From an Adaptation Rule to a Calibrated Test
An adaptive sample-size rule answers a design question: how much additional information will be collected after the interim data are observed? It does not, by itself, specify a valid final test. To obtain a complete procedure, the rule must be paired with a final statistic, a rejection boundary, and assumptions that justify the null distribution of that statistic.
This workshop constructs one such procedure. The sample size for stage two depends on the first-stage p-value, while the final test combines the two stagewise results using fixed inverse-normal weights. The key calibration question is whether the data-dependent choice of stage-two sample size changes the probability of falsely rejecting the null hypothesis.
Model, Rule, and Assumptions
Suppose the outcome is continuous, the treatment and control arms have a common known standard deviation \(\sigma\), and each stage has independent observations in the two arms. Let \(n_j\) be the number of participants per arm in stage \(j\), and let \(\bar{x}_{Tj}\) and \(\bar{x}_{Cj}\) be the corresponding sample means. We test for a positive treatment effect, \(H_0:\mu_T-\mu_C=0\) against \(H_a:\mu_T-\mu_C>0\). Under \(H_0\), define the stagewise statistics and p-values by
Here \(\Phi\) is the standard normal cumulative distribution function. Under the null model, each \(Z_j\) is standard normal. The statistics from the two stages are independent because stage-two observations are new and independent of stage one. We also assume that \(\sigma\) is known, the normal model applies, the one-sided direction was selected in advance, and the stagewise p-values are computed as specified. These assumptions give continuous, exactly uniform stagewise p-values under \(H_0\).
For a concrete rule, take \(n_1=40\) participants per arm and choose the stage-two sample size as follows: if \(p_1\leq 0.20\), set \(n_2=40\) per arm; otherwise set \(n_2=80\) per arm. Thus, a less promising interim result triggers the larger stage-two sample. The rule is prespecified, and the final combination weights will not change with \(p_1\).
Use fixed weights \(w_1=0.60\) and \(w_2=0.80\). Since \(w_1^2+w_2^2=0.36+0.64=1\), the inverse-normal statistic and final decision rule at one-sided level \(\alpha=0.025\) are
Equivalently, the combined p-value is \(1-\Phi(C)\), and rejection occurs when that value is at most \(0.025\). The weights are fixed in advance; the sample size rule changes how much stage-two data are collected, not how the two stagewise results are combined.
Worked Example: A Promising Interim Result
Suppose a hypothetical trial observes \(p_1=0.100\). The rule selects \(n_2=40\) participants per arm. Suppose the stage-two data then give \(p_2=0.005\). Calculate the final test.
The standard normal quantiles are \(\Phi^{-1}(1-0.100)=1.2816\) and \(\Phi^{-1}(1-0.005)=2.5758\). Therefore,
The combined p-value is \(1-\Phi(2.8296)\approx 0.00233\), which is below \(0.025\). The procedure rejects \(H_0\) at the prespecified one-sided level. The conclusion follows from the combined test, not from treating either stage’s p-value as the final p-value.
Worked Example: A Larger Stage Two After a Weaker Interim Result
In another hypothetical trial, suppose \(p_1=0.300\), so the rule selects \(n_2=80\) participants per arm. Let the stage-two p-value be \(p_2=0.010\). Determine whether the final test rejects.
The required normal quantiles are \(\Phi^{-1}(0.700)=0.5244\) and \(\Phi^{-1}(0.990)=2.3263\). The combined statistic is
Thus, \(1-\Phi(2.1756)\approx 0.0148\). Since \(2.1756>1.960\), or equivalently \(0.0148<0.025\), the test rejects \(H_0\). The increased sample size enabled a new stage-two result, but it did not change the rejection boundary or the weights.
Why the Null Rejection Probability Is Calibrated
The crucial fact is conditional: under \(H_0\), for every possible first-stage result and every stage-two sample size selected by the rule, the independently collected stage-two statistic remains standard normal. In particular, conditioning on the stage-one data determines \(n_2\), but it does not select or alter the independent stage-two observations.
The independence assertion follows directly from the conditional distribution. For any sets \(A\) and \(B\),
The conditional probability in the middle is constant because the conditional distribution of \(Z_2\) is standard normal for every stage-one outcome. Thus the two statistics are independent. Their weighted sum is normal with mean zero and variance \(w_1^2+w_2^2=1\), which gives the stated rejection probability.
The same result can be seen by conditioning directly on \(Z_1=z\). The test rejects when \(Z_2>(1.960-0.60z)/0.80\). Under the null, its conditional rejection probability is therefore
Averaging this quantity over the standard normal distribution of \(Z_1\) gives the overall Type I error. Equivalently, because \(C\sim N(0,1)\),
Worked Example: Checking the Calibration at Two Interim Values
Under the null, consider the conditional rejection probability when \(Z_1=0\) and when \(Z_1=2\). These are illustrations of conditional error, not the overall Type I error.
