Statistical Methods for Psychology

Sum of Squares for Contrasts

One of the advantages of linear contrasts is that they can be converted to sums of squares

very easily and can represent the sum of squared differences between the means of sets of

treatments. If we write

it can be shown that

is a component of the overall on 1 df, where nrepresents the number of scores per

treatment.^4

Suppose we have three treatments such that

n 510

For the overall analysis of variance,

Suppose we wanted to compare the average of treatments 1 and 2 with treatment 3. Let

5 1 2, a 25 1 2, 52 1. Then

This sum of squares is a component of the overall on 1 df. We have 1 dfbecause we

are really comparing two quantities (the mean of the first two treatments with the mean of

the third treatment).

Now suppose we obtain an additional linear contrast comparing treatment 1 with treat-

ment 2. Let 5 1, 52 1, and. Then

This is also a component of on 1 df. In addition, because of the particular

contrasts that we chose to run,

the two contrasts account for all of the and all of the dfattributable to treatments. We

say that we have completelypartitioned SStreat.

SStreat

11.667=10.417 1 1.25

SStreat=SScontrast 11 SScontrast 2

SScontrast SStreat

SScontrast=

nc^2

aa

2

j

=

10(-0.5)^2

2

=

2.5

2

=1.25

c= aajXj=(1)(1.5) 1 (-1)(2.0) 1 (0)(3.0)=-0.5

a 1 a 2 a 3 = 0

SStreat

SScontrast=

nc^2

aa

2

j

=

10(-1.25)^2

1.5

=

15.625

1.5

=10.417

c= aajXj=A^12 B(1.5) 1 A^12 BA2.0B 1 (-1)(3.0)=-2.5

a 1 > > a 3

= 103 0.4449 1 0.0278 1 0.6939 4 =11.667

SStreat=na(Xj 2 X..)^2 = 103 (1.5 2 2.167)^21 (2 2 2.167)^21 (3 2 2.167)^24

X 1 =1.5 X 2 =2.0 X 3 =3.0

SStreat

SScontrast=

nc^2

aa

2

j

=

nAaajXjB^2

aa

2

j

c=a 1 X 11 a 2 X 2 1 Á 1 akXk=aajXj

372 Chapter 12 Multiple Comparisons Among Treatment Means

(^4) For unequal sample sizes, SScontrast= c
2
g(aj^2 >nj)
partitioned