Changing The Maximum Sample Size

Why a Changed Maximum Sample Size Needs a New Boundary

At an interim analysis, investigators may decide that the trial should continue to a different maximum sample size than originally planned. More patients can provide more information, but the ordinary final critical value from the original fixed design does not, by itself, preserve the original type I error after a data-dependent change. The relevant question is how much rejection probability the original test had left, conditional on the interim result, and how to allocate exactly that probability under the revised design.

This tutorial works with a one-sided, two-stage normal test. The interim information is fixed; the adaptation changes the planned total information, not the interim look or the interim statistic. This isolates the effect of changing the maximum sample size. The introductory material on conditional error functions defines the general idea; here it is used to derive and check the revised continuation boundary.

Model and assumptions: Under the null hypothesis \(H_0\), let \(Z_1\) be the standardized interim statistic and \(Z_2\) the standardized statistic from the independent stage-two increment. Assume \(Z_1\) and \(Z_2\) are independent standard normal random variables under \(H_0\). The planned final statistic is \(Z=\sqrt{t}Z_1+\sqrt{1-t}Z_2\), where \(t=I_1/I_0\) is the original interim information \(I_1\) divided by the original planned final information \(I_0\), with \(0<t<1\). The original test rejects for \(Z>c\), where \(c\) is its one-sided critical value.

The normal independent-increments model is appropriate, for example, for a known-variance normal endpoint with information proportional to sample size. It also describes the canonical large-sample normal model when its approximation is adequate. The adaptation must be based only on information available at interim, such as \(Z_1\); the future stage-two data must not influence the choice. Once a new maximum information level is selected, the stage-two statistic must have the stated null distribution conditional on that choice.

Original Conditional Error and the Revised Boundary

For a realized interim statistic \(Z_1=z\), the original test rejects if \(\sqrt{t}z+\sqrt{1-t}Z_2>c\). Solving this inequality for \(Z_2\) gives the original stage-two threshold. Because \(Z_2\) is standard normal under the null, the probability of crossing that threshold is the original conditional error.

Definition: For the specified one-sided normal test, the original conditional error at interim value \(z\) is \(e(z)=P_0(Z>c\mid Z_1=z)\), where \(P_0\) denotes probability under \(H_0\). Equivalently, with \(\Phi\) the standard normal cumulative distribution function, \(e(z)=1-\Phi(q(z))\), where \(q(z)=(c-\sqrt{t}z)/\sqrt{1-t}\).

Now suppose the trial changes its planned final information to \(I_m\), while keeping \(I_1\) fixed. Write \(t_m=I_1/I_m\). The revised final statistic, under the same independent-increments model, is \(Z_m=\sqrt{t_m}Z_1+\sqrt{1-t_m}Z_2\). A continuation boundary \(B_m(z)\) means that the revised test rejects when \(Z_m>B_m(z)\), after observing \(Z_1=z\).

Key result: To make the revised rejection probability conditional on \(Z_1=z\) equal the original conditional error, set \(B_m(z)=\sqrt{t_m}z+\sqrt{1-t_m}q(z)\). Thus \(P_0(Z_m>B_m(z)\mid Z_1=z)=e(z)\), provided the stated conditional normal model holds.

The formula preserves the conditional error by keeping the stage-two threshold at \(q(z)\), even though the final statistic now combines interim and stage-two information using different weights. The boundary is generally a function of the interim result and of the selected information level. It need not equal the original fixed cutoff \(c\).

Derivation and Error-Rate Justification

The derivation is a direct inversion of a conditional normal tail. Given \(Z_1=z\), the revised final statistic has mean \(\sqrt{t_m}z\) and standard deviation \(\sqrt{1-t_m}\), because \(Z_2\) is standard normal and independent of the interim data. Therefore the probability of crossing a proposed boundary \(B_m(z)\) is the standard normal upper tail at the corresponding stage-two threshold.

\begin{aligned} P_0(Z_m>B_m(z)\mid Z_1=z) &=1-\Phi\left(\frac{B_m(z)-\sqrt{t_m}z}{\sqrt{1-t_m}}\right).\\ \text{Set this equal to } e(z)=1-\Phi(q(z)) &\quad\Longrightarrow\quad \frac{B_m(z)-\sqrt{t_m}z}{\sqrt{1-t_m}}=q(z),\\ &\quad\Longrightarrow\quad B_m(z)=\sqrt{t_m}z+\sqrt{1-t_m}q(z). \end{aligned}

The second line follows because \(\Phi\) is strictly increasing, so equal upper-tail probabilities imply equal thresholds. This gives an exact pointwise match under the model, rather than only a match averaged over interim outcomes. If the selected information fraction \(t_m=t_m(Z_1)\) depends on the interim result, the same calculation applies separately at each observed \(z\), using the information level selected there.

