10.3. CRITERION-BASED PROCEDURES
129
- Income
1
0.0044
23.3 -24.2
- Illiteracy
1
0.0047
23.3 -24.2
<none>
23.3 -22.2
- Population
1
1.7
25.0 -20.6
- Frost
1
1.8
25.1 -20.4
- HS.Grad
1
2.4
25.7 -19.2
- Murder
1
23.1
46.4
10.3
Step:
AIC= -24.18
Life.Exp ˜ Population + Income + Illiteracy + Murder + HS.Grad +
Frost
.. intermediate steps omitted ..
Step:
AIC= -28.16
Life.Exp ˜ Population + Murder + HS.Grad + Frost
Df Sum of Sq
RSS
AIC
<none>
23.3 -28.2
- Population
1
2.1
25.4 -25.9
- Frost
1
3.1
26.4 -23.9
- HS.Grad
1
5.1
28.4 -20.2
- Murder
1
34.8
58.1
15.5
Coefficients:
(Intercept)
Population
Murder
HS.Grad
Frost
7.10e+01
5.01e-05
-3.00e-01
4.66e-02
-5.94e-03
The sequence of variable removal is the same as with backward elimination. The only difference is
the the Population variable is retained.
2. Adjusted R2 — called R2a. Recall that R2
1
RSS T SS. Adding a variable to a model can only
¤
decrease the RSS and so only increase the R2 so R2 by itself is not a good criterion because it would
always choose the largest possible model.
§
¡
RSS n
p
n
1
ˆ
σ2
¡
R2
¢
model
a
1
1
1
R2
1
¡
¤
¤
¢
¤
T SS n
1
n
p
ˆ
σ2
¢
¨
null
Adding a predictor will only increase R2a if it has some value. Do you see the connection to ˆσ2?
Minimizing the standard error for prediction means minimizing ˆ
σ2 which in term means maximizing
R2a.
3. Predicted Residual Sum of Squares (PRESS) is defined as ∑i ˆε2 where the ˆε
are the residuals
i
i¡
¡
calculated without using case i in the fit. The model with the lowest PRESS criterion is then selected.
This tends to pick larger models (which may be desirable if prediction is the objective).
4. Mallow’s Cp Statistic. A good model should predict well so average MSE of prediction might be a
good criterion:
1
¡
2
σ ∑E ˆy Ey
2
i
i
¢
i
10.3. CRITERION-BASED PROCEDURES
130
which can be estimated by the Cp statistic
RSSp
Cp
2p
n
¤
ˆ
σ2
where ˆ
σ2 is from the model with all predictors and RSSp indicates the RSS from a model with p
parameters.
(a) Cp is easy to compute
(b) It is closely related to R2a and the AIC.
(c) For the full model Cp
p exactly.
¤
¡
¡
¡
(d) If a p predictor model fits then E RSS
σ2
p
n
p
and then E Cp
p. A model with a bad
¢
¤
¢
¢
fit will have Cp much bigger than p.
It is usual to plot Cp against p. We desire models with small p and Cp around or less than p.
Now we try the Cp and R2a methods for the selection of variables in the State dataset. The default for the
leaps() function is the Mallow’s Cp criterion:
> library(leaps)
> x <- model.matrix(g)[,-1]
> y <- statedata$Life
> g <- leaps(x,y)
> Cpplot(g)
146
12346
234567
123457
246
1467
2346
13467
1246
8
1234567
1457
1245
23456
24567
13457
34567
12345
6
123456
134567
124567
Cp
145
2456
3456
4567
1345
4
456
12456
13456
14567
2
1456
4
5
6
7
8
p
Figure 10.1: The Cp plot for the State data
The models are denoted by indices for the predictors. The competition is between the “456” model i.e.
the Frost, HS graduation and Murder model and the model also including Population. Both models are on
or below the Cp
p line, indicating good fits. The choice is between the smaller model and the larger model
¤
10.3. CRITERION-BASED PROCEDURES
131
which fits a little better. Some even larger models fit in the sense that they are on or below the C p
p line but
¤
we would not opt for these in the presence of smaller models that fit. Smaller models with 1 or 2 predictors
are not shown on this plot because their Cp plots are so large.
Now let’s see which model the adjusted R2 criterion selects.
> adjr <- leaps(x,y,method="adjr2")
> maxadjr(adjr,8)
1456
12456
13456
14567 123456 134567 124567
456
0.713
0.706
0.706
0.706
0.699
0.699
0.699
0.694
We see that the Population, Frost, HS graduation and Murder model has the largest R2a. The best three
predictor model is in eighth place but the intervening models are not attractive since they use more predictors
than the best model.
Variable selection methods are sensitive to outliers and influential points. Let’s check for high leverage
points:
> h <- hat(x)
> names(h) <- state.abb
> rev(sort(h))
AK
CA
HI
NV
NM
TX
NY
WA
0.809522 0.408857 0.378762 0.365246 0.324722 0.284164 0.256950 0.222682
Which state sticks out? Let’s try excluding it (Alaska is the second state in the data).
> l <- leaps(x[-2,],y[-2],method="adjr2")
> maxadjr(l)
12456
1456 123456
0.710
0.709
0.707
We see that area now makes it into the model. Transforming the predictors can also have an effect: Take
a look at the variables:
> par(mfrow=c(3,3))
> for(i in 1:8) boxplot(state.x77[,i],main=dimnames(state.x77)[[2]][i])
In Figure 10.3, we see that Population, Illiteracy and Area are skewed - we try transforming them:
> nx <- cbind(log(x[,1]),x[,2],log(x[,3]),x[,4:6],log(x[,7]))
And now replot:
> par(mfrow=c(3,3))
> apply(nx,2,boxplot)
which shows the appropriately transformed data.
Now try the adjusted R2 method again.
> a <- leaps(nx,y,method="adjr2")
> maxadjr(a)
1456 12456 13456
0.717 0.714 0.712
This changes the ”best” model again to log(Population), Frost, HS graduation and Murder.
The adjusted R2 is the highest models we have seen so far.
10.3.
CRITERION-B
0
50
100
150
68
69
70
71
72
73
0
5000
10000
15000
20000
Population
Life Exp
Frost
ASED
PR
OCEDURES
Figure
10.2:
0e+00
2e+05
4e+05
2
4
6
8
10
12
14
3000
4000
5000
6000
Boxplots
Murder
Income
Area
of
the
State
data
40
45
50
55
60
65
0.5
1.0
1.5
2.0
2.5
HS Grad
Illiteracy
132
10.4. SUMMARY
133
10.4
Summary
Variable selection is a means to an end and not an end itself. The aim is to construct a model that predicts well
or explains the relationships in the data. Automatic variable selections are not guaranteed to be consistent
with these goals. Use these methods as a guide only.
Stepwise methods use a restricted search through the space of potential models and use a dubious hy-
pothesis testing based method for choosing between models. Criterion-based methods typically involve a
wider search and compare models in a preferable manner. For this reason, I recommend that you use a
criterion-based method.
Accept the possibility that several models may be suggested which fit about as well as each other. If this
happens, consider:
1. Do the models have similar qualitative consequences?
2. Do they make similar predictions?
3. What is the cost of measuring the predictors?
4. Which has the best diagnostics?
If you find models that seem roughly equally as good but lead to quite different conclusions then it is
clear that the data cannot answer the question of interest unambiguously. Be alert to the danger that a model
contradictory to the tentative conclusions might be out there.
Chapter 11
Statistical Strategy and Model Uncertainty
11.1
Strategy
Thus far we have learnt various tactics
1. Diagnostics: Checking of assumptions: constant variance, linearity, normality, outliers, influential
points, serial correlation and collinearity.
2. Transformation: Transforming the response — Box-Cox, transforming the predictors — tests and
polynomial regression.
3. Variable selection: Stepwise and criterion based methods
What order should these be done in? Should procedures be repeated at later stages? When should we
stop?
I would recommend Diagnostics
Transformation
Variable Selection
Diagnostics as a rudimen-
§
§
§
tary strategy. However, regression analysis is a search for structure in data and there are no hard-and-fast
rules about how it should be done. Regression analysis requires some skill. You must be alert to unexpected
structure in the data. Thus far, no one has implemented a computer program for conducting a complete
analysis. Because of the difficulties in automating the assessment of regression graphics in an intelligent
manner, I do not expect that this will accomplished soon. The human analyst has the ability to assess plots
in light of contextual information about the data.
There is a danger of doing too much analysis. The more transformations and permutations of leaving
out influential points you do, the better fitting model you will find. Torture the data long enough, and sooner
or later it will confess. Remember that fitting the data well is no guarantee of good predictive performance
or that the model is a good representation of the underlying population. So
1. Avoid complex models for small datasets.
2. Try to obtain new data to validate your proposed model. Some people set aside some of their existing
data for this purpose.
3. Use past experience with similar data to guide the choice of model.
Data analysis is not an automatic process. Analysts have personal preferences in their choices of method-
ology, use software with varying capabilities and will interpret the same graphical display differently. In
comparing the competing analyses of two statisticians, it may sometimes be possible to determine that one
134
11.2. EXPERIMENT
135
analysis is clearly superior. However, in most cases, particularly when the analysts are experienced and
professionally trained, a universally acceptable judgment of superiority will not be possible.
The same data may support different models. Conclusions drawn from the models may differ quanti-
tatively and qualitatively. However, except for those well-known datasets that circulate endlessly through
textbooks and research articles, most data is only analyzed once. The analyst may be unaware that a sec-
ond independent look at the data may result in quite different conclusions. We call this problem model
multiplicity. In the next section, we describe an experiment illustrating the depth of this problem.
11.2
Experiment
In Fall 1996, I taught a semester length masters level course in applied regression analysis to 28 students.
Towards the end of the semester, I decided to set an assignment to test the students ability in building a
regression model for the purposes of prediction. I generated regression data with a response y and five
uncorrelated predictors and n
50 from a model known only to me which was:
¤
¡
1 y
x
ε
1
0 57x2
4x1x2
2 1 exp x4
¤
1
¢
¡
¡
¡
¡
¡
¡
¡
¡
¡
where x1
U 0 1 ,x2
N 0 1 ,1 x3
U 0 1 ,x4
N 1 1 , x5
U 1 3 and ε
N 0 1 .
¢
¢
¢
¢
¢
¢
I asked students to predict the mean response at 10 values of the predictors which I specified. I also asked
them to provide a standard error for each of their predictions. The students understood and were reminded
of the distinction between the standard error for the mean response and for a future observed value. The
students were told that their score on the assignment would depend only on the closeness of their predicted
values and the true values and on how closely their standard errors reflected the difference between these
two quantities. Students were told to work independently.
For a given student’s input, let pi be their prediction, ti be the true value and si be the standard error
where i
1
10. To assess their prediction accuracy, I used
¤
¢¡¢¡¢¡
10 §
2
∑ pi ti
t
i 1
i
¨
§
whereas to measure the “honesty” of their standard errors, I used
1 10
∑ p t
i
i
10
¡
s
i 1
i
§
We’d expect the predicted value to differ from the true value by typically about one standard error if the
latter has been correctly estimated. Therefore, the measure of standard error honesty should be around one.
1.12
1.20
1.46
1.46
1.54
1.62
1.69
1.69
1.79
3.14
4.03
4.61
5.04
5.06
5.13
5.60
5.76
5.76
5.94
6.25
6.53
6.53
6.69
10.20
34.45
65.53
674.98
37285.95
Table 11.1: Prediction accuracy
The prediction accuracy scores for the 28 students are shown in Table 11.1. We see that one student
did very poorly. An examination of their model and some conversation revealed that they neglected to
backtransform their predictions to the original scale when using a model with a transform on the response.
11.3. DISCUSSION
136
Three pairs of scores are identical in the table but an examination of the models used and more digits
revealed that only one pair was due to the students using the same model. This pair of students were known
associates. Thus 27 different models were found by 28 students.
The scores for honesty of standard errors are shown in Table 11.2. The order in which scores are shown
correspond to that given in Table 11.1.
0.75
7.87
6.71
0.59
4.77
8.20
11.74
10.70
1.04
17.10
3.23
14.10
84.86
15.52
80.63
17.61
14.02
14.02
13.35
16.77
12.15
12.15
12.03
68.89
101.36
18.12
2.24
40.08
Table 11.2: Honesty of standard errors - order of scores corresponds to that in Table 11.1.
We see that the students’ standard errors were typically around an order of magnitude smaller than they
should have been.
11.3
Discussion
Why was there so much model multiplicity? The students were all in the same class and used the same
software but almost everyone chose a different model. The course covered many of the commonly un-
derstood techniques for variable selection, transformation and diagnostics including outlier and influential
point detection. The students were confronted with the problem of selecting the order in which to apply
these methods and choosing from several competing methods for a given purpose.
The reason the models were so different was that students applied the various methods in different
orders. Some did variable selection before transformation and others the reverse. Some repeated a method
after the model was changed and others did not. I went over the strategies that several of the students
used and could not find anything clearly wrong with what they had done. One student made a mistake in
computing their predicted values but there was nothing obviously wrong in the remainder. The performance
on this assignment did not show any relationship with that in the exams.
The implications for statistical practice are profound. Often a dataset is analyzed by a single analyst
who comes up with a single model. Predictions and inferences are based on this single model. The analyst
may be unaware that the data support quite different models which may lead to very different conclusions.
Clearly one won’t always have a stable of 28 independent analysts to search for alternatives, but it does point
to the value of a second or third independent analysis. It may also be possible to automate the components
of the analysis to some extent as in Faraway (1994) to see whether changes in the order of analysis might
result in a different model.
Another issue is raised by the standard error results. Often we use the data to help determine the model.
Once a model is built or selected, inferences and predictions may be made. Usually inferences are based
on the assumption that the selected model was fixed in advance and so only reflect uncertainty concerning
the parameters of that model. Students took that approach here. Because the uncertainty concerning the
model itself is not allowed for, these inferences tend to be overly optimistic leading to unrealistically small
standard errors. Methods for realistic inference when the data is used to select the model have come under the
heading of Model Uncertainty — see Chatfield (1995) for a review. The effects of model uncertainty often
overshadow the parametric uncertainty and the standard errors need to be inflated to reflect this. Faraway
(1992) developed a bootstrap approach to compute these standard errors while Draper (1995) is an example
of a Bayesian approach. These methods are a step in the right direction in that they reflect the uncertainty in
11.3. DISCUSSION
137
model selection. Nevertheless, they do not address the problem of model multiplicity since they proscribe a
particular method of analysis that does not allow for differences between human analysts.
Sometimes the data speak with a clear and unanimous voice — the conclusions are incontestable. Other
times, differing conclusions may be drawn depending on the model chosen. We should acknowledge the
possibility of alternative conflicting models and seek them out.
Chapter 12
Chicago Insurance Redlining - a complete
example
In a study of insurance availability in Chicago, the U.S. Commission on Civil Rights attempted to examine
charges by several community organizations that insurance companies were redlining their neighborhoods,
i.e. canceling policies or refusing to insure or renew. First the Illinois Department of Insurance provided
the number of cancellations, non-renewals, new policies, and renewals of homeowners and residential fire
insurance policies by ZIP code for the months of December 1977 through February 1978. The companies
that provided this information account for more than 70% of the homeowners insurance policies written in
the City of Chicago. The department also supplied the number of FAIR plan policies written an renewed
in Chicago by zip code for the months of December 1977 through May 1978. Since most FAIR plan
policyholders secure such coverage only after they have been rejected by the voluntary market, rather than
as a result of a preference for that type of insurance, the distribution of FAIR plan policies is another measure
of insurance availability in the voluntary market.
Secondly, the Chicago Police Department provided crime data, by beat, on all thefts for the year 1975.
Most Insurance companies claim to base their underwriting activities on loss data from the preceding years,
i.e. a 2-3 year lag seems reasonable for analysis purposes. the Chicago Fire Department provided similar
data on fires occurring during 1975. These fire and theft data were organized by zip code.
Finally the US Bureau of the census supplied data on racial composition, income and age and value
of residential units for each ZIP code in Chicago. To adjust for these differences in the populations size
associated with different ZIP code areas, the theft data were expressed as incidents per 1,000 population and
the fire and insurance data as incidents per 100 housing units.
The variables are
race racial composition in percent minority
fire
fires per 100 housing units
theft
theft per 1000 population
age
percent of housing units built before 1939
volact
new homeowner policies plus renewals minus cancellations and non renewals per 100 housing
units
involact new FAIR plan policies and renewals per 100 housing units
138
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
139
income median family income
The data comes from the book by Andrews and Herzberg (1985). We choose the involuntary market
activity variable (the number getting FAIR plan insurance) as the response since this seems to be the best
measure of those who are denied insurance by others. It is not a perfect measure because some who are
denied insurance may give up and others still may not try at all for that reason. The voluntary market
activity variable is not as relevant.
Furthermore, we do not know the race of those denied insurance. We only know the racial composition
in the corresponding zip code. This is an important difficulty and brings up the following topic:
Ecological Correlation
When data is collected at the group level, we may observe a correlation between two variables. The
ecological fallacy is concluding that the same correlation holds at the individual level. For example, in
countries with higher fat intakes in the diet, higher rates of breast cancer have been observed. Does this imply
that individuals with high fat intakes are at a higher risk of breast cancer? Not necessarily. Relationships seen
in observational data are subject to confounding but even if this is allowed for, bias is caused by aggregating
data. We consider an example taken from US demographic data:
> data(eco)
> plot(income ˜ usborn, data=eco, xlab="Proportion US born",
ylab="Mean Annual Income")
In the first panel of Figure 12.1, we see the relationship between 1998 per capita income dollars from all
sources and the proportion of legal state residents born in the United States in 1990 for each of the 50 states
plus the District of Columbia. We can see a clear negative correlation.
