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Question 13 of 20 Fundamentals Practical

What inputs determine the sample size of a trial?

Short answer

Sample size follows from the primary endpoint and its analysis, the type I error (usually 5 percent two sided), the power (usually 80 to 90 percent), the treatment difference worth detecting, the variability or event rate, and the expected loss of participants. Each assumption should be sourced and its sensitivity tested.

Why

ICH E9 section 3.5 lists what a sample size calculation must specify: the primary variable, the test statistic, the null and alternative hypotheses, the type I error, the type II error, and how withdrawals and protocol violations are handled. Type I error is the chance of declaring an effect that is not there. Type II error, conventionally 10 to 20 percent, is the chance of missing a real effect, and power is one minus that.

The difference to detect can be the smallest clinically relevant effect, or the expected effect where that is larger. Variability, response rates or event rates should come from published data or earlier trials, and the protocol must say where. For time to event endpoints, power depends on the number of events, so the sample size is worked back from events.

E9 also asks for a range of sample sizes under plausible deviations, because a variance guessed too low is the most common way a trial ends underpowered. Noninferiority trials are sized on a confidence interval excluding the margin, not on detecting a difference.

How to work it through

  1. Name the primary endpoint and the planned analysis.
  2. Fix the type I error and the target power.
  3. Choose the treatment difference and justify it clinically or from prior data.
  4. Take the variability or event rate from published or earlier trial data.
  5. Calculate the number needed, then inflate for expected dropout.
  6. Repeat the calculation across plausible values to show sensitivity.

An example

A trial compares change in a symptom score between two arms. The minimal clinically important difference is 5 points and earlier data give a standard deviation of 10. With two sided alpha 0.05 and 90 percent power, n per arm is 2 x (1.96 + 1.28) squared x 10 squared divided by 5 squared, which is about 84.1, so 85.

Allowing 10 percent dropout gives 85 divided by 0.9, about 95 per arm, or 190 in total. If the standard deviation were really 12, the same 85 completers per arm would give only about 77 percent power.

Where the answer stops holding

The normal approximation formula suits large samples and continuous outcomes. Binary, count and time to event endpoints need their own methods. Dropout inflation assumes completers are like those who drop out, which may not hold.

Assumptions from small earlier trials are often optimistic, so a blinded sample size re-estimation may be worth planning.

A strong answer covers

  • Type I error, power, difference to detect and variability
  • Events, not participants, drive time to event power
  • Dropout allowance
  • Sourced assumptions in the protocol
  • Sensitivity across plausible deviations

Common mistakes

  • Choosing the difference to detect to fit the budget rather than clinical relevance.
  • Using an optimistic standard deviation from a small pilot without sensitivity analysis.
  • Forgetting that a multiplicity adjustment lowers the alpha used in the calculation.