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Sample Size and Expected Sample Size · Tutorial 223 of 499

Solving for Sample Size at a Target Power

Translate a target power into stage sizes under a fixed MAMS allocation and decision rule, and check that the result meets the target in context.

Advanced 13 min read

What You'll Learn

  • Define scenario-specific power as the probability of meeting a design’s success rule by any permitted stage.
  • Write the power equation for a fixed two-stage rule with independent stagewise tests.
  • Invert that equation for a stage size when allocation and boundaries are held fixed.
  • Check feasibility, integer rounding, assumed effects, and variance before finalising a size.
  • Distinguish marginal arm power from a joint multi-arm success probability.
  • Recognise when the simple inversion does not represent the actual MAMS decision rule.

From a Power Target to Stage Sizes

A sample-size calculation for a MAMS design is not just a request for a total number of participants. The design has a sequence of analyses, and each stage size affects the evidence available at its analysis. The planning question here is precise: under a stated treatment-effect scenario, what stage sizes attain a specified power while the allocation and interim decision rules remain fixed?

Power is scenario-specific. It is the probability that the design declares success under the specified treatment effects and outcome model. In a multi-arm design, that probability may refer to success for one named treatment arm, success for every active arm, or success for at least one arm. Those are different criteria. The target must identify which event is being powered.

Definition: For a specified parameter scenario \(\theta\) and a fixed MAMS design, power is \(P_\theta(\text{the design declares the specified success})\). A stage-size calculation finds stage sizes for which this probability reaches or exceeds a prespecified target, while retaining the stated allocation and decision rules.

The calculation depends on the effect size, outcome variability, allocation, analysis rule, and any fixed stopping boundaries. A target such as 80% is not an unconditional property of a design: it is power against a particular scenario. The introductory material on expected sample size concerns how much enrollment is expected under different scenarios; here the quantity being targeted is the probability of success.

A Power Equation for a Fixed Two-Stage Rule

To make the inversion explicit, begin with a defined, simplified two-stage rule. For one treatment arm and its control, suppose each stage enrolls \(n_j\) participants per group, \(j=1,2\). The continuous outcome has common known variance \(\sigma^2\), observations are independent across participants and stages, and allocation is equal within each stage. Under a treatment effect \(\delta\), the stagewise difference-in-means statistic is

$$ Z_j=\frac{\bar{X}_{T,j}-\bar{X}_{C,j}}{\sigma\sqrt{2/n_j}} \sim N\left(\delta\sqrt{\frac{n_j}{2\sigma^2}},1\right). $$

The two stagewise statistics are independent because they use different participants. Assume the prespecified rule declares success at stage 1 if \(Z_1\ge c_1\); otherwise it continues to stage 2 and declares success if \(Z_2\ge c_2\). The constants \(c_1\) and \(c_2\) are fixed decision boundaries. In a real MAMS design they must come from the design’s prespecified error-control procedure; the values below are illustrative and are not asserted to control a particular familywise error rate.

Let \(q_j=P_\theta(Z_j\ge c_j)\), the probability of crossing the boundary at stage \(j\) under the stated scenario. Independence gives the overall probability of success as the chance of crossing at stage 1 plus the chance of not crossing at stage 1 and crossing at stage 2:

$$ \pi(n_1,n_2;\theta) =q_1+(1-q_1)q_2, \qquad q_j=\Phi\left(\delta\sqrt{\frac{n_j}{2\sigma^2}}-c_j\right). $$

This expression is specific to this rule: stage 2 tests use independent stage-2 data, rather than a cumulative statistic combining both stages. It is useful for showing how a fixed-rule power target can be inverted. For another MAMS rule, the success event and therefore the power equation must match that rule.

Why the Inversion Works

Suppose \(n_1\) is fixed and the target power is \(\pi^\star\). Rearrange the power equation to find the stage-2 crossing probability required to meet the target:

$$ q_2^\star=\frac{\pi^\star-q_1}{1-q_1}. $$

If \(0<q_2^\star<1\) and \(\delta>0\), the normal model gives \(q_2=\Phi\left(\delta\sqrt{n_2/(2\sigma^2)}-c_2\right)\). Because \(\Phi\) is strictly increasing, apply its inverse to both sides, then isolate \(n_2\):

$$ \begin{aligned} \Phi^{-1}(q_2^\star) &=\delta\sqrt{\frac{n_2}{2\sigma^2}}-c_2,\\ n_2 &=\frac{2\sigma^2}{\delta^2} \left[c_2+\Phi^{-1}(q_2^\star)\right]^2. \end{aligned} $$

This is the continuous-valued solution. Stage sizes must ultimately be integers, so round up and recompute power at the rounded size. If the bracketed quantity is negative, the algebraic expression is not a valid positive-size solution to the assumed upper-tail test: it signals that the requested crossing probability is already below the limiting crossing probability as \(n_2\) approaches zero. Feasibility and the interpretation of a zero or minimum stage size must be checked against the actual design constraints.

