Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 19H (Statistics)

Assume the explanatory variables are not all equal, so $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2>0$. In the Gaussian linear model, the log-likelihood isFor fixed ${\sigma }^2$, maximum likelihood estimation therefore minimizes the residual sum of squares. Differentiating in $\alpha ,\beta$ gives the least-squares normal equationsThe first gives $\hat{\alpha }=\bar{Y}-\hat{\beta }\bar{x}$; substituting it into the second gives $\hat{\beta }=S_{xy}/S_{xx}$, with $S_{xy}={\sum }_i(x_i-\bar{x})(Y_i-\bar{Y})$. Let $\mathrm{SSE}={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$. Differentiating the profiled log-likelihood in ${\sigma }^2$ givesFor the variance estimator this is the interior maximum when $\mathrm{SSE}>0$, which occurs almost surely under the model when $n>2$.

The coefficient estimators are linear functions of the independent normal errors. In particular,Their multivariate normal distribution has mean $(\alpha ,\beta {)}^T$ and covariance matrixThe fitted vector is the orthogonal projection onto the two-dimensional span of $(1,\dots ,1{)}^T$ and $(x_1,\dots ,x_n{)}^T$. The residual vector is the complementary projection of the spherical normal error vector. Orthogonal normal projections are independent, and in an orthonormal basis of that complementary space its $n-2$ coordinates are independent $N(0,{\sigma }^2)$. Consequentlyindependently of $(\hat{\alpha },\hat{\beta })$. This also gives the unbiased estimator $s^2=\mathrm{SSE}/(n-2)$ rather than the variance MLE.

Under the null hypothesis $\alpha =0$, for $n>2$ the Student t test for a simple-regression intercept usesHere $\eta$ is the prescribed significance level. The normal numerator divided by its true standard deviation is independent of $(n-2)s^2/{\sigma }^2$, which proves the stated Student t-distribution. If $S_{xx}=0$, the two coefficients are not separately identifiable; if $n\le 2$, this residual-variance-based test is unavailable.

Set $x_i=\sin (2{\theta }_i)$. The maximum likelihood estimation formulas for simple linear regression giveUsing the supplied rounded summaries gives $\hat{\beta }\approx 21082.7$ metres and $\hat{\alpha }\approx 312.0$ metres. Recomputing directly from the printed table gives approximately $21082.4$ and $312.0$ metres, respectively, so the small summary rounding differences do not affect the conclusion.

I would not accept the model merely because the Student t-test fails to reject a zero regression intercept. That hypothesis test tests one parameter restriction within the assumed model, not the adequacy of its error assumptions or its functional form. In particular, complementary elevations $\theta$ and $90^{\circ }-\theta$ have exactly the same $x_i$ and hence the same fitted mean, but their observed low-angle minus high-angle ranges arefor the four corresponding pairs. All four are positive, suggesting an elevation-dependent effect omitted by the mean model, such as air resistance, rather than random scatter determined only by $\sin (2\theta )$. This is a regression diagnostics concern, not a claim that these finite data logically rule out the normal-error model. The correct conclusion is nonrejection of the intercept restriction does not validate the regression model; further checks of the residual pattern and error assumptions are needed.