Past exam of the options I course of the University of Oxford / 2008 / O / O1

For a two-way contingency table, Pearson's statistic isHere $O_{i+},O_{+j}$ are the observed margins and $N$ their grand total. Under the independence null, its usual large-sample reference distribution has $(r-1)(c-1)$ degrees of freedom for an $r$-by-$c$ table.

Under the null, the common fitted CC frequency is its pooled proportion $133/3874$. Multiplying by the fixed number of cases givesThis estimates the common type frequency from both groups, rather than using the case frequency alone.

With both column totals fixed, each column is a three-category multinomial sample. Under the null the two columns share one probability vector, which has two free parameters because its three probabilities sum to one. Under the alternative each has its own vector, giving four free parameters. Their difference is two, the reference degrees of freedom for the association test.

Use two degrees of freedom from part (iii). Since $8.42>5.99$, the statistic exceeds the supplied five-percent threshold. Thus reject no association at the five-percent level: the observed genotype frequencies provide evidence of a difference between cases and controls. The approximate p-value is $e^{-8.42/2}\approx 0.0148$ under the ${\chi }_2^2$ reference law.

The recessive alternative distinguishes CC from the other two types while assigning CT and TT the same disease-risk category. Pooling those two types therefore removes a distinction that this alternative does not use. The resulting counts are $83,1856$ for cases and $50,1885$ for controls, arranged as CC versus non-CC.

The resulting two-by-two table has $\boxed{(2-1)(2-1)=1}$ reference degree of freedom. Since $8.41>3.84$, reject no association at the five-percent level using the supplied chi-squared test. The observed CC proportion is higher among cases, $83/1939$, than among controls, $50/1935$, so the direction agrees with the proposed increased-risk category.