Choosing Sample Size Floors And Caps

Why Put Limits on a Re-Estimated Sample Size?

A sample-size re-estimation rule can respond to interim information about a nuisance parameter, such as outcome variability. But an unconstrained rule may recommend an impractically small trial when an estimate is unusually favorable, or an unaffordable trial when it is unfavorable. A floor prevents the final sample size from dropping below a chosen minimum; a cap prevents it from exceeding a chosen maximum. These limits make the rule operational, but they can also change its power and its expected enrollment.

The key is to distinguish the target calculated by the re-estimation formula from the sample size the trial will actually use. The target is an input to the rule, not a guarantee that the final design reaches its nominal power. The calculations below use equal allocation and express sample size per arm; total enrollment is twice that number. Unless stated otherwise, sample sizes refer to evaluable participants, with any allowance for attrition handled separately.

Definition: Let $R$ be the integer sample size per arm recommended by a re-estimation rule before limits are applied. Let $N_{\min}$ be the minimum and $N_{\max}$ the maximum permitted sample size per arm, with $1\leq N_{\min}\leq N_{\max}$. The implemented final sample size per arm is $$ N=\min\{N_{\max},\max(N_{\min},R)\}. $$ Thus $N=N_{\min}$ when $R<N_{\min}$, $N=R$ when $N_{\min}\leq R\leq N_{\max}$, and $N=N_{\max}$ when $R>N_{\max}$. The corresponding total enrollment is $2N$.

If the re-estimation formula first gives a non-integer target $n_{\mathrm{target}}$, a common conservative convention is to round upward, $R=\lceil n_{\mathrm{target}}\rceil$, and then apply the floor and cap. The limits must be specified on the same scale as the target: a per-arm cap cannot be compared directly with a total-enrollment target. If the floor exceeds the cap, the design specification is inconsistent and must be revised before enrollment begins.

From a Target to an Implemented Sample Size

For a simple planning benchmark, suppose the outcome is normally distributed with known common standard deviation $\sigma$, the allocation is equal, and the treatment effect of interest is a difference in means $\delta>0$. For a two-sided z test at significance level $\alpha$, the approximate required sample size per arm for power $1-\beta$ is $$ n_{\mathrm{target}}= \frac{2\sigma^2(z_{1-\alpha/2}+z_{1-\beta})^2}{\delta^2}. $$ Here $z_q$ is the $q$-quantile of the standard normal distribution. This planning expression is useful for understanding the consequence of a cap, but an actual re-estimation rule may use an interim estimate of a nuisance parameter rather than the true $\sigma$.

Worked Example: A Cap Below the Planned Target

Suppose a hypothetical, equally randomized trial plans to detect a 4-point difference in a continuous outcome. Assume a common standard deviation of 15 points, a two-sided significance level of 0.05, and 90% power. The design has a floor of 200 and a cap of 280 evaluable participants per arm. Find the target, the implemented sample size, and the approximate power at the cap.

For these assumptions, $z_{0.975}=1.960$ and $z_{0.90}=1.282$. The continuous target is $$ n_{\mathrm{target}} = \frac{2(15^2)(1.960+1.282)^2}{4^2} = \frac{450(3.242)^2}{16} = \frac{450(10.510564)}{16} \approx 295.6. $$

Rounding upward gives $R=296$ per arm. Since $296>280$, the cap binds, so $N=\min\{280,\max(200,296)\}=280$ per arm, or 560 evaluable participants total. The floor does not affect this result.

At $N=280$, the noncentrality parameter for the difference in means is $$ \lambda=\frac{\delta}{\sigma\sqrt{2/N}} =\frac{4\sqrt{280}}{15\sqrt{2}} =4\sqrt{\frac{280}{450}} \approx 3.155. $$ The approximate two-sided power is $$ P(Z>1.960-\lambda)+P(Z<-1.960-\lambda) = \Phi(3.155-1.960)+\Phi(-3.155-1.960) \approx \Phi(1.195)+\Phi(-5.115) \approx 0.884. $$

Thus the capped design has about 88.4% power under the stated model, below the 90% target. The target calculation is internally consistent, but the cap prevents the trial from reaching that target under these assumptions.

