Statistical Methods for Psychology

Now let us approach the statistical treatment of these data by means of least-squares

multiple linear regression. We will take as our criterion (Y) the raw data in Table 16.1. For

the predictors we will use a design matrix of the form

A 1 A 2 A 3

Here the elements of any one row of the design matrix are taken to apply to all the sub-

jects in the treatment. The multiple-regression solution using the design matrix Xas the

matrix of predictors is presented in Exhibit 16.1. Here the dependent variable (Y) is the

first column of the data matrix. The next three columns together form the matrix X. SPSS

was used to generate this solution, but any standard program would be suitable. (I have

made some very minor changes in the output to simplify the discussion.)

Notice the patterns of intercorrelations among the Xvariables in Exhibit 16.1. This

type of pattern with constant off-diagonal correlations will occur whenever there are equal

numbers of subjects in the various groups. (The fact that we don’t have constant off-diagonal

correlations with unequal-nfactorial designs is what makes our life more difficult in those

situations.)

Notice that the regression coefficients are written in a column. This column can be

called a vector, and is the vector b, or, in analysis of variance terms, the vector t. Notice

that b 15 2.50, which is the same as the estimated treatment effect of Treatment 1 shown in

Table 16.1. In other words, b 1 5t 1. This also happens for b 2 and b 3. This fact necessarily

X=

Treament 1

Treament 2

Treament 3

Treament 4

D

100

010

001

21 21 21

T

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

Table 16.1 Illustrative calculations for simple one-way design with equal ns

(a) Data

Treatment 1 Treatment 2 Treatment 3 Treatment 4

8536

9744

7319

8 5 2.667 6.333

5 5.500

(b) Summary Table

Source df SS MS F

Treatments 3 45.667 15.222 4.46 .626

Error 8 27.333 3.417

Total 11 73.000

(c) Estimated Treatment Effects

tN 3 =X 32 X..=2.67 2 5.5= 2 2.83

tN 2 =X 22 X..=5.0 2 5.5= 2 0.5

tN 1 =X 12 X..=8.0 2 5.5=2.5

h^2

X..