e-Statistics

Testing a Standard Deviation

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. 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 statistic

$\chi^2 = \frac{(n-1)S^2}{\sigma_0^2}$ =

is likely observed around the mean value of chi-square distribution under the respective null hypothesis “ $\sigma = \sigma_0$ .” The opposite of such observation is expressed by

the p-value =

being less than $\alpha$ , suggesting an evidence against the null hypothesis in favor of .

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

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