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

Under the recessive alternative, CT and TT have the same status as non-CC individuals, so their separate labels do not distinguish the proposed risk groups. Pooling them makes the intended comparison CC versus non-CC. The table becomesThe independence test on this two-by-two table has one degree of freedom. Since $8.41>3.84$, reject no association at the 5% level; its chi-squared upper-tail probability is approximately $0.0037$. CC is more frequent among cases, consistent with the stated directional alternative. The supplied Pearson statistic is conventionally compared with its one-degree-of-freedom chi-squared reference.