I have a linear mixed effects model (say AxBxC), where all of the 2-way interactions are significant but the 3 way interaction is not, and I want to perform post hoc contrasts on the 2 way interactions (e.g. AxB). I've found several recommended methods: 1) create a new linear model y=AxB and perform contrasts on AxB using glht. 2) create a new linear model y=AxB and perform contrast on AxB using emmeans. 3) use the original model (AxBxC) and perform contrast on AxB using emmeans.

So far they all seem to give very similar results, but I'm wondering how they are different, and which method is preferred (or if some other method not covered is better).


  • $\begingroup$ could we have a reproducible example please? $\endgroup$
    – Ben Bolker
    Jul 9, 2018 at 23:43
  • $\begingroup$ By "create a new linear model y=AxB" do you mean create a new model with the 3-way interaction excluded, or create a model that doesn't have C at all? $\endgroup$ Jul 10, 2018 at 12:47

1 Answer 1


In your first two options, as I understand it, you are proposing to fit a new model with predictors A and B, but not C. Thus, the variations due to C and its interactions with A and B are left unexplained, so those fixed variations become part of the error variance. That would be a bad model, and so I don’t think those are good options. Emmeans and glht results are model-based and they are only good if the model is good.

So I think your third option is a far better choice. Another possibility would be to fit a new model where all the main effects and two-way interactions involving A, B, and C are included, but the three-way interaction is excluded. Then use emmeans or glht as you see fit. (If glht, using emmeans::emm() is a lot easier than multcomp::mcp() when more than one factor is involved.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.