chi square distribution probability

Question

I am having a problem with this.

Suppose a stock's returns are normally distributed with mean $m$ and variance $\alpha^2$ and we compute the sample variance from a sample of $41$ periods and find that $s^2 = 95$.

If the true variance $\alpha^2 = 64$, would our estimate be in the right most $5$ percent tail of the sampling distribution?

I got that I should use chi square distribution but it's confusing whether it is $\textrm{P}[(41-1)95/64]$ $>$ or $< \dfrac{55.758}{40}$

Can someone clarify this for me?

Thanks

Answer 1

The random variable $(n-1)\frac{s^2}{\alpha^2}$ is $\chi^2(n-1)$ distributed and $\mathbb{P}(\chi^2(n-1)>55.758)=0.05$

$(n-1)\frac{s^2}{\alpha^2}=59.375>55.758$. So $\mathbb{P}(s^2=95)<0.05$.