Hi David,

In the 2008 Quants Study Notes, under section Sampling distribution of means, the following is written.

Variance of Sample Mean = [Variance of Population/n]*[(N-n)/(N-1)]

Also there is a mention that if sample size n <=N then this is the formula.

If we consider a scenario of Sample size n = Population size N, won’t the Sample Variance become Zero as per the above formula due to the multiple of N-n.

I might be missing something here. Could you please clarify?

Thank you,
Shiva

Hi Shiva,

Yes, note it is “without replacement” so if n = N, the sample equals the population and the sample mean is always the population mean; i.e., there will be no variation in the sample mean, it is always the same result. So, indeed as you say this does converge to zero, but appropriately so. In fact, it occurs to me this is a nice illustration of the “variance of sample mean” concept, which is a bit hard to grasp initially.

However, the testability of this variation is quite low, IMO.
Rather, the test almost always has 1 question on the “regular” sample mean “for infinite population or sampling WITH REPLACEMENT” in which case we are back to the commonly used:

variance of sample mean = population variance/N
std dev of sample mean = population standard deviation/SQRT(N)

These two are the ones i would especially be familiar with. I hope that’s helpful.

David

Thanks David for the quick response. Yes as you said, if Sample size is equal to the population size, then there is not going to be any variance in the sample mean. Hence the formula does hold up.

I understand it better now.

Thanks,
Shiva