[R] contrast matrix for aov
Prof Brian Ripley
ripley at stats.ox.ac.uk
Thu Mar 10 09:24:44 CET 2005
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 7
Residual 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 21
Residual 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 1.3033
Coefficients:
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 0.816
where the F values are the squares of the t values.
--
Brian D. Ripley, ripley at stats.ox.ac.uk
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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