Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 32 / 5 / i / Solution

At day 165 there are two A and two B individuals at risk. The event is in A, with expected A count $2/4=1/2$, so its score contribution is $1-1/2=1/2$ and its variance contribution is $1/4$. The remaining A individual is censored at 173 and leaves the risk set before either later event. The risk-set calculations are

ThusThe positive score indicates higher A event hazard in this late contribution. Although B has more observed events, its two events occur with no A individual at risk, so those times do not compare the groups. This is why raw event totals alone are insufficient for the log-rank test.