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Unblinded Re-Estimation and the Promising Zone · Tutorial 400 of 699

Lower and Upper Sample-Size Constraints

You will learn to apply lower and upper sample-size limits to a re-estimation target and distinguish a feasible target from one that the constraints cannot attain.

Advanced 10 min read

What You'll Learn

  • Define minimum and maximum sample-size constraints on the same scale as the re-estimation target.
  • Apply the clamp rule to targets below, within, and above the permitted range.
  • Check whether a constrained sample size can meet the conditional-power objective.
  • Recognize why a cap may leave the target power unattained.
  • Diagnose infeasible bounds and ambiguous sample-size definitions.
  • Distinguish sample-size constraints from guarantees of type I error control.

Why Constrain a Re-Estimated Sample Size?

A conditional-power rule can produce a sample-size target that is operationally unacceptable. A target below the planned minimum may leave too little information for the intended analysis; a target above the maximum may exceed available resources or a prespecified enrollment limit. The rule therefore needs a clear answer to a practical question: what sample size will be used when the unconstrained target is outside the allowed range?

This tutorial treats the minimum and maximum as prespecified limits applied to the output of a deterministic re-estimation rule. The limits do not change the interim data or the conditional-power calculation that produced the unconstrained target. They determine which sample size is selected after that calculation.

Definition: Let \(n_0\) be the unconstrained sample-size target from a prespecified re-estimation rule, and let \(n_{\min}\) and \(n_{\max}\) be the allowed lower and upper bounds, respectively. The constrained target is $$ n_{\mathrm{c}}=\min\!\left\{n_{\max},\ \max\!\left(n_{\min},n_0\right)\right\}, $$ provided \(n_{\min}\leq n_{\max}\). Thus \(n_{\mathrm{c}}=n_{\min}\) if \(n_0<n_{\min}\), \(n_{\mathrm{c}}=n_0\) if \(n_{\min}\leq n_0\leq n_{\max}\), and \(n_{\mathrm{c}}=n_{\max}\) if \(n_0>n_{\max}\).

For clarity, examples below use \(n\) to mean the number randomized per treatment group in a two-group trial with equal allocation. The total planned enrollment is then \(2n\). A design that defines its limits as total enrollment must instead apply the constraint formula to total enrollment; mixing these scales can double or halve the intended cap. At an interim analysis, \(n_{\min}\) must also be at least the number already randomized per group, so the selected final sample size does not discard participants already enrolled.

Conditional Power and the Constrained Target

The clamp rule is straightforward, but its interpretation depends on the calculation that supplies \(n_0\). Consider a one-sided test of \(H_0:\delta=0\) against \(H_a:\delta>0\) for a normally distributed endpoint. Assume equal allocation, a common known standard deviation \(\sigma\), independent observations, and a fixed interim sample size \(n_1\) per group. Let \(D_1\) be the observed treatment-minus-control difference at interim, and let \(Z_1=D_1/(\sigma\sqrt{2/n_1})\) be its standardized statistic. For the conditional-power calculation, suppose future observations are modeled with a specified effect \(\delta^*\), and let the final sample size per group be \(n>n_1\).

Under the independent-increments normal model, the final standardized statistic, conditional on the interim data, has mean \(m(n)\) and standard deviation \(s(n)\):

$$ m(n)=\sqrt{\frac{n_1}{n}}\,Z_1+ \frac{(n-n_1)\delta^*}{\sigma\sqrt{2n}}, \qquad s(n)=\sqrt{\frac{n-n_1}{n}}. $$

For a one-sided level-\(\alpha\) final test, write \(z_{1-\alpha}\) for the \(1-\alpha\) quantile of the standard normal distribution. The conditional power at sample size \(n\) is

$$ \operatorname{CP}(n\mid Z_1,\delta^*)= \Phi\!\left(\frac{m(n)-z_{1-\alpha}}{s(n)}\right), $$

where \(\Phi\) is the standard normal cumulative distribution function. The unconstrained target \(n_0\) may, for example, be obtained by solving \(\operatorname{CP}(n\mid Z_1,\delta^*)=q\) for a prespecified target \(q\). Applying a cap or floor afterward does not alter this equation; it can, however, make its target unattainable.

