For observed counts in an by contingency table, Pearson's chi-squared test statistic for independence is
Under the null model and the usual large-expected-count conditions, its reference distribution is with degrees of freedom.
The pooled CC count is , the case total is 1939, and the grand total is 3874. Under the null, the common estimated CC proportion is . Multiplication by the fixed case total gives
This is the fitted expected count, rather than a separate observation.
Under the null, the two groups share one vector of three category probabilities. The sum-to-one constraint leaves two free parameters. Under the unrestricted alternative, each group has its own such probability vector, leaving four free parameters. The fixed case and control totals introduce no further parameters; the difference is two, giving the chi-squared reference degrees of freedom.
The null reference is , because the alternative has two additional parameters. The supplied statistic satisfies , so reject no association at the 5% level. For two degrees of freedom the upper-tail probability is .
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.
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
The independence test on this two-by-two table has one degree of freedom. Since , reject no association at the 5% level; its chi-squared upper-tail probability is approximately . 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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