Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 20H (Statistics)

The sample mean isWrite the Gaussian sample vector as its orthogonal projection onto the span of $(1,\dots ,1)$ plus its projection onto the orthogonal complement. These two Gaussian projections are independent. Their squared standardized lengths give Cochran's theorem:

Combining the independent standard normal variable $\sqrt{n}(\overline{X}-\mu )/\sigma$ with the independent chi-squared variable $S_{XX}/{\sigma }^2$ givesthe Student's t-distribution with $n-1$ degrees of freedom.

Let $z_p$ denote the $p$-quantile of the standard normal distribution. Since the variance is known,so a $100(1-\alpha )\%$ confidence interval is

Let $t_{\nu ,p}$ denote the $p$-quantile of Student's t-distribution. Part (b) gives the Student t confidence interval

Let ${\chi }_{\nu ,p}^2$ denote the $p$-quantile of the chi-squared distribution. Since $S_{XX}/{\sigma }^2\sim {\chi }_{n-1}^2$, inversion of the central probability event givesas a $100(1-\alpha )\%$ confidence interval for ${\sigma }^2$.

For an independent second sample, letUnder $H_0:{\sigma }^2={\tilde{\sigma }}^2$, independence and the two chi-squared laws give the F-distributionFor the one-sided alternative ${\sigma }^2>{\tilde{\sigma }}^2$, reject for large $F$, using the upper $\alpha$ critical value.

No values of the unknown means enter this statistic: centering by the sample means removes them. If both means are known, one may instead use sums about the known means; the corresponding degrees of freedom are $n$ and $\tilde{n}$, rather than $n-1$ and $\tilde{n}-1$. Thus knowledge of the means changes the degrees of freedom and critical value, while the variance-ratio principle is unchanged.