This method is generally considered "old-fashioned" so while it may be possible, the syntax is difficult and I suspect fewer people know how to manipulate the anova commands to get what you want. The more common method is using glht with a likelihood-based model from nlme or lme4. (I'm certainly welcome to be proved wrong by other answers though.)
That said, if I needed to do this, I wouldn't bother with the anova commands; I'd just fit the equivalent model using lm, pick out the right error term for this contrast, and compute the F test myself (or equivalently, t test since there's only 1 df). This requires everything to be balanced and have sphericity, but if you don't have that, you should probably be using a likelihood-based model anyway. You might be able to somewhat correct for non-sphericity using the Greenhouse-Geiser or Huynh-Feldt corrections which (I believe) use the same F statistic but modify the df of the error term.
If you really want to use car, you might find the heplot vignettes helpful; they describe how the matrices in the car package are defined.
Using caracal's method (for the contrasts 1&2 - 3 and 1&2 - 4&5), I get
psiHat tStat F pVal
1 -3.0208333 -7.2204644 52.1351067 2.202677e-09
2 -0.2083333 -0.6098777 0.3719508 5.445988e-01
This is how I'd get those same p-values:
Reshape the data into long format and run lm to get all the SS terms.
library(reshape2)
d <- OBrienKaiser
d$id <- factor(1:nrow(d))
dd <- melt(d, id.vars=c(18,1:2), measure.vars=3:17)
dd$hour <- factor(as.numeric(gsub("[a-z.]*","",dd$variable)))
dd$phase <- factor(gsub("[0-9.]*","", dd$variable),
levels=c("pre","post","fup"))
m <- lm(value ~ treatment*hour*phase + treatment*hour*phase*id, data=dd)
anova(m)
Make an alternate contrast matrix for the hour term.
foo <- matrix(0, nrow=nrow(dd), ncol=4)
foo[dd$hour %in% c(1,2) ,1] <- 0.5
foo[dd$hour %in% c(3) ,1] <- -1
foo[dd$hour %in% c(1,2) ,2] <- 0.5
foo[dd$hour %in% c(4,5) ,2] <- -0.5
foo[dd$hour %in% 1 ,3] <- 1
foo[dd$hour %in% 2 ,3] <- 0
foo[dd$hour %in% 4 ,4] <- 1
foo[dd$hour %in% 5 ,4] <- 0
Check that my contrasts give the same SS as the default contrasts (and the same as from the full model).
anova(lm(value ~ hour, data=dd))
anova(lm(value ~ foo, data=dd))
Get the SS and df for just the two contrasts I want.
anova(lm(value ~ foo[,1], data=dd))
anova(lm(value ~ foo[,2], data=dd))
Get the p-values.
> F <- 73.003/(72.81/52)
> pf(F, 1, 52, lower=FALSE)
[1] 2.201148e-09
> F <- .5208/(72.81/52)
> pf(F, 1, 52, lower=FALSE)
[1] 0.5445999
Optionally adjust for sphericity.
pf(F, 1*.48867, 52*.48867, lower=FALSE)
pf(F, 1*.57413, 52*.57413, lower=FALSE)
Best Answer
Your error term is not quite correct.
Assuming you've got a balanced design, the repeated measures ANOVA model using
aovisThe term
Error(Subject/Time)models the subject-specific variance; or in the language of mixed-effect models:Subjectis your random andTimeyour fixed effect, andTimeis nested (i.e. repeatedly measured) withinSubject.Two more comments:
Keep in mind that
aovonly allows for random effects if the design is balanced (which is the case for the sample data you provide). If you've got an unbalanced design (e.g. due to missing observations), you will need to use a mixed-effect model.It's a bit odd to use ANOVA to model a time dependence. I am assuming here that
Timeis a categorical variable.