[R] Hotelling T-Squared vs Two-Factor Anova

Peter Dalgaard p.dalgaard at biostat.ku.dk
Sun Apr 15 10:20:53 CEST 2007

Sean Scanlan wrote:
> Hi,
> I am a graduate student at Stanford University and I have a general
> statistics question.  What exactly is the difference between doing a
> two-factor repeated measures ANOVA and a Hotelling T-squared test for
> a paired comparison of mean vectors?
> Given:
> Anova: repeated measures on both factors, 1st factor = two different
> treatments, 2nd factor = 4 time points, where you are measuring the
> blood pressure at each of the time points.
> Hotelling T^2: You look at the difference in the 4x1 vector of blood
> pressure measurements for the two different treatments, where the four
> rows in the vector are the four time points.
> I am mainly interested in the main effects of the two treatments.  Can
> someone please explain if there would be a difference in the two
> methods or any advantage in using one over the other?
In a few words (the full story takes a small book), the difference is in 
the assumptions, and in the hypothesis being tested. In the most common 
incarnation, T^2 tests for *any* difference in the means, whereas ANOVA 
removes the average before comparing the shapes of the time course. If 
you look at intra-individual differences (e.g. x2-x1, x3-x2, x4-x3, but 
other choices are equivalent), then T^2 on these three variables will 
test the same hypothesis about the means. The remaining difference is 
then that ANOVA assumes a particular pattern of the covariance matrix, 
whereas T^2 allows a general covariance structure. In particular, T^2 
applies even when your response variables are not of the same quantity, 
say if you had simultaneous measurements of heart rate and blood pressure.

The standard assumption for ANOVA is "compound symmetry" (one value on 
the diagonal, another off-diagonal), which can be weakened to 
"sphericity" (covariance of differences behave as they would under 
comp.symm.). On closer inspection, sphericity actually means that the 
covariance matrix for differences is proportional to a known matrix.

Since T^2 has more parameters to estimate it will have less power if 
both methods are applicable. Even if the assumptions are not quite 
right, procedure based on the ANOVA F may still be stronger, but this 
requires correction terms to be applied (these are known as 
Greenhouse-Geisser and Huynh-Feldt epsilons).
> Thanks,
> Sean
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