The chi squared test for the sample variance is:

$\chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}}$

Say that we do not know $\sigma^{2}$. In these cases it is common to use an estimator $s^{2}$, since $s^{2} \rightarrow \sigma^{2}$ in probability. Let us then swap $\sigma^{2}$ with $s^{2}$

$\chi^{2} = \frac{(n-1)s^{2}}{s^{2}}$

And with simple algebra, this leaves us with:

$ \chi^{2} = n-1 $

So, when we don't now the population variance, then why go through all the effort of calculating this chi square value when we can just use n-1?

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    $\begingroup$ Your second line gives $n-1$, and you have no need for the rest $\endgroup$
    – Henry
    Jul 29, 2021 at 23:59
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    $\begingroup$ If you are using this to test the variance, the null hypothesis is typically that $\sigma^2$ is a specified value, and you are testing how likely the observed sample variance (or something more extreme) would be seen if the null hypothesis were true. It makes no sense to estimate $\sigma^2$ from $s^2$ in such a test. $\endgroup$
    – Henry
    Jul 29, 2021 at 23:59
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    $\begingroup$ Your initial formula is wrong; it should have denominator $\sigma^2$ (population variance), not $S^2$ (sample variance. $\endgroup$
    – BruceET
    Jul 30, 2021 at 0:32
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    $\begingroup$ $s^2/s^2$ is obviously not identical to $s^2/\sigma^2$ - no mystery remains. $\endgroup$
    – Glen_b
    Jul 30, 2021 at 3:26
  • $\begingroup$ It is my understanding that $s^{2}$ is an uncontroversial way to estimate $\sigma^{2}$. Thanks for @Henry for clarifying that when using this test the population variance should be known. As seen in when calculating a t-statistic, it is commonly accepted to use $\bar{x}$ to estimate $\mu$, so I thought that the same would apply here. Is there a test that is used when the population variance is unknown? When I search for hypothesis tests for a difference of variances I can only find $\chi^{2}$ $\endgroup$ Jul 30, 2021 at 4:50

1 Answer 1


Perhaps an example will help. Let's test if the population variance, which we do not know, is equal to $1$.

$$ H_0: \sigma^2 = 1\\H_a: \sigma^2\ne 1 $$

Under this null hypothesis, we posit that $\sigma^2=1$, and we calculate the $\chi^2$ statistic that we can use to calculate p-value or compare to reference values.

$$\chi^2 = \dfrac{(n-1)s^2}{1} $$

The key is that we must posit that the population variance has some value. In my example, I posited that the population variance was $1$. Perhaps you have a different null hypothesis you want to test. As an exercise, you might want to calculate the $\chi^2$ value for a null hypothesis of $H_0: \sigma^2 = 4$ for the following data: $1, 2, 3$.

  • $\begingroup$ Oh gosh I see I just made a very fundamental mistake in the role of $\sigma^{2}$ in the Chi square test, I hadn't realized it was hypothesized in this context. Thank you for clearing that up $\endgroup$ Aug 2, 2021 at 21:37
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    $\begingroup$ It's the same as how you hypothesize some null value in other tests, like the t-test or a proportion test (to name two with which you might be familiar). $\endgroup$
    – Dave
    Aug 2, 2021 at 21:41

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