Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / O (Statistics)

For observed counts $O_{ij}$ in an $r$ by $c$ contingency table, Pearson's chi-squared test statistic for independence isUnder the null model and the usual large-expected-count conditions, its reference distribution is ${\chi }^2$ with $(r-1)(c-1)$ degrees of freedom.

The pooled CC count is $83+50=133$, the case total is 1939, and the grand total is 3874. Under the null, the common estimated CC proportion is $133/3874$. Multiplication by the fixed case total givesThis is the fitted expected count, rather than a separate observation.

Under the null, the two groups share one vector of three category probabilities. The sum-to-one constraint leaves two free parameters. Under the unrestricted alternative, each group has its own such probability vector, leaving four free parameters. The fixed case and control totals introduce no further parameters; the difference is two, giving the chi-squared reference degrees of freedom.

The null reference is ${\chi }_2^2$, because the alternative has two additional parameters. The supplied statistic satisfies $8.42>5.99$, so reject no association at the 5% level. For two degrees of freedom the upper-tail probability is $e^{-8.42/2}\approx 0.0148$.

The result is evidence that genotype proportions differ between cases and controls. It is a statistical association test; it does not by itself identify a causal mechanism.

Under the recessive alternative, CT and TT have the same status as non-CC individuals, so their separate labels do not distinguish the proposed risk groups. Pooling them makes the intended comparison CC versus non-CC. The table becomesThe independence test on this two-by-two table has one degree of freedom. Since $8.41>3.84$, reject no association at the 5% level; its chi-squared upper-tail probability is approximately $0.0037$. CC is more frequent among cases, consistent with the stated directional alternative. The supplied Pearson statistic is conventionally compared with its one-degree-of-freedom chi-squared reference.

Differentiating the least-squares residual norm gives the normal equationsIf $X$ has full column rank, $\hat{\theta }=(X^{T}X{)}^{-1}{X}^{T}Y$, and thereforeHere $\bar{x}=0$ and $\bar{z}=2$, so the design matrix and its Gram matrix areThe centered predictor columns are perpendicular, givingBoth estimates are linear combinations of the jointly normal errors, hence jointly normal. Zero covariance therefore also gives independence. The normality assumption is what permits this last step; orthogonality alone would only prove uncorrelatedness.

The Neyman-Pearson lemma says that for testing a simple null density $f_0$ against a simple alternative $f_1$, a size-$\alpha$ test that rejects for the largest values of $f_1/f_0$ is most powerful. Randomization on a likelihood-ratio boundary can be used to achieve the exact size.

The source writes the same ${\lambda }_0$ in both hypotheses but immediately introduces ${\lambda }_1>{\lambda }_0$. Use the intended alternative $\lambda ={\lambda }_1$. If read literally as two identical hypotheses, there is no such distinct alternative to test.

For Weibull shape two, the sample likelihood is proportional toThuswhich is increasing in $S={\sum }_i{X}_i^2$. The most powerful critical region is thereforeIndeed, $2S/{\lambda }_0^2\sim {\chi }_{2n}^2$, so $C=({\lambda }_0^2/2)q_{2n,1-\alpha }$. The critical value depends only on the null and size, not on the chosen larger alternative. Hence the same test is uniformly most powerful for $\lambda >{\lambda }_0$, and in particular for the printed restricted range $\lambda >{\lambda }_1$.