Why the Clamp Rule Is the Natural Constraint

The clamp selects the feasible value closest to the unconstrained target on the sample-size scale. This is the projection of \(n_0\) onto the interval \([n_{\min},n_{\max}]\). The short case analysis below also shows why there is no fourth outcome when the bounds are valid.

1
Check feasibility.
Confirm \(n_{\min}\leq n_{\max}\), and confirm both bounds use the same unit as \(n_0\), such as participants per group.
2
Apply the lower limit.
If \(n_0<n_{\min}\), then \(\max(n_{\min},n_0)=n_{\min}\); since this is no larger than \(n_{\max}\), the outer minimum leaves \(n_{\mathrm{c}}=n_{\min}\).
3
Keep an interior target.
If \(n_{\min}\leq n_0\leq n_{\max}\), the inner maximum is \(n_0\), and the outer minimum also leaves \(n_{\mathrm{c}}=n_0\).
4
Apply the upper limit.
If \(n_0>n_{\max}\), the inner maximum is \(n_0\), and the outer minimum gives \(n_{\mathrm{c}}=n_{\max}\).

The rule guarantees that the selected target is in the allowed interval. It does not by itself guarantee that a conditional-power target is attained, that a particular actual power will result, or that the final test has type I error \(\alpha\). Those are separate design and inference questions.

Worked Example: A Minimum Raises a Target Below the Floor

Suppose \(n_1=100\) participants per group have been observed, with \(Z_1=0\). Assume \(\sigma=10\), \(\delta^*=2\) outcome units, and a one-sided final critical value \(z_{1-\alpha}=1.960\). For this hypothetical conditional-power calculation, the unconstrained target is about \(n_0=552\) per group. A prespecified minimum requires at least \(600\) per group, and the maximum is \(900\).

At \(n=552\), the conditional mean and standard deviation are $$ m(552)=\frac{2(552-100)}{10\sqrt{2(552)}}= \frac{904}{10\sqrt{1104}}\approx 2.721, \qquad s(552)=\sqrt{\frac{452}{552}}\approx 0.9049. $$

Thus the standardized distance above the critical value is approximately $$ \frac{2.7207-1.960}{0.904898}\approx 0.8407, $$ so the conditional power is \(\Phi(0.8407)\approx 0.800\). The unconstrained target is below the floor, and the clamp gives $$ n_{\mathrm{c}}=\min\{900,\max(600,552)\}=600 $$ participants per group, or \(1{,}200\) in total.

The floor changes the selected sample size; it does not mean the conditional power is still exactly \(0.800\). At \(n=600\), the same model gives \(m(600)=1000/(10\sqrt{1200})\approx2.887\) and \(s(600)=\sqrt{500/600}\approx0.9129\). Therefore, conditional power is \(\Phi((2.887-1.960)/0.9129)\approx0.845\). This calculation checks the conditional-power implication of the floor under the stated model.

Worked Example: An Interior Target Is Unchanged

In a second hypothetical design, a prespecified conditional-power procedure has produced an unconstrained target of \(n_0=700\) per group. The allowed range is \(500\) to \(800\) per group. Both bounds exceed the interim enrollment of \(120\) per group, so they are feasible for the accrued sample.

Because \(500\leq700\leq800\), the target is already allowed: $$ n_{\mathrm{c}}=\min\{800,\max(500,700)\} =\min\{800,700\}=700. $$

The final target remains \(700\) per group, giving total enrollment of \(1{,}400\). No further conditional-power calculation is needed merely to apply the bounds: neither limit is active. If the target had been stated as total enrollment instead, the corresponding interval and target would need to be expressed in total-enrollment units before applying the rule.

Worked Example: A Maximum Prevents Reaching the Unconstrained Target

Suppose a third hypothetical re-estimation gives \(n_0=1{,}200\) per group, while the prespecified limits are \(400\) and \(1{,}000\) per group. The interim enrollment is \(300\) per group, so neither the minimum nor the maximum conflicts with participants already enrolled.

The unconstrained target exceeds the maximum. Hence $$ n_{\mathrm{c}}=\min\{1{,}000,\max(400,1{,}200)\} =\min\{1{,}000,1{,}200\}=1{,}000 $$ per group, or \(2{,}000\) participants in total. The cap prevents the design from using the requested target of \(1{,}200\) per group.

