I am reading meta analysis such as this one in page 89.
It says the weight assigned to each study is $w_i=\frac{1}{v_i}$ where $v_i$ is the within-study variance for study $(i)$.
The weighted mean $\bar{T_•}$is then computed as
$$\bar{T_•}=\frac{\sum_{i=1}^kw_i T_i}{\sum_{i=1}^kw_i}$$
I want to do a meta analysis on means with known 95% CIs on serval studies.
Since means are the sample means and variance of the sample mean is $\frac{\sigma^2}{n}$ or (approximately $\frac{s^2}{n}$). So I think the weights should be square of the standard errors(i.e $\frac{\sigma^2}{n})$. i.e I use standard errors from 95% CIs directly without multiply by sample size.
But someone says the weight should be square of standard deviarion, i.e $n\times(se)^2=s^2$. I am confused, since we are weighting means which are the random varialbes, therefore, we should use variance of sample mean, but not the variance directly.
Read this post , I think Glen_b's answer is quite related to Meta-analysis, if the $\hat{\mu}$ is $\bar{X}$ should the weight be $\frac{\sigma^2}{n}$?
Thank you very much.