LAUSR

The article by In and Lee entitled “Survival analysis: Part I — analysis of time-to-event”, contained an error in the sample size calculation example. The calculated numbers are incorrect; here, we present the corrected calculation process and results. Supposing that these five patients were observed for four weeks on average, the hazard rate (λ) is 2/(5 × 4 weeks) = 0.1/person-week. The value of the 4-week survival function for conventional drug A, estimated using the relationship between the survival function and hazard function, is SA(4) = exp (–0.1 × 4) = 0.670. Since new drug B decreases recurrence by 30%, the hazard ratio is 0.7, and the value of the 4-week survival function for new drug B is SB(4) = exp ((–0.1 × 0.7) × 4) = 0756. If both groups have the same sample size, π1 = π2 = 0.5, the probability of an event, which is the denominator of the sample size calculation formula, is 1 –(π1S1(t) + π2S2(t)) = 1 – (0.5 × 0.670 + 0.5 × 0.756) = 0.287. The total event count, which is the numerator of the sample size calculation formula, can be obtained from Equation 3. zα/2 and zβ, which represent the values of probability in a standard normal distribution, are 1.96 and 0.842, respectively, for a significance level of 0.05 and statistical power of 80%. With the values of π1 and π2 set to 0.5 each and the hazard ratio set at 0.7, the total event count required is (1.96 + 0.842)/{0.5 × 0.5 × (log0.7)} = 246.9, i.e., 247 events. Substituting this value and the incidence rate into Equation 4, 247/0.287 = 860.6, i.e., 861 is obtained. Applying the generally assumed withdrawal rate of 10% to the value obtained, 861/(1 – 0.1) = 956.7, i.e., a total of 957 subjects, is set as the required sample size. With the group size ratio set at 0.5, 479 subjects are to be assigned to each group.