How population size affects the variance of the distribution of sample mean We know the variane of the distribution of the sample mean is $\sigma_{m}^2=\sigma^2/n$, where $n$ is the sample size and $\sigma^2$ is the population variance.
Say we have two populations -- all male in your univeristy department (Po1, 120 males) and all male in your city (Po2, 120K males).
Assuming the population variance of Po1 is equal to Po2, and we fix the sample size $n=100$ for both populations. Then by $\sigma_{m}^2=\sigma^2/n$, the two variances of sample mean are also identical.
But this is quite counter-intutive, isn't it? Since the sample size $n$ is much closer to the size of Po1 than Po2, but still they have the same varaince in thoese point estimates. In other words, the sample size should be somehow determined by the population size -- the larger the population size the more samples you need to have a good estimate.
Is there any intutive reasons behind this or is it simply becasue the assumption ``population variance of Po1 is equal to Po2'' is counter-intutive? Thanks.

If your sample is without replacement from a finite population size $N$, then the variance of the sample mean is not $\dfrac{\sigma^2}{n}$ but $\dfrac{N-n}{N-1}\dfrac{\sigma^2}{n}$
So in your example of $n=100$, $N_1=120$ and $N_2=120000$, then instead of a variance of $\dfrac{\sigma^2}{100}$,
we would get $\dfrac{\sigma^2}{595}$ for the smaller population and about $\dfrac{\sigma^2}{100.083}$ for the variance population, with the variance for the smaller population substantially smaller, in line with your intuition
