 Solution

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.