The overall null rejection probability is then preserved by conditioning on the interim statistic. If the original test has size \(\alpha\), its conditional error averages to \(\alpha\). The revised conditional rejection probability equals that same conditional error at every interim value, so its average is also \(\alpha\).

\begin{aligned} P_0(\text{reject under revised design}) &=E_0\left[P_0(\text{reject}\mid Z_1)\right]\\ &=E_0[e(Z_1)]\\ &=P_0(\text{reject under original design})=\alpha. \end{aligned}

This argument depends on the conditional distribution used in the boundary calculation being correct. It does not establish protection against model misspecification, nor does it license an adaptation based on future data. It shows why matching the conditional error is sufficient for preserving the original size under the stated assumptions.

Worked Example: Increasing the Maximum Information

Suppose a hypothetical trial has original information fraction \(t=0.50\), one-sided critical value \(c=1.960\), and interim result \(z=1.500\). The trial changes its maximum information so that the revised fraction is \(t_m=0.25\). Find the original conditional error and the revised final-statistic boundary.

First calculate the stage-two threshold under the original test:

q(1.500)=\frac{1.960-\sqrt{0.50}(1.500)}{\sqrt{0.50}} =\frac{1.960-1.060660\ldots}{0.7071068\ldots} =1.271858\ldots

The original conditional error is \(1-\Phi(1.271858\ldots)\approx0.1017\). Thus, conditional on this interim result, the original test had about a 10.17% probability of rejecting under the null.

For the revised information fraction, the boundary is:

\begin{aligned} B_m(1.500) &=\sqrt{0.25}(1.500)+\sqrt{0.75}(1.271858\ldots)\\ &=0.750000+1.101460\ldots\\ &=1.851460\ldots\approx1.8515. \end{aligned}

Check the result by recovering the stage-two threshold: \((1.851460\ldots-0.750000)/0.8660254\ldots=1.271858\ldots\). The revised conditional rejection probability is therefore again about 0.1017.

Worked Example: A Larger Increase in Planned Information

Consider a second hypothetical calculation with original \(t=0.50\), \(c=1.960\), and interim result \(z=1.200\). Suppose the revised information fraction is \(t_m=0.20\). Calculate the conditional error and the corresponding boundary.

The original stage-two threshold is \(q(1.200)=(1.960-\sqrt{0.50}(1.200))/\sqrt{0.50}\). Since \(\sqrt{0.50}(1.200)=0.8485281\ldots\), this gives:

q(1.200)=\frac{1.960-0.8485281\ldots}{0.7071068\ldots} =1.571858\ldots

Consequently, \(e(1.200)=1-\Phi(1.571858\ldots)\approx0.0580\). The revised boundary is:

\begin{aligned} B_m(1.200) &=\sqrt{0.20}(1.200)+\sqrt{0.80}(1.571858\ldots)\\ &=0.5366563\ldots+1.405914\ldots\\ &=1.942570\ldots\approx1.9426. \end{aligned}

The revised standardized stage-two threshold is \((1.942570\ldots-0.5366563\ldots)/0.8944272\ldots=1.571858\ldots\), giving the same conditional error, approximately 0.0580. The smaller information fraction means the trial plans more total information relative to the interim information; it does not mean the boundary can be chosen without accounting for the original conditional error.

Why Reusing the Old Cutoff Can Fail

A common shortcut is to change the maximum sample size but keep rejecting whenever the revised final statistic exceeds the original critical value \(c\). That rule generally has a different conditional rejection probability. It therefore does not satisfy the pointwise condition used in the error-preservation argument. The size of the difference depends on the interim result and on the new information fraction; it is not safe to infer that a larger sample automatically makes the old cutoff valid.

