Meta analysis for means, should the weight be variance of the sample mean?

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.

• You are right, the square of the standard error is what you need. – mdewey Sep 24 '17 at 12:52
• Do you have interest in meta-analysis of effect-sizes or "sample means" ? Alternatively do you want to challenge inverse-variance weighing procedure ? Please clarify your goal(s). – Subhash C. Davar Sep 28 '17 at 12:09
• I am not sure what your meaning, A [standard error][1] of a statistic (or estimator) is the (estimated) standard deviation of the statistic. I would think if you want to summarize any statistics (or estimators, such as coefficiets, ORs, RR), probably you only use square of standard errors as weight. [1]: web.eecs.umich.edu/~fessler/papers/files/tr/stderr.pdf – Deep North Sep 28 '17 at 12:44