estimating the population standard deviation from the sample standard deviation Is there a way to estimating the population standard deviation, $\sigma$, from the sample standard deviation, $s$, if the size of the finite population is $N$ and the size of the sample is $n$?
Is there expression true: $$ \sigma = s \cdot \sqrt{n} \cdot \sqrt{\dfrac{N-1}{N-n}}$$
 A: Let's take the components one at a time.


*

*The $\sqrt{n}$ term would be appropriate if $s$ was the sample estimate of the standard error of the mean.
However, you state $s$ to be the standard deviation, so that term doesn't belong.

*When the population is finite and the sample fraction isn't really small ($n/N$ isn't very small), the finite population correction factor is used to adjust the standard error of a sample mean for the fact that as $n$ samples (without replacement) a larger fraction of the population, the variance of the estimate reduces from the infinite-population form:
$\hat{\sigma}_{\bar{X}}=\sqrt{\frac{N-n}{N-1}} \frac{s}{\sqrt{n}}$
(So (i) you have it upside down and (ii) it shouldn't be there, since you're not esitmating $\hat{\sigma}_{\bar{X}}$). So we drop that term as well.

*If you want to estimate $\sigma$ itself - the standard deviation of the distribution of values rather than of means - you'd usually still just use $s$. If you want it to exactly equal $\sigma$ when $N=n$ you'd need to make an adjustment, but not that one (dropping the Bessel correction would work, for example).
