An observed-to-expected estimate of the low-versus-high relative failure risk is
$$
\frac{O_A/E_A}{O_B/E_B}
=\frac{3/1.6}{2/3.4}
=3.1875.
$$
Thus the low-dose failure hazard is estimated to be roughly $3.2$ times the high-dose hazard, with great uncertainty in this tiny dataset.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 207 / 4 / a / iv / Solution

 Solution

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