35000
60000
30000
40000
25000
Mean Annual Income
Mean Annual Income
20000
20000
0.75
0.85
0.95
0.0
0.2
0.4
0.6
0.8
Proportion US born
Proportion US born
Figure 12.1: 1998 annual per capita income and proportion US born for 50 states plus DC. Plot on the right
is the same data as on the left but with an extended scale and the least squares fit shown
We can fit a regression line and show the fitted line on an extended range:
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
140
> g <- lm(income ˜ usborn, eco)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
68642
8739
7.85
3.2e-10
usborn
-46019
9279
-4.96
8.9e-06
Residual standard error: 3490 on 49 degrees of freedom
Multiple R-Squared: 0.334,
Adjusted R-squared: 0.321
F-statistic: 24.6 on 1 and 49 degrees of freedom, p-value: 8.89e-06
> plot(income ˜ usborn, data=eco, xlab="Proportion US born",
ylab="Mean Annual Income",xlim=c(0,1),ylim=c(15000,70000),xaxs="i")
> abline(g$coef)
We see that there is a clear statistical significant relationship between per capita annual income and the
proportion who are US born. What does this say about the average annual income of people who are US
born and those who are naturalized citizens? If we substitute, usborn=1 into the regression equation, we
get 68642-46019=$22,623, while if we put usborn=0, we get $68,642. This suggests that on average,
naturalized citizens are three times wealthier than US born citizens. In truth, information US Bureau of the
Census indicates that US born citizens have an average income just slightly larger than naturalized citizens.
What went wrong with our analysis?
The ecological inference from the aggregate data to the individuals requires an assumption of constancy.
Explicitly, the assumption would be that the incomes of the native-born do not depend on the proportion of
native born within the state (and similarly for naturalized citizens). This assumption is unreasonable for this
data because immigrants are naturally attracted to wealthier states.
This is also relevent to the analysis of the Chicago insurance data since we have only aggregate data.
We must keep in mind that the results for the aggregated data may not hold true at the individual level.
We will focus on the relationship between race and the response although similar analyses might be
done for the income variable.
Start by reading the data in and examining it:
> data(chicago)
> chicago
race fire theft
age volact involact income
60626 10.0
6.2
29 60.4
5.3
0.0
11744
60640 22.2
9.5
44 76.5
3.1
0.1
9323
etc.
60645
3.1
4.9
27 46.6
10.9
0.0
13731
Rescale the income variable and omit volact
> ch <- data.frame(chicago[,1:4],involact=chicago[,6],income=chicago[,7]/1000)
> ch
race fire theft
age involact income
60626 10.0
6.2
29 60.4
0.0 11.744
60640 22.2
9.5
44 76.5
0.1
9.323
etc.
60645
3.1
4.9
27 46.6
0.0
13.731
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
141
Summarize:
> summary(ch)
race
fire
theft
age
Min.
: 1.00
Min.
: 2.00
Min.
:
3.0
Min.
: 2.0
1st Qu.: 3.75
1st Qu.: 5.65
1st Qu.: 22.0
1st Qu.:48.6
Median :24.50
Median :10.40
Median : 29.0
Median :65.0
Mean
:35.00
Mean
:12.30
Mean
: 32.4
Mean
:60.3
3rd Qu.:57.60
3rd Qu.:16.10
3rd Qu.: 38.0
3rd Qu.:77.3
Max.
:99.70
Max.
:39.70
Max.
:147.0
Max.
:90.1
involact
income
Min.
:0.000
Min.
: 5.58
1st Qu.:0.000
1st Qu.: 8.45
Median :0.400
Median :10.70
Mean
:0.615
Mean
:10.70
3rd Qu.:0.900
3rd Qu.:12.00
Max.
:2.200
Max.
:21.50
We see that there is a wide range in the race variable with some zip codes being almost entirely
minority or non-minority. This is good for our analysis since it will reduce the variation in the regression
coefficient for race, allowing us to assess this effect more accurately. If all the zip codes were homogenous,
we would never be able to discover an effect from this aggregated data. We also note some skewness in the
theft and income variables. The response involact has a large number of zeroes. This is not good
for the assumptions of the linear model but we have little choice but to proceed.
Now make some graphical summaries:
> par(mfrow=c(2,3))
> for(i in 1:6) hist(ch[,i],main=names(ch)[i])
> for(i in 1:6) boxplot(ch[,i],main=names(ch)[i])
> pairs(ch)
Only the boxplots are shown in Figure 12.
An examination of the data using xgobi would also be worthwhile.
Now look at the relationship between involact and race:
> summary(lm(involact ˜ race,data=ch))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
0.12922
0.09661
1.34
0.19
race
0.01388
0.00203
6.84
1.8e-08
Residual standard error: 0.449 on 45 degrees of freedom
Multiple R-Squared: 0.509,
Adjusted R-squared: 0.499
F-statistic: 46.7 on 1 and 45 degrees of freedom,
p-value: 1.78e-08
We can clearly see that homeowners in zip codes with high % minority are being denied insurance at
higher rate than other zip codes. That is not in doubt. However, can the insurance companies claim that the
discrepancy is due to greater risks in some zip-codes? For example, we see that % minority is correlated
CHAPTER
0.0
0.5
1.0
1.5
2.0
0
20
40
60
80
100
120
140
0
20
40
60
80
100
12.
CHICA
GO
Figure
involact
theft
race
INSURANCE
12.2:
Boxplots
REDLINING
of
the
Chicago
-
5
10
15
20
0
20
40
60
80
10
20
30
40
A
COMPLETE
Insurance
data
EXAMPLE
income
age
fire
142
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
143
with the fire rate from the plots. The insurance companies could say that they were denying insurance in
neighborhoods where they had sustained large fire-related losses and any discriminatory effect was a by-
product of (presumably) legitimate business practice. What can regression analysis tell us about this claim?
The question of which variables should also be included in the regression so that their effect may be
adjusted for is difficult. Statistically, we can do it, but the important question is whether it should be done at
all. For example, it is known that the incomes of women in the US are generally lower than those of men.
However, if one adjusts for various factors such as type of job and length of service, this gender difference is
reduced or can even disappear. The controversy is not statistical but political - should these factors be used
to make the adjustment?
In this example, suppose that if the effect of adjusting for income differences was to remove the race
effect? This would pose an interesting but non-statistical question. I have chosen to include the income
variable here just to see what happens.
I use log(income) partly because of skewness in this variable but also because income is better considered
on a multiplicative rather than additive scale. In other words, $1,000 is worth a lot more to a poor person
than a millionaire because $1,000 is a much greater fraction of the poor person’s wealth.
We start with the full model:
> g <- lm(involact ˜ race + fire + theft + age + log(income), data = ch)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.18554
1.10025
-1.08
0.28755
race
0.00950
0.00249
3.82
0.00045
fire
0.03986
0.00877
4.55
4.8e-05
theft
-0.01029
0.00282
-3.65
0.00073
age
0.00834
0.00274
3.04
0.00413
log(income)
0.34576
0.40012
0.86
0.39254
Residual standard error: 0.335 on 41 degrees of freedom
Multiple R-Squared: 0.752,
Adjusted R-squared: 0.721
F-statistic: 24.8 on 5 and 41 degrees of freedom,
p-value: 2.01e-11
Before we start making any conclusions, we should check the model assumptions.
> plot(g$fit,g$res,xlab="Fitted",ylab="Residuals",
main="Residual-Fitted plot")
> abline(h=0)
> qqnorm(g$res)
These two diagnostic plots are shown in Figure 12.
The diagonal streak in the residual-fitted plot is caused by the large number of zero response values in
the data. When y
, the residual ˆε
ˆ
y
xT ˆβ, hence the line. Turning a blind eye to this feature, we see
¤
¤
¤
no particular problem. The Q-Q plot looks fine too.
Now let’s look at influence - what happens if points are excluded? We’ll use a function qqnorml()
that I wrote that labels the points in a Q-Q plot with the case numbers Plot not shown but cases 6 and 24
seem to stick out.
> gi <- lm.influence(g)
> for(i in 1:5) qqnorml(gi$coef[,i+1],main=names(ch)[-5][i])
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
144
Residual−Fitted plot
Normal Q−Q Plot
0.5
0.5
0.0
0.0
Residuals
−0.5
Sample Quantiles
−0.5
0.0
0.5
1.0
1.5
−2
−1
0
1
2
Fitted
Theoretical Quantiles
Jacknife Residuals
Cook−Statistics
3
35
6
3
2
0.3
1
24
0
0.2
−1
17
36
2
21
45
1
34
19
13
46
29
15
43
39
18
33
16
47
23
12
9
20
28
27
32
7
38
26
42
44
10
4
14
24
37
41
22
511
25
3035
831
40
30
0.1
Sample Quantiles
Sample Quantiles
11
−3
6
0.0
20
12
9
28
47
16
33
23
18
27
32
46
15
43
39
38
26
1
44
4
19
21
13
41
29
42
10
72
34
45
14
17
36
8
22
5
37
25
40
313
−2
−1
0
1
2
0.0
0.5
1.0
1.5
2.0
Theoretical Quantiles
Theoretical Quantiles
Figure 12.3: Diagnostic plots of the Chicago Insurance data
Check out the jacknife residuals:
> qqnorml(rstudent(g),main="Jacknife Residuals")
> qt(0.05/(2*47),47-6-1)
[1] -3.529468
Nothing too extreme - now look at the Cook statistics using the halfnorm() function that I wrote:
> halfnorm(cooks.distance(g),main="Cook-Statistics")
Cases 6 and 24 stick out again. Let’s take a look at these two cases:
> ch[c(6,24),]
race fire theft
age involact income
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
145
60610 54.0 34.1
68 52.6
0.3
8.231
60607 50.2 39.7
147 83.0
0.9
7.459
These are high theft and fire zip codes. See what happens when we exclude these points:
> g <- lm(involact ˜ race + fire + theft + age + log(income),ch,
subset=(1:47)[-c(6,24)])
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.57674
1.08005
-0.53
0.596
race
0.00705
0.00270
2.62
0.013
fire
0.04965
0.00857
5.79
1e-06
theft
-0.00643
0.00435
-1.48
0.147
age
0.00517
0.00289
1.79
0.082
log(income)
0.11570
0.40111
0.29
0.775
Residual standard error: 0.303 on 39 degrees of freedom
Multiple R-Squared: 0.804,
Adjusted R-squared: 0.779
F-statistic:
32 on 5 and 39 degrees of freedom,
p-value: 8.19e-13
theft and age are no longer significant at the 5% level. We now address the question of transfor-
mations - because the response has some zero values and for interpretational reasons we will not try to
transform it. Similarly, since the race variable is the primary predictor of interest we won’t try transforming
it either so as to avoid interpretation difficulties. We try fitting a polynomial model with quadratic terms in
each of the predictors:
> g2 <- lm(involact ˜ race + poly(fire,2) + poly(theft,2) + poly(age,2)
+ poly(log(income),2), ch, subset=(1:47)[-c(6,24)])
> anova(g,g2)
Analysis of Variance Table
Model 1: involact ˜ race + fire + theft + age + log(income)
Model 2: involact ˜ race + poly(fire, 2) + poly(theft, 2) + poly(age,
Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
1
39
3.59
2
35
3.20
4
0.38
1.04
0.4
Seems that we can do without the quadratic terms. A check of the partial residual plots also reveals no
need to transform. We now move on to variable selection. We are not so much interested in picking one
model here because we are mostly interested in the dependency of involact on the race variable. So ˆβ1 is the
thing we want to focus on. The problem is that collinearity with the other variables may cause ˆβ1 to vary
substantially depending on what other variables are in the model. We address this question here. leaps()
is bit picky about its input format so I need to form the x and y explicitly:
> y <- ch$inv[cooks.distance(g) < 0.2]
> x <- cbind(ch[,1:4],linc=log(ch[,6]))
> x <- x[cooks.distance(g) < 0.2,]
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
146
Removing all points with Cook’s Statistics greater than 0.2 takes out cases 6 and 24.
We make the Cp plot.
> library(leaps)
> a <- leaps(x,y)
> Cpplot(a)
See Figure 12.
25
1235
7.0
1245
123
6.0
12345
Cp
125
5.0
12
124
1234
4.0
3.0
3.5
4.0
4.5
5.0
5.5
6.0
p
Figure 12.4: Cp plot of the Chicago Insurance data
The best model seems to be this one:
> g <- lm(involact ˜ race + fire + theft + age, ch, subset=(1:47)[-c(6,24)])
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.26787
0.13967
-1.92
0.0623
race
0.00649
0.00184
3.53
0.0011
fire
0.04906
0.00823
5.96
5.3e-07
theft
-0.00581
0.00373
-1.56
0.1271
age
0.00469
0.00233
2.01
0.0514
Residual standard error: 0.3 on 40 degrees of freedom
Multiple R-Squared: 0.804,
Adjusted R-squared: 0.784
F-statistic: 40.9 on 4 and 40 degrees of freedom,
p-value: 1.24e-13
The fire rate is also significant and actually has higher t-statistics. Thus, we have verified that there is a
positive relationship between involact and race while controlling for a selection of the other variables.
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
147
How robust is the conclusion? Would other analysts have come to the same conclusion? One alternative
model is
> galt <- lm(involact ˜ race+fire+log(income),ch,subset=(1:47)[-c(6,24)])
> summary(galt)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
0.75326
0.83588
0.90
0.373
race
0.00421
0.00228
1.85
0.072
fire
0.05102
0.00845
6.04
3.8e-07
log(income) -0.36238
0.31916
-1.14
0.263
Residual standard error: 0.309 on 41 degrees of freedom
Multiple R-Squared: 0.786,
Adjusted R-squared: 0.77
F-statistic: 50.1 on 3 and 41 degrees of freedom,
p-value: 8.87e-14
In this model, we see that race is not statistically significant. The previous model did fit slightly better
but it is important that there exists a reasonable model in which race is not significant since although the
evidence seems fairly strong in favor of a race effect, it is not entirely conclusive. Interestingly enough, if
log(income) is dropped:
> galt <- lm(involact ˜ race+fire,ch,subset=(1:47)[-c(6,24)])
> summary(galt)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.19132
0.08152
-2.35
0.0237
race
0.00571
0.00186
3.08
0.0037
fire
0.05466
0.00784
6.97
1.6e-08
Residual standard error: 0.31 on 42 degrees of freedom
Multiple R-Squared: 0.779,
Adjusted R-squared: 0.769
F-statistic: 74.1 on 2 and 42 degrees of freedom,
p-value: 1.7e-14
we find race again becomes significant which raises again the question of whether income should be
adjusted for since it makes all the difference here.
We now return to the two left-out cases. Observe the difference in the fit when the two are re-included
on the best model. The quantities may change but the qualitative message is the same. It is better to include
all points if possible, especially in a legal case like this where excluding points might lead to criticism and
suspicion of the results.
> g <- lm(involact ˜ race + fire + theft + age, data=ch)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.24312
0.14505
-1.68
0.10116
race
0.00810
0.00189
4.30
0.00010
fire
0.03665
0.00792
4.63
3.5e-05
theft
-0.00959
0.00269
-3.57
0.00092
age
0.00721
0.00241
2.99
0.00460
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
148
Residual standard error: 0.334 on 42 degrees of freedom
Multiple R-Squared: 0.747,
Adjusted R-squared: 0.723
F-statistic:
31 on 4 and 42 degrees of freedom,
p-value: 4.8e-12
The main message of the data is not changed - we should check the diagnostics. I found no trouble.
(Adding back in the two points to the race+fire+log(income) model made race significant again.
So it looks like there is some good evidence that zip codes with high minority populations are being “red-
lined” - that is improperly denied insurance. While there is evidence that some of the relationship between
race and involact can be explained by the fire rate, there is still a component that cannot be attributed to the
other variables.
However, there is some doubt due to the response not being a perfect measure of people being denied
insurance. It is an aggregate measure which raises the problem of ecological correlations. We have implicitly
assumed that the probability that a minority homeowner would obtain a FAIR plan after adjusting for the
effect of the other covariates is constant across zip-codes. This is unlikely to be true. If the truth is simply
variation about some constant, then our conclusions will still be reasonable but if this probability varies in
a systematic way, then our conclusions may be off the mark. It would be a very good idea to obtain some
individual level data.
Another point to be considered is the size of the effect. The largest value of the response is only 2.2%
and most cases are much smaller. Even assuming the worst, the number of people affected is small.
There is also the problem of a potential latent variable that might be the true cause of the observed
relationship, but it is difficult to see what that variable might be. Nevertheless, this always casts a shadow
of doubt on our conclusions.
There are some special difficulties in presenting this during a court case. With scientific enquiries, there
is always room for uncertainty and subtlety in presenting the results, but this is much more difficult in the
court room. The jury may know no statistics and lawyers are clever at twisting words. A statistician giving
evidence as an expert witness would do well to keep the message simple.
Another issue that arises in cases of this nature is how much the data should be aggregated. For example,
I divided the data using a zip code map of Chicago into north and south. Fit the model to the south of
Chicago:
> data(chiczip)
> g <- lm(involact ˜ race + fire + theft +age, subset=(chiczip == "s"), ch)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.23441
0.23774
-0.99
0.338
race
0.00595
0.00328
1.81
0.087
fire
0.04839
0.01689
2.87
0.011
theft
-0.00664
0.00844
-0.79
0.442
age
0.00501
0.00505
0.99
0.335
Residual standard error: 0.351 on 17 degrees of freedom
Multiple R-Squared: 0.743,
Adjusted R-squared: 0.683
F-statistic: 12.3 on 4 and 17 degrees of freedom,
p-value: 6.97e-05
and now to the north.
CHAPTER 12. CHICAGO INSURANCE REDLINING - A COMPLETE EXAMPLE
149
> g <- lm(involact ˜ race + fire + theft +age, subset=(chiczip == "n"), ch)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.31857
0.22702
-1.40
0.176
race
0.01256
0.00448
2.81
0.011
fire
0.02313
0.01398
1.65
0.114
theft
-0.00758
0.00366
-2.07
0.052
age
0.00820
0.00346
2.37
0.028
Residual standard error: 0.343 on 20 degrees of freedom
Multiple R-Squared: 0.756,
Adjusted R-squared: 0.707
F-statistic: 15.5 on 4 and 20 degrees of freedom,
p-value: 6.52e-06
What differences do you see? By dividing the data into smaller and smaller subsets it is possible to
dilute the significance of any predictor. On the other hand it is important not to aggregate all data without
regard to whether it is reasonable. Clearly a judgment has to be made and this often a point of contention in
legal cases.
