Example 1.7. Survival times are recorded for ten patients. Test \[H_0 : \theta = 200 \qquad \text {against}\qquad H_1 : \theta \neq 200\] at the \(5\%\) level, where \(\theta \) is the population median. Suppose three of the ten observations exceed \(200\).

The statistic. Count the observations above \(200\), recording each as a “\(+\)”. By the definition of the median, under \(H_0\) each observation is above \(200\) with probability \(\tfrac 12\), independently. So the number \(R\) of “\(+\)” signs satisfies \[R \sim B\!\left (10, \tfrac 12\right ), \qquad P(R = r) = \binom {10}{r}\left (\tfrac 12\right )^{10}.\]

The null distribution.

The distribution is symmetric about \(r = 5\), as it must be when \(P = \tfrac 12\).

The test. With \(R = 3\) observed, the evidence against \(H_0\) in the lower tail is \[P(R \leq 3) = \sum ^{3}_{r=0}\binom {10}{r}\left (\tfrac 12\right )^{10} = \dfrac {1+10+45+120}{1024} = \dfrac {176}{1024} = 0.1719 .\] The alternative is two-sided, and by symmetry \(P(R \geq 7) = P(R \leq 3)\), so \[P\text {-value} = 2(0.1719) = 0.3438 .\]

Decision. Since \(0.3438 > 0.05\) we do not reject \(H_0\). There is no evidence at the \(5\%\) level that the median differs from \(200\).