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
$$
\min_{n_0,n_1>0}F(n_0,n_1)
\quad\text{subject to}\quad
\frac a{n_0}+\frac b{n_1}=C.
$$
The <Lagrange multiplier> stationary equations determine the ratio; the second-order condition requires the constrained stationary point to be a local minimum.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-207/2/b/i/solution