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The problem is

Find the variance $S^2$ for random sample of size 21 from a normal population with variance 5.

(Hint: Use the fact that the statistic $\frac{(n-1)S^2}{\sigma^2}$ has a Chi-squared distribution with n-1 degrees of freedom, for a normal population with variance $\sigma^2$)

So I get that the statistic $\frac{(21-1)S^2}{5}$ has a Chi-squared distribution with 20 degrees of freedom. But I'm not sure how to proceed from there since I don't get how the Chi-squared distribution is related to getting the particular variance value.

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  • $\begingroup$ $S^2$ is random, but perhaps they want its mean using $X\sim\chi_\nu^2\implies\Bbb EX=\nu$. $\endgroup$
    – J.G.
    Commented Apr 30, 2020 at 11:36
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    $\begingroup$ Assuming you want $Var(S^2)$, use the variance of chi-square distribution. $\endgroup$
    – StubbornAtom
    Commented Apr 30, 2020 at 12:45

1 Answer 1

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Hint:

It is asked for $Var\left(S^2 \right)$. And we know that $\frac{(n-1)\cdot S^2}{\sigma^2} \sim \chi_{n-1}^2$ with $Var\left(\chi_{n-1}^2\right)=2(n-1)$. Thus

$$Var\left(\frac{(n-1)\cdot S^2}{\sigma^2} \right)=2\cdot (n-1),$$

where $(n-1)$ and $\sigma^2$ are constants.

$$\frac{(n-1)^2}{\sigma^4}Var\left( S^2 \right)=2\cdot (n-1)$$

Go on and solve the equation for $Var\left( S^2 \right)$.

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