Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 2 / b / i

Table of contents
- Solution i

Solution

The expected number of failures is

F (n_{0}, n_{1}) = n_{0} (1 - p_{0}) + n_{1} (1 - p_{1}) .

Holding the alternative and test size fixed, constant power is equivalent to fixing

V (n_{0}, n_{1}) = C

. The ethical allocation problem is therefore

min_{n_{0}, n_{1} > 0} F (n_{0}, n_{1}) subject to \frac{a}{n _{0}} + \frac{b}{n _{1}} = C .

The Lagrange multiplier stationary equations determine the ratio; the second-order condition requires the constrained stationary point to be a local minimum.

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