Variance of a Normal distribution and the F-testEdexcel A-Level Further Maths: Revision notes

If $X_1,\ldots,X_n$ is a random sample from $\mathrm{N}(\mu,\sigma^2)$ and $S^2$ is the sample variance, then$\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}.$The chi-squared distribution $\chi^2_\nu$ is positively skewed, takes only positive values and depends on its degrees of freedom $\nu=n-1$. Critical values come from $\chi^2$ tables or a calculator, for example the upper $5\%$ point of $\chi^2_{15}$ is $24.996$. This result needs the sample to come from a Normal population.

To test $H_0:\sigma^2=\sigma_0^2$, calculate $\frac{(n-1)s^2}{\sigma_0^2}$ and compare it with $\chi^2_{n-1}$. For $H_1:\sigma^2>\sigma_0^2$ reject if the statistic exceeds the upper critical value; for $H_1:\sigma^2<\sigma_0^2$ reject if it is below the lower critical value; for $H_1:\sigma^2\ne\sigma_0^2$ split the significance level between both tails. Example: $n=20$, $s^2=6.8$, $\sigma_0^2=4$: statistic $=\frac{19\times6.8}{4}=32.3$. The upper $5\%$ point of $\chi^2_{19}$ is $30.144$, so reject $H_0$ at $5\%$ in favour of $\sigma^2>4$.

A $(100-\alpha)\%$ confidence interval for $\sigma^2$ uses the lower and upper $\frac{\alpha}{2}\%$ points of $\chi^2_{n-1}$:$\frac{(n-1)s^2}{\chi^2_{\text{upper}}}<\sigma^2<\frac{(n-1)s^2}{\chi^2_{\text{lower}}}.$The upper $\chi^2$ point gives the lower limit and vice versa, because $\chi^2$ is in the denominator. Example: $n=10$, $s^2=8.4$, $90\%$: $\frac{75.6}{16.919}<\sigma^2<\frac{75.6}{3.325}$, i.e. $(4.47,\ 22.7)$. The interval is not symmetrical about $s^2$. A $(100-\alpha)\%$ interval matches a two-tailed test at the $\alpha\%$ level.

For independent samples from two Normal populations with equal variances, $\frac{S_1^2}{S_2^2}\sim F_{n_1-1,\,n_2-1}$, the F-distribution with numerator and denominator degrees of freedom. To test $H_0:\sigma_1^2=\sigma_2^2$, calculate $F=\frac{s_1^2}{s_2^2}$.

• For $H_1:\sigma_1^2>\sigma_2^2$ (one-tailed) compare with the upper $\alpha\%$ point of $F_{n_1-1,\,n_2-1}$.
• For $H_1:\sigma_1^2\ne\sigma_2^2$ (two-tailed at $\alpha\%$), put the larger variance on top (so $F>1$) and use the upper $\frac{\alpha}{2}\%$ point with the degrees of freedom in the matching order. Example: $s_1^2=14.4$ ($n_1=12$), $s_2^2=5.6$ ($n_2=9$): $F=2.57$ against the upper $5\%$ point of $F_{11,8}=3.313$. This is a two-tailed $10\%$ test, so $H_0$ is not rejected.