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
Bill.Venables at csiro.au
2007-Apr-15 08:03 UTC
[R] Hotelling T-Squared vs Two-Factor Anova
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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 > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >
Sean
Both Bill V and Peter D are right regarding traditional
repeated-measures ANOVA that assumes equality of the variance-covariance
matrices across groups and compound symmetry of the covariance matrix.
However, there is a multivariate approach to repeated measures that does
not require the assumption of compound symmetry. This approach is given
in the reference below. The difference between multivariate repeated
measures and general MANOVA is that the former imposes a model matrix
(contrasts) on the measures to reflect the repeated measures design.
General MANOVA does not have this extra matrix. The Hotelling T-squared
approach with a measures model matrix would be equivalent to
multivariate repeated measures.
In your case, you need to specify the measure model matrix for the 4
time points (polynomial, differences, Helmert), which usually cannot be
done in a standard T-square program. However, you could compute the
model matrix results (contrasts) beforehand and submit them to the
T-square program. Your taking the difference scores on the 1st factor
is doing just that.
Joe
@BOOK{Bock1975,
author = {Bock, R. D.},
title = {Multivariate statistical methods in behavioral research},
year = {1975},
publisher = {McGraw-Hill},
address = {New York},
keywords = {regression; analysis of covariance; analysis of variance;
log-linear
analysis; matrices; distribution theory;},
}
-----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, April 14, 2007 7:38 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
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Lucke, Joseph F wrote:> Sean > > Both Bill V and Peter D are right regarding traditional > repeated-measures ANOVA that assumes equality of the variance-covariance > matrices across groups and compound symmetry of the covariance matrix. > However, there is a multivariate approach to repeated measures that does > not require the assumption of compound symmetry. This approach is given > in the reference below. The difference between multivariate repeated > measures and general MANOVA is that the former imposes a model matrix > (contrasts) on the measures to reflect the repeated measures design. > General MANOVA does not have this extra matrix. The Hotelling T-squared > approach with a measures model matrix would be equivalent to > multivariate repeated measures. >I mentioned in my reply that one might look at successive differences to remove the mean level, I think this is effectively the same general idea. It is actually implemented in anova.mlm(), which allows you to specify a transformation matrix, possibly using formula notation to compute the relevant model matrix/matrices. There is, btw, an alternative approach which involves _conditioning_ on contrasts that at known to have mean zero (second order differences for a linear within-subjects trend, e.g.). This is not implemented systematically anywhere though, as far as I know.> In your case, you need to specify the measure model matrix for the 4 > time points (polynomial, differences, Helmert), which usually cannot be > done in a standard T-square program. However, you could compute the > model matrix results (contrasts) beforehand and submit them to the > T-square program. Your taking the difference scores on the 1st factor > is doing just that. > > Joe > > @BOOK{Bock1975, > author = {Bock, R. D.}, > title = {Multivariate statistical methods in behavioral research}, > year = {1975}, > publisher = {McGraw-Hill}, > address = {New York}, > keywords = {regression; analysis of covariance; analysis of variance; > log-linear > analysis; matrices; distribution theory;}, > } > > -----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, April 14, 2007 7:38 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 > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >