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

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.