After all this analysis, the reader may be feeling somewhat dissatisfied. It seems we are unable to come
to any truly definite conclusions and everything we say has been hedged with “ifs” and “buts”. Winston
Churchill once said
Indeed, it has been said that democracy is the worst form of Government except all those other
forms that have been tried from time to time.
We might say the same thing about Statistics in relation to how it helps us reason in the face of uncer-
tainty.
Chapter 13
Robust and Resistant Regression
When the errors are normal, least squares regression is clearly best but when the errors are nonnormal, other
methods may be considered. A particular concern is long-tailed error distributions. One approach is to
remove the largest residuals as outliers and still use least squares but this may not be effective when there
are several large residuals because of the leave-out-one nature of the outlier tests. Furthermore, the outlier
test is an accept/reject procedure that is not smooth and may not be statistically efficient for the estimation
of β. Robust regression provides an alternative.
There are several methods. M-estimates choose β to minimize
n
§
∑
β
ρ yi xT
i
σ
i 1
¨
§
Possible choices for ρ are
1. ρ¡ x
x2 is just least squares
¢
¤
¡
¡
2. ρ¡ x
x is called least absolute deviations regression (LAD). This is also called L1 regression.
¢
¤
3.
¡
¡
ρ
x2 2
if x
c
¡
x
¡
¡
¢
¤
c x
c2 2
otherwise
is called Huber’s method and is a compromise between least squares and LAD regression. c can be
¡
¡
an estimate of σ but not the usual one which is not robust. Something ∝ median ˆεi for example.
Robust regression is related to weighted least squares. The normal equations tell us that
X T ¡ y
X ˆβ
0
¢
¤
¡
With weights and in non-matrix form this becomes:
n
p
∑
¡
w
β
ixi j yi
∑ xij j 0 j 1 p
¢
¤
¤
¢¡¢¡¢¡
i 1
j 1
§
§
Now differentiating the M-estimate criterion with respect to β j and setting to zero we get
n
∑
y
∑p x β
ρ
i
i j
j
j 1
§
σ
xi j
0
j
1
p
¤
¤
¢¡¢¡¢¡
i 1
¡
§
150
CHAPTER 13. ROBUST AND RESISTANT REGRESSION
151
Now let u
β
i
yi
∑p xij j to get
¤
j 1
§
n
p
∑ ρ ¡ ui ¡
¢
x
β
i j yi
∑ xij j 0 j 1 p
¢
¤
¤
u
¢¡¢¡¢¡
i 1
i
j 1
§
§
so we can make the identification of
¡
w u
ρ ¡ u u
¢
¤
¢
¡
and we find for our choices of ρ above:
¡
1. LS: w u is constant.
¢
¡
¡
¡
2. LAD: w u
1 u - note the asymptote at 0 - this makes a weighting approach difficult.
¢
¤
3. Huber:
¡
¡
1
if u
c
¡
w u
¡
¡
¢
¤
c u
otherwise
There are many other choices that have been used. Because the weights depend on the residuals, an
iteratively reweighted least squares approach to fitting must be used. We can sometimes get standard errors
by vˆar ˆβ
ˆ
σ2 ¡ XTW X 1 (use a robust estimate of σ2 also).
¤
¢
We demonstrate the methods on the Chicago insurance data. Using least squares first.
> data(chicago)
> g <- lm(involact ˜ race + fire + theft + age + log(income),chicago)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -3.57398
3.85729
-0.93
0.35958
race
0.00950
0.00249
3.82
0.00045
fire
0.03986
0.00877
4.55
4.8e-05
theft
-0.01029
0.00282
-3.65
0.00073
age
0.00834
0.00274
3.04
0.00413
log(income)
0.34576
0.40012
0.86
0.39254
Residual standard error: 0.335 on 41 degrees of freedom
Multiple R-Squared: 0.752,
Adjusted R-squared: 0.721
F-statistic: 24.8 on 5 and 41 degrees of freedom,
p-value: 2.01e-11
Least squares works well when there are normal errors but can be upset by long-tailed errors. A conve-
nient way to apply the Huber method is to apply the rlm() function which is part of the MASS (see the
book Modern Applied Statistics in S+) which also gives standard errors. The default is to use the Huber
method but there are other choices.
> library(MASS)
> g <- rlm( involact ˜ race + fire + theft + age + log(income), chicago)
Coefficients:
Value
Std. Error t value
(Intercept) -2.926
3.397
-0.861
race
0.008
0.002
3.583
CHAPTER 13. ROBUST AND RESISTANT REGRESSION
152
fire
0.046
0.008
5.940
theft
-0.010
0.002
-3.912
age
0.006
0.002
2.651
log(income)
0.283
0.352
0.803
Residual standard error: 0.249 on 41 degrees of freedom
The R2 and F-statistics are not given because they cannot be calculated (at least not in the same way).
The numerical values of the coefficients have changed a small amount but the general significance of the
variables remains the same and our substantive conclusion would not be altered. Had we seen something
different, we would need to find out the cause. Perhaps some group of observations were not being fit well
and the robust regression excluded these points.
Another method that can be used is Least Trimmed Squares(LTS). Here one minimizes ∑q ˆε2 where
i 1
i
§
¡
¡
q is some number less than n and i indicates sorting. This method has a high breakdown point because
¢
it can tolerate a large number of outliers depending on how q is chosen. The Huber and L1 methods will
still fail if some ε
∞
i
. LTS is an example of a resistant regression method. Resistant methods are good at
§
dealing with data where we expect there to be a certain number of “bad” observations that we want to have
no weight in the analysis.
> library(lqs)
> g <- ltsreg(involact ˜ race + fire + theft + age + log(income),chicago)
> g$coef
(Intercept)
race
fire
theft
age log(income)
-1.6950187
0.0037348
0.0549117
-0.0095883
0.0018549
0.1700325
> g <- ltsreg(involact ˜ race + fire + theft + age + log(income),chicago)
> g$coef
(Intercept)
race
fire
theft
age log(income)
2.2237795
0.0050697
0.0423565
-0.0084868
0.0008755
-0.2398183
¡
The default choice of q is n 2
p
1 2 where x indicates the largest integer less than or equal
¡
¢
¢
¡
¡
¡
to x. I repeated the command twice and you will notice that the results are somewhat different. This is
because the default genetic algorithm used to compute the coefficients is non-deterministic. An exhaustive
search method can be used
> g <- ltsreg(involact ˜ race + fire + theft + age + log(income),chicago,
nsamp="exact")
> g$coef
(Intercept)
race
fire
theft
age log(income)
-1.12093591
0.00575147
0.04859848 -0.00850985
0.00076159
0.11251547
This takes about 20 minutes on a 400Mhz Intel Pentium II processor. For larger datasets, it will take
much longer so this method might be impractical.
The most notable difference from LS for the purposes of this data is the decrease in the race coefficient
- if the same standard error applied then it would verge on insignificance. However, we don’t have the
standard errors for the LTS regression coefficients. We now use a general method for inference which is
especially useful when such theory is lacking - the Bootstrap.
To understand how this method works, think about how we might empirically determine the distribution
of an estimator. We could repeatedly generate artificial data from the true model, compute the estimate each
CHAPTER 13. ROBUST AND RESISTANT REGRESSION
153
time and gather the results to study the distribution. This technique, called simulation, is not available to us
for real data because we don’t know the true model. The Bootstrap emulates the simulation procedure above
except instead of sampling from the true model, it samples from the observed data itself. Remarkably, this
technique is often effective. It sidesteps the need for theoretical calculations that may be extremely difficult
or even impossible. The Bootstrap may be the single most important innovation in Statistics in the last 20
years.
To see how the bootstrap method compares with simulation, let’s spell out the steps involved. In both
cases, we consider X fixed.
Simulation
In general the idea is to sample from the known distribution and compute the estimate, repeating many
times to find as good an estimate of the sampling distribution of the estimator as we need. For the regression
case, it is easiest to start with a sample from the error distribution since these are assumed to be independent
and identically distributed:
1. Generate ε from the known error distribution.
2. Form y
X β
ε from the known β.
¤
3. Compute ˆβ.
We repeat these three steps many times. We can estimate the sampling distribution of ˆβ using the em-
pirical distribution of the generated ˆβ, which we can estimate as accurately as we please by simply running
the simulation for long enough. This technique is useful for a theoretical investigation of the properties of
a proposed new estimator. We can see how its performance compares to other estimators. However, it is of
no value for the actual data since we don’t know the true error distribution and we don’t know the true β.
The bootstrap method mirrors the simulation method but uses quantities we do know. Instead of sam-
pling from the population distribution which we do not know in practice, we resample from the data itself.
Bootstrap
1. Generate ε by sampling with replacement from ˆε1
ˆεn.
¢¡¢¡¢¡£
2. Form y
X ˆβ
ε
¤
¡
3. Compute ˆβ from X y
¢
This time, we use only quantities that we know. For small n, it is possible to compute ˆβ for every
possible sample from ˆε1
ˆεn, but usually we can only take as many samples as we have computing power
¢¡¢¡¢¡£
available. This number of bootstrap samples can be as small as 50 if all we want is an estimate of the
variance of our estimates but needs to be larger if confidence intervals are wanted.
To implement this, we need to be able to take a sample of residuals with replacement. sample() is
good for generating random samples of indices:
> sample(10,rep=T)
[1] 7 9 9 2 5 7 4 1 8 9
and hence a random sample (with replacement) of RTS residuals is:
> g$res[sample(47,rep=T)]
60639
60641
60634
60608
60608
60612
60651
60620
0.091422 -0.039899
0.013526
0.342344
0.342344 -0.022214
0.255031
0.333714
CHAPTER 13. ROBUST AND RESISTANT REGRESSION
154
(rest deleted
You will notice that there is a repeated value even in this small snippet. We now execute the bootstrap
- first we make a matrix to save the results in and then repeat the bootstrap process 1000 times: (This takes
about 6 minutes to run on a 400Mhz Intel Pentium II processor)
> x <- model.matrix(˜ race+fire+theft+age+log(income),chicago)[,-1]
> bcoef <- matrix(0,1000,6)
> for(i in 1:1000){
+ newy <- g$fit + g$res[sample(47,rep=T)]
+ brg <- ltsreg(x,newy,nsamp="best")
+ bcoef[i,] <- brg$coef
+ }
It is not convenient to use the nsamp="exact" since that would require 1000 times the 20 minutes it
takes to make original estimate. That’s about two weeks, so I compromised and used the second best option
of nsamp="best". This likely means that our bootstrap estimates of variability will be somewhat on the
high side. This illustrates a common practical difficulty with the bootstrap — it can take a long time to
compute. Fortunately, this problem recedes as processor speeds increase. It is notable that this calculation
was the only one in this book that did not take a negligible amount of time. You typically do not need the
latest and greatest computer to do statistics on the size of datasets encountered in this book.
To test the null hypothesis that H
¥
0 : βrace
0 against the alternative H1 : βrace
0 we may figure what
¤
fraction of the bootstrap sampled βrace were less than zero:
> length(bcoef[bcoef[,2]<0,2])/1000
[1] 0.019
So our p-value is 1.9% and we reject the null at the 5% level.
We can also make a 95% confidence interval for this parameter by taking the empirical quantiles:
> quantile(bcoef[,2],c(0.025,0.975))
2.5%
97.5%
0.00099037 0.01292449
We can get a better picture of the distribution by looking at the density and marking the confidence
interval:
> plot(density(bcoef[,2]),xlab="Coefficient of Race",main="")
> abline(v=quantile(bcoef[,2],c(0.025,0.975)))
See Figure 13.1. We see that the distribution is approximately normal with perhaps so longish tails.
This would be more accurate if we took more than 1000 bootstrap resamples. The conclusion here
would be that the race variable is significant but the effect is less than that estimated by least squares. Which
is better? This depends on what the ”true” model is which we will never know but since the QQ plot did
not indicate any big problem with non-normality I would tend to prefer the LS estimates. However, this
does illustrate a general problem that occurs when more than one statistical method is available for a given
dataset.
Summary
CHAPTER 13. ROBUST AND RESISTANT REGRESSION
155
250
150
Density
50
0
−0.01
0.00
0.01
0.02
Coefficient of Race
Figure 13.1: Bootstrap distribution of ˆβrace with 95% confidence intervals
1. Robust estimators provide protection against long-tailed errors but they can’t overcome problems with
the choice of model and its variance structure. This is unfortunate because these problems are more
serious than non-normal error.
2. Robust estimates just give you ˆβ and possibly standard errors without the associated inferential meth-
ods. Software and methodology for this inference is not easy to come by. The bootstrap is a general
purpose inferential method which is useful in these situations.
3. Robust methods can be used in addition to LS as a confirmatory method. You have cause to worry if
the two estimates are far apart.
Chapter 14
Missing Data
Missing data is the situation where some values of some cases are missing. This is not uncommon. Deal-
ing with missing data is time consuming. In my experience, fixing up problems caused by missing data
sometimes takes longer than the analysis itself.
What can be done? Obviously, finding the missing values is the best option but this is not always
possible. Next ask why the data are missing. If the reason for a datum being missing is non-informative,
then a fix is easier. For example, if a data point is missed because it was large then this could cause some
bias and a simple fix is not possible. Patients may drop out of a drug study because they feel their treatment
is not working - this would cause bias.
Here are several fix-up methods to use when data are missing for noninformative reasons:
1. Delete the case with missing observations. This is OK if this only causes the loss of a relatively small
number of cases. This is the simplest solution.
2. Fill-in or impute the missing values. Use the rest of the data to predict the missing values. Simply
replacing the missing value of a predictor with the average value of that predictor is one easy method.
Using regression on the other predictors is another possibility. It’s not clear how much the diagnos-
tics and inference on the filled-in dataset is affected. Some additional uncertainty is caused by the
imputation which needs to be allowed for.
¡
3. Missing observation correlation. Consider just xi yi pairs with some observations missing. The
¢
means and SDs of x and y can be used in the estimate even when a member of a pair is missing. An
analogous method is available for regression problems.
4. Maximum likelihood methods can be used assuming the multivariate normality of the data. The EM
algorithm is often used here. We will not explain the details but the idea is essentially to treat missing
values as nuisance parameters.
Suppose some of the values in the Chicago Insurance dataset were missing. I randomly declared some
the observations missing in this modified dataset. Read it in and take a look:
> data(chmiss)
> chmiss
race fire theft
age involact income
60626 10.0
6.2
29 60.4
NA 11.744
60640 22.2
9.5
44 76.5
0.1
9.323
60613 19.6 10.5
36
NA
1.2
9.948
156
CHAPTER 14. MISSING DATA
157
60657 17.3
7.7
37
NA
0.5 10.656
--- etc ---
60645
3.1
4.9
27
NA
0.0 13.731
There are 20 missing observations denoted by NA here. It’s important to know what the missing value
code is for the data and/or software you are using. What happens if we try to fit the model?
> g <- lm(involact ˜ .,chmiss)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.11648
0.60576
-1.84
0.07947
race
0.01049
0.00313
3.35
0.00302
fire
0.04388
0.01032
4.25
0.00036
theft
-0.01722
0.00590
-2.92
0.00822
age
0.00938
0.00349
2.68
0.01390
income
0.06870
0.04216
1.63
0.11808
Residual standard error: 0.338 on 21 degrees of freedom
Multiple R-Squared: 0.791,
Adjusted R-squared: 0.741
F-statistic: 15.9 on 5 and 21 degrees of freedom,
p-value: 1.59e-06
Any case with at least one missing value is omitted from the regression. You can see there are now only
21 degrees of freedom - almost half the data is lost. We can fill in the missing values by their variable means
as in:
> cmeans <- apply(chmiss,2,mean,na.rm=T)
> cmeans
race
fire
theft
age involact
income
35.60930 11.42444 32.65116 59.96905
0.64773 10.73587
> mchm <- chmiss
> for(i in c(1,2,3,4,6)) mchm[is.na(chmiss[,i]),i] <- cmeans[i]
We don’t fill in missing values in the response because this is the variable we are trying to model. Now
refit:
> g <- lm(involact ˜ ., data=mchm)
> summary(g)
Coefficients:
Value Std. Error t value Pr(>|t|)
(Intercept)
0.0707
0.5094
0.1387
0.8904
race
0.0071
0.0027
2.6307
0.0122
fire
0.0287
0.0094
3.0623
0.0040
theft -0.0031
0.0027
-1.1139
0.2723
age
0.0061
0.0032
1.8954
0.0657
income -0.0271
0.0317
-0.8550
0.3979
Residual standard error: 0.3841 on 38 degrees of freedom
Multiple R-Squared: 0.6819
F-statistic: 16.3 on 5 and 38 degrees of freedom, the p-value is 1.41e-08
CHAPTER 14. MISSING DATA
158
Compare with the previous results - what differences do you see? Different statistical packages have
different ways of handling missing observations. For example, the default behavior in S-PLUS would
refuse to fit the model at all.
The regression coefficients are now all closer to zero. The situation is analogous to the error in variables
case. The bias introduced by the fill-in method can be substantial and may not be compensated by the
attendant reduction in variance.
