I've calculated Cohen's d for regression coefficients (from the t statistic), odds ratios and means differences, hoping to pool the results in a meta-analysis and see how it works. However, in Stata, it doesn't seem you're able to pool these results without confidence intervals for Cohen's d, so my question is how do I get around this? Is there a way of calculating it, or is there a way of pooling the results in Stata without this information?

I know there are several negative sides to this sort of meta-analysis, but am intrigued to see how this works in comparison to several smaller analyses of specific effect sizes.


2 Answers 2


According to p238 of standard text on meta-analysis in social science The Handbook of Research Synthesis, the variance of Cohen's $d$ is $$\left( \frac{n_1 + n_2}{n_1 n_2} + \frac{d^2}{2(n_1+n_2-2)}\right) \left(\frac{n_1 + n_2}{n_1+n_2-2} \right), $$ where $n_1$ and $n_2$ are the sample sizes of the two groups being compared and $d$ is Cohen's $d$.

Taking the square-root of this variance gives the standard error of $d$, needed as input by several of the user-written meta-analysis packages for Stata. (Some of them also accept confidence intervals as input, but they simply convert them to standard errors internally anyway.)

  • 2
    $\begingroup$ Why don't you get the standard deviation by taking the square-root of that variance? I thought variance was defined as the standard deviation squared. Or are we dealing with a different definition of "variance" here? $\endgroup$
    – Speldosa
    Sep 22, 2014 at 8:40

This is an old question, but someone may be looking for a quick answer (this is in R, but quite quick). The package MBESS provides a straightforward conversion tool.



ci.smd(smd=SOMEVAR,n.1=X, n.2=Y) 
  • $\begingroup$ Why did you use .69 as t value? $\endgroup$
    – user248711
    Nov 1, 2020 at 15:50
  • $\begingroup$ I was converting a very specific data set 6 years ago. I wiped that data a long time ago. $\endgroup$
    – d-cubed
    Nov 2, 2020 at 16:10

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