Worked Example: Checking the Fixed-Cutoff Shortcut

Use the first example’s hypothetical values: original \(t=0.50\), \(c=1.960\), \(z=1.500\), and revised \(t_m=0.25\). If the original cutoff is reused, the revised conditional rejection probability becomes:

\begin{aligned} P_0(Z_m>1.960\mid Z_1=1.500) &=1-\Phi\left(\frac{1.960-\sqrt{0.25}(1.500)}{\sqrt{0.75}}\right)\\ &=1-\Phi\left(\frac{1.210000}{0.8660254\ldots}\right)\\ &=1-\Phi(1.397187\ldots)\approx0.0812. \end{aligned}

This is about 8.12%, rather than the original conditional error of about 10.17%. The fixed-cutoff shortcut is conservative at this particular interim value, but that single conditional comparison does not establish the overall behavior for all values or adaptation rules. The conditional-error boundary instead reproduces the original conditional probability by construction.

A combination test is a related alternative: it combines stagewise evidence using a prespecified valid rule, rather than translating the original conditional error into a revised final-statistic boundary. Both approaches can address design changes, but they are not interchangeable calculations. In this tutorial’s normal setup, the conditional-error route makes the relationship between the original and revised rejection regions explicit. Whichever method is used, the adaptation rule and its assumptions must be specified and the resulting test’s null error justified.

Diagnostics, Scope, and Common Pitfalls

Several checks help detect calculation or implementation errors:

Recover the stage-two threshold. Substitute the calculated \(B_m(z)\) into \((B_m(z)-\sqrt{t_m}z)/\sqrt{1-t_m}\). It should equal \(q(z)\), up to numerical rounding.
Check the direction of the information fraction. With fixed interim information, increasing final information decreases \(t_m=I_1/I_m\). Do not confuse the information fraction with the total sample size.
Check the null distribution after adaptation. The derivation requires a standard normal stage-two increment conditional on the interim result and the selected information. If that conditional model fails, the boundary may not preserve the claimed error.
Check the adaptation rule. The new information level must be selected using interim information only. A decision that depends on future stage-two outcomes is not covered by this argument.
Use adequate numerical precision. Retain unrounded quantities in the boundary and tail calculations, then round the reported result. Rounding \(q(z)\) too early can make a boundary appear inconsistent with its implied stage-two threshold.

The calculation is limited to the stated one-sided two-stage normal setting. It assumes no change to the interim information or statistic, and it does not by itself handle a changed correlation structure, a new interim look, a discrete test statistic, or an uncertain nuisance parameter. In those settings the conditional distribution and the relevant conditional error must be established for the actual design rather than borrowed unchanged from this formula.

Common mistake: “The sample size increased, so the original critical value still controls the type I error.” That conclusion does not follow. The adaptation changes how the final statistic depends on the interim and stage-two data. The correction is to calculate the original conditional error at the observed interim result and choose the revised boundary so that its conditional rejection probability matches it.

Calculate the original threshold.
For interim value \(z\), find \(q(z)=(c-\sqrt{t}z)/\sqrt{1-t}\).

Calculate the original conditional error.
Evaluate \(e(z)=1-\Phi(q(z))\), retaining precision for subsequent steps.

Apply the revised information fraction.
Use \(t_m=I_1/I_m\) for the selected maximum information and set \(B_m(z)=\sqrt{t_m}z+\sqrt{1-t_m}q(z)\).

Verify the conditional tail.
Confirm that the revised boundary gives stage-two threshold \(q(z)\), and hence conditional rejection probability \(e(z)\).

Check Your Understanding

Answer each question, then compare with the brief solutions below.

In the two-stage model, what does \(t_m\) represent when the interim information is held fixed?
For an original test with \(t=0.50\), \(c=1.960\), and \(z=0\), write the original stage-two threshold and conditional error.
Why does setting \(B_m(z)=\sqrt{t_m}z+\sqrt{1-t_m}q(z)\) preserve the conditional rejection probability?
State one reason the conditional-error argument may not apply if the sample-size adaptation is implemented differently from the setup here.

Solutions. (1) \(t_m=I_1/I_m\), the interim information divided by the revised planned final information. (2) \(q(0)=1.960/\sqrt{0.50}=2.771859\ldots\), so the conditional error is \(1-\Phi(2.771859\ldots)\), approximately 0.0028. (3) The boundary gives revised stage-two threshold \((B_m(z)-\sqrt{t_m}z)/\sqrt{1-t_m}=q(z)\); the same standard normal tail is therefore used. (4) For example, if the selected sample size depends on future stage-two outcomes, the stated conditional model and pointwise argument do not apply.