On Wed, 9 Mar 2005, Darren Weber wrote:
How do we specify a contrast interaction matrix for an ANOVA model?
We have a two-factor, repeated measures design, with
Where does `repeated measures' come into this? You appear to have repeated a 2x2 experiment in each of 8 blocks (subjects). Such a design is usually analysed with fixed effects. (Perhaps you averaged over repeats in the first few lines of your code?)
Cue Direction (2) x Brain Hemisphere(2)
Each of these has 2 levels, 'left' and 'right', so it's a simple 2x2 design matrix. We have 8 subjects in each cell (a balanced design) and we want to specify the interaction contrast so that:
CueLeft>CueRght for the Right Hemisphere CueRght>CueLeft for the Left Hemisphere.
Here is a copy of the relevant commands for R:
######################################## lh_cueL <- rowMeans( LHroi.cueL[,t1:t2] ) lh_cueR <- rowMeans( LHroi.cueR[,t1:t2] ) rh_cueL <- rowMeans( RHroi.cueL[,t1:t2] ) rh_cueR <- rowMeans( RHroi.cueR[,t1:t2] ) roiValues <- c( lh_cueL, lh_cueR, rh_cueL, rh_cueR )
cuelabels <- c("CueLeft", "CueRight") hemlabels <- c("LH", "RH")
roiDataframe <- data.frame( roi=roiValues, Subject=gl(8,1,32,subjectlabels), Hemisphere=gl(2,16,32,hemlabels), Cue=gl(2,8,32,cuelabels) )
roi.aov <- aov(roi ~ (Cue*Hemisphere) + Error(Subject/(Cue*Hemisphere)), data=roiDataframe)
I think the error model should be Error(Subject). In what sense are `Cue' and `Cue:Hemisphere' random effects nested inside `Subject'?
Let me fake some `data':
set.seed(1); roiValues <- rnorm(32)
subjectlabels <- paste("V"1:8, sep = "")
options(contrasts = c("contr.helmert", "contr.poly"))
roi.aov <- aov(roi ~ Cue*Hemisphere + Error(Subject), data=roiDataframe)roi.aov
Call: aov(formula = roi ~ Cue * Hemisphere + Error(Subject), data = roiDataframe)
Grand Mean: 0.1165512
Stratum 1: Subject
Terms:
Residuals
Sum of Squares 4.200946
Deg. of Freedom 7Residual standard error: 0.7746839
Stratum 2: Within
Terms:
Cue Hemisphere Cue:Hemisphere Residuals
Sum of Squares 0.216453 0.019712 0.057860 21.896872
Deg. of Freedom 1 1 1 21Residual standard error: 1.021131 Estimated effects are balanced
Note that all the action is in one stratum, and the SSQs are the same as
aov(roi ~ Subject + Cue * Hemisphere, data = roiDataframe)
(and also the same as for your fit).
print(summary(roi.aov))
It auto-prints, so you don't need print().
########################################
I've tried to create a contrast matrix like this:
cm <- contrasts(roiDataframe$Cue)
which gives the main effect contrasts for the Cue factor. I really want to specify the interaction contrasts, so I tried this:
######################################## # c( lh_cueL, lh_cueR, rh_cueL, rh_cueR ) # CueRight>CueLeft for the Left Hemisphere. # CueLeft>CueRight for the Right Hemisphere
cm <- c(-1, 1, 1, -1) dim(cm) <- c(2,2)
(That is up to sign what Helmert contrasts give you.)
roi.aov <- aov( roi ~ (Cue*Hemisphere) + Error(Subject/(Cue*Hemisphere)), contrasts=cm, data=roiDataframe) print(summary(roi.aov)) ########################################
but the results of these two aov commands are identical. Is it the case that the 2x2 design matrix is always going to give the same F values for the interaction regardless of the contrast direction?
Yes, as however you code the design (via `contrasts') you are fitting the same subspaces. Not sure what you mean by `contrast direction', though.
However, you have not specified `contrasts' correctly:
contrasts: A list of contrasts to be used for some of the factors in
the formula.and cm is not a list, and an interaction is not a factor.
OR, is there some way to get a summary output for the contrasts that is not available from the print method?
For more than two levels, yes: see `split' under ?summary.aov.
Also, see se.contrasts which allows you to find the standard error for any contrast.
For the fixed-effects model you can use summary.lm:
fit <- aov(roi ~ Subject + Cue * Hemisphere, data = roiDataframe) summary(fit)
Df Sum Sq Mean Sq F value Pr(>F) Subject 7 4.2009 0.6001 0.5756 0.7677 Cue 1 0.2165 0.2165 0.2076 0.6533 Hemisphere 1 0.0197 0.0197 0.0189 0.8920 Cue:Hemisphere 1 0.0579 0.0579 0.0555 0.8161 Residuals 21 21.8969 1.0427
summary.lm(fit)
Call: aov(formula = roi ~ Subject + Cue * Hemisphere, data = roiDataframe)
Residuals:
Min 1Q Median 3Q Max
-1.7893 -0.4197 0.1723 0.5868 1.3033Coefficients:
Estimate Std. Error t value Pr(>|t|)
[...]
Cue1 -0.08224 0.18051 -0.456 0.653
Hemisphere1 0.02482 0.18051 0.137 0.892
Cue1:Hemisphere1 -0.04252 0.18051 -0.236 0.816where the F values are the squares of the t values.
-- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595
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