Worked Example: Solving for the Second-Stage Size

Consider a hypothetical two-stage trial with a continuous outcome of known standard deviation \(\sigma=1\). The design allocates \(n_j\) participants to each group at stage \(j\), uses fixed boundaries \(c_1=2.50\) and \(c_2=1.96\), and follows the independent stagewise rule just defined. Suppose the scenario of interest has \(\delta=0.50\) outcome units, \(n_1=16\) per group, and the target power is 0.80. Find the second-stage size.

First calculate the stage-1 crossing probability. Its mean statistic under the scenario is \(0.50\sqrt{16/2}=1.4142\), so \(q_1=\Phi(1.4142-2.50)=\Phi(-1.0858)\approx0.1389\). Thus the probability of continuing is \(1-q_1=0.8611\).

The stage-2 crossing probability required is \(q_2^\star=(0.80-0.1389)/0.8611\approx0.7678\). The corresponding standard-normal quantile is approximately \(\Phi^{-1}(0.7678)\approx0.7316\). Substitution into the inverted formula gives

$$ n_2 =\frac{2(1)^2}{(0.50)^2} \left[1.96+0.7316\right]^2 =8(2.6916)^2 \approx57.96. $$

Round up to \(n_2=58\) per group. At that size, the stage-2 mean statistic is \(0.50\sqrt{58/2}=2.6926\), giving a boundary-subtracted value \(2.6926-1.96=0.7326\) and \(q_2\approx\Phi(0.7326)\approx0.7681\). The resulting power is \(0.1389+0.8611(0.7681)\approx0.8003\). The rounded design therefore meets the 0.80 target under the stated scenario and assumptions.

Worked Example: Solving for the First-Stage Size

Use the same hypothetical outcome model, boundaries, and fixed-rule design, with \(\sigma=1\) and \(\delta=0.50\). This time, stage 2 is fixed at \(n_2=40\) per group, and the goal is to find a stage-1 size that gives overall power 0.80.

The fixed second-stage crossing probability is \(q_2=\Phi(0.50\sqrt{40/2}-1.96)=\Phi(2.2361-1.96)=\Phi(0.2761)\approx0.6087\). Solving \(\pi^\star=q_1+(1-q_1)q_2\) for \(q_1\) gives

$$ q_1=\frac{\pi^\star-q_2}{1-q_2} =\frac{0.8000-0.6087}{1-0.6087} \approx0.4889. $$

Therefore \(\Phi^{-1}(q_1)\approx-0.0278\). The stage-1 mean statistic must satisfy \(0.50\sqrt{n_1/2}-2.50=-0.0278\), so

$$ n_1 =2\left(\frac{2.50-0.0278}{0.50}\right)^2 \approx48.89. $$

Round up to \(n_1=49\) per group. Then the mean statistic is \(0.50\sqrt{49/2}=2.4749\), and \(q_1=\Phi(2.4749-2.50)=\Phi(-0.0251)\approx0.4900\). Rechecking the full rule gives \(0.4900+(1-0.4900)(0.6087)\approx0.8004\). This is a stage-size solution for the specified rule, not a recommendation to choose \(n_1\) without considering recruitment, stage timing, or other design constraints.

Multi-Arm Targets and Assumption Checks

For multiple experimental arms, calculate power for the success event that matters. If the requirement is at least 80% marginal power for each active arm under a specified scenario, each arm’s marginal success probability must meet 0.80. With a shared control, arm-specific statistics are generally correlated because they reuse control observations. That correlation does not alter each arm’s marginal power calculation, but it does affect joint probabilities such as the chance of at least one rejection or of rejecting all active arms. Such joint targets require the joint distribution and cannot be obtained by treating the arms as independent.

Before using a direct inversion, verify the assumptions that support it:

  • The effect \(\delta\), variance \(\sigma^2\), allocation, and success criterion are specified for the scenario.
  • The stated normal model and standard error are appropriate; here the outcome variance is known and observations are independent.
  • The stagewise statistics are independent and the rule really uses the separate stagewise tests and fixed boundaries described above.
  • The boundaries are the ones that will be used in the design, including any multiplicity adjustment across arms.
  • The calculated size obeys operational constraints, and power is recomputed after rounding.

For example, if allocation is unequal, the standard error is not \(\sigma\sqrt{2/n_j}\). With \(n_{T,j}\) treatment participants and \(n_{C,j}\) control participants, it is \(\sigma\sqrt{1/n_{T,j}+1/n_{C,j}}\). The corresponding mean of the standardized statistic must be used in the power equation. Likewise, if the outcome variance is estimated, binary outcomes are used, or a covariate-adjusted analysis is planned, the assumed distribution and standard error must reflect that analysis rather than being borrowed uncritically from the known-variance example.

