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re: st: Re: Comparisons in repeated measures anova


From   David Airey <[email protected]>
To   [email protected]
Subject   re: st: Re: Comparisons in repeated measures anova
Date   Tue, 24 Jun 2008 15:02:57 -0500

.
Seems like you have one between sample factor (2 levels) and one within sample factor (11 levels) and because you have an interaction, you now want to test the simple effects, or the levels of system at each time, or the levels of time within each system. For testing different levels of the within sample factor (time) across the between sample factor (system), you can just use a oneway ANOVA (or even ttest since you have just two factor levels for system). You do have a choice of errors for this test, one for just the time you are interested in, or a pooled term, but I think most packages like SPSS and SAS that automate this kind of ANOVA post estimation analysis (ie give to you without asking) will use the MS at time k.
So, if you did
anova y a/s|a b a*b, repeated(b)
then if b = 1, 2, or 3, and a = 1 or 2
anova y a if b == 1
is fine, and applying a FWER as needed.
Also see the Stata FAQ of ANOVA.

I am stuck with evaluation of a comparison. I have carried out a repeated measures anova of two systems (labeled 3 & 8) over 11 time periods, and I want to see if the two systems differ at each time period.

. anova deltae sys / sample|sys time sys*time, repeated(time)

Repeated variable: time
Huynh-Feldt epsilon = 0.2594
Greenhouse-Geisser epsilon = 0.1914
Box's conservative epsilon = 0.1000

------------ Prob > F ------------
Source | df F Regular H-F G- G Box
----------- +----------------------------------------------------
time | 10 20.30 0.0000 0.0000 0.0000 0.0011
sys*time | 10 5.27 0.0000 0.0076 0.0160 0.0446
Residual | 100
----------- +----------------------------------------------------
The design matrix has 48 columns and I tried to compare the systems at time 1 as follows:
. test _coef[time[1]*sys[3]] = _coef[time[1]*sys[8]]

( 1) sys[3]*time[1] - sys[8]*time[1] = 0

F( 1, 100) = 10.59
Prob > F = 0.0016

I interpret this as the 2 systems being significantly different. The problem I have is that the means of the two systems are 0.618 and 0.642. When the data is plotted there appears to be no difference at this time and a t-test (I appreciate that it is not valid in this design) has a probability 0.668 which seems to agree with the plot.

I would be grateful if somebody could tell what I am doing wrong in this comparison and how to carry it out.
Jayne



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