How to compute Cohen's d from means difference in T-Score (or Z-Scores) and its CI

Question

I am performing a meta analysis on longitudinal cohort studies reporting outcomes in cognitive performance as a function of adherence to a nutritional variable. Most of them provide "standardized means differences" or "standardized beta regression coefficients" which I can easily convert to a correlation coefficient.

However, there are some studies that only report the results of their regression analysis as raw means differences (95% CI, p value) derived by using the highest tertile (vs. lowest and middle) as reference for comparison. These means differences are expressed in transformed scores, either T or Z scores.

How can I compute a Cohen's d coefficient with this information?

My first thought was to calculate SE/SD from the CI but given the fact that they are asymmetrical I don't think that makes any sense. Should I just divide the value by the SD of T-Scores (SD = 10) or Z-Scores (SD = 1)?

Answer 1

Divide the mean difference by 10 for the T-scores and 1 for the Z-scores. The only issue here is that you are effectively dividing by the overall standard deviation rather than the within groups pooled standard deviation. For small values of $d$ this won't make much difference. For larger values it will. You can compute the pooled standard deviation for the T-Score values as $$ s_{pooled} = \sqrt{ \frac{10^2(N-1) - \frac{\left(\overline{X}_1^2 + \overline{X}_2^2 - 2 \overline{X}_1 \overline{X}_2 \right)} {N}}{N}} . $$ For the z-score, drop the $10^2$ from the equation.