Hypothesis testing and confidence interval The log return $X$ on a certain stock investment is an $N(\mu,\sigma^2)$ random variable.
A financial analyst has claimed that the volatility $\sigma$ of the log return on this stock is less than $3$ units. A random sample of $11$ returns on this stock gave an estimated variance of the log-returns as $s^2 = 16$.

*

*Assess the analyst's claim by using a significance test at level $\alpha = 0.05$ to test
$H_0: \sigma^2 \leq 9 \text{ against } H_1: \sigma^2 > 9.$


*Find a two-sided $95\%$ confidence interval for $\sigma^2$.
 A: Since log returns is normally distributed the sample variance $S^2$ has a distribution that is proportional to a chi-square with $n-1$ degrees of freedom.  So precisely stated 
$(n-1)S^2/σ^2$ has a chi square distribution with $n-1$ degrees of freedom.  So for the above test use 
$(n-1)S^2/9$ as the test statistic and use the chi square distribution with $n-1$ degrees of freedom to get the test results.
In your case n-1 =10.
A: Since
$$
\frac{(n-1)S^2}{\sigma^2} = \frac{10 S^2}{\sigma^2} \sim \chi^2_{11-1} = \chi^2_{10},
$$
you have
$$
\Pr\left( A < \frac{(n-1)S^2}{\sigma^2} < B\right) = 0.95
$$
where $A$ and $B$ are so chosen that
$$
\Pr(\chi^2_{10}<A) = 0.025 = \Pr(\chi^2_{10}>B).
$$
(You can get $A$ and $B$ from tables or software.)
Via simple algebra it follows that
$$
\Pr\left( \frac 1 B < \frac{\sigma^2}{10S^2} < \frac 1 A   \right) = 0.95,
$$
whence
$$
\Pr\left( \frac {10S^2}{B} < \sigma^2 < \frac{10S^2}{A}   \right) =0.95.
$$
So there's your confidence interval.
If $9$ is less than the lower end of this confidence interval, you'd reject the null hypothesis at the 2.5% level.  So instead of half of 5% that we used above, use half of 10%, and that will tell you whether to reject the null hypothesis at the 5% level.
