Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / O / 3

For a normal sample with unknown mean and variance, the sample mean and sample variance are independent, with $\sqrt{n}(\overline{X}-\mu )/\sigma \sim N(0,1)$ and $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$. Therefore the ratio defining the Studentized normal sample mean hasusing $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$.

Let $t_{\nu ,\beta }$ denote a quantile of the Student t-distribution with $\nu =n-1$ degrees of freedom. Symmetry gives $\mathbb{P}(|Y|\le t_{n-1,1-\alpha /2})=1-\alpha$. Solving for $\mu$ gives the exact central confidence interval

Under $H_0:\mu =0$, the statistic $T=\sqrt{n}\overline{X}/S$ has Student t-distribution with $n-1$ degrees of freedom. Reject in either tail, splitting the null error probability equally. Thus the critical region isThe distribution is continuous, so the boundary convention does not change the size $\alpha$.

Let $F_{n-1}$ be the cumulative distribution function of the Student t-distribution. The two-sided p-value is $2[1-F_{n-1}(|t_0|)]$. If it is less than $\alpha$, strict monotonicity gives $|t_0|>t_{n-1,1-\alpha /2}$, henceThe magnitude of the interval's centre is larger than its half-width, so zero lies outside the confidence interval. This is the duality between confidence intervals and hypothesis tests.