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I would like to calculate Cohen's d for 2x2 ANOVA interaction (nationality: Germany, France; gender: male, female).

Someone asked similar question earlier (Cohen's d for 2x2 interaction), and got an advise to calculate Cohen's d using (a1 - b1) - (a2 - b2) as a numerator and the square root of the MSE from the ANOVA as a denominator.

Could someone please tell me a reference, where I could find more information how to calculate Cohen's d using this formula? I really cannot find any. What would a1, a2, b1 and b2 be in my case?


marked as duplicate by John, Silverfish, gung, whuber Nov 4 '15 at 13:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You'd have to actually add details about your case for anyone to answer the second part of your question. There's no such thing as a reference for your first part since that's just the trivial mathematical equation for your interaction effect. To be more clear, a and b are the variables and the numbers are levels of the variables. $\endgroup$ – John Nov 4 '15 at 12:15
  • $\begingroup$ Now that @John has edited his answer at the possible duplicate to clarify what the a1 etc refer to, I can't see anything in this question which isn't answered there. $\endgroup$ – Silverfish Nov 4 '15 at 12:27

Edit: I see that I missed a key part of the question, which is your confusion about what a1, a2, b1, and b2 actually are. These refer to the 4 cell means in the 2x2 design. Imagine that subjects get randomly assigned twice: first to either condition a or condition b, and then to either condition 1 or condition 2. So each subjects ends up in either group a1, a2, b1, or b2. In what I wrote below, I use these labels as shorthand to refer to the group means rather than the groups themselves (as in the previous advice you received).

Actually I think it is more appropriate to use $$ d=\frac{(a1-b1)-(a2-b2)}{2\sigma} $$ rather than the definition you mentioned. I covered this on my blog a few months back (LINK) but I'll cover the basic argument again here.

If we take your numerator and distribute the implicit $-1$, we see that it equals $$ a1-b1-a2+b2=(+1)a1 + (-1)a2 + (-1)b1 + (+1)b2. $$ The key here is to realize that this is still a comparison between two group means, just like in the classical definition of Cohen's d. We are comparing the a1 and b2 groups (which have coefficients of +1 in the above sum) against the a2 and b1 groups (which have coefficients of -1). So the two relevant means to use in computing d are the mean of the a1 and b2 means, $\mu_1=\frac{a1+b2}{2}$, and the mean of the a2 and b1 means, $\mu_2=\frac{a2+b1}{2}$. This gives us $$ d=\frac{\mu_1-\mu_2}{\sigma}=\frac{\frac{a1+b2}{2}-\frac{a2+b1}{2}}{\sigma}=\frac{(a1-b1)-(a2-b2)}{2\sigma}. $$ I think that this is the most natural extension of the classical Cohen's d to a 2x2 interaction effect.

If you're not convinced yet, see my blog comment (HERE) for some further arguments for why this should be preferred over the effect size definition that you mentioned.


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