Post-hoc test after 2-factor repeated measures ANOVA in R?

I have problems finding a solution regarding how to run a post-hoc test (Tukey HSD) after a 2-factor (both within-subjects) repeated-measures ANOVA in R. For the ANOVA, I’ve used the aov -function:

summary(aov(dv ~ x1 * x2 + Error(subject/(x1*x2)), data=df1))

After reading answers to other questions, I gathered that I would first have to re-run the ANOVA using some other fuction (e.g., lme). This is what I came up with.

Lme.mod <- lme(dv ~ x1*x2, random=list(subject=pdBlocked(list(~1, pdIdent(~x1-1), pdIdent(~x2-1)))), data=df1)

Both main effects were significant, but there were no interaction effects. Then, I used these functions for the post-hoc comparisons:

summary(glht(Lme.mod, linfct=mcp(x1="Tukey")))
summary(glht(Lme.mod, linfct=mcp(x2="Tukey")))

However, there were some problems:

First of all, the R Help file says that “The mcp function must be used with care when defining parameters of interest in two-way ANOVA or ANCOVA models (…) multcomp version 1.0-0 and higher generates comparisons for the main effects only, ignoring covariates and interactions (older versions automatically averaged over interaction terms). A warning is given.” And sure enough, I received the following warning message:

Warning message:
In mcp2matrix(model, linfct = linfct) :
covariate interactions found -- default contrast might be inappropriate

Another puzzling thing was that although both main effects were significant, there were no significant differences in the post-hoc comparisons for one of the factors (x1). I’ve never encountered this before. Are the scripts/analyses correct/appropriate, or is there something that I’m missing? Any help would be most appreciated!



Lme.mod <- lme(dv ~ x1x2, random=~1|subject,
summary(glht(Lme.mod, linfct=mcp(x1x2="Tukey")))

be what you are after, i.e. do posthoc tests among all combinations of measurements levels of both factors x1 and x2? (I’ve also imposed compound symmetry, to make the lme result match that of the repeated measures aov call)

Source : Link , Question Author : Jonna , Answer Author : Tom Wenseleers

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