 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-41/2/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

24/99

and

12/100

OR = \frac{24 \cdot 88}{75 \cdot 12} = 2.34667, δ = lo g OR = 0.852996.

Var (δ) = \frac{1}{24} + \frac{1}{75} + \frac{1}{12} + \frac{1}{88} = 0.149697, SE (δ) = 0.386907.

The stated normal approximation and quantile approximately two give the confidence interval

δ \in 0.852996 \pm 2 (0.386907) = (0.07918, 1.62681) .

Exponentiation gives the corresponding odds ratio interval

(1.0824, 5.0876)

. Its lower endpoint exceeds one.

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