Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / O / 1 / iv / Solution

The null reference is ${\chi }_2^2$, because the alternative has two additional parameters. The supplied statistic satisfies $8.42>5.99$, so reject no association at the 5% level. For two degrees of freedom the upper-tail probability is $e^{-8.42/2}\approx 0.0148$.

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.