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Sample Size and Expected Sample Size · Tutorial 221 of 499

Using Covariate Adjustment in Sample-Size Scenarios

Translate covariate-adjustment efficiency into outcome variance, stage-specific sample sizes, and scenario-based expected enrollment.

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What You'll Learn

  • Define the efficiency assumption using the fraction of outcome variance explained by baseline covariates.
  • Derive the adjusted variance of a treatment-control contrast and its sample-size implication.
  • Translate per-comparison sample sizes into cumulative MAMS stage sizes.
  • Calculate expected enrollment when treatment arms and shared control have different continuation probabilities.
  • Diagnose the consequences of an optimistic efficiency assumption.
  • Distinguish the effect of smaller stage sizes from the effect of changed continuation probabilities.

Why Covariate Adjustment Matters for MAMS Planning

A prespecified baseline covariate can explain part of the outcome variation among participants. If the planned analysis adjusts for such covariates, the treatment-control comparison may be more precise than an unadjusted comparison of the same size. In sample-size planning, that potential gain can translate into fewer participants at each stage, a smaller maximum enrollment, or both.

The gain is not automatic. It depends on how much outcome variation the covariates explain, whether the planning assumption is credible in the trial population, and whether the analysis model is specified appropriately. In a multi-arm multi-stage (MAMS) design, the adjustment also affects interim statistics and therefore can change the probabilities that arms continue. Consequently, reducing stage sizes and recalculating expected enrollment are related but distinct tasks.

Definition: Let \(R^2\) denote the prespecified proportion of outcome variance explained by the baseline covariates used in the planned adjusted analysis. The corresponding variance-efficiency factor is \(1-R^2\). For a continuous outcome under the assumptions below, the planning residual variance is \(\sigma^2_{\mathrm{res}}=\sigma^2(1-R^2)\), where \(\sigma^2\) is the unadjusted outcome variance.

Assumptions and the Variance Calculation

The simple \(1-R^2\) variance reduction is a planning approximation for a continuous outcome. It is appropriate when the covariates are measured before randomization, are prespecified, and are prognostic for the outcome; the adjusted model is correctly specified; and the assumed \(R^2\) is relevant to the trial population. The usual calculation also assumes a common residual variance across randomized groups and a treatment effect that is not changed by adjustment. In practice, a model may include treatment and baseline covariates, with treatment contrasts estimated conditional on those covariates.

For a treatment group with \(n_t\) participants and a shared control group with \(n_0\), the approximate variance of an adjusted difference in means is \(\sigma^2(1-R^2)(1/n_t+1/n_0)\). This is a marginal variance for one comparison. Because MAMS comparisons share control observations, their test statistics are correlated; a joint design calculation must retain that shared-control structure rather than treating the comparisons as independent.

$$ \operatorname{Var}(\widehat{\Delta}_{\mathrm{adj}}) \approx \sigma^2(1-R^2)\left(\frac{1}{n_t}+\frac{1}{n_0}\right). $$

To see how this enters a sample-size calculation, suppose a one-sided comparison has a positive target effect \(\delta\), a normal critical value \(c\), and target power \(1-\beta\). With a normal approximation, the standardized mean under the alternative is \(\delta/SE\). Requiring power \(1-\beta\) means \(\delta/SE-c=z_{1-\beta}\), where \(z_{1-\beta}\) is the corresponding standard-normal quantile. If \(n_0=r n_t\), substitution and rearrangement give the required treatment-group size.

$$ \frac{\delta}{\sigma\sqrt{(1-R^2)(1+1/r)/n_t}}=c+z_{1-\beta} \quad\Longrightarrow\quad n_t= \frac{\sigma^2(1-R^2)(1+1/r)(c+z_{1-\beta})^2}{\delta^2}. $$

This derivation shows the key scaling result: with the effect, allocation ratio, critical value, and power target held fixed, the unrounded sample size is multiplied by \(1-R^2\) relative to the unadjusted calculation. It does not establish that a complete MAMS design's maximum or expected enrollment is reduced by exactly that factor. Stage boundaries, integer rounding, shared-control allocation, and continuation probabilities also matter.

Worked Example: Translating Efficiency Into a Sample Size

Consider a hypothetical continuous outcome measured in points. Suppose the unadjusted standard deviation is 10 points, the target treatment effect is 5 points, equal numbers are assigned to treatment and control, the one-sided critical value is \(c=2.00\), and target power is 80%. Assume a prespecified baseline adjustment explains \(R^2=0.36\) of outcome variance. For this illustration, use the normal approximation and the same critical value for the adjusted and unadjusted calculations.

