Testing a Standard Deviation

Here we are interested in the plausibility of the statement regarding the population standard deviation $ \sigma$ of a single variable. The null hypothesis $ H_0$ and the alternative hypothesis

$ H_A:\hspace{0.05in}\sigma$ $ \sigma_0 =$
forms a hypothesis test problem on whether we can reject ``$ H_0$ in favor of $ H_A$.''

The test procedure is based upon the sample standard deviation $ S =$ and the sample size $ n =$ . The normality assumption is essential for the appropriateness of the test. That is, sample size $ n$ is adequately large ($ n \ge 30$), or the sample distribution has a small sample size but is approximately normal. Then the test statistic $ \chi^2 = \frac{(n-1)S^2}{\sigma_0^2}$ is likely observed around, greater than, or less than the mean value $ (n-1)$ of chi-square distribution under the respective null hypothesis `` $ \sigma = \sigma_0$,'' `` $ \sigma \ge \sigma_0$,'' or `` $ \sigma \le \sigma_0$.'' The opposite of such observation is expressed by p-value < $ \alpha$, suggesting an evidence against the null hypothesis $ H_0$.

When the null hypothesis is rejected it is reasonable to find out the confidence interval for the population standard deviation $ \sigma$.

$ \displaystyle\left(
\sqrt{\frac{(n-1) S^2}{\chi_{\alpha/2,n-1}}}
\sqrt{\frac{(n-1) S^2}{\chi_{1-\alpha/2,n-1}}}
\right) =$ ( , )