I am analyzing a data set using a one-way, within-subjects anova with 12 levels. Each of the levels represents a unique trial type in a reaction time experiment. However, the trials can be reasonably grouped together into 4 equal groups, let us call them A through D. My primary interest is in showing that A differs from the controls (B-D). Following a significant Anova, linear contrasts show that this is the case. However, it seems that I should also demonstrate that B-D are not different from each other. Parsimony suggests I could run another Anova (rather than pairwise tests) to demonstrate this. However, I have never seen this done. Is this simply convention or is there a more rigorous reason that I am missing? Thanks!
1 Answer
Doing a second one-way ANOVA will be more statistically parsimonious if the null hypothesis (B-D have no significant difference in means) is accepted. But it will be less parsimonious if it is rejected, because then you will do the linear contrasts anyway.
Personally, I would just do the linear contrasts with the first ANOVA so I only have to calculate and present a single ANOVA table. I think that is the reason that it seems to be the convention.