I have a significant 2x2 interaction in a between-subjects design. I would normally report omega-squared as my effect-size measure, but I've been asked to provide Cohen's d instead? Can Cohen's d be computed for an interaction? If so, how?


You can translate pretty much any of the effect sizes to Cohen's d, so start with omega and work your way over. Equations are available for them all online... the wikipedia page for effect size is good for this.

In your particular case though you can get a numerator for cohen's d as (a1 - b1) - (a2 - b2). That's the simple effect of a 2x2 interaction with, a and b as the variables (e.g. sex and handedness) and the numbers are levels of the variables (e.g. male / female and left / right). Your denominator is just the square root of the MSE from the ANOVA*.

*assuming homogeneity of variance

  • 1
    $\begingroup$ I think that a more appropriate effect size is $d=((a1-b1)-(a2-b2))/2\sigma$, where $\sigma$ is the RMSE, as I explain in my answer here: stats.stackexchange.com/questions/179098/… $\endgroup$ – Jake Westfall Nov 4 '15 at 5:27
  • $\begingroup$ Your answer might be handy for power calculations but that's not what Cohen's d is for. It's for equivalent interpretation of effects. Cohen never specified the math, only that it be counted as number of standard deviations and your version of it does not do that. Your d for an interaction will violate the consistency of interpretation that is the entire purpose for the standardized measure. You might want to look at the arguments for and against different ds for effect sizes in repeated measures design to get a flavour of the problem you're introducing. $\endgroup$ – John Nov 4 '15 at 12:24
  • $\begingroup$ Actually I think you have it exactly backwards: the effect size that I define is comparable in size/interpretation to the classical definition of d, as well as to how d is typically defined for main effects in the 2x2 context (where one typically ignores the other factor in computing d), while yours is not. See my arguments that I linked to at the bottom of my post. One of the big issues is that, unless you divide by an additional factor of 2 to account for there being twice as many means, then the variance of this measure differs from the classical d, rendering it on a different scale. $\endgroup$ – Jake Westfall Nov 4 '15 at 14:48
  • $\begingroup$ It was never about the variance of d. You might want to vet this argument a different way than the comments on an answer. Perhaps construct your own question about it. That seems the best idea. $\endgroup$ – John Nov 4 '15 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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