0

Is it (N-n)/N or (N-n)/(N-1)? Provided that they can be used interchangeably, why would you use one instead of the other?

user208557
  • 33
  • 2
  • 7
  • To the extent that chl's answer [here](https://stats.stackexchange.com/questions/5158/explanation-of-finite-correction-factor) covers this question (even though the question is a little different), I guess this counts as a duplicate – Glen_b May 17 '18 at 00:50

1 Answers1

2

The term $\frac{N-n}{N-1}$ is a term in the variance of a hypergeometric.

If you use the binomial variance instead (substituting $K/N$ for $p$), then the terms are the same aside from this term.

Thus you could think of this term as a way of adjusting a variance calculated under an assumption of binomial sampling (in effect, sampling without replacement, or sampling from an infinite population) for the fact that you're actually sampling with replacement from a finite population.

Under the assumptions relevant for these formulas to apply in their respective cases, the $\frac{N-n}{N}$ version could be seen as an approximation, but if $N$ and $n$ are reasonably large then that will make very little difference (and even less difference again once you convert to a standard deviation).

(However there are other ways to argue for the N denominator besides as an approximation.)

Glen_b
  • 257,508
  • 32
  • 553
  • 939