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

Bill.Venables at csiro.au Bill.Venables at csiro.au
Sun Apr 15 10:03:14 CEST 2007

```I take it all subjects are measured at the same time points, or
Hotelling's T^2 becomes rather messy.

The essential difference lies in the way the variance matrix is
modelled.  The usual repeated measures model would model the variance
matrix as equal variances and equal covariances, i.e. with two
parameters, (though you can vary this using, e.g. lme).  Hotelling's T^2
would model the variance matrix as a general symmetric matrix, i.e. for
the 4x4 case using 4+3+2+1 = 10 parameters.  If it is appropriate, the
repeated measures model is much more parsimonious.

Bill Venables.

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Sean Scanlan
Sent: Saturday, 14 April 2007 5:38 PM
To: r-help at stat.math.ethz.ch
Subject: [R] Hotelling T-Squared vs Two-Factor Anova

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?

Thanks,
Sean

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help