2 sample z test - Standardized Effect Size


comparing the average of 2 groups when we know each group's standard deviation: σ1, σ2

Cohen’s effect Size = | avg(x1)- avg(x2)| / σ pooled

How do you calculate the σ pooled for 2 sample z test?

σ pooled^2= (n1σ1^2+n2σ2^2)/(n1+n2) {similar to t-test}

or just a simple average σ pooled =(σ1+σ1)/2 ?
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Hi hlsmith,

Per my understanding, the book describes the: 2 sample t test - Standardized Effect Size.
When you use the sample standard deviation:

S pooled ^2= ( (n1-1)S1^2+(n2-1)S2^2)/(n1+n2-2)

The question is how to calculate the 2 sample z test Cohens effect size, when you know the standard deviation of the two groups are: σ1,σ2.

The book treats only the case that σ1=σ2: effect size=(μ1-μ2)/σ

I know the basic assumption of "pool" idea is that the standard deviations are the same but only the sample standard deviations are not the same, and in this case, we know that the standard deviations are not the same, so probably the average of the standard deviations? or the average of the variances is more appropriate?
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No, I agree - if you need to address that it needs some type of weighted pooling. I would imagine that should be a formula out there. You didn't see anything in that book? I don't recall ever doing that process myself. If I was you I would continue combing the web to confirm your approach or find a formula. It seems like it would be a common process in meta-analyses.

I think the satterthwaite formula for calculating a t-test with unequal variances might be useful because the gist is to pool and use the appropriate (approximate) degrees of freedom. This is just of the top off my head.
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