Quantifying Expected Enrollment and Power

If the re-estimation recommendation is random, the implemented sample size is random too. For possible recommendations $r_1,\ldots,r_k$, with probabilities $q_j=P(R=r_j)$, define $c(r)=\min\{N_{\max},\max(N_{\min},r)\}$. Then expected enrollment is $$ E(2N)=2\sum_{j=1}^{k}q_jc(r_j). $$ This calculation is about the average across repeated trials under the specified model for the re-estimation recommendation; it does not mean an individual trial will enroll that average. The floor can raise expected enrollment by lifting low recommendations, while the cap can lower it by truncating high recommendations. Their combined effect depends on how often each limit binds.

For a fixed sample size $n$ per arm, the normal-model power for the two-sided z test under a true treatment difference $\delta$ is $$ \pi(n)=\Phi(\lambda_n-z_{1-\alpha/2})+\Phi(-\lambda_n-z_{1-\alpha/2}), \qquad \lambda_n=\frac{\delta\sqrt{n}}{\sigma\sqrt{2}}. $$ When the rule's selected size is independent of the efficacy statistic conditional on the planning scenario, the scenario-averaged power is $\sum_j q_j\pi(c(r_j))$. This mixture is a useful way to compare policies. If the adaptation uses interim trial data that are associated with the final test statistic, this simple averaging is not generally an exact power calculation. In that setting, the joint behavior of the adaptation and test must be evaluated, typically by deriving or simulating the full design.

Worked Example: A Floor and Cap Trade Power for Enrollment

Consider a hypothetical planning model with a true difference of 4 points, common standard deviation 15 points, and a two-sided z test at $\alpha=0.05$. Suppose the re-estimation recommendation $R$, in evaluable participants per arm, is 160, 240, or 360, with probabilities 0.25, 0.50, and 0.25. Treat these scenario probabilities as independent of the efficacy statistic for this illustrative calculation. Compare no limits with a floor of 180 and a cap of 240.

Without limits, expected enrollment is $$ E(2R)=2\{0.25(160)+0.50(240)+0.25(360)\} =2(250)=500. $$ With the limits, the recommendations become 180, 240, and 240 per arm. Therefore, $$ E(2N)=2\{0.25(180)+0.50(240)+0.25(240)\} =2(225)=450. $$ The limits reduce expected enrollment by 50 participants in this scenario distribution.

To compare power, the relevant noncentrality parameter is $\lambda_n=4\sqrt{n/450}$. Applying the two-sided power formula gives approximate powers of 0.665 at $n=160$, 0.716 at $n=180$, 0.832 at $n=240$, and 0.947 at $n=360$. Without limits, the scenario-weighted power is $$ 0.25(0.665)+0.50(0.832)+0.25(0.947)\approx 0.819. $$ With the floor and cap it is $$ 0.25(0.716)+0.50(0.832)+0.25(0.832)\approx 0.803. $$

In this example, the floor improves power in the low-recommendation scenario, but the cap removes the extra power associated with the 360-per-arm recommendation. The net average power is lower, while expected enrollment also falls. Neither direction is automatic: the result depends on the recommendation distribution and the amount of truncation at each limit.

Feasibility as a Design Constraint

A cap is often chosen because the trial must fit a recruitment, budget, drug-supply, or calendar constraint. If the operational capacity is $M$ evaluable participants total under equal allocation, a cap satisfying $2N_{\max}\leq M$ ensures that the planned maximum does not exceed that capacity. A recruitment plan should also distinguish evaluable sample size from the number who must be randomized. For example, if a fraction $a$ is expected to provide an evaluable outcome, a rough total randomized target is $2N/a$; this allowance does not change the definition of the per-arm floor or cap.

Worked Example: Choosing a Cap to Fit Recruitment Capacity

A hypothetical trial has recommendations of 180, 240, or 360 evaluable participants per arm, each occurring with probability 0.25, 0.50, or 0.25. The investigators require a floor of 200 per arm and have capacity for at most 520 evaluable participants total. Compare a cap of 300 with a cap of 260, and use the same normal model as the previous example: $\delta=4$, $\sigma=15$, and two-sided $\alpha=0.05$.

A cap of 300 permits up to 600 participants total, exceeding the 520-person capacity. After applying the floor and that cap, the three recommendations become 200, 240, and 300 per arm. Expected total enrollment is $$ 2\{0.25(200)+0.50(240)+0.25(300)\} =2(245)=490. $$ Its maximum total enrollment is 600, so this policy is not feasible under the stated capacity.

