Statistical Methods for Psychology

suggesting a choice. If the covariate naturally varies in the population (as it does in this

case, where we expect different animals to vary in their pretest score, then it makes the

most sense to divide the SStreatfrom the analysis of covariance by the SStotal(unadjusted)

from that analysis. This will produce a value of h^2 which is the percentage of “normal vari-

ation” accounted for by the independent variable.^10 Then

An alternative approach, which will produce the same answer, is to take h^2 as the dif-

ference between the R^2 from a model predicting the dependent variable from only the co-

variate (the pretest) and one predicting the dependent variable from both the covariate and

the treatment. The increase in explained variation from the first of these models to the sec-

ond represents what the treatment contributes after controlling for the covariate. For our

example R^2 using just the covariate is .714. (You can obtain this by an analysis of variance

using the covariate as the independent variable, or by a regression of the independent vari-

able on the covariate.) When you add in the treatment effect the R^2 is .804. These values

are shown in the following table.

Step Predictors R^2 Change in R^2 Ffor change

1 Pretest .714

2 Pretest, Treatment .804 .090 4.689

h^2 is the difference between these two values of R^2 , which is the contribution to explained

variation of the treatment after controlling for the covariate. This is the same value we ob-

tained by the first approach.

d-Family Measure

Measures from the d-family often are more interpretable, and they are most often used for

specific contrasts between two means. The example we have been using is not a very good

one for a contrast of two means because the independent variable is a continuum. But I will

use the contrast between the control group and the .5 mg group as an example, because

these are the two conditions that Conti and Musty’s theory would have expected to show

the greatest mean difference. Because we are working with an analysis of covariance, the

appropriate means to compare are the adjusted means ( ) from that analysis. In this case

they are 3.1719 for the .5 mg condition and 1.7153 for the control condition. (You may

recall that we performed a test on the difference between these adjusted means in the pre-

vious section, and it was significant.)

You should recall that we have generally used

as our effect size estimate. When we are comparing two group means, is simply the dif-

ference between the two means because the coefficients are [ 2 1 0 1 0 0]. For our example

is 3.1719 – 1.7153 5 1.4566. But the question of most importance is what we will use

for the estimate of the standard deviation. One of several choices would be the square root

of MSerrorfrom an analysis of variance, because this would be an estimate of the average

variability within each group, and would thus standardize the mean difference in the metric

cN

cN

d=

cN

sN

Yi

h^2 =

SStreat(adj)

SStotal

=

9.2239

102.7689

=.09

610 Chapter 16 Analyses of Variance and Covariance as General Linear Models

(^10) If you were interested in the for the quadratic relationship between dose and the activity level, controlling
for the pretest activity level, you could just divide the SSquadraticby SStotal.
h^2