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I'm doing a meta-analysis of one-sample t-tests and I want to find a variance standardizing transformation for Cohen's d for one sample t-test and the variance of the transformed effect size (to verify that the PET-PEESE results are not affected by the induced correlation between effect sizes and standard errors due to using Cohen's d for one-sample t-tests). In other words, an equivalent to Fisher's z transformation for correlation coefficient and its variance.

From reading the Wikipedia page on variance standardizing transformations, I understand that I need a formula h(u) for the variance of one-sample t-test. Then apply the formula

enter image description here

to obtain the transformation. Finally, I need to get a distribution of Cohen's d for one-sample t-test and find out how the change of variables from the derived formula changes the variance.

I found 3 different suggestions for the variance of Cohen's d for one-sample t-test here and a suggestion to use 1/n in the reply to the answer here. If 1/n was indeed the variance for Cohen's d of one-sample t-test, it would have greatly simplified the whole process -- it would already be variance stabilized. However, I do not believe it to be true since the uncertainty in the variance of the original observations (which is probably Chi^2 distributed) should propagate into the distribution of Cohen's d (as similarly explained for two-sample t-test here).

Assuming that either one of the proposed formulas in 3 holds, the variance stabilizing transformations (dropping the constants) would be:

enter image description here for the second variance defined in 3 and computed by wolfram alpha

or

enter image description here for the third variance defined in 3 and computed by wolfram alpha

(I did not used the first formula since the post shows that it aproximates the true variance poorly).

Then, I "just" need to apply the transformations to the distribution of Cohen's d for one-sample t-test and obtain its variance after transformation, however, I could not find it anywhere, and I am not sure that I would be able to make the final step correctly.

Thanks, Frantisek

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  • $\begingroup$ One problem with the formulas derived in this way is that the effect size made dependent on the sample size. I run a few simulations and it seems like that the variance of the first one is slightly above 1 and approaching it as N increases. The variance of the second formula seems to be more dependent on N. $\endgroup$ Commented Feb 23, 2021 at 10:16

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