Statistical Methods for Psychology

Salary than we do with a model that does not include either PTratio or Salary but does in-
clude the other two predictors. In the case of nested models, it is relatively easy to test
whether one is significantly better than another. We can just compare their R^2 values or the
sums of squares for regression.
For example, suppose that we start with a model that contains Expend, LogPctSAT,
PTratio, and Salary. (I chose this model because it has two more predictors than the sim-
pler one that we will look at.) The multiple R^2 is .888 and the analysis of variance summary
table is

15.10 Regression Diagnostics 545

Sum of

Squares

243689.5

30618.141

274307.7

Model

1 Regression

Residual

Total

df

4

45

49

Mean

Square

60922.385

680.403

F

89.539

Sig.

.000a

ANOVAb

aPredictors: (Constant), Salary, PTratio, LogPctSAT, Expend

bDependent Variable: SAT

Sum of

Squares

243069.3

31238.381

274307.7

Model

1 Regression

Residual

Total

df

2

47

49

Mean

Square

121534.649

664.646

F

182.856

Sig.

.000a

ANOVAb

aPredictors: (Constant), LogPctSAT, Expend

bDependent Variable: SAT

Next we drop PTratio and Salary and just use Expend and LogPctSAT. Now the R^2 is
.886 and the analysis of variance summary table is

Notice that the first model explained slightly more variation than the second. If we
compute the difference in SSregressionwe have 243,689.5 – 243,069.3 5 620.2 5 SSdifference.
This difference in the sum of squares can be converted to a mean square by dividing by the
degrees of freedom, but what are the degrees of freedom? They are simply the difference in
the number of predictors, which is 2. Therefore MSdifference 5 SSdifferencedf 5 620.2 2 5
310.1. Moreover, this mean square can be tested by dividing by the residual mean square
from the fuller model. So

This is an Fon 2 and 45 degrees of freedom and is clearly not significant. We do not do a
better job of predicting SAT scores with the additional two predictors.
If we had just compared the model with Expend, LogPctSAT, and PTratio against the
model without PTratio, our resulting Fwould be .175, and its square root would be .418,
which is exactly the tfor the test of PTratio in the fuller model. In other words if we only
want to drop one predictor we know whether that drop will be significant simply by look-
ing at the t-test on the predictor in the fuller model.
But what do we do if we do not have nested models? That question arises in this exam-
ple when we ask if I made a wise choice to use LogPctSAT rather than PctSAT as my pre-
dictor. Because the models are not nested we cannot simply test the difference in SSregression

F=

SSreg(full) 2 SSreg(reduced)

dfreg(full) 2 dfreg(reduced)

MSresidual(full)

=

(243,689.5 2 243,069.3)

2

664.646

=

310.1

664.646

=0.467

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