Why Choose the Maximum Sample Size Deliberately?
A promising-zone rule can increase the planned sample size when interim results fall in a prespecified range. The maximum sample size is therefore not just an administrative limit: it determines how much additional information the design can acquire in that range, and how much enrollment the trial may require. A cap that is too low can leave power inadequate where expansion is intended to help. A cap that is unnecessarily high can increase expected enrollment without a worthwhile gain in the scenarios that matter.
The selection problem is to compare candidate caps using operating characteristics under a set of plausible true treatment effects. In particular, examine both the power at each effect and the expected total enrollment. Building on the earlier material on expected sample size, the key is to evaluate these quantities over the whole adaptive rule, not just to compute the sample size associated with one interim estimate.
Model and Operating Characteristics
Consider a two-stage design with one interim decision. All participants first contribute to a base sample size \(N_0\). If the interim estimate falls in a prespecified promising zone, enrollment increases to a candidate maximum \(M\), where \(M \geq N_0\). Otherwise, the trial ends at \(N_0\). Assume no early stopping for futility in this example. The same logic extends to designs with multiple interim looks, but their region probabilities and expected enrollment must account for all paths through the looks.
The power expression follows by partitioning the possible interim results into the promising zone and its complement, then applying the law of total probability. The enrollment expression follows in the same way: the total is \(M\) with probability \(q(\delta)\), and \(N_0\) otherwise. These are unconditional operating characteristics at a specified true effect; they are not conditional-power targets.
For the conditional probabilities to describe a valid design, specify the endpoint model, allocation, interim rule, final test, and any variance assumptions. In particular, the rejection rule must have its type I error calibrated for the adaptation. Simply changing sample size based on an unblinded interim effect and then applying an ordinary fixed-sample critical value does not, in general, guarantee that the overall type I error is \(\alpha\). The values below are stipulated outputs from a hypothetical, fully specified design whose final rule has been calibrated under \(H_0\) to the planned one-sided level \(\alpha=0.025\). They illustrate cap selection; they are not data from a real study.
Suppose the endpoint is continuous, allocation is equal, and the interim estimate \(\hat{\delta}\) has distribution \(N(\delta,1)\), with effects and standard errors expressed in standardized-effect units. Let the promising zone be \(4.5<\hat{\delta}<5.5\), let \(N_0=100\), and consider candidate caps \(M=110,120,130\). Under this model, $$ q(\delta)=P(4.5<\hat{\delta}<5.5\mid\delta) =\Phi(5.5-\delta)-\Phi(4.5-\delta). $$ Thus \(q(4)=0.2417\), \(q(5)=0.3829\), and \(q(6)=0.2417\), to four significant figures. For example, at \(\delta=6\), \(q(6)=\Phi(-0.5)-\Phi(-1.5)\). The conditional rejection probabilities used below come from evaluating the same calibrated design under each effect and cap.
Worked Examples: Calculating and Comparing Caps
Worked Example: Power at a Cap of 130
At \(\delta=6\), suppose the calibrated design has \(\pi_O(6)=0.9994\) and \(\pi_P(6;130)=0.99996\). Calculate power at \(M=130\).
The promising-zone probability is \(q(6)=0.2417\), so the probability of being outside the zone is \(1-0.2417=0.7583\). Substitution into the total-probability formula gives $$ \pi(6;130)=0.2417(0.99996)+0.7583(0.9994) =0.241690332+0.757845020 =0.999535352. $$ The power is approximately \(0.9995\) (four significant figures). Under the stated model and calibrated rule, the probability of rejecting \(H_0\) when the true standardized effect is 6 is about 99.95%. This is not a statement about the probability that \(H_0\) is true.
Expected enrollment is $$ E_6[N]=100+0.2417(130-100) =100+7.251 =107.251. $$ Thus the expected total is about 107.3 participants at this effect, although the realized total is either 100 or 130 under this rule.
Worked Example: Selecting the Smallest Cap That Meets Planning Criteria
Suppose the planning team regards \(\delta=5\) as the nominal scenario and \(\delta=4\) as a stress scenario. It requires power of at least 0.90 at \(\delta=5\) and at least 0.85 at \(\delta=4\). The hypothetical calibrated evaluation gives the following powers:
| True effect \(\delta\) | Cap \(M=110\) | Cap \(M=120\) | Cap \(M=130\) |
|---|---|---|---|
| 4 | 0.820 | 0.850 | 0.870 |
| 5 | 0.900 | 0.925 | 0.940 |
| 6 | 0.9988 | 0.9992 | 0.9995 |
A cap of 110 meets the nominal requirement exactly, but fails the stress requirement because \(0.820<0.85\). A cap of 120 meets both: its powers are \(0.925\geq0.90\) at \(\delta=5\) and \(0.850\geq0.85\) at \(\delta=4\). A cap of 130 also meets both, but it is not the smallest qualifying cap.
