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The correction factor for finite populations is $\sqrt{1-\frac{n}{N}}$ (for known variance) where $n$ is the sample size and $N$ is the population size.

I wonder where the square root comes from. Let me explain with an example for the aritmetic average used to estimate the mean. I compute $\bar{x}=\sum_{i=1} ^n x_i$ from the sample and use this to find the population total as $T=N\bar{x}$.

I can write $T$ as $T=n\bar{x}+(N-n)\bar{x}$. The first term in the right hand side is the sample total, where I have no estimation error, so it is not estimated and there is no estimation error on it. For the second term, the variance is $(N-n)^2\frac{\sigma^2}{n}$. So the let's call it 'correct variance' is $0+(N-n)^2\frac{\sigma^2}{n}$.

If I would compute the variance for the 'infinite population' extrapolation then I would use $N^2\frac{\sigma^2}{n}$, in order to find the ''finite population correction factor'' that reduces the latter variance to the former I have to multiply the latter by $\frac{(N-n)^2}{N^2}$ and for the standard deviation I find that the correction factor is $FPC=\frac{N-n}{N}=1-\frac{n}{N}$.

So I wonder where the square root comes from ?

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    $\begingroup$ Hint: exactly what statistic is this correction factor being applied to? Is there a square root anywhere in its formula? $\endgroup$
    – whuber ♦
    Aug 20, 2015 at 14:59
  • $\begingroup$ @whuber: it is applied to the standard deviation and yes, there is a square root in its definition? Is that the reason? It is not derived from one or other property or theorem than? $\endgroup$
    – user83346
    Aug 20, 2015 at 15:17

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