We illustrate the variable selection methods on some data on the 50 states - the variables are
population 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)
> help(state)
> statedata<data.frame(state.x77,row.names=state.abb,check.names=T)
> g <- lm(Life.Exp~., data=statedata)
> summary(g)
##### outputs ######
Coefficients:
Estimate
(Intercept)
7.09e+01
Population
5.18e-05
Income
-2.18e-05
Illiteracy
3.38e-02
Murder
-3.01e-01
HS.Grad
4.89e-02
Frost
-5.74e-03
Area
-7.38e-08
Std. Error
1.75e+00
2.92e-05
2.44e-04
3.66e-01
4.66e-02
2.33e-02
3.14e-03
1.67e-06
t value
40.59
1.77
-0.09
0.09
-6.46
2.10
-1.82
-0.04
Pr(>|t|)
< 2e-16
0.083
0.929
0.927
8.7e-08
0.042
0.075
0.965
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 expectancythat’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
(Intercept)
7.10e+01
1.39e+00
Population
5.19e-05
2.88e-05
Income
-2.44e-05
2.34e-04
Illiteracy
2.85e-02
3.42e-01
Murder
-3.02e-01
4.33e-02
HS.Grad
4.85e-02
2.07e-02
Frost
-5.78e-03
2.97e-03
> g <- update(g, .~. - Illiteracy)
> summary(g)
t value
51.17
1.80
-0.10
0.08
-6.96
2.35
-1.94
Pr(>|t|)
< 2e-16
0.079
0.917
0.934
1.5e-08
0.024
0.058
Coefficients:
(Intercept)
Population
Income
Murder
HS.Grad
Frost
Estimate
7.11e+01
5.11e-05
-2.48e-05
-3.00e-01
4.78e-02
-5.91e-03
Std. Error
1.03e+00
2.71e-05
2.32e-04
3.70e-02
1.86e-02
2.47e-03
t value
69.07
1.89
-0.11
-8.10
2.57
-2.39
Pr(>|t|)
< 2e-16
0.066
0.915
2.9e-10
0.014
0.021
> g <- update(g, .~. - Income)
> summary(g)
Coefficients:
Estimate
(Intercept) 7.10e+01
Population
5.01e-05
Murder
-3.00e-01
HS.Grad
4.66e-02
Frost
-5.94e-03
Std.Error
9.53e-01
2.51e-05
3.66e-02
1.48e-02
2.42e-03
t value
74.54
2.00
-8.20
3.14
-2.46
Pr(>|t|)
< 2e-16
0.0520
1.8e-10
0.0030
0.0180
> g <- update(g, .~. - Population)
> summary(g)
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.
Best subset: Cp
Note: Here you might need two packages: leaps and faraway. How to install packages?
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 adjR2 methods for the selection of variables in the State dataset. The default for
the leaps() function is the Mallow’s Cp criterion:
> library(leaps)
> library(faraway)
> x <- model.matrix(g)[,-1]
> y <- statedata$Life
> g <- leaps(x,y)
> Cpplot(g)
If we get the following Cpplot, what should we do?
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 which fits a little better. Some even larger models fit in the
sense that they are on or below theCp=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.
_
> 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 adj R2. 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.
ANOVA in R
Consider a model with only one variable, income:
>g1 <- lm(Life.Exp~Income, data=statedata)
>summary(g1) ## Take a look
>anova(g1) ## Analysis of Variance
Call:
lm(formula = Life.Exp ~ Income, data = statedata)
Residuals:
Min
1Q
Median
-2.96547 -0.76381 -0.03428
3Q
0.92876
Max
2.32951
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.758e+01 1.328e+00 50.906
<2e-16 ***
Income
7.433e-04 2.965e-04
2.507
0.0156 *
--Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.275 on 48 degrees of freedom
Multiple R-Squared: 0.1158,
Adjusted R-squared: 0.09735
F-statistic: 6.285 on 1 and 48 DF, p-value: 0.01562
Analysis of Variance Table
Response: Life.Exp
Df Sum Sq Mean Sq F value Pr(>F)
Income
1 10.223 10.223 6.2847 0.01562 *
Residuals 48 78.076
1.627
-3
-1.5
-1.0
-2
-0.5
0.0
residuals
0
-1
residuals
0.5
1
1.0
2
1.5
Residual plots in R
g <- lm(Life.Exp~., data=statedata)
g1<-lm(Life.Exp~Income, data=statedata)
par(mfrow=c(1,2)) ## draw two figures
plot(g1$fit,g1$res,xlab="Fitted",ylab="residuals")
plot(g$fit,g$res,xlab="Fitted",ylab="residuals")
### What is the difference?
70.0
70.5
71.0
Fitted
71.5
72.0
69
70
71
72
Fitted
Plot against ŷ . This is the most important diagnostic plot that you can make. If all is well, you
should see constant variance in the vertical ( ˆ) direction and the scatter should be symmetric
vertically about 0. Things to look for are heteroscedascity (non-constant variance) and
nonlinearity (which indicates some change in the model is necessary).
Look at the following figures:
Residuals vs Fitted plots - the first suggests no change to the current model while the second
shows non-constant variance and the third indicates some nonlinearity which should prompt some
change in the structural form of the model.
You should also plot ˆagainst xi (for predictors that are both in and out of the model). Look for
the same things except in the case of plots against predictors not in the model, look for any
relationship which might indicate that this predictor should be included.
2
1
0
-3
-2
-1
residuals
0
-1
-2
-3
residuals
1
2
For example:
plot(statedata$Popula,g1$res,xlab="Population",ylab="residuals")
plot(statedata$Murder,g1$res,xlab="Murder",ylab="residuals")
0
5000
10000
Population
15000
20000
2
4
6
8
Murder
10
12
14