Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 4 / 19C (Statistics)

Write the common variance as ${\sigma }^2$. Independence and the normal distribution giveMoreover $\bar{X}$ and $S_{XX}$ are independent. To see both assertions, standardize the observations into an isotropic Gaussian vector and choose an orthonormal basis with first vector $(1,\dots ,1)/\sqrt{m}$. Its first coordinate is $\sqrt{m}(\bar{X}-{\mu }_X)/\sigma$; the remaining independent standard normal coordinates have squared sum $S_{XX}/{\sigma }^2$. The same argument applies to the $Y$ sample, independently of the first sample.

Under the null hypothesis, the difference of sample means is $N(0,{\sigma }^2(1/m+1/n))$. The residual sums of squares add to an independent chi-squared distribution with $\nu =m+n-2$ degrees of freedom. Define the pooled sample variance and test statistic byFor $\nu >0$, under $H_0$ this is $Z/\sqrt{U/\nu }$ with independent $Z\sim N(0,1)$ and $U\sim {\chi }_{\nu }^2$. Hence $T\sim t_{m+n-2}$ under the null, the Student t-distribution.

Let $t_{\nu ,0.99}$ be its 99th percentile. For the specified one-sided alternative, reject $H_0$ precisely whenBecause this distribution is continuous, the rejection probability under $H_0$ is exactly $0.01$, irrespective of the nuisance mean and variance. Large positive values supply evidence for ${\mu }_X>{\mu }_Y$; a two-sided or 0.995 cutoff would test a different alternative. The common-variance and independence assumptions are essential to this exact Student t-test. If both samples have size one, there are no residual degrees of freedom and the proposed unknown-variance test cannot be formed.