11.4 The Log-Rank Test

We now continue our analysis of the BrainCancer data introduced in Section 11.3. We wish to compare the survival of males to that of females. Figure 11.3 shows the Kaplan–Meier survival curves for the two groups. Females seem to fare a little better up to about 50 months, but then the two curves both level off to about 50%. How can we carry out a formal test of equality of the two survival curves?

At first glance, a two-sample t -test seems like an obvious choice: we could test whether the mean survival time among the females equals the mean survival time among the males. But the presence of censoring again creates a complication. To overcome this challenge, we will conduct a log-rank test ,[4]

log-rank test

11.4 The Log-Rank Test 475

	Group 1 Group 2	Total
Died Survived	q_1_k q_2_k r_1_k −q_1_k r_2_k −q_2_k	qk rk −qk
Total	r_1_k r_2_k	rk

TABLE 11.1. Among the set of patients at risk at time dk, the number of patients who died and survived in each of two groups is reported.

which examines how the events in each group unfold sequentially in time.

Recall from Section 11.3 that d 1 < d 2 < · · · < dK are the unique death times among the non-censored patients, rk is the number of patients at risk at time dk , and qk is the number of patients who died at time dk . We further define r 1 k and r 2 k to be the number of patients in groups 1 and 2, respectively, who are at risk at time dk . Similarly, we define q 1 k and q 2 k to be the number of patients in groups 1 and 2, respectively, who died at time dk . Note that r 1 k + r 2 k = rk and q 1 k + q 2 k = qk .

At each death time dk , we construct a 2 × 2 table of counts of the form shown in Table 11.1. Note that if the death times are unique (i.e. no two individuals die at the same time), then one of q 1 k and q 2 k equals one, and the other equals zero.

The main idea behind the log-rank test statistic is as follows. In order to test H 0 : E( X ) = µ for some random variable X , one approach is to construct a test statistic of the form

\[W = \frac{ \sum_{k=1}^K (q_{1k} - \operatorname{E}(q_{1k})) }{ \sqrt{ \sum_{k=1}^K \operatorname{Var}(q_{1k}) } } \quad (11.5)\]

To construct the log-rank test statistic, we compute a quantity that takes exactly the form (11.4), with X =[�] [K] k =1 [q][1] [k][,][where] [q][1] [k][is][given][in][the][top] left of Table 11.1.

In greater detail, if there is no difference in survival between the two groups, and conditioning on the row and column totals in Table 11.1, the expected value of q 1 k is

\[\operatorname{E}(q_{1k}) = \frac{q_k}{n_k} n_{1k} \quad (11.6)\]

So the expected value of X =[�] [K] k =1 [q][1] [k][is] [µ][=][�] k[K] =1 rr 1 kk[q][k][.][Furthermore,] it can be shown[5] that the variance of q 1 k is

\[\operatorname{Var}(q_{1k}) = \frac{q_k (n_{1k} / n_k) (1 - n_{1k}/n_k) (n_k - q_k)}{n_k - 1} \quad (11.7)\]

Though q 11 , . . . , q 1 K may be correlated, we nonetheless estimate

\[\operatorname{Var} \left( \sum_{k=1}^K q_{1k} \right) = \sum_{k=1}^K \operatorname{Var}(q_{1k})\]

4The log-rank test is also known as the Mantel–Haenszel test or Cochran–Mantel– Haenszel test .

5For details, see Exercise 7 at the end of this chapter.

476 11. Survival Analysis and Censored Data

Therefore, to compute the log-rank test statistic, we simply proceed as in (11.4), with X =[�] [K] k =1 [q][1] [k][, making use of][(][11.5][) and (][11.7][). That is, we calculate

\[W = \frac{ \sum_{k=1}^K \left( q_{1k} - \frac{q_k}{n_k} n_{1k} \right) }{ \sqrt{ \sum_{k=1}^K \frac{q_k (n_{1k} / n_k) (1 - n_{1k}/n_k) (n_k - q_k)}{n_k - 1} } } \quad (11.8)\]

When the sample size is large, the log-rank test statistic W has approximately a standard normal distribution; this can be used to compute a p -value for the null hypothesis that there is no difference between the survival curves in the two groups.[6]

Comparing the survival times of females and males on the BrainCancer data gives a log-rank test statistic of W = 1 . 2, which corresponds to a twosided p -value of 0 . 2 using the theoretical null distribution, and a p -value of 0 . 25 using the permutation null distribution with 1,000 permutations. Thus, we cannot reject the null hypothesis of no difference in survival curves between females and males.

The log-rank test is closely related to Cox’s proportional hazards model, which we discuss in Section 11.5.2.