Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 19H / b

Put ${\mathrm{RSS}}_1={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$ for the unrestricted fit. Under $H_0$, the best slope is still $\hat{\beta }$ because the design is centered. Orthogonality of the residuals to the constant vector givesAt either fitted mean, the maximum-likelihood estimator of variance is $\mathrm{RSS}/n$. The ratio used by the generalized likelihood-ratio test is consequentlySmall likelihood ratios correspond to large $|T_n|$. Under $H_0$, part (a) gives independent $\sqrt{n}\overline{Y}/\sigma \sim N(0,1)$ and ${\mathrm{RSS}}_1/{\sigma }^2\sim {\chi }_{n-2}^2$. Thus $T_n$ has the Student's t-distribution with $n-2$ degrees of freedom. The exact size-5% critical region iswhere $t_{\nu ,0.975}$ is the 97.5th percentile of that distribution. The denominator uses the unbiased residual variance estimate, not the likelihood variance denominator $n$.

Under an alternative, the same statistic has the representationThe joint law of $Z,W$ is parameter-free, and $\beta$ has disappeared. Thereforewhere $g(d)=\mathbb{P}(|(Z+\sqrt{n}d)/\sqrt{W/(n-2)}|>t_{n-2,0.975})$. This is also the noncentral t-distribution description of the power. As explained in part (a), an exact residual-based test requires $n>2$.