At \(Z_1=0\), the rejection condition requires \(Z_2>1.960/0.80=2.450\). The conditional rejection probability is \(1-\Phi(2.450)\approx 0.00714\).
At \(Z_1=2\), the boundary for stage two is \(Z_2>(1.960-0.60(2))/0.80=0.950\). The conditional rejection probability is \(1-\Phi(0.950)\approx 0.1711\).
These conditional probabilities differ because a larger first-stage statistic contributes more toward rejection. They do not imply that the overall Type I error exceeds \(0.025\): the overall error averages the conditional probabilities across all possible \(Z_1\) values, weighted by their null distribution. The normal-combination result establishes that this average is \(0.025\).
Diagnostics, Alternatives, and Limitations
A calibration check should follow the actual algorithm from beginning to end: generate stage one under \(H_0\), apply the sample-size rule, generate independent stage-two data at the selected size, compute both stagewise p-values, and apply the fixed combination test. In simulation, the estimated rejection proportion should be compatible with \(0.025\), allowing for Monte Carlo error. For example, with one million simulated trials, the standard error at a true rejection probability of \(0.025\) is approximately \(\sqrt{0.025(0.975)/1{,}000{,}000}=0.000156\). A discrepancy much larger than ordinary simulation error warrants checking the implementation and assumptions.
The exact guarantee depends on the procedure actually matching its specification. Common failure points include using stage-two observations in the adaptation decision, changing the weights after seeing the interim result, reusing stage-one observations in the stage-two statistic, or using p-values that are not conditionally valid for the selected stage-two design. Any of these can invalidate the independence or conditional-distribution argument. If variance is estimated, outcomes are discrete, or nuisance parameters affect the stagewise null distribution, the simple exact normal argument may no longer apply; calibration must then be justified for that setting rather than assumed.
The earlier material on conditional error and combination tests provides two broad ways to protect Type I error after adaptation. This workshop uses a combination test with fixed weights: it is straightforward when independent, valid stagewise p-values are available. Conditional-error recalibration instead preserves a specified conditional rejection probability and can accommodate a suitably recalibrated second-stage test. A naive pooled test is not a safe substitute: its fixed-design reference distribution need not account for how the sample size was selected. These approaches are alternatives, not interchangeable calculations; the final rule must be calibrated as a whole.
The result also has limits. It guarantees Type I error under the stated null model and assumptions; it does not guarantee a particular power, make a sample-size rule scientifically sensible, or correct bias in the data collection. Nor does the rule ensure that a larger sample is always chosen when evidence is weaker unless that relationship is explicitly built into the rule. Keep the adaptation algorithm, stagewise analysis, weights, and decision boundary prespecified and reproducible.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For the rule in this tutorial, what stage-two sample size is selected when \(p_1=0.18\)? What must remain fixed for the stated calibration argument to apply?
- Under the null, why does selecting \(n_2\) using \(p_1\) not by itself make \(Z_2\) dependent on \(Z_1\)?
- Suppose \(w_1=0.60\) and \(w_2=0.70\). Is the stated \(N(0,1)\) calibration valid without changing anything? Explain.
- For \(p_1=0.050\) and \(p_2=0.020\), calculate \(C\) using the tutorial’s weights and decide whether to reject at \(\alpha=0.025\).
- Give one example of a change to the procedure that could invalidate the calibration argument.
Solutions. (1) Since \(0.18\leq0.20\), choose \(n_2=40\) per arm. The weights, combination statistic, and boundary must remain as specified, and the stage-two p-value must retain its conditional null distribution. (2) The stage-two observations are new and independent; conditional on the stage-one data and selected \(n_2\), \(Z_2\) is still standard normal. Its conditional distribution does not depend on the first-stage result, which gives independence. (3) No. The weight variance sum is \(0.60^2+0.70^2=0.85\), not 1, so the combination statistic has null variance 0.85 under independence. The stated standard normal boundary is not calibrated for that statistic. (4) The quantiles are \(\Phi^{-1}(0.950)=1.6449\) and \(\Phi^{-1}(0.980)=2.0537\). Thus \(C=(0.60)(1.6449)+(0.80)(2.0537)=0.9869+1.6430=2.6299\). Since \(2.6299>1.960\), reject \(H_0\); the combined p-value is approximately \(0.00427\). (5) For example, choosing the combination weights after inspecting \(p_1\) changes the final statistic based on interim data and is not covered by the fixed-weight argument. That altered procedure would need its own calibration.