If the target \(n_0\) was defined to attain conditional power \(q\), the cap alone does not establish that \(\operatorname{CP}(1{,}000\mid Z_1,\delta^*)=q\). That conditional power must be evaluated at \(1{,}000\) using the specified model. If the value is below \(q\), the right conclusion is that the constrained design cannot attain the target under that calculation—not that the cap has somehow preserved it.

What the Bounds Do—and Do Not—Change

A lower limit and an upper limit play different operational roles. A floor can prevent the target from falling below a design minimum, even when the interim result would otherwise lead to a smaller recommendation. A cap can make an otherwise indicated expansion impossible. In either case, the selected sample size is a design decision governed by the prespecified algorithm, not an estimate of the effect itself.

ApproachRuleConsequence
Unconstrained re-estimationUse \(n_0\) directly.Preserves the original target calculation, but may produce an operationally unacceptable size.
Clamp to boundsUse \(\min\{n_{\max},\max(n_{\min},n_0)\}\).Always selects a value within valid limits; a boundary may not attain the conditional-power target.
Do not expand beyond a limitKeep the sample size at a prespecified ceiling or stop expansion.Controls enrollment, but may leave the design with less conditional power than requested.

These approaches are not interchangeable after observing interim results. The minimum, maximum, units, and action at a binding limit should be specified in the design algorithm before the adaptation is used. A sample-size cap is not a substitute for a valid final analysis: unblinded sample-size adaptation can affect operating characteristics, and the constraint formula alone does not provide type I error control.

Diagnostics and Common Pitfalls

  • Inconsistent units. A target of 600 per group cannot be compared directly with a maximum of 1,000 total participants. Convert both to a common scale first.
  • Impossible bounds. If \(n_{\min}>n_{\max}\), no value satisfies both constraints. The algorithm is not defined until the conflict is resolved.
  • A floor below accrued enrollment. If \(n_{\min}<n_1\), clarify whether the rule is intended to allow a final size below the number already observed. In an ordinary fixed-final-size design, the final size must be at least \(n_1\).
  • Assuming a cap preserves target power. A cap selects a feasible size, not necessarily a size that meets the conditional-power target. Recalculate conditional power at the selected boundary if that implication matters.
  • Confusing a target with a guarantee. Conditional power is calculated under its stated effect and model assumptions. It is not a guarantee of the trial’s eventual rejection or of power under every possible true effect.
  • Forgetting inferential consequences. A deterministic clamp does not automatically preserve the nominal type I error rate. The final testing procedure and adaptation must be assessed as part of the complete design.

A practical diagnostic is to record, at each hypothetical interim result used to check the algorithm, the unconstrained target, which bound is active, the constrained target, and the conditional power at that constrained target. This makes it possible to detect a cap that frequently prevents the intended target from being reached, or a floor that makes the re-estimation behave differently from the design rationale.

Key takeaway: Apply the minimum and maximum to the unconstrained target on a clearly defined sample-size scale. The clamp guarantees feasibility within the interval; it does not guarantee attainment of the conditional-power target or control of the final test’s error rate.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

  1. An unconstrained target is \(n_0=450\) per group, with \(n_{\min}=500\) and \(n_{\max}=900\). What is the constrained target?
  2. An unconstrained target is \(n_0=750\) total participants. The limits are 600 and 900 total participants. What is the result, and why must the units be stated?
  3. If \(n_0=1{,}050\), \(n_{\min}=400\), and \(n_{\max}=1{,}000\), is the conditional-power target necessarily attained at the constrained sample size? Explain.
  4. At interim, \(n_1=320\) per group have been enrolled, but the specified minimum is \(n_{\min}=300\). What issue should be checked before applying the clamp?
  5. Does applying a sample-size cap, by itself, prove that the final test has type I error \(\alpha\)?

Solutions. (1) \(n_{\mathrm{c}}=\min\{900,\max(500,450)\}=500\) per group. (2) The target \(750\) is within the total-enrollment interval, so it remains \(750\) total; the units matter because per-group and total counts differ by a factor of two under equal allocation. (3) No. The constrained target is \(1{,}000\), but conditional power must be evaluated there; the cap may prevent attainment of the target. (4) Check whether the algorithm permits a final size below the number already enrolled. In an ordinary fixed-final-size design it must not; the minimum should be compatible with accrued enrollment. (5) No. Type I error control depends on the complete adaptive design and final analysis, not on the clamp formula alone.

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