2
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ You are right, the square of the standard error is what you need. $\endgroup$
    – mdewey
    Sep 24, 2017 at 12:52
  • $\begingroup$ 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). $\endgroup$
    – user10619
    Sep 28, 2017 at 12:09
  • $\begingroup$ 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 $\endgroup$
    – Deep North
    Sep 28, 2017 at 12:44

1 Answer 1

Reset to default
1
$\begingroup$

I already answered this as a comment but since they are potentially ephemeral I give a proper answer here.

The appropriate weight in meta-analysis is the inverse of variance as stated by the OP and this is the sampling variance of the estimator being meta-analysed, not the variance of the data involved. If the standard error is available instead it just needs to be squared.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.