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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: $$ \sigma = s \cdot \sqrt{n} \cdot \sqrt{\dfrac{N-1}{N-n}}$$

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  • $\begingroup$ The equality is false in general, and it's incorrect in several different ways at once. You're trying to produce an estimate, so the left hand side should not be $\sigma$. The scaling factor won't do what you need. What properties do you seek for an estimator of $\sigma$ to have? Are you simply trying to give a finite population correction? $\endgroup$ – Glen_b Nov 24 '14 at 23:46
  • $\begingroup$ That still looks wrong in at least two ways ... but it now looks even more like you're trying to do some kind of finite population correction. Please describe more accurately what you're trying to achieve. Why is $\sqrt n$ there? $\endgroup$ – Glen_b Nov 24 '14 at 23:48
  • $\begingroup$ Please, see again. I had changed the expression. $\endgroup$ – Jorge Nov 24 '14 at 23:48
  • $\begingroup$ What's $\sqrt n$ for? Are you confusing $s$ with the usual sample estimate of the standard error of the mean? $\endgroup$ – Glen_b Nov 24 '14 at 23:50
  • $\begingroup$ My expression is derivated from the $$ \sigma_{\bar{X}} = \dfrac{\sigma}{\sqrt{n}} \cdot \sqrt{\dfrac{N-n}{N-1}} $$ where i make $s =\sigma_{\bar{X}} $ $\endgroup$ – Jorge Nov 24 '14 at 23:53
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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).

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