1 More about contrasts and interactions
Refer to the computer code for today’s lectures to see how to perform these calculations in SAS.
1.1 Doing 1 way ANOVA as a set of contrasts
The ANOVA null hypothesis for 4 groups H0:µ1=µ2=µ3=µ4is equivalent to the following set of
contrasts:
L1:µ1−µ2= 0
L2:µ1−µ3= 0
L3:µ1−µ4= 0
For these three contrasts to be simultaneously equal to zero is equivalent to having all four means equal
to each other.
1.2 Contrasts with interactions
For an experiment with factorial treatment structure, with 2 factors each having three levels, we can look at
the treatment-combination means in the following arrangement:
B
1 2 3
1µ11 µ12 µ13
A2µ21 µ22 µ23
3µ31 µ32 µ33
1.3 Simple main effects tests
We can express questions of interest regarding an AxB interaction in the form of contrasts. The first example
is simple-main-effect tests. If we wish to test the equality of factor A when factor B is held at the first level
(in our police training example, we are testing the three patrols when training time is fixed at 5 hours) we
wish to test H0:µ11 =µ21 =µ31, which is equivalent to the following set of contrasts:
L1:µ11 −µ21 = 0
L2:µ11 −µ31 = 0
1.4 Treatment-contrast interactions
The problem with simple-main-effect tests is that they involve a combination of main effects and interaction
effects. If you repeat the above question for when training time is held at 10 and then 15 hours, the sum of
these three simple-main-effect test sum of squares equals SSA + SSAB from the factorial model. Thus these
tests are mixing main effects and interaction effects together. An alternative approach is to take a contrast
in one factor, and see if it interacts with the other factor. In the police example, it may be of interest to see
if the upper-class patrol - middle-class patrol contrast (call it ψ1(A)=µ1·
−µ2·) interacts with the training
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