Single Sample Inferences about the Population Variance: $H_{0}: \sigma^{2}=\sigma_{0}^{2}$

In the previous sections we were only concerned about population means exclusively. In some cases, we, however would be interested to make claims about the dispersion (i.e. the variance) of the populations instead.

For instance, some medications or consumer products (e.g. alcohol) might make some people much better at performing a certain task, while others worse. The opposing difference in the performance between these two subpopulations could end up cancelling each other out, thus resulting in no change in the mean; the variance, on the other hand, might change significantly. In this section we will introduce a method for testing hypotheses regarding the population variance.

Mathematically, we could suggest the following experimental set up:

Now suppose that we have a sample $X_{1}, X_{2}, \ldots, X_{n}$ drawn from a Normal population with distribution $N\left(\mu, \sigma^{2}\right)$, where we know neither of the parameters. We defer from providing a formal proof here here, but a well-defined result in statistics claims that the test statistic $T=\frac{(n-1) S^{2}}{\sigma_{0}^{2}}$, where $S^{2}$ is the sample variance, follows the $\chi^{2}$ (read this as: "chi-square") distribution with $n-1$ degrees of freedom, given the assumed null hypothesis is true:

The critical values for $\chi^{2}$ distribution can also be looked up in the statistical tables, just like for any other statistical test described in this chapter. Note for the two-tailed test case (i.e. simple alternative hypothesis) that the critical values are not symmetrical!

For instance, for a test with $\alpha=0.01$ and the sample size $n=17$, the appropriate critical values for $\chi_{16}^{2}$ are $c_{l}=5.14$ and $c_{r}=34.27$, and therefore we would use the following decision rule: "reject $H_{0}$ if $T<c_{l}$ or $T>c_{r} "$.