11. Sample Size II: Time-to-Event, Non-Inferiority, Cluster, Loss
After this lesson you can
- Compute the events needed for a hazard ratio and convert events to participants
- Compute a non-inferiority sample size from the margin
- Turn a per-arm size into a number of clinics with the design effect
- Inflate for loss by dividing by the proportion retained
- Explain why non-adherers are never excluded to fix dilution
Video coming soon
Lessons publish daily from November 17, 2026.
This lesson sizes time-to-event endpoints by events, non-inferiority comparisons by the margin and cluster versions by the design effect. It then inflates for loss to follow-up, explains non-adherence dilution and allocation ratios, and assembles the final fictional Oakhaven plan of 125 per arm.
Questions and answers
Questions a learner would ask about this lesson. Think through your own answer before you open each one.
- Why does a time-to-event comparison count events instead of people?
Show answer
Power depends on the number of events, and a participant who never has an event contributes only a partly known time. In the fictional Oakhaven trial, the events formula is 4 times 7.84 over the squared log hazard ratio: ln 0.5 is minus 0.693, squared 0.480, so 31.36 over 0.480 is about 65.3, rounded to 66 events. At a 30 percent event proportion, 66 divided by 0.30 is about 220 participants.
- Why does a non-inferiority trial need more participants than a superiority trial?
Show answer
The margin replaces the expected difference in the denominator, and when the two regimens are expected to be equal only the margin is left. For the shorter-regimen side case, 16 times 0.16 (0.80 times 0.20) over 0.01 (the ten-point margin squared) is 256 per arm, 512 in total. Halving the margin quadruples the sample.
- How is a cluster version sized?
Show answer
Multiply the individually randomized per-arm size by the design effect, 1 plus 49 times 0.02, about 2, so 84 per arm becomes 168. Then round in clinics: 168 divided by 50 is 3.36, so four clinics of 50 per arm, eight clinics and 400 participants. More clusters give more power than bigger ones.
- What is the right way to inflate for loss to follow-up?
Show answer
Divide by the proportion retained. With 10 percent loss, 111 divided by 0.9 is 123.3, so 124, rounded to 125 per arm and 250 in total. Multiplying by 1.1 gives 122.1, so 123, which undershoots, and the error grows at higher loss. Inflation is a backstop; the retention plan comes first.
- How should non-adherence in the program arm be handled?
Show answer
Non-adherers stay in the program arm, and their outcomes look more like usual care, which dilutes the intention-to-treat difference. The only two cures are to size for the diluted difference or to raise adherence by design. Excluding non-adherers breaks randomization, so dilution is a cost but exclusion is a bias.
Practice quiz
- Question 1For a hazard ratio of 0.5 at eighty percent power, about how many events are needed?
- Question 2With 66 events needed and 30 percent of participants expected to have an event, about how many participants are needed?
- Question 3Which assumption does the events formula NOT make?
- Question 4The shorter-regimen non-inferiority trial (success 80 percent, margin ten points) needs about how many per arm?
- Question 5If the non-inferiority margin is halved, the sample size:
- Question 6The individually randomized size is 84 per arm and the design effect is about 2, with 50 participants per clinic. How many clinics does the cluster version need in total?
- Question 7Inflating 111 per arm for 10 percent loss gives:
- Question 8Why is multiplying 111 by 1.1 the wrong way to inflate for loss?
- Question 9Non-adherence in the program arm:
- Question 10What is Oakhaven's final sample-size plan for the primary endpoint?
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Design of Clinical Studies is education only. Every lesson follows the fictional Oakhaven Adherence Trial with invented numbers; the course uses no real data and cites no real studies.