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This question already has an answer here:

I have been reading http://www.polyu.edu.hk/mm/effectsizefaqs/effect_size_equations2.html and the only difference I can find between the two is that they have ever so slightly different pooled standard deviation values but I don't understand how changing these SD values is meant to improve the accuracy of the estimate of effect size.

The differences in standard deviation are the following ($SD_{pooled}$ is the pooled SD for Cohen's d and $SD^{*}_{pooled}$ is for Hedges' g):

$SD_{pooled}=\sqrt{\frac{n_1 SD_1^2+n_2 SD_2^2}{n_1+n_2-2}}$

$SD^{*}_{pooled}=\sqrt{\frac{(n_1-1)SD_1^2+(n_2-1)SD_2^2}{n_1+n_2-2}}$

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marked as duplicate by gung, COOLSerdash, Matt Krause, Nick Cox, Peter Flom Aug 10 '13 at 17:22

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You are right, the difference between them is very small and with large $N$ will disappear. In fact most people (at least in my experience) are not aware of any of this; "Cohen's $d$" is often used generically, many people have not heard of Hedges' $g$, but they use the latter formula and call it by the former name. The difference is that Cohen used the maximum likelihood estimator for the variance, which is biased with small $N$, whereas Hedges used Bessel's correction to estimate the variance. (For more on this topic, it may help you to read this CV thread: What is the difference between N and N-1 in calculating population variance?) The corresponding formulas are often known as the population formula for the variance and the sample formula. Recall that these are:
\begin{align} \text{Var}(X)_\text{population} &= \frac{\sum (x_i-\bar x)^2}{N} \\ ~ \\ ~ \\ \text{Var}(X)_\text{sample} &= \frac{\sum (x_i-\bar x)^2}{N-1} \end{align} As $N$ increases indefinitely, these two estimates will converge to the same value. However, with small samples, the population formula will underestimate the variance because it does not take into account the fact that the mean, $\bar x$, was estimated from the same dataset. When these estimates are subsequently used to estimate the standardized mean difference, that implies that the former will overestimate the effect size.

Thus, with small samples, Hedges' $g$ provides a superior estimate of the standardized mean difference, but the superior performance fades as the sample size increases.

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