e-Statistics

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 and the alternative hypothesis

$H_A:\hspace{0.05in}\sigma$ $\sigma_0 =$

forms a hypothesis test problem on whether we can reject ``

in favor of

.''

The test procedure is based upon the sample standard deviation and the sample size . The normality assumption is essential for the appropriateness of the test. That is, sample size 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 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 .

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

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