What is the justification for weighting meta-analysis by standard error instead of sample size? I have observed in many meta-analyses, and in tutorials on the matter, that the analyzed studies are commonly weighted by the reciprocal of the standard error. This seems to have a face validity because a lower standard error suggests a better estimate of the effect in question.
However, a standard error could be low for two reasons: (1) it could be because there is a larger sample size $n$ in a study; or (2) it could be because there is a lower variance in the study sample(s). Weighting by $n$ makes sense because one should weight the study not just by the more accurate estimate but also because the study is providing more evidence. And weighting by standard error would tend to be highly correlated with weighting by $n$. But why should low sample variance also matter? It could simply be a very unrepresentative sample in a study that causes the variance to be low. In that case one should not be weighting the study heavily.
Also, when doing a fixed effects meta-analysis it seems to violate the assumptions of the analysis to weight by study standard errors since one would expect there is really only one variance and all studies deviate from that variance by some amount. Once study variances are assumed equal weighting by standard error should be weighting by $n$.
So, why is weighting by standard error so often favoured over weighting by $n$?
 A: I suspect they are not weighting by the standard error, which would mean giving higher weight to less certain results, but weighting by the reciprocal of the square of the standard error, i.e. inverse-variance weighting, because this has the least variance among all weighted averages.
For example the Cochrane Statistical Methods Group suggests the generic inverse-variance weighted average
$$\frac{\sum Y_i (1/SE_i^2)}{\sum (1/SE_i^2)}$$ with a reference to Rice K, Higgins JPT, Lumley T. A re-evaluation of fixed effect(s) meta-analysis. Journal of the Royal Statistical Society Series A (Statistics in Society) 2018; 181: 205-227.
Since $SE_i^2$ is an estimator of $\frac1{n_i}\sigma_i^2$, you might hope that its reciprocal was approximately proportional to $n_i$ if all the $\sigma_i^2$ are the same and so you might get something similar to your suggestion of weighting by $n_i$.  But there are two additional reasons for this particular method:

*

*the individual studies may have taken different approaches to estimating the same parameter, some more accurate than others, and this difference in precision can be taken into account

*most studies may report a standard error for their estimate even when they are more confused about the actual $n$ used (especially if parts of the original sample are dropped for one reason or another)