We can also use regression methods to predict the missing values of the covariates. Let’s try to fill-in
the missing race values:
> gr <- lm(race ˜ fire+theft+age+income,chmiss)
> chmiss[is.na(chmiss$race),]
race fire theft
age involact income
60646
NA
5.7
11 27.9
0.0 16.250
60651
NA 15.1
30 89.8
0.8 10.510
60616
NA 12.2
46 48.0
0.6
8.212
60617
NA 10.8
34 58.0
0.9 11.156
> predict(gr,chmiss[is.na(chmiss$race),])
60646
60651
60616
60617
-17.847
26.360
70.394
32.620
Can you see a problem with filling these values in? Obviously we would need to put more work into
the regression models used to fill-in the missing values. One trick that can be applied when the response is
bounded between 0 and 1 is the logit transformation:
¡
¡
y
log y 1
y
¢
¢
§
This transformation maps to the whole real line. We define the logit function and its inverse:
> logit <- function(x) log(x/(1-x))
> ilogit <- function(x) exp(x)/(1+exp(x))
We now fit the model with a logit-transformed response and then back-transform the predicted values re-
membering to convert our percentages to proportions and vice versa at the appropriate times:
> gr <- lm(logit(race/100) ˜ fire+theft+age+income,chmiss)
> ilogit(predict(gr,chmiss[is.na(chmiss$race),]))*100
60646
60651
60616
60617
0.41909 14.73202 84.26540 21.31213
We can see how our predicted values compare to the actual values:
> data(chicago)
> chicago$race[is.na(chmiss$race)]
[1]
1.0 13.4 62.3 36.4
So our first two predictions are good but the other two are somewhat wide of the mark.
Like the mean fill-in method, regression fill-in will also introduce a bias towards zero in the coefficients
while tending to reduce the variance also. The success of the regression method depends somewhat on the
collinearity of the predictors - the filled-in values will be more accurate the more collinear the predictors
are.
CHAPTER 14. MISSING DATA
159
For situations where there is a substantial proportion of missing data, I recommend that you investigate
more sophisticated methods, likely using the EM algorithm. Multiple imputation is another possibility. The
fill-in methods described above will be fine when only a few cases need to be filled but will become less
reliable as the proportion of missing cases increases.
Chapter 15
Analysis of Covariance
Predictors that are qualitative in nature, like for example eye color, are sometimes called categorical or
factors. How can these predictors be incorporated into a regression analysis? Analysis of Covariance refers
to regression problems where there is a mixture of quantitative and qualitative predictors.
Suppose we are interested in the effect of a medication on cholesterol level - we might have two groups
- one of which receives the medication and the other which does not. However, we could not treat this as a
simple two sample problem if we knew that the two groups differed with respect to age and this would affect
the cholesterol level. See Figure 15 for a simulated example. For the patients who received the medication,
the mean reduction in cholesterol level was 0% while for those who did not the mean reduction was 10%.
So superficially it would seem that it would be better not to be treated. However, the treated group ranged in
age from 50 to 70 while those who were not treated ranged in age between 30 and 50. We can see that once
age is taken into account, the difference between treatment and control is again 10% but this time in favor
of the treatment.
20
10
5
0
Percentage reduction
−10
30
40
50
60
70
Age
Figure 15.1: Simulated example showing the confounding effect of a covariate. The patients who took the
medication are marked with a solid dot while those who did not are marked with an empty dot
Analysis of covariance is a method for adjusting the groups for the age difference and then seeing the
160
15.1. A TWO-LEVEL EXAMPLE
161
effect of the medication. It can also be used when there are more than two groups and more than one
covariate.
Our strategy is to incorporate the qualitative predictors within the y
X β
ε framework. We
¤
can then use the estimation, inferential and diagnostic techniques that we have already learnt.
This avoids have to learn a different set of formulae for each new type of qualitative predictor configu-
ration which is the approach taken by many texts. To put qualitative predictors into the y
X β
ε form we
¤
need to code the qualitative predictors. Let’s consider a specific example:
y = change in cholesterol level
x = age
0
did not take medication
d ¤
1
took medication
A variety of linear models may be considered here:
1. The same regression line for both groups — y
β
β
ε
0
1x
or in R y ˜ x
¤
2. Separate regression lines for each group but with the same slope — y
β
β
β
ε
0
1x
2d
or in R y
¤
˜ x + d. In this case β2 represents the distance between the regression lines i.e. the effect of the
drug.
3. Separate regression lines for each group y
β
β
β
β
ε
0
1x
2d
3x d
or in R y ˜ x + d + d:x
¤
¡
or y ˜ x*d. Any interpretation of the effect of the drug will now depend on age also. To form the
slope interaction term x d in the X-matrix, simply multiply x by d elementwise.
¡
Estimation and testing works just as it did before. Interpretation is much easier if we can eliminate the
slope interaction term.
Other codings of d are possible, for instance
1
did not take medication
d
¤
1
took medication
is used by some. This coding enables β2 and β3 to be viewed as differences from a response averaged
over the two groups. Any other coding that assigned a different number to the two groups would also work
but interpretation of the estimated parameters might be more difficult.
15.1
A two-level example
The data for this example consist of x= nave height and y = total length in feet for English medieval
cathedrals. Some are in the Romanesque (r) style and others are in the Gothic (g) style. Some cathedrals
have parts in both styles and are listed twice. We wish to investigate how the length is related to height for
the two styles. Read in the data and make a summary on the two styles separately.
15.1. A TWO-LEVEL EXAMPLE
162
> data(cathedral)
> cathedral
style
x
y
Durham
r
75 502
Canterbury
r
80 522
....etc....
Old.St.Paul
g 103 611
Salisbury
g
84 473
> lapply(split(cathedral,cathedral$style),summary)
$g
style
x
y
g:16
Min.
: 45.0
Min.
:182
r: 0
1st Qu.: 60.8
1st Qu.:299
Median : 73.5
Median :412
Mean
: 74.9
Mean
:397
3rd Qu.: 86.5
3rd Qu.:481
Max.
:103.0
Max.
:611
$r
style
x
y
g:0
Min.
:64.0
Min.
:344
r:9
1st Qu.:70.0
1st Qu.:425
Median :75.0
Median :502
Mean
:74.4
Mean
:475
3rd Qu.:80.0
3rd Qu.:530
Max.
:83.0
Max.
:551
Now plot the data — see Figure 15.1.
> plot(cathedral$x,cathedral$y,type="n",xlab="Nave height",ylab="Length")
> text(cathedral$x,cathedral$y,as.character(cathedral$s))
Now fit the separate regression lines model. y ˜ x*style is equivalent.
> g <- lm(y ˜ x+style+x:style, data=cathedral)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
37.11
85.68
0.43
0.6693
x
4.81
1.11
4.32
0.0003
style
204.72
347.21
0.59
0.5617
x.style
-1.67
4.64
-0.36
0.7227
Residual standard error: 79.1 on 21 degrees of freedom
Multiple R-Squared: 0.541,
Adjusted R-squared: 0.476
F-statistic: 8.26 on 3 and 21 degrees of freedom,
p-value: 0.000807
Because style is non-numeric, R automatically treats it as a qualitative variables and sets up a coding -
but which coding?
15.1. A TWO-LEVEL EXAMPLE
163
g
600
r
r r r
g g
r
g
500
g
r
r
g
g
g
g
g
r
400
g
Length
r
g
300
g
g
g
200
g
50
60
70
80
90
100
Nave Height
Figure 15.2: A comparison of Romanesque (r) and Gothic (g) Cathedrals
> model.matrix(g)
(Intercept)
x style x.style
Durham
1
75
1
75
Canterbury
1
80
1
80
...etc...
Old.St.Paul
1 103
0
0
Salisbury
1
84
0
0
We see that the model can be simplified to
> g <- lm(y ˜ x+style, cathedral)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
44.30
81.65
0.54
0.5929
x
4.71
1.06
4.45
0.0002
style
80.39
32.31
2.49
0.0209
Residual standard error: 77.5 on 22 degrees of freedom
Multiple R-Squared: 0.538,
Adjusted R-squared: 0.496
F-statistic: 12.8 on 2 and 22 degrees of freedom,
p-value: 0.000203
Put the two parallel regression on the plot:
> abline(44.30,4.71)
> abline(44.30+80.39,4.71,lty=2)
15.2. CODING QUALITATIVE PREDICTORS
164
A check on the diagnostics reveals no particular problems.
Our conclusion is that for cathedrals of the same height, Romanesque ones are 80.39 feet longer. For
each extra foot in height, both types of cathedral are about 4.7 feet longer. Gothic cathedrals are treated as
the reference level because “g” comes before “r” in the alphabet. We can change this:
> cathedral$style <- relevel(cathedral$style,ref="r")
> g <- lm(y ˜ x+style, cathedral)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
124.69
82.92
1.50
0.1469
x
4.71
1.06
4.45
0.0002
style
-80.39
32.31
-2.49
0.0209
Residual standard error: 77.5 on 22 degrees of freedom
Multiple R-Squared: 0.538,
Adjusted R-squared: 0.496
F-statistic: 12.8 on 2 and 22 degrees of freedom,
p-value: 0.000203
Although the coefficients have different numerical values, this coding leads to the same conclusion as
before.
Notice that in this case the two groups have about the same average height — about 74 feet. The
difference in the lengths is 78 feet on average which is similar to the 80 feet from the fit. This is in contrast
to the cholesterol example above where the two groups had very different means in their predictors.
15.2
Coding qualitative predictors
There is no unique coding for a two-level factor — there are even more choices with multi-level predictors.
For a k-level predictor, k
1 dummy variables are needed for the representation. One parameter is used to
represent the overall mean effect or perhaps the mean of some reference level and so only k
1 variables
are needed rather than k.
These dummy variables cannot be exactly collinear but otherwise there is no restriction. The choice
should be based on convenience.
Treatment coding
Consider a 4 level factor that will be coded using 3 dummy variables. This table describes the coding:
Dummy coding
1
2
3
1
0
0
0
levels
2
1
0
0
3
0
1
0
4
0
0
1
This treats level one as the standard level to which all other levels are compared so a control group, if
one exists, would be appropriate for this level. R assigns levels to a factor in alphabetical order by default.
The columns are orthogonal and the corresponding dummies will be too. The dummies won’t be orthogonal
to the intercept. Treatment coding is the default choice for R
Helmert Coding
15.3. A THREE-LEVEL EXAMPLE
165
Dummy coding
1
2
3
1
-1
-1
-1
levels
2
1
-1
-1
3
0
2
-1
4
0
0
3
If there are equal numbers of observations in each level (a balanced design) then the dummy variables
will be orthogonal to the each other and the intercept. This coding is not so nice for interpretation. It is the
default choice in S-PLUS.
There are other choices of coding — anything that spans the k
1 dimensional space will work. The
choice of coding does not affect the R2, ˆ
σ2 and overall F-statistic. It does effect the ˆβ and you do need to
know what the coding is before making conclusions about ˆβ.
15.3
A Three-level example
Here’s an example with a qualitative predictor with more than one level. The data for this example come
from a 1966 paper by Cyril Burt entitled ”The genetic determination of differences in intelligence: A study
of monozygotic twins reared apart”. The data consist of IQ scores for identical twins, one raised by foster
parents, the other by the natural parents. We also know the social class of natural parents (high, middle or
low). We are interested in predicting the IQ of the twin with foster parents from the IQ of the twin with the
natural parents and the social class of natural parents. Let’s read in and take a look at the data:
> data(twins)
> twins
Foster Biological Social
1
82
82
high
2
80
90
high
etc.
26
107
106
low
27
98
111
low
> plot(twins$B,twins$F,type="n",xlab="Biological IQ",ylab="Foster IQ")
> text(twins$B,twins$F,substring(as.character(twins$S),1,1))
See Figure 15.3 — what model seems appropriate? The most general model we’ll consider is the seper-
ate lines model:
> g <- lm(Foster ˜ Biological*Social, twins)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-1.8720
17.8083
-0.11
0.92
Biological
0.9776
0.1632
5.99
6e-06
Sociallow
9.0767
24.4487
0.37
0.71
Socialmiddle
2.6881
31.6042
0.09
0.93
Biological.Sociallow
-0.0291
0.2446
-0.12
0.91
Biological.Socialmiddle
-0.0050
0.3295
-0.02
0.99
Residual standard error: 7.92 on 21 degrees of freedom
15.3. A THREE-LEVEL EXAMPLE
166
h
130
h
h
l
m
110
h
l l
l
l
m
l l m
l
90
Foster IQ
h l
m
lh l h
l
m
70
m
l
70
80
90
100
110
120
130
Biological IQ
Figure 15.3: Burt twin data, l=low class, m=middle class and h= high class, regression fit shown
Multiple R-Squared: 0.804,
Adjusted R-squared: 0.757
F-statistic: 17.2 on 5 and 21 degrees of freedom,
p-value: 8.31e-07
The reference level is high class, being first alphabetically. We see that the intercept for low class line
would be -1.872+9.0767 while the slope for the middle class line would be 0.9776-0.005. Check the design
matrix for the gory details
> model.matrix(g)
Now see if the model can be simplified to the parallel lines model:
> gr <- lm(Foster ˜ Biological+Social, twins)
> anova(gr,g)
Analysis of Variance Table
Model 1: Foster ˜ Biological + Social
Model 2: Foster ˜ Biological + Social + Biological:Social
Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
1
23
1318
2
21
1317
2
1
0.01
1.0
Yes it can. The sequential testing can be done in one go:
> anova(g)
Analysis of Variance Table
15.3. A THREE-LEVEL EXAMPLE
167
Response: Foster
Df Sum Sq Mean Sq F value
Pr(>F)
Biological
1
5231
5231
83.38 9.3e-09
Social
2
175
88
1.40
0.27
Biological:Social
2
1 4.7e-01
0.01
0.99
Residuals
21
1317
63
We see that a further reduction to a single line model is possible:
> gr <- lm(Foster ˜ Biological, twins)
Plot the regression line on the plot:
> abline(gr$coef)
A check of the diagnostics shows no cause for concern. The (almost) final model:
> summary(gr)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
9.2076
9.2999
0.99
0.33
Biological
0.9014
0.0963
9.36
1.2e-09
Residual standard error: 7.73 on 25 degrees of freedom
Multiple R-Squared: 0.778,
Adjusted R-squared: 0.769
F-statistic: 87.6 on 1 and 25 degrees of freedom,
p-value: 1.2e-09
The icing on the cake would be a further simplification of this model to the line y=x (the IQ’s are equal).
¡
The model has no parameters at all so it has RSS
∑ y
2
i
xi
and degrees of freedom equal to the sample
¤
i
¢
size. We compute the F-test and p-value:
> sum(gr$resˆ2)
[1] 1493.5
> sum((twins$F-twins$B)ˆ2)
[1] 1557
> ((1557-1493.5)/2)/(1493.5/25)
[1] 0.53147
> 1-pf(0.53147,2,25)
[1] 0.59423
So the null is not rejected.
Burt was interested in demonstrating the importance of heredity over environment in intelligence and
this data certainly point that way. (Although it would be helpful to know the social class of the foster parents)
However, before jumping to any conclusions, you may be interested to know that there is now consid-
erable evidence that Cyril Burt invented some of his data on identical twins. In light of this, can you see
anything in the above analysis that might lead one to suspect this data?
Chapter 16
ANOVA
Predictors are now all categorical/ qualitative. The name ANOVA stands for Analysis of Variance is used
because the original thinking was to try to partition the overall variance in the response to that due to each
of the factors and the error. Predictors are now typically called factors which have some number of levels.
The parameters are now often called effects. We shall first consider only models where the parameters are
considered fixed but unknown — called fixed-effects models but random-effects models are also used where
parameters are taken to be random variables.
16.1
One-Way Anova
16.1.1
The model
Given a factor α occurring at i
1
I levels, with j
1
Ji observations per level. We use the model
¤
¤
¢¡¢¡¢¡
¢¡¢¡¢¡
y
α
ε
i j
µ
i
i j
i
1
I
j
1
Ji
¤
¤
¤
¢¡¢¡¢¡
¢¡¢¡¢¡
As it stands not all the parameters are identifiable and some restriction is necessary:
1. Set µ
0 and use I different dummy variables.
¤
2. Set α1
0 — this corresponds to treatment contrasts
¤
3. Set ∑ α
i Ji
i
0 which leads to the least squares estimates
¤
ˆµ
¯
y
ˆ
αi
¯
yi
¯
y
¤
¤
where indicates which index or indices the mean is taken over.
This last method is the most commonly recommended for manual calculation in older textbooks although
it is harder to represent within in the y
X β
ε framework. The first two are easier to implement for
¤
computations. As usual, some preliminary graphical analysis is appropriate before fitting. A side-by-side
boxplot is often the most useful plot. Look for equality of variance, transformations, outliers (influence is
not relevant here since leverages won’t differ unless the design is very unbalanced).
16.1.2
Estimation and testing
The effects can be estimated using direct formulae as above or by using the least squares approach (the
outcome is the same). The first test of interest is whether there is a difference in the levels of the factor. We
compare
168
16.1. ONE-WAY ANOVA
169
H0 : αi
0
i
¤
Ha : at least one αi is non zero.
We use the same F-test as we have used for regression. The outcome of this test will be the same no
matter what coding/restriction we use. If the null is accepted then we are done (subject to an investigation
of transformation and outliers). If we reject the null, we must investigate which levels differ.
16.1.3
An example
The example dataset we will use is a set of 24 blood coagulation times. 24 animals were randomly assigned
to four different diets and the samples were taken in a random order. This data comes from Box, Hunter,
and Hunter (1978).
> data(coagulation)
> coagulation
coag diet
1
62
A
2
60
A
...etc...
23
63
D
24
59
D
The first step is to plot the data - boxplots are useful:
> plot(coag ˜ diet, data=coagulation)
See the first panel of Figure 16.1.3. We are hoping not to see
1. Outliers — these will be apparent as separated points on the boxplots. The default is to extend the
whiskers of the boxplot no more than one and half times the interquartiles range from the quartiles.
Any points further away than this are plotted separately.
2. Skewness — this will be apparent from an asymmetrical form for the boxes.
3. Unequal variance — this will be apparent from clearly unequal box sizes. Some care is required
because often there is very little data be used in the construction of the boxplots and so even when the
variances truly are equal in the groups, we can expect a great deal of variability
In this case, there are no obvious problems. For group C, there are only 4 distinct observations and one
is somewhat separated which accounts for the slightly odd looking plot.
Now let’s fit the model.
