A suggestion for the standard error for Cohen's d

Since from Taylor's series the ratio of two estimates a and b is given as

V(a/b) = $$\frac{\mbox{a}^{text{2} \mbox{V(b) - 2ab Cov(a,b) + } \mbox{b}}text{2} \mbox{V(a)}}{\mbox{b}^text{4}}$$

with, for Cohen's d, a = difference in two group means, $$\bar{x}_text{1}$$ and $$\bar{x}_text{2}$$ with respective sizes n1 and n2 and standard deviations, s1 and s2 and b the pooled standard deviation, sd.

then V(a) = $$ \frac{s1^{text{2}}{n1} + \frac{s2}text{2}}{n2}$$ and V(b) = the square of $$\frac{0.71sd}{sqrt{n1+n2}}$$ (see [http://davidmlane.com/hyperstat/A19196.html here]). If Cov(a,b) is equal to zero as the groups are independent

we have

s.e. (Cohen's d) = $$\frac{ (\bar{x}_text{1}-\bar{x}_text{2})^{text{2} \(\frac{0.71sd}{sqrt{n1+n2}})}text{2} + sd^{text{2} (\frac{s1}text{2}}{n1}+\frac{s2^{text{2}}{n2}) }{sd}text{4}}$$

In the example of the hospital waiting times $$\bar{x}_text{1}$$=12.08, s1=2.89, n1=480, $$\bar{x}_text{2}$$=16.07, s2=2.98, n2=493 and the (pooled) sd=3.55. Inserting these into the above equation we then get Cohen's d =1.36, s.e.(Cohen's d) = 0.059 giving 95% CI of -1.24, -1.48) compared to (-1.22, -1.50) using Smithson's and the bootstrap.