# Finding the value of a sample statistic using chi-squared distribution?

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.

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

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 $$Var\left( S^2 \right)$$.