> g <- lm(coag ˜ diet, coagulation)
> summary(g)
Coefficients:
Estimate Std. Error
t value Pr(>|t|)
(Intercept)
6.10e+01
1.18e+00
51.55
< 2e-16
dietB
5.00e+00
1.53e+00
3.27
0.00380
dietC
7.00e+00
1.53e+00
4.58
0.00018
dietD
-1.00e-14
1.45e+00 -7.4e-15
1.00000
16.1. ONE-WAY ANOVA
170
Normal Q−Q Plot
70
4
2
65
0
coag
Sample Quantiles
−2
60
−4
A
B
C
D
−2
−1
0
1
2
diet
Theoretical Quantiles
Residual−Fitted plot
Jittered plot
4
4
2
2
0
0
Residuals
Residuals
−2
−2
−4
−4
61
63
65
67
62
64
66
68
Fitted
Fitted
Figure 16.1: One way anova plots
16.1. ONE-WAY ANOVA
171
Residual standard error: 2.37 on 20 degrees of freedom
Multiple R-Squared: 0.671,
Adjusted R-squared: 0.621
F-statistic: 13.6 on 3 and 20 degrees of freedom,
p-value: 4.66e-05
We conclude from the small p-value for the F-statistic that there is some difference between the groups?
Group A is the reference level and has a mean of 61, groups B, C and D are 5, 7 and 0 seconds larger on
average. Examine the design matrix to understand the coding:
> model.matrix(g)
We can fit the model without an intercept term as in
> gi <- lm(coag ˜ diet -1, coagulation)
> summary(gi)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
dietA
61.000
1.183
51.5
<2e-16
dietB
66.000
0.966
68.3
<2e-16
dietC
68.000
0.966
70.4
<2e-16
dietD
61.000
0.837
72.9
<2e-16
Residual standard error: 2.37 on 20 degrees of freedom
Multiple R-Squared: 0.999,
Adjusted R-squared: 0.999
F-statistic: 4.4e+03 on 4 and 20 degrees of freedom,
p-value:
0
We can directly read the level means but the tests are not useful since they involve comparisons with zero.
Note the miscalculation of R2.
16.1.4
Diagnostics
Remember to plot the residuals/fitted values and do the QQ plot. Influential points and transforming the
predictors are not an issue although it is reasonable to consider transforming the response if the situation
demands it.
See the last three panels of Figure 16.1.3.
> qqnorm(g$res)
> plot(g$fit,g$res,xlab="Fitted",ylab="Residuals",
main="Residual-Fitted plot")
> plot(jitter(g$fit),g$res,xlab="Fitted",ylab="Residuals",
main="Jittered plot")
Because the data are integers and the fitted values turn out to integers also, some discreteness is obvious
in the Q-Q plot. Of course, discrete data can’t be normally distributed. However, here it is approximately
normal and so we can go ahead with the inference without any qualms. The discreteness in the residuals and
fitted values shows up in the residual-fitted plot because we can see fewer points than the sample size. This
is because of overplotting of the point symbols. There are several ways round this problem. One simple
solution is to add a small amount of noise to the data. This is called jittering. Sometimes you have to tune
the amount of noise but the default setting is adequate here.
16.1. ONE-WAY ANOVA
172
16.1.5
Multiple Comparisons
After detecting some difference in the levels of the factor, interest centers on which levels or combinations
of levels are different. Note that it does not make sense to ask whether a particular level is significant since
this begs the question, “significantly different from what”. Any meaningful test must involve a comparison
of some kind.
It is important to ascertain whether the comparison made were decided on before or after examining the
data. After fitting a model, one might decide to test only those differences that look large. To make such
a decision, you also have to examine the small differences. Even if you do not actually test these small
differences, it does have an effect on the inference.
If the comparisons were decided on prior to examining the data, there are three cases:
1. Just one comparison — use the standard t-based confidence intervals that we have used before.
2. Few comparisons — use the Bonferroni adjustment for the t. If there are m comparisons, use α m for
the critical value.
3. Many comparisons — Bonferroni becomes increasingly conservative as m increases. At some point
it is better to use the Tukey or Scheff´e or related methods described below.
It is difficult to be honest and be seen to be honest when using pre-data comparisons. Will people
really believe that you only planned to make certain comparisons? Although some might make a distinction
between pre and post-data comparisons, I think it is best to consider all comparisons as post-data.
If the comparisons were decided on after examining the data, you must adjust the CI to allow for the
possibility of all comparisons of the type to be made.
There are two important cases:
1. Pairwise comparisons only. Use the Tukey method.
2. All contrasts i.e. linear combinations. Use the Scheff´e method.
We consider pairwise comparisons first. A simple C.I. for α
α
i
j is
α
1
ˆ
α
2
¢
i
ˆ
αj t
ˆ
σ 1
n I
Ji
Jj
A test for α
α
i
j amounts to seeing whether zero lies in this interval or not. This is fine for just one test but
¤
suppose we do all possible pairwise tests when α
5% and the null hypothesis is in fact true. In Table 16.1,
¤
we see effects of multiple comparisons on the true error rates.
I
2
3
4
5
6
Nominal Type I error
5%
5%
5%
5%
5%
Actual overall Type I error
5%
12.2%
20.3%
28.6%
36.6%
Table 16.1: True error rates for multiple comparisons
We see that the true type I error can get quite high. Using the t-based CI for multiple comparisons is
called least significant differences or LSD but this one is a bad trip. When comparisons are only made after
the overall F-test shows a difference, it’s called Fisher’s LSD — this one isn’t quite so bad but the type I
error will still be too big.
16.1. ONE-WAY ANOVA
173
We can use a simulation to illustrate the issues involved in multiple comparisons. Because random
numbers are random, your results may differ but the message will be the same.
Suppose we have a factor of 6 levels with 4 observations per level:
> x <- factor(rep(LETTERS[1:6],rep(4,6)))
> x
[1] A A A A B B B B C C C C D D D D E E E E F F F F
Levels:
A B C D E F
and suppose the response has no relationship to the factor (i.e. the null hypothesis holds):
> g <- lm(rnorm(24) ˜ x)
> gs <- summary(g)
Here are the coefficients:
> g$coef
(Intercept)
xB
xC
xD
xE
xF
0.221638
0.331200
0.058631
-0.536102
0.295339
0.067889
¡
¡
¡
¡
The t-statistic for testing whether level A = level B is ˆ
αA ˆαB se ˆαA ˆαB where se ˆαA ˆαB
¢
¡
¢
¢
¢
¤
¡
ˆ
σ 1 4 1 4
ˆ
σ
2
¤
¢
> g$coef[2]*sqrt(2)/gs$sig
xB
0.41881
This would (in absolute value) need exceed this t-critical value for significance at the 5% level:
> qt(0.975,24-6)
[1] 2.1009
Out of all the possible pairwise comparisons, we may compute the maximum t-statistic as
> range(c(0,g$coef[-1]))
[1] -0.5361
0.3312
> rg <- range(c(0,g$coef[-1]))
> (rg[2]-rg[1])*sqrt(2)/gs$sig
[1] 1.0967
which just fails to meet significance. Now let’s repeat the experiment 1000 times.
> res <- matrix(0,1000,2)
> for(i in 1:1000){
g <- lm(rnorm(24) ˜ x)
gs <- summary(g)
res[i,1] <- abs(g$coef[2]*sqrt(2)/gs$sig)
rg <- range(c(0,g$coef[-1]))
res[i,2] <- (rg[2]-rg[1])*sqrt(2)/gs$sig
}
16.1. ONE-WAY ANOVA
174
Now see how many of the test statistics for comparing level A and level B were significant at the 5%
level:
> sum(res[,1] > 2.1)/1000
[1] 0.045
Just a shade under the 5% it should be. Now see how many times the maximum difference is significant
at the 5% level.
> sum(res[,2] > 2.1)/1000
[1] 0.306
About 1/3 of the time. So in cases where there is no difference between levels of the factor, about 1/3
of the time, an investigator will find a statistically significant difference between some levels of the factor.
Clearly there is a big danger that one might conclude there is a difference when none truly exists.
We need to make the critical value larger so that the null is rejected only 5% of the time. Using the
simulation results we estimate this value to be:
> quantile(res[,2],0.95)
95%
3.1627
It turns out that this value may be calculated using the ”Studentized Range distribution”:
> qtukey(0.95,6,18)/sqrt(2)
[1] 3.1780
which is close to the simulated value.
Now let’s take a look at the densities of our simulated t-statistics:
> dmax <- density(res[,2],from=0,to=5)
> d2 <- density(res[,1],from=0,to=5)
> matplot(d2$x,cbind(dmax$y,d2$y),type="l",xlab="Test statistic",
ylab="Density")
> abline(h=0)
> abline(v=2.1,lty=2)
> abline(v=3.178)
We see the result in Figure 16.2. We see that the distribution of the maximum t-statistic has a much heavier
tail than the distribution for a prespecified difference. The true density for the prespecified difference is a the
upper half of a t-distribution — the maximum in the estimated distribution does not occur at zero because
boundary error effects in the density estimator.
Now we return to our real data. We’ve found that there is a significant difference among the diets but
which diets can be said to be different and which diets are not distinguishable. Let’s do the calculations for
the difference between diet B and diet C which is 2. First we do the LSD calculation:
> qt(1-.05/2,20)*2.366*sqrt(1/6+1/6)
[1] 2.8494
> c(2-2.85,2+2.85)
[1] -0.85
4.85
16.1. ONE-WAY ANOVA
175
0.6
0.4
Density
0.2
0.0
0
1
2
3
4
5
Test statistic
Figure 16.2: Estimated densities of the maximum t-statistic (solid line) and t-statistic for a prespecified
difference (dashed line). The corresponding theoretical 95% quantiles are marked with vertical lines
An alternative way we can make the same calculation is to recode the factor with B as the reference level
and refit the model:
> coagulation$diet <- relevel(coagulation$diet,ref="B")
> g <- lm(coag ˜ diet, coagulation)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
66.000
0.966
68.32
< 2e-16
dietA
-5.000
1.528
-3.27
0.00380
dietC
2.000
1.366
1.46
0.15878
dietD
-5.000
1.278
-3.91
0.00086
Residual standard error: 2.37 on 20 degrees of freedom
Multiple R-Squared: 0.671,
Adjusted R-squared: 0.621
F-statistic: 13.6 on 3 and 20 degrees of freedom,
p-value: 4.66e-05
We can read the B vs. C difference directly from the output now as 2 and compute the width of the
confidence band using the corresponding standard error:
> qt(0.975,20)*1.366
[1] 2.8494
We can verify that the standard error of 1.366 can also be obtained directly as
> 2.366*sqrt(1/6+1/6)
[1] 1.366
16.1. ONE-WAY ANOVA
176
As can be seen, the result is the same as before.
Suppose two comparisons were pre-planned, then critical value is now this, using the Bonferroni cor-
rection.
> qt(1-.05/4,20)*2.37*sqrt(1/6+1/6)
[1] 3.3156
> c(2-3.32,2+3.32)
[1] -1.32
5.32
Tukey’s Honest Significant Difference (HSD) is designed for all pairwise comparisons and depends
¡
on the studentized range distribution. Let X
σ2
1
Xn be i.i.d. N µ
and let R
maxi Xi
mini Xi be the
¢
¤
¢¡¢¡¢¡£
range. Then R ˆ
σ has the studentized range distribution qn ν where ν is the number of degrees of freedom
¢
used in estimating σ.
The Tukey C.I.’s are
ˆ
σ
1
1
ˆ
α
¢
i
ˆ
αj qI n I
¢
J
J
2
¢
i
j
When the sample sizes Ji are very unequal, Tukey’s HSD may be too conservative but in general they are
narrower than those produced by Scheff´e’s theorem. There are several other methods for multiple compar-
isons — the Tukey method tends to be more conservative than most because it takes the rather pessimistic
approach based on the maximum difference. Not all the differences will be as large as the maximum and so
some competing methods take advantage of this to get tighter intervals.
For future reference, a more general form for the Tukey intervals is
¡
¡
¡
difference ¢
q
l d f
¢
2
se of difference
¢
¢
¢
¢
where l is the number of levels of the factor on which we are making multiple comparisons and d f is the
degrees of freedom for the error.
We compute the Tukey HSD bands for the diet data. First we need the critical value from the studentized
range distribution.
> qtukey(0.95,4,20)
[1] 3.9583
and the interval is:
> (3.96/sqrt(2))*2.37*sqrt(1/6+1/6)
[1] 3.8315
> c(2-3.83,2+3.83)
[1] -1.83
5.83
A convenient way to obtain all the intervals is
> TukeyHSD(aov(coag ˜ diet, coagulation))
Tukey multiple comparisons of means
95% family-wise confidence level
$diet
diff
lwr
upr
16.1. ONE-WAY ANOVA
177
A-B -5.0000e+00
-9.2754 -0.72455
C-B
2.0000e+00
-1.8241
5.82407
D-B -5.0000e+00
-8.5771 -1.42291
C-A
7.0000e+00
2.7246 11.27545
D-A -1.4211e-14
-4.0560
4.05604
D-C -7.0000e+00 -10.5771 -3.42291
The Bonferroni based bands would have been just slightly wider:
> qt(1-.05/12,20)*2.37*sqrt(1/6+1/6)
[1] 4.0052
We divide by 12 here because there are 6 possible pairwise differences and we want a two-sided con-
fidence interval: 6 2
12. With a bit of work we find that only the A-D and B-C differences are not
¤
significant.
The Tukey method assumes the worst by focusing on the largest difference. There are other competitors
like the Newman-Keuls, Duncan’s multiple range and the Waller-Duncan procedure. For a detailed descrip-
tion of the many availabe alternatives see Hsu (1996). Some other pairwise comparison tests may be found
in the R package ctest.
16.1.6
Contrasts
A contrast among the effects α
α
α
1
I is a linear combination ∑i ci i where the ci are known and ∑i ci
0.
¤
¢¡¢¡¢¡£
For example
1. α
α
1
2 is a contrast with c1
1 c2
1 and the other ci
0. All pairwise differences are contrasts.
¤
¤
¤
¡
¡
2. α
α
α
α
1
2
2
3
4
2 with c1
c2
1 2 and c3
c4
1 2 and the other ci
0. This contrast
¢
¡
¢
¡
¤
¤
¤
¤
¤
is directly interpretable.
16.1.7
Scheff´e’s theorem for multiple comparisons
An estimable function of the parameters is one that can be estimated given the data we have. More precisely,
a linear combination ψ
cT β is estimable if there exists an aT y such that EaT y
cT β. Contrasts are
¤
¤
estimable but something like αi is not because it will depend on the coding used. Now ˆ
ψ aT y and var ˆψ
¤
¤
σ2aT a which can be estimated by ˆσ2aT a. Suppose we let the dimension of the space of possible c be q and
¡
the rank X
r. (r
p if we have complete identifiability.)
¢
¤
¤
Scheff´e’s theorem
¡
A 100 1
α % simultaneous confidence interval for all estimable ψ is
¢
ˆ
ψ ¢
qFα
q n r
vˆar ˆ
ψ
¢
Example: Simultaneous confidence interval for the regression surface:
xT ˆβ ¢
pFα
ˆ
σ
1
p n p
xT ¡ X T X
x
¢
¢
We can illustrate this with corrosion data used in the lack of fit chapter. We compute the usual t-based
pointwise bands along with the simultaneous Scheff´e bands:
16.1. ONE-WAY ANOVA
178
> data(corrosion)
> gf <- lm(loss ˜ Fe, corrosion)
> grid <- seq(0,3,by=0.1)
> p <- predict(gf,data.frame(Fe=grid),se=T)
> fmult <- sqrt(2*qf(0.95,2,11))
> tmult <- qt(0.975,11)
> matplot(grid,cbind(p$fit,p$fit-fmult*p$se,p$fit+fmult*p$se,
p$fit-tmult*p$se,p$fit+tmult*p$se),type="l",lty=c(1,2,2,5,5),
ylab="loss",xlab="Iron Content")
> points(corrosion$Fe,corrosion$loss)
The plot is shown in Figure 16.3. The bands form a 95% simultaneous confidence region for the true
regression line. These bands are slightly wider than the t-based pointwise confidence bands described in
Chapter 3. This is because they hold over the whole real line and not just a single point.
120
100
loss
80
60
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Iron Content
Figure 16.3: Scheff´e 95% simultaneous confidence bands are shown as dotted lines surrounding the least
squares fit. The interior (dashed) lines represent the pointwise confidence intervals.
Example: One-way anova: Consider ψ
∑ αi, a contrast which is therefore estimable. We compute
¤
i ci
I
c2
v ˆ
ar ˆ
ψ
ˆ
σ2 ∑ i
¤
J
i 1
i
§
and the SCI for ψ is then
I
∑
c2
¡
c
¢
i
i ˆ
αi
I
1 Fα
ˆ
σ ∑
¢
I 1 n I
¢
J
i
i
i 1
§
Here we apply the Scheff´e method for (B+C)/2 -(A+D)/2 so that c2
c3
1 2 and c1
c4
1 2
¤
¤
¤
¤
> sqrt(3*qf(0.95,3,20))*2.37*sqrt(1/4+1/6+1/6+1/8)/2
[1] 3.0406
16.2. TWO-WAY ANOVA
179
> (5+7)/2-(0+0)/2
[1] 6
> c(6-3.04,6+3.04)
[1] 2.96 9.04
We see that this difference is significantly different from 0 and so we may conclude that there is signif-
icant difference between the average of B and C and the average of A and D despite the fact that we may
have chosen to test this difference after seeing the data.
16.1.8
Testing for homogeneity of variance
This can be done using Levene’s test. Simply compute the absolute values of the residuals and use these as
the response in a new one-way anova. A significant difference would indicate non constant variance.
There are other tests but this one is quite insensitive to non-normality and is simple to execute. Most
tests and CI’s are relatively insensitive to non-constant variance so there is no need to take action unless the
Levene test is significant at the 1% level.
