Hi Stuart,
> Say these are the conditions entered:
>
> A B C A B D E
>
> And we are interested in the first A only versus average of C, D, E,
> then the contrast is:
>
> [3 0 -1 0 0 -1 -1]
>
> Yes?
Yup; maybe easier to see it as [1 0 -1/3 0 0 -1/3 -1/3], so the
negatives are clearly an average of C, D, & E.
> Assuming that is correct, let us say that, in the example above,
> C=D=E and we now enter the conditions as:
>
> A B C A B C C
>
> is the subsequent contrast [1 0 -1] now equivalent to the contrast
> above?
The contrast is expressing the equivalent effect, but the model is
different. In this later model, you are assuming that C=D=E; you save
yourself estimating another two parameters (D & E) and hence have 2
more df. But to the extent that C=D=E is not true, you will have an
inflated error variance.
Hence, if you have the df to spare, you're generally always better off
fitting the larger model.
-Tom
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