# Cohen’s d for 2×2 anova interaction [duplicate]

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

Someone asked similar question earlier (Cohen’s d for 2×2 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?

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 2×2 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 2×2 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.