Applying this to the diet data, we find:
> summary(lm( abs(g$res) ˜ coagulation$diet))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
1.500
0.716
2.10
0.049
coagulation$dietB
0.500
0.924
0.54
0.594
coagulation$dietC
-0.500
0.924
-0.54
0.594
coagulation$dietD
0.500
0.877
0.57
0.575
Residual standard error: 1.43 on 20 degrees of freedom
Multiple R-Squared: 0.0956,
Adjusted R-squared: -0.0401
F-statistic: 0.705 on 3 and 20 degrees of freedom,
p-value: 0.56
Since the p-value is large, we conclude that there is no evidence of a non-constant variance.
16.2
Two-Way Anova
Suppose we have two factors, α at I levels and β at J levels. Let ni j be the number of observations at level
i of α and level j of β and let those observations be yi j1 yi j2
. A complete layout has ni j
1 for all i j.
¡
¢¡¢¡¢¡
The most general model that may be considered is
¡
y
α
β
αβ
ε
i jk
µ
i
j
i j
i jk
¤
¢
¡
The interaction effect αβ i j is interpreted as that part of the mean response not attributable to the additive
¢
effect of αi and β j. For example, you may enjoy strawberries and cream individually, but the combination
is superior. In contrast, you may like fish and ice cream but not together.
A balanced layout requires that ni j
n. Not all the parameters are identifiable but if the main effects α
¤
and β are coded appropriately and the interaction effects coding is then derived from the product of these
codings, then every contrast of parameters can be estimated.
16.2. TWO-WAY ANOVA
180
16.2.1
One observation per cell
When ni j
1 we would have as many observation as parameters if we tried to fit the full model as above.
¤
The parameters could be estimated but no further inference would be possible.
¡
We can assume αβ i j
0 to free up degrees of freedom to make some tests and CI’s. This assumption
¢
¤
can be checked graphically using an interaction plot - plot the cell means on the vertical axis and the factor
α on the horizontal. Join points with same level of β. The role of α and β can be reversed. Parallel lines on
the plot are a sign of a lack of interaction. Tukey’s non-additivity test provides another way of investigating
an interaction - the model
y
α
β
φα β
ε
i j
µ
i
j
i
j
i jk
¤
is fit to the data and then we test if φ
0. This is a nonlinear model and that it makes the assumption that
¤
the interaction effect is multiplicative in a form which seems somewhat tenuous.
Barring any trouble with interaction, because of the balanced design, the factors are orthogonal and their
significance can be tested in the usual way.
16.2.2
More than one observation per cell
When ni j
n i.e. the same number of observations per cell, we have orthogonality. Orthogonality can
¤
also occur if the row/column cell numbers are proportional. Orthogonality is desirable and experiments are
usually designed to ensure it.
With more than one observation per cell we are now free to fit and test the model:
¡
y
α
β
αβ
ε
i jk
µ
i
j
i j
i jk
¤
¢
The interaction effect may be tested by comparison to the model
y
α
β
ε
i jk
µ
i
j
i jk
¤
and computing the usual F-test. If the interaction effect is found to be significant, do not test the main effects
even if they appear not to be significant. The estimation of the main effects and their significance is coding
dependent when interactions are included in the model.
If the interaction effect is found to be insignificant, then test the main effects but use RSS d f from the
full model in the denominator of the F-tests — this has been shown to maintain the type I error better. So
the F-statistic used is
¡
F
RSSsmall
RSSlarge d fsmall
d flarge
¤
¢
ˆ
σ2full
16.2.3
Interpreting the interaction effect
No interactions You can do pairwise comparisons on α without regard to β and vice versa.
Interaction present A comparison of the levels of α will depend on the level of β. Interpretation is not
simple. Consider the following two layouts of ˆµi j in a 2x2 case:
Male
Female
Male
Female
drug 1
3
5
2
1
drug 2
1
2
1
2
16.2. TWO-WAY ANOVA
181
The response is a measure of performance. In the case on the left, we can say that drug 1 is better than
drug 2 although the interaction means that its superiority over drug 2 depends on the gender. In the case on
the right, which drug is best depends on the gender. We can also plot this as in Figure 16.4. We see that
neither case are the lines parallel indicating interaction but the superiority of drug 1 is clear in the first plot
and the ambiguous conclusion is clear in the second. I recommend making plots like this when you want to
understand an interaction effect.
5
2.0
4
1.8
1.6
3
Response
Response
1.4
2
1.2
1
1.0
Male
Female
Male
Female
Sex
Sex
Figure 16.4: Two 2x2 tables with the response plotted by the factors, sex on the horizontal axis and drug 1
as the solid line and drug 2 as the dotted line.
When the interaction is significant, the main effects cannot be defined in an obvious and universal way.
For example, we could the gender effect as the effect for females, the effect for males, the effect for the
average males and females or something else. If there was no interaction effect, the gender effect could be
defined unambiguously.
When you have a significant inteaction, you can fit a model
y
ε
i jk
µi jk
i jk
¤
and then treat the data as a one-way anova with IJ levels. Obviously this makes for more complex compar-
isons but this is unavoidable when interactions exist.
Here is a two-way anova design where there are 4 replicates. As part of an investigation of toxic agents,
48 rats were allocated to 3 poisons (I,II,III) and 4 treatments (A,B,C,D). The response was survival time in
tens of hours. The Data:
A
B
C
D
I
0.31
0.82
0.43
0.45
0.45
1.10
0.45
0.71
0.46
0.88
0.63
0.66
0.43
0.72
0.76
0.62
II
0.36
0.92
0.44
0.56
16.2. TWO-WAY ANOVA
182
0.29
0.61
0.35
1.02
0.40
0.49
0.31
0.71
0.23
1.24
0.40
0.38
III
0.22
0.30
0.23
0.30
0.21
0.37
0.25
0.36
0.18
0.38
0.24
0.31
0.23
0.29
0.22
0.33
We make some plots:
> data(rats)
> plot(time ˜ treat + poison, data=rats)
Some evidence of skewness can be seen, especially since it appears that variance is in some way related
to the mean response. We now check for an interaction using graphical methods:
> interaction.plot(rats$treat,rats$poison,rats$time)
> interaction.plot(rats$poison,rats$treat,rats$time)
Do these look parallel? The trouble with interaction plots is that we expect there to be some random
variation regardless so it is difficult to distinguish true interaction from just noise. Fortunately, in this case,
we have replication so we can directly test for an interaction effect.
Now fit the full model and see the significance of the factors:
> g <- lm(time ˜ poison*treat, rats)
> anova(g)
Analysis of Variance Table
Response: time
Df Sum Sq Mean Sq F value
Pr(>F)
poison
2
1.033
0.517
23.22 3.3e-07
treat
3
0.921
0.307
13.81 3.8e-06
poison:treat
6
0.250
0.042
1.87
0.11
Residuals
36
0.801
0.022
We see that the interaction effect is not significant but the main effects are. We check the diagnostics:
> qqnorm(g$res)
> plot(g$fitted,g$res,xlab="Fitted",ylab="Residuals")
Clearly there’s a problem - perhaps transforming the data will help. Try logs first:
> g <- lm(log(time) ˜ poison*treat,rats)
> plot(g$fitted,g$res,xlab="Fitted",ylab="Residuals",main="Log response")
Not enough so try the reciprocal:
> g <- lm(1/time ˜ poison*treat,rats)
> plot(g$fitted,g$res,xlab="Fitted",ylab="Residuals",
main="Reciprocal response")
16.2. TWO-WAY ANOVA
183
Interaction plot
0.9
1.2
1.2
I
0.8
II
1.0
1.0
0.7
III
0.8
0.8
0.6
tim
tim
0.5
0.6
0.6
Cell means
0.4
0.4
0.4
0.3
0.2
0.2
0.2
A
B
C
D
I
II
III
A
B
C
D
treat
poison
Treatment
Interaction plot
Normal Q−Q Plot
0.9
0.4
0.4
A
0.8
B
0.7
C
0.2
0.2
0.6
D
Cell means
0.5
0.0
Residuals
0.0
Sample Quantiles
0.4
0.3
−0.2
−0.2
0.2
I
II
III
−2
−1
0
1
2
0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Poison
Theoretical Quantiles
Fitted
Log response
Reciprocal response
Normal Q−Q Plot
1.0
1.0
0.4
0.2
0.5
0.5
0.0
Residuals
Residuals
0.0
0.0
Sample Quantiles
−0.2
−0.5
−0.5
−0.4
−1.6
−1.2
−0.8
−0.4
2
3
4
−2
−1
0
1
2
Fitted
Fitted
Theoretical Quantiles
Figure 16.5: Two way anova plots
16.2. TWO-WAY ANOVA
184
Looks good - the reciprocal can be interpreted as the rate of dying. Better check the Q-Q plot again:
> qqnorm(g$res)
This looks better than the first Q-Q plot. We now check the ANOVA table again, find the interaction is
not significant, simplify the model and examine the fit:
> anova(g)
Analysis of Variance Table
Response: 1/time
Df Sum Sq Mean Sq F value
Pr(>F)
poison
2
34.9
17.4
72.63 2.3e-13
treat
3
20.4
6.8
28.34 1.4e-09
poison:treat
6
1.6
0.3
1.09
0.39
Residuals
36
8.6
0.2
> g <- lm(1/time ˜ poison+treat, rats)
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
2.698
0.174
15.47
< 2e-16
poisonII
0.469
0.174
2.69
0.0103
poisonIII
1.996
0.174
11.45
1.7e-14
treatB
-1.657
0.201
-8.23
2.7e-10
treatC
-0.572
0.201
-2.84
0.0069
treatD
-1.358
0.201
-6.75
3.3e-08
Residual standard error: 0.493 on 42 degrees of freedom
Multiple R-Squared: 0.844,
Adjusted R-squared: 0.826
F-statistic: 45.5 on 5 and 42 degrees of freedom,
p-value: 6.66e-16
Let’s construct pairwise confidence intervals for the treatment factor using the Tukey method. Because
of the balance, the CI’s will all be the same width. First the standard error for a pairwise difference may be
obtained from the output as 0.201. We then compute the width of the interval as
> qtukey(0.95,4,42)*0.201/sqrt(2)
[1] 0.53767
So the bands will be difference plus or minus 0.54. All the bands except the B-D do not include 0 so
we can conclude that all these other pairs of treatments are significantly different. The treatment reduces the
rats survival time the most is A since it is the reference level and all other treatments reduce the response
(rate of dying). Can you distinguish between the poisons?
16.2.4
Replication
It’s important that the observations observed in each cell are genuine replications. If this is not true, then the
observations will be correlated and the analysis will need to be adjusted. It is a common scientific practice
16.3. BLOCKING DESIGNS
185
to repeat measurements and take the average to reduce measurement errors. These repeat measurements
are not independent observations. Data where the replicates are correlated can be handled with repeated
measures models.
For example, imagine that the experiment above involved the reaction times of human subjects under two
factors. We need to distinguish between an experiment that uses 48 subjects and one that uses 12 subjects
where each subject repeats their assigned factor combination 4 times. In the latter case, the responses will
not be independent and a repeated measures style of analysis will be necessary.
16.3
Blocking designs
In completely randomized designs (CRD) like the one and two-way anova, the treatments are assigned to
the experimental units at random. This is appropriate when the units are homogenous. Sometimes, we may
suspect that the units are heterogenous, but we can’t describe the form it takes - for example, we may know
a group of patients are not identical but we may have no further information about them. In this case, it is
still appropriate to use a CRD. Of course, the randomization will tend to spread the heterogeneity around
to reduce bias, but the real justification lies in the randomization test discussed earlier for regression. The
usual testing argument may be applied. Under the null hypothesis, there is no link between a factor and
the response. In other words, the responses have been assigned to the units in a way that is unlinked to the
factor. This corresponds to the randomization used in assigning the levels of the factor to the units. This
is why the randomization is crucial because it allows us to make this argument. Now if the difference in
the response between levels of the factor seems too unlikely to have occurred by chance, we can reject the
null hypothesis. The normal-based inference is approximately equivalent to the permutation-based test. The
normal-based inference is much quicker so we might prefer to use that.
When the experimental units are heterogenous in a known way and can be arranged into blocks where
the intrablock variation is ideally small but the interblock variation is large, a block design can be more
efficient than a CRD.
The contrast between the two designs is shown in Figure 16.6.
Examples:
Suppose we want to compare 4 treatments and have 20 patients available. We might be able divide the
patients in 5 blocks of 4 patients each where the patients in each block have some relevant similarity. We
would then randomly assign the treatments within each block.
Suppose we want to test 3 crop varieties on 5 fields. Divide each field into 3 strips and randomly assign
the crop variety.
Note: We prefer to have block size equal to the number of treatments. If this is not done or possible, an
incomplete block design must be used.
Notice that under the randomized block design the randomization used in assigning the treatments to
the units is restricted relative to the full randomization used in the CRD. This has consequences for the
inference.
16.3.1
Randomized Block design
We have one factor (or treatment) at t levels and one blocking variable at r levels. The model is
y
τ ρ
ε
i j
µ
i
j
i j
¤
The analysis is then very similar to the two-way anova with one observation per cell. We can check for
interaction and check for a treatment effect. We can also check the block effect but this is only useful for
future reference. Blocking is a feature of the experimental units and restricts the randomized assignment of
16.3. BLOCKING DESIGNS
186
Completely randomized design
Randomized Block Design
y
Block 1
¢¡
y
Block 2
£
All
Treatments
units
randomly
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randomly
assigned
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assigned
X
to the units
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with each
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block
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¥
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y
Block r
Figure 16.6: Completely randomized design vs. Randomized Block design
the treatments. This means that we cannot regain the degrees of freedom devoted to blocking even if the
blocking effect turns out not to be significant. The randomization test-based argument means that we must
judge the magnitude of the treatment effect within the context of the restricted randomization that has been
used.
We illustrate with an experiment to compare 4 processes, A,B,C,D for the production of penicillin.
These are the treatments. The raw material, corn steep liquor, is quite variable and can only be made in
blends sufficient for 4 runs. Thus a randomized complete block design is definitely suggested by the nature
of the experimental units. The data is:
A
B
C
D
Blend 1
89
88
97
94
Blend 2
84
77
92
79
Blend 3
81
87
87
85
Blend 4
87
92
89
84
Blend 5
79
81
80
88
We start with some graphical checks:
> data(penicillin)
> plot(yield ˜ blend+treat,data=penicillin)
See the first two panels of Figure 16.3.1
Did you see any problems? Now check for interactions:
16.3. BLOCKING DESIGNS
187
95
95
90
90
y
y
85
85
80
80
Blend1
Blend2
Blend3
Blend4
Blend5
A
B
C
D
blend
treat
Interaction Plot
Interaction Plot
95
95
90
90
Response
85
Response
85
80
80
A
B
C
D
1
2
3
4
5
Treatment
Blend
Normal Q−Q Plot
6
6
4
4
2
2
0
0
Residuals
Sample Quantiles
−2
−2
−4
−4
80
85
90
95
−2
−1
0
1
2
Fitted
Theoretical Quantiles
Figure 16.7: RCBD plots for the penicillin data
16.3. BLOCKING DESIGNS
188
> interaction.plot(penicillin$treat,penicillin$blend,penicillin$yield)
> interaction.plot(penicillin$blend,penicillin$treat,penicillin$yield)
What do you think? It is hard to tell — interaction plots are only suggestive, not definitive.
Regardless, we now fit the model:
> g <- lm(yield ˜ treat+blend,penicillin)
> anova(g)
Analysis of Variance Table
Response: yield
Df Sum Sq Mean Sq F value Pr(>F)
treat
3
70.0
23.3
1.24
0.339
blend
4
264.0
66.0
3.50
0.041
Residuals 12
226.0
18.8
We see no significant treatment effect but the block effect is, as suspected, significant. The analysis of
variance table corresponds to a sequential testing of models, here corresponding to the sequence
y ˜ 1
y ˜ treat
y ˜ treat + blend
So here the p-value 0.339 corresponds to a comparison of the first two models in this list, while the p-value
of 0.041 corresponds to the test comparing the second two. One small point to note is that the denominator
in both F-test is the mean square from the full model, here 18.8
Notice that if we change the order of the terms in the ANOVA, it makes no difference because of the
orthogonal design:
> anova(lm(yield ˜ blend+treat,penicillin))
Analysis of Variance Table
Response: yield
Df Sum Sq Mean Sq F value Pr(>F)
blend
4
264.0
66.0
3.50
0.041
treat
3
70.0
23.3
1.24
0.339
Residuals 12
226.0
18.8
By way of comparison, see what happens if we omit the first observation in the dataset — this might
happen in practice if this run was lost:
> anova(lm(yield ˜ blend+treat,penicillin[-1,]))
Analysis of Variance Table
Response: yield
Df Sum Sq Mean Sq F value Pr(>F)
blend
4
266.5
66.6
3.27
0.054
treat
3
59.7
19.9
0.98
0.439
Residuals 11
224.3
20.4
16.3. BLOCKING DESIGNS
189
> anova(lm(yield ˜ treat+blend,penicillin[-1,]))
Analysis of Variance Table
Response: yield
Df Sum Sq Mean Sq F value Pr(>F)
treat
3
91.8
30.6
1.50
0.269
blend
4
234.4
58.6
2.87
0.075
Residuals 11
224.3
20.4
Notice that now the order does matter. If we want to test for a treatment effect, we would prefer the first
table since in that version the blocking factor blend is already included when we test the treatment factor.
Since the blocking factor is an unalterable feature of the chosen design, this is as it should be.