A cap of 260 gives implemented sizes of 200, 240, and 260 per arm. The maximum is 520 participants total, and expected total enrollment is $$ 2\{0.25(200)+0.50(240)+0.25(260)\} =2(235)=470. $$ The approximate powers at 200, 240, and 260 per arm are 0.760, 0.832, and 0.860, respectively. Hence scenario-averaged power is $$ 0.25(0.760)+0.50(0.832)+0.25(0.860)\approx 0.821. $$

The 260-per-arm cap satisfies the stated capacity and yields about 82.1% power under this illustrative scenario model. The 300 cap would yield slightly more power in the highest-recommendation scenario, but its maximum enrollment makes it infeasible. A design decision should state this trade-off explicitly rather than report expected enrollment alone: a low average does not guarantee that the maximum will fit the available capacity.

Justification, Diagnostics, and Limitations

The effect of a floor or cap follows directly from the truncation rule. For every recommendation $R$, $N_{\min}\leq c(R)\leq N_{\max}$, so averaging both sides gives $N_{\min}\leq E(N)\leq N_{\max}$. Also, if the fixed-size test's power $\pi(n)$ increases with $n$, then raising a floor cannot reduce the size selected in any scenario, and lowering a cap cannot increase it. Under the independent-scenario mixture calculation, these pointwise comparisons also order the corresponding average powers. A floor and cap applied together have competing effects, so their net effect on average power cannot be determined from the limits alone.

This argument relies on a fixed-size power curve that increases with sample size, as it does for the stated normal test with a fixed effect and significance level. It does not by itself establish the exact type I error or power of an adaptive test. When the interim information used to choose $N$ is related to the final test statistic, conditioning on the selected sample size can change the distribution of that statistic. The mixture calculation is then only an approximation unless that dependence has been handled in the operating-characteristic analysis.

Before adopting limits, check at least the following:

Scale and rounding: confirm that floor, cap, and target are all per arm or all total; document whether non-integer targets are rounded upward before truncation.
Binding frequency: estimate how often the recommendation falls below the floor or above the cap. A frequently binding cap is a substantive power constraint, not a rare contingency.
Achievable target: calculate power at the cap under clinically relevant assumptions. If that power is below the design goal, either increase the cap, revise the goal transparently, or accept the limitation with justification.
Capacity at the maximum: test the worst-case total enrollment, not only its expectation, against recruitment and resource constraints.
Dependence and model fit: if interim data determine the sample size, evaluate the joint adaptation-and-test procedure. Also examine whether the nuisance-parameter estimate and assumed effect scenarios are plausible.

Floors and caps are not substitutes for a sound re-estimation rule. A high floor can waste resources when the target is usually small; a low cap can materially reduce power when high sample sizes are often needed. Conversely, no cap may make the rule impossible to deliver. A defensible specification reports the floor, cap, rounding convention, expected and maximum enrollment, and power implications across relevant scenarios. It also distinguishes an operationally feasible design from one that attains a desired power under every plausible nuisance-parameter value.

Common pitfall: “The re-estimation target is 296 per arm, so the capped design still has 90% power.” This is incorrect when the cap is 280: under the stated model, the implemented size is 280 and its approximate power is 88.4%. Calculate operating characteristics using the implemented sample size, and account for adaptation dependence when the re-estimation data and final test statistic are not independent.

Check Your Understanding

Answer each question using the floor-and-cap rule. Solutions follow the questions.

A rule recommends 175 per arm. The floor is 190 and the cap is 300. What sample size per arm and total enrollment are implemented?
Recommendations of 100, 200, and 400 per arm occur with probabilities 0.2, 0.5, and 0.3. With a floor of 150 and cap of 300, find expected total enrollment.
Why is it not generally sufficient to average fixed-size power values by the probabilities of selected sample sizes when the re-estimation uses interim trial data?
A trial can recruit at most 480 evaluable participants total under equal allocation. What is the largest feasible per-arm cap?

Solutions. (1) The floor binds: $N=\min\{300,\max(190,175)\}=190$ per arm, giving $2(190)=380$ participants total. (2) The implemented sizes are 150, 200, and 300 per arm. Thus $E(2N)=2\{0.2(150)+0.5(200)+0.3(300)\}=2(220)=440$ participants. (3) Interim information can be associated with the final efficacy statistic, so the statistic's distribution conditional on the selected size need not be its fixed-size distribution. Exact power must account for their joint behavior. (4) Equal allocation gives at most $480/2=240$ evaluable participants per arm, so the cap cannot exceed 240.