Worked Example: Sizing for the Less Responsive of Two Arms

Suppose a hypothetical two-arm MAMS design uses the same fixed rule and equal per-group sizes for each treatment-control comparison. The scenario specifies effects \(\delta_A=0.50\) and \(\delta_B=0.40\) outcome units, \(\sigma=1\), \(n_1=16\) per group, boundaries \(c_1=2.50\) and \(c_2=1.96\), and a requirement that each arm have at least 0.80 marginal power. Find a common stage-2 size sufficient for both arms.

The weaker effect, \(\delta_B=0.40\), has the smaller stage-1 mean statistic: \(0.40\sqrt{16/2}=1.1314\). Thus \(q_{1B}=\Phi(1.1314-2.50)=\Phi(-1.3686)\approx0.0855\). The required stage-2 crossing probability is \((0.80-0.0855)/(1-0.0855)\approx0.7813\), with quantile approximately \(0.7766\). Hence

$$ n_2 =\frac{2}{(0.40)^2}(1.96+0.7766)^2 \approx93.61. $$

Round up to 94 per group. For arm B, the stage-2 mean is \(0.40\sqrt{94/2}=0.40\sqrt{47}\approx2.7423\), so \(q_{2B}=\Phi(2.7423-1.96)=\Phi(0.7823)\approx0.7830\). Its overall power is approximately \(0.0855+0.9145(0.7830)=0.8016\). For arm A, the same size gives a larger mean statistic at both stages, so its marginal crossing probabilities, and therefore its power, are larger. This meets the stated per-arm marginal criterion. It does not by itself calculate the probability that both arms succeed jointly.

Diagnostics, Alternatives, and Limitations

A result should be checked by substituting the rounded stage sizes back into the power equation. If the recalculated power falls below the target, the rounded size is inadequate. Also inspect the required probability \(q_2^\star\): a value outside \([0,1]\) means the target cannot be achieved by the proposed inversion as stated, or is already attained before the stage being solved for. If the assumed treatment effect is zero, increasing sample size does not make a fixed upper-tail test powerful against a positive effect in the same way as under \(\delta>0\); the effect scenario and direction of the test must be checked.

The simple inversion is an alternative to using a conventional single-stage formula for total sample size. A single-stage formula does not account for the probability of early success or the continuation rule, so it generally does not yield the required MAMS stage sizes. Conversely, the displayed closed-form inversion is not a substitute for calculating power under a cumulative-data rule, futility stopping, arm-specific stopping, or a joint multi-arm success criterion. For those designs, write the actual success event under the fixed rules and calculate its probability under the specified model; a direct closed-form inversion may not be available.

A common mistake is to treat the stage-2 crossing probability as the overall power, or to multiply it by the chance of continuation and forget early success. The correct calculation for the defined rule is \(q_1+(1-q_1)q_2\), not \(q_2\) and not \((1-q_1)q_2\) alone. Another pitfall is to change a boundary while adjusting sample size: that changes the decision rule and can change type I error, so it is no longer the fixed-rule problem considered here.

Key takeaway: Hold the allocation and interim rule fixed, express power as the probability of the design’s full success event under the stated scenario, solve for the unknown stage size, then round and recompute. A correct answer is conditional on the model, effect scenario, and decision rules used.

Check Your Understanding

Use the independent two-stage rule in this tutorial where applicable. The solutions give the calculation steps as well as the answers.

  1. In the displayed rule, explain why \(q_1+q_2\) is not the overall power.
  2. For \(\delta=0.50\), \(\sigma=1\), \(n_1=16\), and \(c_1=2.50\), calculate \(q_1\). What is the continuation probability?
  3. If \(q_1=0.20\) and the target power is 0.80, what stage-2 crossing probability is required?
  4. Why does meeting an 80% marginal power target for each of two arms not establish an 80% probability that both arms succeed?

Solutions. (1) The stage-2 test is reached only when stage 1 does not cross, and stage-1 success already counts. Thus power is \(q_1+(1-q_1)q_2\), not the sum of two unconditional crossing probabilities. (2) The stage-1 mean is \(0.50\sqrt{16/2}=1.4142\), so \(q_1=\Phi(1.4142-2.50)=\Phi(-1.0858)\approx0.1389\). Continuation probability is \(1-0.1389=0.8611\). (3) \(q_2^\star=(0.80-0.20)/(1-0.20)=0.60/0.80=0.75\). (4) The joint event that both arms succeed depends on their joint distribution. With a shared control, their statistics are generally correlated, so two marginal probabilities alone do not determine the probability of joint success.

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