Check the diagnostics:
> plot(g$fitted,g$res,xlab="Fitted",ylab="Residuals")
> qqnorm(g$res)
And that might be the end of the story except for that worrying interaction effect possibility. We execute
the Tukey non-additivity test:
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
90.00
2.74
32.79
4.1e-13
treatB
1.00
2.74
0.36
0.7219
treatC
5.00
2.74
1.82
0.0935
treatD
2.00
2.74
0.73
0.4802
blendBlend2
-9.00
3.07
-2.93
0.0125
blendBlend3
-7.00
3.07
-2.28
0.0416
blendBlend4
-4.00
3.07
-1.30
0.2169
blendBlend5
-10.00
3.07
-3.26
0.0068
Residual standard error: 4.34 on 12 degrees of freedom
Multiple R-Squared: 0.596,
Adjusted R-squared: 0.361
F-statistic: 2.53 on 7 and 12 degrees of freedom,
p-value: 0.0754
> alpha <- c(0,g$coef[2:4])
> alpha
treatB treatC treatD
0
1
5
2
> beta <- c(0,g$coef[5:8])
> beta
blendBlend2 blendBlend3 blendBlend4 blendBlend5
0
-9
-7
-4
-10
> ab <- rep(alpha,5)*rep(beta,rep(4,5))
> h <- update(g,.˜.+ab)
> anova(h)
Analysis of Variance Table
16.3. BLOCKING DESIGNS
190
Response: yield
Df Sum Sq Mean Sq F value Pr(>F)
treat
3
70.0
23.3
1.15
0.374
blend
4
264.0
66.0
3.24
0.055
ab
1
2.0
2.0
0.10
0.760
Residuals 11
224.0
20.4
Because the p-value of the treat times block effect is .76 we accept the null hypothesis of no interaction.
Of course, the interaction may be of a non-multiplicative form, but there is little we can do about that.
We can do multiple comparisons for the treatment effects, using the Tukey or the Scheff´e method as
appropriate:
ˆτ
¢
i
ˆτ j
q
ˆ
σ
r
¢
t t 1 r 1
¢
¡
¡
or
∑
¡
c
¢
i ˆ
τi
t
1 F
ˆ
σ ∑c2 r
¢
t 1 t 1 r 1
i
¢
¡
¡
i
i
Now, just for the sake of the example, we compute the Tukey pairwise confidence intervals for the
treatment effects:
> qtukey(0.95,4,12)
[1] 4.1987
The standard errors for the differences are
> 4.34*sqrt(1/5+1/5)
[1] 2.7449
Can you find this in the output above? The bands are difference plus or minus this:
> 4.2*2.745/sqrt(2)
[1] 8.1522
How does this compare with the observed difference in the treatment effects?
16.3.2
Relative advantage of RCBD over CRD
We can measure precision by considering var ˆτ (or equivalently ˆσ2). Compare the ˆσ2 for designs with the
same sample size. We define relative efficiency as
ˆ
σ2CRD
ˆ
σ2RCBD
where the quantities can be computed by fitting models with and without the blocking effect. For example,
¡
¡
suppose ˆ
σ2
226
264
12
4
30 6 and ˆ
σ2
18 8, as it is in the example above, then the
CRD ¤
¢
¡
¢
¤
RCBD ¤
¡
¡
relative efficiency is 1.62. The ˆ
σ2
numbers come from combining the sums of squares and degrees of
CRD
freedom for the residuals and the blend in the first anova table we made for this data. An alternative method
would be to simply fit the model yield ˜ treat and read off ˆ
σCRD from the output.
The interpretation is that a CRD would require 62% more observations to obtain the same level of
precision as a RCBD.
The efficiency is not guaranteed to be greater than one. Only use blocking where there is some hetero-
geneity in the experimental units. The decision to block is a matter of judgment prior to the experiment.
There is no guarantee that it will increase precision.
16.4. LATIN SQUARES
191
16.4
Latin Squares
These are useful when there are two blocking variables. For example, in a field used for agricultural exper-
iments, the level of moisture may vary across the field in one direction and the fertility in another. In an
industrial experiment, suppose we wish to compare 4 production methods (the treatment) — A, B, C, and
D. We have available 4 machines 1, 2, 3, and 4, and 4 operators, I, II, III, IV. A Latin square design is
1
2
3
4
I
A
B
C
D
II
B
D
A
C
III
C
A
D
B
IV
D
C
B
A
Table 16.2: Latin Square
Each treatment is assigned to each block once and only once.
The design and assignment of treatments and blocks should be random.
We use the model
y
τ
β
γ
ε
i jk
µ
i
j
k
i jk
i j k
1
t
¤
¤
¢¡¢¡¢¡
To test for a treatment effect simply fit a model without the treatment effect and compare using the F-test.
The Tukey pairwise CI’s are
ˆτ
¢
l
ˆτm
q
ˆ
σ 1 t
t t 1 t 2
¢
¡
¡
The Latin square can be even more efficient than the RCBD provided that the blocking effects are
sizable.
We need to have both block sizes to be equal to the number of treatments. This may be difficult to
achieve. Latin rectangle designs are possible by adjoining latin squares.
The Latin square can be used for comparing 3 treatment factors. Only t 2 runs are required compared to
the t3 required if all combinations were run. (The downside is that you can’t estimate the interactions
if they exist). This is an example of a fractional factorial.
The Latin square can be replicated if more runs are available.
When there are 3 blocking variables, a Graeco-Latin square may be used but these rarely arise in
practice.
An engineer wants to compare the qualities of raw materials from four suppliers, A, B, C, D. The raw
material is used to produce a component whose breaking strength is measured. It takes an operator a whole
day to make one component and there are 4 operators and 4 days on which the experiment will take place.
A Latin square design is appropriate here where the operator and the day are the blocking effects.
> data(breaking)
> breaking
y operator
day supplier
16.4. LATIN SQUARES
192
1
810
op1 day1
B
2 1080
op1 day2
C
...etc...
15 1025
op4 day3
D
16
900
op4 day4
C
We can check the Latin square structure:
> matrix(breaking$supplier,4,4)
[,1] [,2] [,3] [,4]
[1,] "B"
"C"
"D"
"A"
[2,] "C"
"D"
"A"
"B"
[3,] "A"
"B"
"C"
"D"
[4,] "D"
"A"
"B"
"C"
Plot the data:
> plot(y ˜ operator + day + supplier, breaking)
Examine the boxplots in Figure 16.4. There appear to be differences in suppliers but not in the two
blocking variables. No outlier, skewness or unequal variance is apparent.
Now fit the Latin squares model:
> g <- lm(y ˜ operator + day + supplier, breaking)
> anova(g)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
operator
3
7662
2554
0.41 0.7510
day
3
17600
5867
0.94 0.4759
supplier
3 371138
123712
19.93 0.0016
Residuals
6
37250
6208
Does it make a difference if we change the order of fitting? Let’s see:
> anova(lm(y ˜ day + supplier + operator, breaking))
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
day
3
17600
5867
0.94 0.4759
supplier
3 371137
123712
19.93 0.0016
operator
3
7662
2554
0.41 0.7510
Residuals
6
37250
6208
They are the same because of the balanced design. We see that there is clear supplier effect but no
evidence of an effect due to day or operator.
Now check the diagnostics
16.4. LATIN SQUARES
193
1000
1000
900
900
y
y
800
800
700
700
600
600
op1
op2
op3
op4
day1
day2
day3
day4
operator
day
100
1000
50
900
y
0
800
Residuals
700
−50
600
A
B
C
D
600
700
800
900
supplier
Fitted
Figure 16.8: Latin square analysis
16.4. LATIN SQUARES
194
> plot(g$fit,g$res,xlab="Fitted",ylab="Residuals")
> qqnorm(g$res,ylab="Residuals")
I show only the residual-fitted plot which is fine as was the Q-Q plot. Now look at the estimates of the
effects:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
667.5
62.3
10.72
3.9e-05
operatorop2
-35.0
55.7
-0.63
0.55302
operatorop3
-58.7
55.7
-1.05
0.33227
operatorop4
-46.2
55.7
-0.83
0.43825
dayday2
-40.0
55.7
-0.72
0.49978
dayday3
40.0
55.7
0.72
0.49978
dayday4
-40.0
55.7
-0.72
0.49978
supplierB
167.5
55.7
3.01
0.02381
supplierC
411.2
55.7
7.38
0.00032
supplierD
291.2
55.7
5.23
0.00196
Residual standard error: 78.8 on 6 degrees of freedom
Multiple R-Squared: 0.914,
Adjusted R-squared: 0.785
F-statistic: 7.09 on 9 and 6 degrees of freedom,
p-value: 0.0135
We see that Supplier C looks best followed by D. Is the difference significant though? Which suppliers
in general are significantly better than others? We need the Tukey pairwise intervals to help determine this.
The width of the bands calculated in the usual manner:
> qtukey(0.95,4,6)*55.7/sqrt(2)
[1] 193
The width of the interval is 193 - what can we say about differences between suppliers? We can make a
handy table of the supplier differences:
> scoefs <- c(0,g$coef[8:10])
> outer(scoefs,scoefs,"-")
supplierB supplierC supplierD
0.00
-167.50
-411.25
-291.25
supplierB 167.50
0.00
-243.75
-123.75
supplierC 411.25
243.75
0.00
120.00
supplierD 291.25
123.75
-120.00
0.00
We see that the (A,B), (B,D) and (D,C) differences are not significant at the 5% level. Notice that it
would not be reasonable to include that A is no different from C by chaining these comparisons together
because each comparison is made using a statistical test where doubt exists about the conclusion and not a
logical and definite assertion of equality.
If maximizing breaking strength is our aim, we would pick supplier C but if supplier D offered a better
price we might have some cause to consider switching to D. The decision would need to be made with
cost-quality trade-offs in mind.
How much more (or less ) efficient is the Latin square compared to other designs? First compare to the
completely randomized design:
16.5. BALANCED INCOMPLETE BLOCK DESIGN
195
> gr <- lm(y ˜ supplier,breaking)
> (summary(gr)$sig/summary(g)$sig)ˆ2
[1] 0.8391
We see that the LS is 16% less efficient than the CRD. Now compare to the blocked designs:
> gr <- lm(y ˜ supplier+operator,breaking)
> (summary(gr)$sig/summary(g)$sig)ˆ2
[1] 0.98166
> gr <- lm(y ˜ supplier+day,breaking)
> (summary(gr)$sig/summary(g)$sig)ˆ2
[1] 0.8038
We see that the Latin square turned out to be a bad choice of design because there was very little if any
difference between the operators and days but we did not know that until after the experiment! Next time
we will know better.
16.5
Balanced Incomplete Block design
For a complete block design, the block size is equal to the number of treatments. When the block size is
less than the number of treatments, an incomplete block design must be used. For example, in the penicillin
example, suppose 6 production processes were to be compared but each batch of material was only sufficient
for four runs.
In an incomplete block design, the treatments and blocks are not orthogonal. Some treatment contrasts
will not be identifiable from certain block contrasts - this is an example of confounding. This means that
those treatment contrasts effectively cannot be examined. In a balanced incomplete block design, all the
pairwise differences are identifiable and have the same standard error. Pairwise differences are more likely
to be interesting than other contrasts. Here is an example:
We have 4 treatments (t=4) A,B,C,D and the block size, k = 3 and there are b
4 blocks. Therefore,
¤
each treatment appears r
3 times in the design. One possible BIB design is
¤
1
A
B
C
2
A
B
D
3
A
C
D
4
B
C
D
Table 16.3: BIB design
Each pair of treatments appears in the same block λ
2 times — this feature enables simple pairwise
¤
comparison. For a BIB design, we require
b
t ¥
k
¡
rt
bk
n
¤
¤
λ¡
¡
t
1
r k
1
¢
¤
¢
¡
This last relation holds because the number of pairs in a block is k k
1 2 so the total number of pairs
¢
¡
¡
¡
must be bk k
1 2. On the other hand the number of treatment pairs is t t
1 2. The ratio of these two
¢
¡
¢
¡
quantities must be λ.
16.5. BALANCED INCOMPLETE BLOCK DESIGN
196
Since λ has to be integer, a BIB design is not always possible even when the first two conditions are
satisfied. For example, consider r
4 t
3 b
6 k
2 then λ
2 which is OK but if r
4 t
4 b
¤
¤
¤
¤
¤
¤
¤
¤
8 k
2 then λ
4 3 so no BIB is possible. (Something called a partially balanced incomplete block design
¤
¤
can then be used). BIB’s are also useful for competitions where not all contestants can fit in the same race.
The model we fit is the same as for the RCBD:
y
τ ρ
ε
i j
µ
i
j
i j
¤
A nutrition specialist studied the effects of six diets, a, b, c, d, e, and f on weight gain of domestic
rabbits. When rabbits reached 4 weeks of age they were assigned to a diet. It was decided to block on litters
of rabbits but from past experience about sizes of litters, it was felt that only 3 uniform rabbits could be
selected from each available litter. There were ten litters available forming blocks of size three. Each pair
of diets appear in the same block twice. Examine the data.
> data(rabbit)
> rabbit
block treat gain
1
b1
f 42.2
2
b1
b 32.6
etc.
30
b10
a 37.3
We can see the BIB structure:
> matrix(rabbit$treat,nrow=3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "f"
"c"
"c"
"a"
"e"
"b"
"d"
"a"
"d"
"f"
[2,] "b"
"a"
"f"
"e"
"c"
"f"
"a"
"e"
"b"
"d"
[3,] "c"
"b"
"d"
"c"
"d"
"e"
"b"
"f"
"e"
"a"
Now plot the data:
> plot(gain ˜ block + treat, rabbit)
See Figure 16.5. What do you conclude? Now fit the model:
> g <- lm(gain ˜ block+treat,data=rabbit)
> anova(g)
Analysis of Variance Table
Response: gain
Df Sum Sq Mean Sq F value
Pr(>F)
block
9
730
81
8.07 0.00025
treat
5
159
32
3.16 0.03817
Residuals 15
151
10
Changing the order of treatment and block:
16.5. BALANCED INCOMPLETE BLOCK DESIGN
197
50
50
45
45
40
40
gain
35
gain
35
30
30
25
25
20
20
b1
b2
b4
b6
b8
a
b
c
d
e
f
block
treat
Normal Q−Q Plot
4
4
2
2
0
0
Residuals
Sample Quantiles
−2
−2
−4
−4
25
30
35
40
45
−2
−1
0
1
2
Fitted
Theoretical Quantiles
Figure 16.9: Balanced incomplete block analysis
16.5. BALANCED INCOMPLETE BLOCK DESIGN
198
> anova(lm(gain ˜ treat+block,data=rabbit))
Analysis of Variance Table
Response: gain
Df Sum Sq Mean Sq F value
Pr(>F)
treat
5
293
59
5.84 0.00345
block
9
596
66
6.59 0.00076
Residuals 15
151
10
Does make a difference because the design is not orthogonal because of the incompleteness. Which
table is appropriate for testing the treatment effect or block effect? The first one, because we want to test for
a treatment effect after the blocking effect has been allowed for.
Now check the diagnostics
> plot(g$fitted,g$res,xlab="Fitted",ylab="Residuals")
> qqnorm(g$res)
Which treatments differ? We need to do pairwise comparisons. Tukey pairwise confidence intervals are
easily constructed:
q
2k
ˆτ
t n b t 1
¢
¢
©
l
ˆτm
ˆ
σ
2
λt
¢
First we figure out the difference between the treatment effects:
> tcoefs <- c(0,g$coef[11:15])
> outer(tcoefs,tcoefs,"-")
treatb
treatc
treatd treate
treatf
0.000000
1.7417 -0.40000 -0.066667 5.2250 -3.3000
treatb -1.741667
0.0000 -2.14167 -1.808333 3.4833 -5.0417
treatc
0.400000
2.1417
0.00000
0.333333 5.6250 -2.9000
treatd
0.066667
1.8083 -0.33333
0.000000 5.2917 -3.2333
treate -5.225000 -3.4833 -5.62500 -5.291667 0.0000 -8.5250
treatf
3.300000
5.0417
2.90000
3.233333 8.5250
0.0000
Now we want the standard error for the pairwise comparisons:
> summary(g)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
36.0139
2.5886
13.91
5.6e-10
blockb10
3.2972
2.7960
1.18
0.2567
blockb2
4.1333
2.6943
1.53
0.1458
blockb3
-1.8028
2.6943
-0.67
0.5136
blockb4
8.7944
2.7960
3.15
0.0067
blockb5
2.3056
2.7960
0.82
0.4225
blockb6
5.4083
2.6943
2.01
0.0631
blockb7
5.7778
2.7960
2.07
0.0565
blockb8
9.4278
2.7960
3.37
0.0042
blockb9
-7.4806
2.7960
-2.68
0.0173
16.5. BALANCED INCOMPLETE BLOCK DESIGN
199
treatb
-1.7417
2.2418
-0.78
0.4493
treatc
0.4000
2.2418
0.18
0.8608
treatd
0.0667
2.2418
0.03
0.9767
treate
-5.2250
2.2418
-2.33
0.0341
treatf
3.3000
2.2418
1.47
0.1617
Residual standard error: 3.17 on 15 degrees of freedom
Multiple R-Squared: 0.855,
Adjusted R-squared: 0.72
F-statistic: 6.32 on 14 and 15 degrees of freedom,
p-value: 0.000518
We see that the standard error for the pairwise comparison is 2.24. This can also be obtained as
2k
λ ˆσ:
t
> sqrt((2*3)/(2*6))*3.17
[1] 2.2415
Notice that all the treatment standard errors are equal because of the BIB. If the roles of blocks and treat-
ments were reversed, we see that the design would not be balanced and hence the unequal standard errors
for the blocks.
Now compute the Tukey critical value:
> qtukey(0.95,6,15)
[1] 4.5947
So the intervals have width
> 4.59*2.24/sqrt(2)
[1] 7.2702
We check which pairs are significantly different:
> abs(outer(tcoefs,tcoefs,"-")) > 7.27
treatb treatc treatd treate treatf
FALSE
FALSE
FALSE
FALSE
FALSE
FALSE
treatb FALSE
FALSE
FALSE
FALSE
FALSE
FALSE
treatc FALSE
FALSE
FALSE
FALSE
FALSE
FALSE
treatd FALSE
FALSE
FALSE
FALSE
FALSE
FALSE
treate FALSE
FALSE
FALSE
FALSE
FALSE
TRUE
treatf FALSE
FALSE
FALSE
FALSE
TRUE
FALSE
Only the e-f difference is significant.
How much better is this blocked design than the CRD? We compute the relative efficiency:
> gr <- lm(gain ˜ treat,rabbit)
> (summary(gr)$sig/summary(g)$sig)ˆ2
[1] 3.0945
Blocking was well worthwhile here.
