*1) one-sample / paired-sample vs. independent samples Cohen's d calculation*

It seems that for calculation of Cohen's d (an estimate of effect size), you should use the difference in means (or in one-sample case, from baseline), divided by SD, if available:

(M1 - M2) / SD

For paired samples, you use the mean of the difference scores divided by the SD of the difference scores.

However, when you only have the t-statistic and the degrees of freedom (df), I cannot figure out how the formula for d is derived. In the one-sample/paired-sample case, it's intuitive how to derive d from t, since t is calculated as the difference in means over standard error:

t = (M1 - M2) / se

which equals:

(M1 - M2) / (SD / sqrt(n))

so d equals:

d = (M1 - M2) / SD = t / sqrt(n)

And that's what you find in online articles here (formula 3) and here.

But for independent samples, the equation is instead:

d = 2t / sqrt(df), as seen here and here

or even

d = t * sqrt(2/n), as seen here

So where does the "2" come from? I think I'm okay regardless, because I have the original data so I can use the actual means and SDs, but I would really like to know why this difference exists, and where this "2" comes from. Is it related to pooled SD's?

*2) For independent samples, which Pooled SD is used for Cohen's d calculations?*

If I have the means, SDs, and n's of two independent groups, I should be able to calculate Cohen's d without using t-statistics and df's, and using the pooled SD in formula (M1 - M2) / SD above. However, I have seen both of these:

SDpooled1 = square root of ( (n1 - 1)s1^2 + (n2 - 1)s2^2 ) / (n1 + n2)

or

SDpooled2 = square root of ( (n1 - 1)s1^2 + (n2 - 1)s2^2 ) / (n1 + n2 -2)

From what I've seen, SDpooled1 is a biased estimator of SD, and SDpooled2 is unbiased, that is, I think it means the biased estimator (SDpooled1) underestimates SD. It seems like Cohen's d uses SDpooled1, whereas Hedges' g uses SDpooled2 (see Effect size on Wikipedia, and this Rosnow et al. 2000 Psych Science article, formula (5). )

Is this true? And if so, why would one Effect size estimator (Hedges' g) use the unbaised estimator, SDpooled2, while the other (Cohen's d) not? Other sites list SDpooled as sqrt((s1^2 + s2^2) / 2), which is the same as SDpooled1 when sample sizes are equal: here and here.

Thanks so much everyone!

Alon Hafri