I've already read this, which isn't very helpful.

Consider Cohen's $d$ for two independent random samples $\{X_{1j}\}_{j=1}^{n_1=N}, \{X_{2j}\}_{j=1}^{n_2 = N}$, given by $$d = \dfrac{\bar{X}_1 - \bar{X}_2}{\text{SD}}$$ where $\text{SD}$ is usually pooled. Obviously, $$d = \dfrac{T}{\sqrt{N}}\text{,}$$ where $T$ is the $T$-test statistic provided when doing a hypothesis test on $\mu_{D} = \mu_{1} - \mu_{2} = 0$ as the null and $\mu_{D} = \mu_{1} - \mu_{2} \neq 0$ as the alternative.

By transformations (letting $D$ be the random variable representing Cohen's $d$): $$f_{D}(d) = f_{T}(d\sqrt{N})\sqrt{N}$$ where $$f_{T}(t) = \dfrac{\Gamma\left(\dfrac{\nu + 1}{2}\right)}{\Gamma\left(\dfrac{\nu}{2}\right)\sqrt{\nu\pi}}\left(1+\dfrac{t^2}{\nu}\right)^{-(\nu+1)/2}\text{, } t \in (-\infty, +\infty)$$ is the pdf of the $T$-distribution with degrees of freedom $\nu = n_1 + n_2 - 2 = 2N - 2 = 2(N-1)$. This implies that $N = 1+\dfrac{\nu}{2}$. So, $$f_{D}(d) = \dfrac{\Gamma\left(\dfrac{\nu + 1}{2}\right)}{\Gamma\left(\dfrac{\nu}{2}\right)\sqrt{\nu\pi}}\left[1+\dfrac{\left(1+\dfrac{\nu}{2}\right)d^2}{\nu}\right]^{-(\nu+1)/2}\sqrt{1+\dfrac{\nu}{2}}\text{, } d \in (-\infty, +\infty)\text{.}$$

It's pretty clear how to calculate a Confidence Interval from this PDF, but this article states the following:

A formula for calculating the confidence interval for an effect size is given by Hedges and Olkin (1985, p86). If the effect size estimate from the sample is $d$, then it is Normally distributed...

I don't think the PDF I have up there is that of a normal distribution. Did I do something wrong here?

  • $\begingroup$ Great question. I too have been looking for an answer. The "easy" approach is to just calculate the distribution of d through numerical simulations. $\endgroup$
    – Jessica
    Commented Dec 3, 2015 at 16:51
  • $\begingroup$ I don't understand your Edit. Why is it "quite obvious"? I think d should be t-distributed (appropriately scaled), as you wrote above. Perhaps the claim that it's normally distributed should be understood asymptotically? $\endgroup$
    – amoeba
    Commented Dec 3, 2015 at 17:46
  • $\begingroup$ @amoeba Oh, wait. I'm wrong. I for some reason thought for a sec that the SD was using the population value $\sigma_{\bar{X}_1 - \bar{X}_2}$. $\endgroup$ Commented Dec 3, 2015 at 17:47
  • $\begingroup$ @amoeba Edit removed $\endgroup$ Commented Dec 3, 2015 at 17:48
  • $\begingroup$ @amoeba If it is asymptotic, I suppose I should see what happens to $f_{D}$ as $\nu \to \infty$... $\endgroup$ Commented Dec 3, 2015 at 17:48


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