16.6. FACTORIAL EXPERIMENTS
200
16.6
Factorial experiments
Suppose we have
Factors α β γ
¡¢¡¢¡
with levels lα lβ lγ
¡¢¡¢¡
A full factorial experiment has at least one run for each combination of the levels. The number of
combinations is lαlβlγ
which could easily be very large. The biggest model for a full factorial contains all
¡¢¡¢¡
possible interaction terms which may be of quite high order.
Advantages of factorial designs
1. If no interactions are significant, we get several one-way experiments for the price of one. Compare
this with doing a sequence of one-way experiments.
2. Factorial experiments are efficient — it is often better to use replication for investigating another factor
instead. For example, instead of doing a 2 factor experiment with replication, it is often better to use
that replication to investigate another factor.
Disadvantage of factorial designs
Experiment may be too large and so cost too much time or money.
Analysis
The analysis of full factorial experiments is an extension of that used for the two-way anova. Typically,
there is no replication due to cost concerns so it is necessary to assume that some higher order interactions
are zero in order to free up degrees of freedom for testing the lower order effects. Not many phenomena
require a precise combination of several factors so this is not unreasonable.
Fractional Factorials
Fractional factorials use only a fraction of the number of runs in a full factorial experiment. This is
done to save the cost of the full experiment or because the experimental material is limited and only a few
runs can be made. It is often possible to estimate the lower order effects with just a fraction. Consider an
experiment with 7 factors, each at 2 levels
mean
main
2-way
3-way
4
5
6
7
no. of pars
1
7
21
35
35
21
7
1
Table 16.4: No. of parameters
If we are going to assume that higher order interactions are negligible then we don’t really need 27
128
¤
runs to estimate the remaining parameters. We could run only a quarter of that, 32, and still be able to
estimate main and 2-way effects. (Although, in this particular example, it is not possible to estimate all the
two-way interactions uniquely. This is because, in the language of experimental design, there is no available
resolution V design, only a resolution IV design is possible.)
A Latin square where all predictors are considered as factors is another example of a fractional factorial.
In fractional factorial experiments, we try to estimate many parameters with as little data as possible.
This means there is often not many degrees of freedom left over. We require that σ2 be small, otherwise there
will be little chance of distinguishing significant effects. Fractional factorials are popular in engineering ap-
plications where the experiment and materials can be tightly controlled. In the social sciences and medicine,
16.6. FACTORIAL EXPERIMENTS
201
the experimental materials, often human or animal, are much less homgenous and less controllable so σ2
tends to be larger. In such cases, fractional factorials are of no value.
Fractional factorials are popular in product design because they allow for the screening of a large number
of factors. Factors identified in a screening experiment can then be more closely investigated.
Speedometer cables can be noisy because of shrinkage in the plastic casing material, so an experiment
was conducted to find out what caused shrinkage. The engineers started with 15 different factors: liner O.D.,
liner die, liner material, liner line speed, wire braid type, braiding tension, wire diameter, liner tension, liner
temperature, coating material, coating die type, melt temperature, screen pack, cooling method and line
speed, labelled a through o. Response is percentage shrinkage per specimen. There were two levels of each
factor. A full factorial would take 215 runs, which is highly impractical so a design with only 16 runs was
used where the particular runs have been chosen specially so as to estimate the the mean and the 15 main
effects. We assume that there is no interaction effect of any kind. The purpose of such an experiment is to
screen a large number of factors to identify which are important. Examine the data. The + indicates the high
level of a factor, the - the low level. The data comes from Box, Bisgaard, and Fung (1988)
Read in and check the data.
> data(speedo)
> speedo
h d l b j f n a i e m c k g o
y
1 - - + - + + - - + + - + - - + 0.4850
2 + - - - - + + - - + + + + - - 0.5750
3 - + - - + - + - + - + + - + - 0.0875
4 + + + - - - - - - - - + + + + 0.1750
5 - - + + - - + - + + - - + + - 0.1950
6 + - - + + - - - - + + - - + + 0.1450
7 - + - + - + - - + - + - + - + 0.2250
8 + + + + + + + - - - - - - - - 0.1750
9 - - + - + + - + - - + - + + - 0.1250
10 + - - - - + + + + - - - - + + 0.1200
11 - + - - + - + + - + - - + - + 0.4550
12 + + + - - - - + + + + - - - - 0.5350
13 - - + + - - + + - - + + - - + 0.1700
14 + - - + + - - + + - - + + - - 0.2750
15 - + - + - + - + - + - + - + - 0.3425
16 + + + + + + + + + + + + + + + 0.5825
Fit and examine a main effects only model:
> g <- lm(y ˜ .,speedo)
> summary(g)
Residuals:
ALL 16 residuals are 0: no residual degrees of freedom!
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
0.582500
h
-0.062188
d
-0.060938
16.6. FACTORIAL EXPERIMENTS
202
l
-0.027188
b
0.055937
j
0.000938
f
-0.074062
n
-0.006562
a
-0.067813
i
-0.042813
e
-0.245312
m
-0.027813
c
-0.089687
k
-0.068438
g
0.140312
o
-0.005937
Residual standard error: NaN on 0 degrees of freedom
Multiple R-Squared:
1,
Adjusted R-squared:
NaN
F-statistic:
NaN on 15 and 0 degrees of freedom,
p-value:
NaN
Why are there no degrees of freedom? Why do we have so many ”NA”’s in the display? Because there
are as many parameters as cases.
It’s important to understand the coding here, so look at the X-matrix.
> model.matrix(g)
(Intercept) h d l b j f n a i e m c k g o
1
1 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0
...etc...
We see that “+” is coded as 0 and “-” is coded as 1. This unnatural ordering is because of their order in
the ASCII alphabet.
We don’t have any degrees of freedom so we can’t make the usual F-tests. We need a different method.
Suppose there were no significant effects and the errors are normally distributed. The estimated effects
would then just be linear combinations of the errors and hence normal. We now make a normal quantile plot
of the main effects with the idea that outliers represent significant effects. The qqnorm() function is not
suitable because we want to label the points.
> coef <- g$coef[-1]
> i <- order(coef)
> plot(qnorm(1:15/16),coef[i],type="n",xlab="Normal Quantiles",
ylab="Effects")
> text(qnorm(1:15/16),coef[i],names(coef)[i])
See Figure 16.6. Notice that ”e” and possibly ”g” are extreme. Since the ”e” effect is negative, the +
level of ”e” increases the response. Since shrinkage is a bad thing, increasing the response is not good so
we’d prefer what ever “wire braid” type corresponds to the - level of e. The same reasoning for g leads us to
expect that a larger (assuming that is +) would decrease shrinkage.
A half-normal plot is better for detecting extreme points. This plots the sorted absolute values against
Φ 1 ¡ ¡
¡
n
i
2n
1 . Thus it compares the absolute values of the data against the upper half of a normal
¢
¡
¢
¢
distribution. We don’t particularly care if the coefficients are not normally distributed, it’s just the extreme
16.6. FACTORIAL EXPERIMENTS
203
g
e
0.1
b
0.20
0.0
no j
ml
g
i
f kahd
Effects
c
Effects
c
−0.1
0.10
bdha k f
i
lm
−0.2
e
jon
0.00
−1.5
−0.5
0.5
1.5
0.0
0.5
1.0
1.5
Normal Quantiles
Half−Normal Quantiles
Figure 16.10: Fractional Factorial analysis
cases we want to detect. Because the half-normal folds over the ends of a QQ plot it “doubles” our resolution
for the detection of outliers.
> coef <- abs(coef)
> i <- order(coef)
> plot(qnorm(16:30/31),coef[i],type="n",xlab="Half-Normal Quantiles",
ylab="Effects")
> text(qnorm(16:30/31),coef[i],names(coef)[i])
We might now conduct another experiment focusing on the effect of ”e” and ”g”.
Appendix A
Recommended Books
A.1
Books on R
There are currently no books written specifically for R , although several guides can be downloaded from
the R web site.
R is very similar to S-plus so most material on S-plus applies immediately to R . I highly recommend
Venables and Ripley (1999). Alternative introductory books are Spector (1994) and Krause and Olson
(2000). You may also find Becker, Chambers, and Wilks (1998) and Chambers and Hastie (1991), useful
references to the S language. Ripley and Venables (2000) is a more advanced text on programming in S or
R .
A.2
Books on Regression and Anova
There are many books on regression analysis. Weisberg (1985) is a very readable book while Sen and
Srivastava (1990) contains more theoretical content. Draper and Smith (1998) is another well-known book.
One popular textbook is Kutner, Nachtschiem, Wasserman, and Neter (1996). This book has everything
spelled out in great detail and will certainly strengthen your biceps (1400 pages) if not your knowledge of
regression.
204
Appendix B
R functions and data
R may be obtained from the R project web site at www.r-project.org.
This book uses some functions and data that are not part of base R . You may wish to download these
functions from the R web site. The additional packages used are
MASS leaps ggobi ellipse nlme
MASS and nlme are part of the “recommended” R installation so depending on what installation option you
choose, you may already have these without additional effort. Use the command
> library()
to see what packages you have. The MASS functions are part of the VR package that is associated with the
book Venables and Ripley (1999). The ggobi data visualization application may also need to be installed.
This may be obtained from www.ggobi.org This is not essential so don’t worry if you can’t install it.
In addition, you will need the splines, mva and lqs packages but these come with basic R installation so no
extra work is necessary.
I have packaged the data and functions that I have used in this book as an R package that you may obtain
from my web site — www.stat.lsa.umich.edu/˜faraway. The functions available are
halfnorm
Half normal plot
Cpplot
Cp plot
qqnorml
Case-labeled Q-Q plot
maxadjr
Models with maximum adjusted Rˆ2
vif
Variance Inflation factors
prplot
Partial residual plot
In addition the following datasets are used:
breaking
Breaking strengths of material by day, supplier, operator
cathedral
Cathedral nave heights and lengths in England
chicago
Chicago insurance redlining
chiczip
Chicago zip codes north/south
chmiss
Chicago data with some missing values
coagulation
Blood coagulation times by diet
corrosion
Corrosion loss in Cu-Ni alloys
eco
Ecological regression example
gala
Species diversity on the Galapagos Islands
205
APPENDIX B. R FUNCTIONS AND DATA
206
odor
Odor of chemical by production settings
pima
Diabetes survey on Pima Indians
penicillin
Penicillin yields by block and treatment
rabbit
Rabbit weight gain by diet and litter
rats
Rat survival times by treatment and poison
savings
Savings rates in 50 countries
speedo
Speedometer cable shrinkage
star
Star light intensities and temperatures
strongx
Strong interaction experiment data
twins
Twin IQs from Burt
Where add-on packages are needed in the text, you will find the appropriate library() command.
However, I have assumed that the faraway library is always loaded. You can add a line reading library(faraway)
to your Rprofile file if you expect to use this package in every session. Otherwise you will need to remember
to type it each time.
I set the following options to achieve the output seen in this book
> options(digits=5,show.signif.stars=FALSE)
The digits=5 reduces the number of digits shown when printing numbers from the default of seven. Note
that this does not reduce the precision with which these numbers are internally stored. One might take this
further — anything more than 2 or 3 significant digits in a displayed table is usually unnecessary and more
important, distracting.
Appendix C
Quick introduction to R
C.1
Reading the data in
The first step is to read the data in. You can use the read.table() or scan() function to read data in
from outside R . You can also use the data() function to access data already available within R .
> data(stackloss)
> stackloss
Air.Flow Water.Temp Acid.Conc. stack.loss
1
80
27
89
42
2
80
27
88
37
... stuff deleted ...
21
70
20
91
15
Type
> help(stackloss)
We can check the dimension of the data:
> dim(stackloss)
[1] 21
4
C.2
Numerical Summaries
One easy way to get the basic numerical summaries is:
> summary(stackloss)
Air.Flow
Water.Temp
Acid.Conc.
stack.loss
Min.
:50.0
Min.
:17.0
Min.
:72.0
Min.
: 7.0
1st Qu.:56.0
1st Qu.:18.0
1st Qu.:82.0
1st Qu.:11.0
Median :58.0
Median :20.0
Median :87.0
Median :15.0
Mean
:60.4
Mean
:21.1
Mean
:86.3
Mean
:17.5
3rd Qu.:62.0
3rd Qu.:24.0
3rd Qu.:89.0
3rd Qu.:19.0
Max.
:80.0
Max.
:27.0
Max.
:93.0
Max.
:42.0
207
C.2. NUMERICAL SUMMARIES
208
We can compute these numbers seperately also:
> stackloss$Air.Flow
[1] 80 80 75 62 62 62 62 62 58 58 58 58 58 58 50 50 50 50 50 56 70
> mean(stackloss$Ai)
[1] 60.429
> median(stackloss$Ai)
[1] 58
> range(stackloss$Ai)
[1] 50 80
> quantile(stackloss$Ai)
0%
25%
50%
75% 100%
50
56
58
62
80
We can get the variance and sd:
> var(stackloss$Ai)
[1] 84.057
> sqrt(var(stackloss$Ai))
[1] 9.1683
We can write a function to compute sd’s:
> sd <- function(x) sqrt(var(x))
> sd(stackloss$Ai)
[1] 9.1683
We might also want the correlations:
> cor(stackloss)
Air.Flow Water.Temp Acid.Conc. stack.loss
Air.Flow
1.00000
0.78185
0.50014
0.91966
Water.Temp
0.78185
1.00000
0.39094
0.87550
Acid.Conc.
0.50014
0.39094
1.00000
0.39983
stack.loss
0.91966
0.87550
0.39983
1.00000
Another numerical summary with a graphical element is the stem plot:
> stem(stackloss$Ai)
The decimal point is 1 digit(s) to the right of the |
5 | 000006888888
6 | 22222
7 | 05
8 | 00
C.3. GRAPHICAL SUMMARIES
209
C.3
Graphical Summaries
We can make histograms and boxplot and specify the labels if we like:
> hist(stackloss$Ai)
> hist(stackloss$Ai,main="Histogram of Air Flow",
xlab="Flow of cooling air")
> boxplot(stackloss$Ai)
Scatterplots are also easily constructed:
> plot(stackloss$Ai,stackloss$W)
> plot(Water.Temp ˜ Air.Flow,stackloss,xlab="Air Flow",
ylab="Water Temperature")
We can make a scatterplot matrix:
> plot(stackloss)
We can put several plots in one display
> par(mfrow=c(2,2))
> boxplot(stackloss$Ai)
> boxplot(stackloss$Wa)
> boxplot(stackloss$Ac)
> boxplot(stackloss$s)
> par(mfrow=c(1,1))
C.4
Selecting subsets of the data
Second row:
> stackloss[2,]
Air.Flow Water.Temp Acid.Conc. stack.loss
2
80
27
88
37
Third column:
> stackloss[,3]
[1] 89 88 90 87 87 87 93 93 87 80 89 88 82 93 89 86 72 79 80 82 91
The 2,3 element:
> stackloss[2,3]
[1] 88
c() is a function for making vectors, e.g.
> c(1,2,4)
[1] 1 2 4
C.5. LEARNING MORE ABOUT R
210
Select the first, second and fourth rows:
> stackloss[c(1,2,4),]
Air.Flow Water.Temp Acid.Conc. stack.loss
1
80
27
89
42
2
80
27
88
37
4
62
24
87
28
The : operator is good for making sequences e.g.
> 3:11
[1]
3
4
5
6
7
8
9 10 11
We can select the third through sixth rows:
> stackloss[3:6,]
Air.Flow Water.Temp Acid.Conc. stack.loss
3
75
25
90
37
4
62
24
87
28
5
62
22
87
18
6
62
23
87
18
We can use ”-” to indicate ”everthing but”, e.g all the data except the first two columns is:
> stackloss[,-c(1,2)]
Acid.Conc. stack.loss
1
89
42
2
88
37
... stuff deleted ...
21
91
15
We may also want select the subsets on the basis of some criterion e.g. which cases have an air flow greater
than 72.
> stackloss[stackloss$Ai > 72,]
Air.Flow Water.Temp Acid.Conc. stack.loss
1
80
27
89
42
2
80
27
88
37
3
75
25
90
37
C.5
Learning more about R
While running R you can get help about a particular commands - eg - if you want help about the stem()
command just type help(stem).
If you don’t know what the name of the command is that you want to use then type:
help.start()
and then browse. You may be able to learn the language simply by example in the text and refering to
the help pages.
You can also buy the books mentioned in the recommendations or download various guides on the web
— anything written for S-plus will also be useful.
Bibliography
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and research worker. New York: Springer-Verlag.
Becker, R., J. Chambers, and A. Wilks (1998). The new S language: A Programing Environment for Data
Analysis and Graphics (revised ed.). CRC.
Belsley, D. A., E. Kuh, and R. E. Welsch (1980). Regression Diagnostics: Identifying Influential Data
and Sources of Collinearity. New York: Wiley.
Box, G. P., S. Bisgaard, and C. Fung (1988). An explanation and critque of taguchi’s contributions to
quality engineering. Quality and reliability engineering international 4, 123–131.
Box, G. P., W. G. Hunter, and J. S. Hunter (1978). Statistics for Experimenters. New York: Wiley.
Carroll, R. and D. Ruppert (1988). Transformation and Weighting in Regression. London: Chapman Hall.
Chambers, J. and T. Hastie (1991). Statistical Models in S. Chapman and Hall.
Chatfield, C. (1995). Model uncertainty, data mining and statistical inference. JRSS-A 158, 419–466.
Draper, D. (1995). Assessment and propagation of model uncertainty. JRSS-B 57, 45–97.
Draper, N. and H. Smith (1998). Applied Regression Analysis (3rd ed.). New York: Wiley.
Faraway, J. (1992). On the cost of data analysis. Journal of Computational and Graphical Statistics 1,
215–231.
Faraway, J. (1994). Order of actions in regression analysis. In P. Cheeseman and W. Oldford (Eds.),
Selecting Models from Data: Artificial Intelligence and Statistics IV, pp. 403–411. Springer Verlag.
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