Calculating variance of one-sample cohen's d effect size I'm trying to perform a meta-analysis where my effects are mainly one-sample t-test against chance level. I've easily found the one-sample cohen's d formula but I'm not able to find the variance formula to weight the estimated effect size.
I've found this post however I would like to have a clear reference or maybe other suggestions.
Thanks!
 A: First off, it is important to clarify the nature of your effect size. There are two ways to standardized the mean difference. The first is with either the time 1 standard deviation or the pooled time 1 and time 1standard deviations. The second is with the standard deviation of the gain scores. The relationship between these two denominators is
$$ s =  \frac{s_{gain}}{\sqrt{2 \left( 1 - r \right)}} $$
where $s$ is the pooled or raw standard deviation and $r$ is the correlation between the time 1 and time 2 scores. I strongly recommend standardizing on $s$ and not $s_{gain}$. It is not clear how to interpret the latter as you are standardizing the mean change on the variability in change.
Assuming you are standardizing the mean difference using $s$, then the variance for the effect size is
$$ v = \frac{2*(1-r)}{n} + \frac{d^2}{2*n} ~.$$
The details for this are worked out in:
Becker, B. J. (1988). Synthesizing standardized mean‐change measures. British Journal of Mathematical and Statistical Psychology, 41(2), 257-278.
The challenge, of course, is that in the context of meta-analysis, the studies often do not report $r$. Some meta-analysts use a guesstimate of $r = .5$ and you can do sensitivity analysis with different values. If the study reports the time 1 and time 2 standard deviations and means and one-sample t-test, then you can recover $r$ using the following:
$$   r = \frac {\left(s_1^2 t^2 + s_2^2 t^2 \right) -
   \left(\overline{x}_2 - \overline{x}_1\right) n} {2 s_1 s_2 t^2} ~.$$
This assumes that the sign of $t$ matches the sign of the mean difference. Flip one or both as needed.
