Questions: Log-Rank Test for Survival Comparison

Question 1

Two Kaplan-Meier survival curves for treatments A and B cross at 12 months — Treatment A is better early but Treatment B is better late. A log-rank test yields p = 0.42. A colleague concludes the treatments are equivalent. What is the problem?

• A. The log-rank test lacks power to detect any difference — a larger sample is needed
• B. The log-rank test averages the difference over time; when curves cross, the early advantage and late advantage cancel out, masking a real difference
• C. The p-value of 0.42 definitively proves equivalence
• D. The log-rank test only works for three or more groups

Answer: (B)

The log-rank test is designed to detect a consistent difference in survival (proportional hazards). When curves cross, one treatment is better early and the other better late — the positive and negative differences cancel, producing a small test statistic even though the treatments behave very differently. Alternative tests like the Wilcoxon (Breslow) test weight early events more heavily, or time-dependent analyses can explicitly model the crossing. A non-significant log-rank test with crossing curves does not mean equivalence — it means the test was inappropriate for that pattern.

Question 2

The log-rank test compares observed to expected events at each event time. Under the null hypothesis, how are expected events computed?

• A. Expected events equal the total events divided equally among groups
• B. Expected events for each group are proportional to the number at risk in that group at each event time
• C. Expected events are computed from a parametric survival distribution fitted to the combined data
• D. Expected events are the historical rates from previous studies

Answer: (B)

Under the null hypothesis of identical survival, the probability of an event in any group is proportional to that group's share of the risk set. If 60 of 100 subjects at risk at time t are in Group A, and 3 events occur at that time, the expected number of events in Group A is 3 × (60/100) = 1.8. This is computed at every event time, and the observed-minus-expected differences are summed to form the test statistic. The approach is analogous to the Mantel-Haenszel method for combining 2×2 tables across strata.

Question 3

The log-rank test requires the proportional hazards assumption — that the ratio of hazard rates between groups remains constant over time.

• T. True
• F. False

Answer: True

The log-rank test is most powerful under proportional hazards because it gives equal weight to all event times. When the hazard ratio is constant, the cumulative difference between observed and expected events grows consistently over time, maximizing the test statistic. If hazards cross (one group has higher early hazard, the other higher late hazard), the contributions at different times cancel, reducing power. The log-rank test is valid under non-proportional hazards (the Type I error rate is still controlled), but it may fail to detect real differences — it is a question of power, not validity.

Question 4

Explain what the log-rank test statistic measures and why it follows a chi-squared distribution.

Model answer: The test statistic sums the squared differences between observed and expected events in each group, divided by the variance of the expected events, across all event times. Under the null hypothesis that survival is identical across groups, the observed-minus-expected differences at each event time are approximately independent with known variance (from the hypergeometric distribution). By the central limit theorem, the sum of many such standardized differences converges to a chi-squared distribution with degrees of freedom equal to the number of groups minus one.

The construction is analogous to a Pearson chi-squared test applied repeatedly at each event time and summed. With two groups, the test statistic has 1 degree of freedom. With k groups, it has k-1 degrees of freedom. The hypergeometric distribution at each event time reflects the fact that the allocation of events to groups, given the total events and the risk set sizes, follows this distribution under the null.