The inverse variance is proportional to the sample size?

Interested in meta-analysis I found the lecture notes http://www.uazuay.edu.ec/cibse/lecture_notes.pdf

I understand that it makes sense in a meta-analysis to assign different weights to different studies. Personally I find it natural to weigh by the sample sizes of the particular studies. However, in the fixed effect model the inverse of the variances is used. As an explanation on page 17 I read the sentence "The inverse variance is roughly proportional to sample size, but has finer distinctions,( and serves to minimize the variance of the combined effect.)"

I do not understand that and have problems with the first claim.

Suppose $X_1,..,X_n$ are random variables iid. Then $\bar{X}=\frac{1}{n} \sum_n X_i$ has variance $\frac{Var(X)}{n}$. This is clear. So one could say that on the level of random variables the inverse variance of the arithmetic mean is proportional to the sample size.

However, on the level of samples, we always compute the sample means and the sample variances, and the sample variance is an estimator for the variance of an individual measurement and not the variance of the (sample) arithmetic mean.

Can somebody clear up my confusion

• So divide $s^2$ by $n$. And if studies $A$ and $B$ each weigh $100$ boxes of corn flakes, $A$ with a professional scale, $B$ with a dime store scale that has greater built in variance, in merging the results it makes sense to give greater weight to results from $A$. – André Nicolas Jul 11 '11 at 22:53
• The first sentence above was a little abrupt. Indeed $s^2$ is an estimator for the population variance. Thus $s^2/n$ is an estimator for the variance of the sample mean. It should be good enough for determining appropriate weights when we do the pooling. – André Nicolas Jul 12 '11 at 6:00
• That is exactly my problem. The weights come from consider the inverse of $s^2$, i.e. the estimator for the population variance and not for the estimator of the variance of the sample mean. $1/s^2$ will not be proportional to $n$, right? – wood Jul 12 '11 at 9:58
• To be more precise. $s^2$ will be an estimator for the population variance. So the sample variance from two studies will approximately be the population variance. So we are weighing by roughly the same number, which does not make sense. Or is something else meant by weighing with variance? i.e. maybe not taking the sample variance but some other quantity? – wood Jul 12 '11 at 10:30
• They probably mean weighting by inverse of $s^2/n$. Or jointly as inverse of $s^2$ and inverse of $n$. Only thing that is consistent with the quoted words. – André Nicolas Jul 12 '11 at 11:18