Solution
= Solution
{c}
The null reference is $\chi^2_2$, because the alternative has two additional parameters. The supplied statistic satisfies $8.42>5.99$, so \b[reject no association at the 5% level]. For two degrees of freedom the upper-tail probability is $e^{-8.42/2}\approx0.0148$.
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.