# Cohen's d for 2x2 anova interaction [duplicate]

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

• 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. – John Nov 4 '15 at 12:15
• 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. – Silverfish Nov 4 '15 at 12:27
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.