= Solution
{c}
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 becomes
$$
\begin{array}{c|rr}
&\text{cases}&\text{controls}\\ \hline
\text{CC}&83&50\\
\text{non-CC}&1856&1885
\end{array}.
$$
The independence test on this two-by-two table has \b[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.
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