Solution
= Solution
Use two <degrees of freedom> from part (iii). Since $8.42>5.99$, the statistic exceeds the supplied five-percent threshold. Thus \b[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 $e^{-8.42/2}\approx0.0148$ under the $\chi^2_2$ reference law.