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This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write

$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}N}\right)$$

Comparing the correction factor with the formula provided in the CV post mentioned in the comment as a possible duplicate-maker of this one, Explanation of finite population correction factor?, there appears as $\left(\sqrt {\frac {N-n}{N-1}}\right)$. Why this difference in the denominator?

@chl answer there mentions

You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.

...which needs to be reconciled with @whuber's answer.

This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write

$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}N}\right)$$

Comparing the correction factor with the formula provided in the CV post mentioned in the comment as a possible duplicate-maker of this one, Explanation of finite population correction factor?, there appears as $\left(\sqrt {\frac {N-n}{N-1}}\right)$. Why this difference in the denominator?

@chl answer there mentions

You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.

...which needs to be reconciled with @whuber's answer.

This result holds specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write

$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}N}\right)$$

Comparing the correction factor with the formula provided in the CV post mentioned in the comment as a possible duplicate-maker of this one, Explanation of finite population correction factor?, where there appears as $\left(\sqrt {\frac {N-n}{N-1}}\right)$, @chl answer there mentions

You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.

This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write

$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}N}\right)$$

Comparing the correction factor with the formula provided in the CV post mentioned in the comment as a possible duplicate-maker of this one, Explanation of finite population correction factor?, there appears as $\left(\sqrt {\frac {N-n}{N-1}}\right)$. Why this difference in the denominator?

@chl answer there mentions

You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.

...which needs to be reconciled with @whuber's answer.

This result holds specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write

$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}N}\right)$$

Comparing the correction factor with the formula provided in the CV post mentioned in the comment as a possible duplicate-maker of this one, Explanation of finite population correction factor?, where there appears as $\frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}{N-1}}\right)$, @chl answer there mentions

You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.

This result holds specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write

$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}N}\right)$$

Comparing the correction factor with the formula provided in the CV post mentioned in the comment as a possible duplicate-maker of this one, Explanation of finite population correction factor?, where there appears as $\left(\sqrt {\frac {N-n}{N-1}}\right)$, @chl answer there mentions

You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.