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I understand that given a sample of data, the following formula can be used to estimate the population variance.

$\displaystyle S^{2} = \frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}$

However, I was asked this question. When a sample is taken from the population, does the sample variance follow that of the population? Is it still more accurate to use $\frac{1}{n-1}$ or we should use $\frac{1}{n}$ instead?

Edit 1: Sorry if the question wasnt clear enough. What I meant is that if we try to find out the sample variance (a random sample from the population) given the population variance, do we take it as such or do we have to adjust it using $\frac{n}{n-1}$?

Thank you.

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    $\begingroup$ What do you mean by "sample is taken from the population"? $\endgroup$ – MAN-MADE Jul 3 '17 at 5:49
  • $\begingroup$ may be the given link help you, if I am understanding the question correctly. math.stackexchange.com/questions/2301265/… $\endgroup$ – MAN-MADE Jul 3 '17 at 5:51
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    $\begingroup$ If you aren't sampling, but instead you are measuring the whole population, then the population variance is defined as $\sigma^{2}=\frac{1}{N}\sum_{i=1}^{N}(x_{i}-\mu)^{2},$ where I'm changing the notation to make it clear that we're talking about a population with measurements $x_{1},\ldots,x_{N}.$ $\endgroup$ – RideTheWavelet Jul 3 '17 at 6:12
  • $\begingroup$ Sorry if the question wasnt clear enough. What I meant is that if we try to find out the sample variance given the population variance, do we take it as such or do we have to adjust it using $\frac{n}{n-1}$? $\endgroup$ – chowsai Jul 3 '17 at 7:53
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If $X_1, X_2, \dots, X_n$ is a random sample from a population with variance $\sigma^2,$ then the sample variance $S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i- \bar X)^2$ has $E(S^2) = \sigma^2,$ so that $S^2$ is an unbiased estimate of $\sigma^2.$ If $n$ is used in the denominator instead of $n - 1$, then $S^2$ is biased, but the bias decreases with increasing $n.$

Furthermore, if the population is $\mathsf{Norm}(\mu,\sigma),$ then $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(df = n-1).$

Thus, if $L$ and $U$ are chosen (from software or printed tables of the chi-squared distribution) so that $P\left(L \le \frac{(n-1)S^2}{\sigma^2} \le U\right) = 0.95,$ then (after manipulation of the inequality) a 95% confidence interval for $\sigma^2$ is of the form $\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L}\right).$

[Note: Technically, there are various ways to define the 'accuracy' of such an estimate, and unbiasedness is not the only criterion for accuracy. It has been argued that according to some criteria for accuracy and for some population distributions, it might be better to use $n$ or, even $n+1,$ in the denominator. However, I'm guessing these are more advanced considerations than you have in mind.]

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  • $\begingroup$ Thank you for your insightful answer. So if we try to find out the sample variance given the population variance, do we take it as such or do we have to adjust it using $\frac{n}{n-1}$? $\endgroup$ – chowsai Jul 3 '17 at 7:58
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    $\begingroup$ If you have data, you would just use the formula to get get $S^2$. If you know the population $\sigma^2$ and are trying to guess in advance of taking data what $S^2$ will be, then $\sigma^2$ is your best guess. For sample variances always use (n-1) in the denominator. // If you need to find $\sigma^2$ for a finite population of $N$ known values, you would use $N$ in the denominator as in the Comment of @RideTheWavelet. $\endgroup$ – BruceET Jul 3 '17 at 8:07

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