It's a between-subject design with a control group and a treatment group with 50 people each. Each participant undergoes 100 trails. (repeated measures). So the only within-subject factor is the number of trails. The independent variable is dummy coded (0 & 1). The dependent variable is continuous, it varies between [3.00 to 6.00]. I want to compare the means between the two. Should I use a repeated-measures ANOVA / mixed model / T-test/manova?

Which is a good approach and why?

I understand that, I can have a mixed model and find anova for that. Is this right?


1 Answer 1


Both an ANOVA and mixed modeling approach will work for your data. I am more familiar with mixed models, and can speak better to that. However, you are correct that you can run your analysis as a mixed model and then get an ANOVA table afterward. If you are using R, then you can use lmer to first estimate the mixed model:

m <- lmer(outcome ~ trtmt + (1|Subject), data=df) 

The above model will give you the mean outcome difference between treatment and non-treatment individuals while properly accounting for the correlated outcome scores from the same individual. If you wanted the ANOVA table based on this model, you can use anova(m) to get that.

It is worth noting that the mixed model, as with any statistical model, is only valid under certain assumptions, many of which are the same assumptions as regression. Among these are the normality of the residuals, which can be examined as such:

residm <- resid(m) #grab L1 residuals
qqnorm(residm) #graphing to check normality assumption

Also important with the residuals is the homogeneity of variance assumption, which you can inspect visually, looking for the absence of any pattern in the residual vs. fitted plot using plot(m).

Mixed models impose a normal distribution on the random intercepts, i.e., (1|Subject) in the lmer syntax above. You can look to see whether the predicted intercepts do a decent enough job of following a normal distribution using the following code: plot(ranef(m)).


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