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When calculating Cohen's $d$ for independent samples, you must use a pooled $SD$. However, I have seen both of these:

$$SD_{\text{pooled1}} = \sqrt{\frac{ (n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2}}$$

vs.

$$SD_{\text{pooled2}} = \sqrt{\frac{ (n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 -2}}$$

Supporting the use of $SD_{\text{pooled1}}$ , some website have $SD_{\text{pooled}}$ listed as $\sqrt{\frac{s_1^2 + s_2^2}{2}}$, which is the same as $SD_{\text{pooled1}}$ when sample sizes are equal.

From online discussions, it sounds like $SD_{\text{pooled1}}$ is a "biased" estimator of $SD$, and $SD_{\text{pooled2}}$ is less biased, that is, I think it means $SD_{\text{pooled1}}$ underestimates $SD$. In addition, some sites suggest that some effect size metrics use $SD_{\text{pooled1}}$ (Cohen's $d$), and others $SD_{\text{pooled2}}$ (Hedges' $g$).

Is this true? And if so, why would one effect size metric (Hedges' $g$) use the unbaised estimator, $SD_{\text{pooled2}}$ , while the other (Cohen's $d$) not?

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  • $\begingroup$ Closely related: stats.stackexchange.com/questions/1850/… That said, I believe the initial definition for Cohen's d didn't specify how to compute the SD, so the de facto formula (with the biased version) might just be a historical accident. $\endgroup$
    – Matt Krause
    Jun 14, 2012 at 8:46
  • $\begingroup$ Thanks for editing my question so the formatting is much more clear, Jeromy. $\endgroup$
    – Alon
    Jun 21, 2012 at 20:48

1 Answer 1

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Both estimates are biased. The square of the second one is an unbiased estimator of the common variance. It is not clear that taking the square root of an unbiased estimator makes the estimate of standard deviation better than taking the square root of a biased estimator of variance.

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  • $\begingroup$ +1 for pointing out the common misconception that the square root of an unbiased estimator of a parameter unbiasedly estimates the square root of the parameter. Btw, Michael, I believe there was a word missing from your second sentence - I've edited it. Please check that I didn't change the content in a substantive way. $\endgroup$
    – Macro
    Jun 14, 2012 at 12:41
  • $\begingroup$ @Macro The change looks good. $\endgroup$
    – Michael R. Chernick
    Jun 14, 2012 at 12:45
  • $\begingroup$ Thank you for pointing that out, Michael. The post above by Matt Krause confirms this. And so it would seem that for small sample sizes, Hedges' g should be used, since it provides a better estimate in that case. Is that reasonable? $\endgroup$
    – Alon
    Jun 21, 2012 at 21:09

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