At \(\delta=5\), the expected enrollments are \(100+0.3829(10)=103.829\) for cap 110, \(100+0.3829(20)=107.658\) for cap 120, and \(100+0.3829(30)=111.487\) for cap 130. Under the stated criteria, cap 120 is the smallest cap that passes both power requirements. This choice makes the tradeoff explicit rather than selecting a cap from the nominal scenario alone.
Worked Example: Summarizing Expected Enrollment Across Scenarios
For a planning summary, suppose the team assigns scenario weights \(0.25, 0.50,\) and \(0.25\) to \(\delta=4,5,\) and \(6\), respectively. These weights define a scenario-weighted average for comparing plans; they do not establish a probability distribution for the true effect. Calculate the average expected enrollment for cap 120.
The expected enrollments at the three effects are: $$ E_4[N]=100+0.2417(20)=104.834, $$ $$ E_5[N]=100+0.3829(20)=107.658, $$ $$ E_6[N]=100+0.2417(20)=104.834. $$ Weighting these values gives $$ 0.25(104.834)+0.50(107.658)+0.25(104.834) =106.246. $$ The scenario-weighted expected enrollment is about 106.2 participants. This summary can be useful alongside the separate scenario results, but it should not replace them: the average can conceal a scenario with unacceptable power or enrollment.
A Practical Selection Procedure
Maximum sample size selection is a constrained planning decision. A cap should be evaluated together with the interim rule and final test because changing the cap can change both the power gain in the promising zone and the resources required. A practical comparison proceeds as follows.
State the endpoint, allocation, variance assumptions, plausible true effects, interim regions, base sample size, and calibrated final rejection rule.
List feasible values of \(M\). State power requirements, resource limits, and any required stress scenarios before comparing results.
For every candidate and effect, calculate the probability of each interim region, the conditional rejection probabilities, and expected total enrollment. Use exact integration or validated simulation under the joint design model.
Exclude caps that fail a prespecified constraint. Among those remaining, select according to the stated efficiency or resource objective, then verify rounding, type I error calibration, and the operational enrollment limit.
Diagnostics, Alternatives, and Limitations
Several checks help detect a flawed cap comparison. First, verify that the region probabilities sum to one and that expected enrollment lies between \(N_0\) and \(M\). In the two-region rule, this requires \(0\leq q(\delta)\leq1\) and \(N_0\leq N_0+q(\delta)(M-N_0)\leq M\). Second, check that the conditional rejection probabilities are between zero and one and that the reported overall power agrees with their weighted average. Third, compare Monte Carlo estimates with numerical integration where available; simulation error can make close candidate caps appear spuriously different. Report simulation uncertainty when it is material to a selection decision.
Power should generally not decrease merely because the cap is larger if the procedure is otherwise unchanged and the additional information is used by a valid, prespecified rule. If a computed comparison shows a substantial decrease, investigate whether the final test, adaptation rule, or calibration changed across caps, or whether simulation noise is large. Do not assume strict monotonicity in every adaptive procedure: changing a cap can alter the rule or its critical value, so the actual design must be evaluated.
The main alternative is a fixed-sample design, which has a single planned total and no interim-dependent increase. It is simpler to describe and operate, but does not vary enrollment in response to interim promise. A promising-zone design with a lower cap limits maximum enrollment but may give less power in the zone; a higher cap may improve power there while increasing expected and maximum enrollment. These alternatives should be compared using the same effect scenarios, endpoint assumptions, and type I error standard.
A cap choice is only as credible as its planning assumptions. If the interim variance, treatment-effect scale, region probabilities, or final-test calibration is wrong, the displayed operating characteristics may not describe the implemented study. Scenario-weighted averages also depend on the chosen weights and can obscure poor performance in an important scenario. Finally, the cap is a maximum, not an expected sample size: report both, and account for integer allocation and operational rounding when translating total enrollment into numbers per group.
Check Your Understanding
Answer each question using the definitions and formulas above. Solutions are included after the questions.
- If \(N_0=80\), \(M=120\), and \(q(\delta)=0.30\), what is the expected total enrollment at that effect?
- Suppose \(q(\delta)=0.20\), \(\pi_P(\delta;M)=0.80\), and \(\pi_O(\delta)=0.70\). What is the overall power?
- Why is a cap that meets a nominal power target not necessarily an adequate choice?
- What should be checked if estimated power decreases substantially when the cap increases?
Solutions. 1. \(E_\delta[N]=80+0.30(120-80)=92\) participants. 2. \(\pi=0.20(0.80)+0.80(0.70)=0.72\), or 72% power at the specified effect. 3. It may fail a prespecified power or enrollment criterion at another plausible effect; the full scenario set and resource constraints should be examined. 4. Check whether the final test or calibration changed, whether the adaptation rule changed, and whether simulation error could explain the difference; then evaluate the specified design rather than assuming the larger cap must improve power.