The adjusted variance factor is \(1-0.36=0.64\), so the planning residual variance is \(10^2(0.64)=64\) points squared. The 80% power quantile is approximately \(z_{0.80}=0.8416\). With equal group sizes, \(r=1\), and the formula gives:

$$ n_t=\frac{64(1+1)(2.00+0.8416)^2}{5^2} =\frac{128(8.0747)}{25} =41.34. $$

Rounding up gives 42 participants per group for this calculation. Without covariate adjustment, the corresponding calculation is \(200(8.0747)/25=64.60\), or 65 per group after rounding up. The unrounded ratio is \(41.34/64.60=0.640\), as predicted by the variance factor. These figures illustrate a single comparison calculation; they do not by themselves guarantee a particular overall MAMS power after interim stopping and multiplicity control.

From a Per-Comparison Calculation to Stage Sizes

A MAMS design uses cumulative sample sizes at successive analyses. For arm \(j\), let \(n_{j,k}\) be the cumulative number assigned to that treatment by stage \(k\), and let \(n_{0,k}\) be the cumulative number assigned to shared control. The stage increment is the new enrollment between consecutive analyses: for example, \(n_{j,k}-n_{j,k-1}\) in treatment arm \(j\). Applying the variance reduction to a per-comparison planning calculation is a starting point for setting these quantities, not a substitute for checking the design's stagewise operating characteristics.

For example, if an illustrative design uses 60% of its maximum per-group sample size at the first analysis, the 42-per-group adjusted calculation above gives \(0.60(42)=25.2\), rounded up to 26 per group at stage 1. The remaining increment is \(42-26=16\) per group. On the unadjusted calculation, \(0.60(65)=39\) per group at stage 1, with a further \(65-39=26\) per group. For two treatment arms and one shared control, these allocations correspond to cumulative stage-1 totals of \(3(26)=78\) and \(3(39)=117\), respectively, and maximum totals of \(3(42)=126\) and \(3(65)=195\). This comparison assumes the same per-group allocation for each arm and control and is an illustration of size translation, not a final calibrated MAMS design.

In an actual design, boundaries and the joint distribution of the test statistics determine power and continuation behavior. If those boundaries are recalibrated, or if the shared-control allocation differs from the simple equal-group illustration, the stage sizes must be checked using those design details. Covariate adjustment does not remove the correlation caused by shared control, and it does not justify scaling every stage quantity blindly.

Worked Example: Building Cumulative Stage Sizes

Suppose a hypothetical two-treatment MAMS design adopts the illustrative adjusted maximum of 42 participants per treatment arm and 42 in shared control. It schedules the first analysis after 60% of each group's maximum enrollment, rounding up as above.

At stage 1, each group has \(26\) participants, so the total enrollment is \(26+26+26=78\). If both treatment arms continue, the next stage recruits \(42-26=16\) additional participants in each group; the stage-2 increment is \(16+16+16=48\). The cumulative maximum is \(78+48=126\), which also equals \(42+42+42\).

If only one treatment arm continues and shared-control recruitment is still required, enrollment after stage 1 depends on the design's rule for adding control participants. For instance, adding 16 more participants to the continuing arm and 16 to shared control would add 32, not 48. Thus, the maximum sample size alone is not enough to obtain expected enrollment: the arm-specific continuation rules and their probabilities are needed.

Expected Enrollment Depends on Continuation Probabilities

Let \(N_1\) be the stage-1 total enrollment. At later stages, a treatment-arm increment is recruited only if that arm remains active, while a shared-control increment may be recruited if at least one treatment remains active. Let \(q_{j,k}\) be the probability that arm \(j\) is active at the start of stage \(k\), and \(q_{0,k}\) the probability that at least one treatment arm is active then. If the increments are fixed whenever their corresponding recruitment rule applies, expected total enrollment is the initial enrollment plus each increment multiplied by its probability of being recruited.

$$ E(N)=N_1+\sum_{k=2}^{K} \left[ (n_{0,k}-n_{0,k-1})q_{0,k} +\sum_{j=1}^{J}(n_{j,k}-n_{j,k-1})q_{j,k} \right]. $$

The expression follows by writing total enrollment as the initial cohort plus each possible later increment times an indicator that the increment is recruited. Taking expectations replaces each indicator with its recruitment probability. The probabilities are not generally interchangeable: the chance that at least one arm continues is not the same as the sum of the arm-specific probabilities, because both arms can continue together.

Worked Example: Expected Enrollment With Shared Control

Consider a hypothetical two-stage design whose stage-1 enrollment is 120 participants. At stage 2, each treatment arm would recruit 30 more participants if it continues, and shared control would recruit 30 more if at least one treatment continues. Suppose, for this planning scenario, the probability that at least one arm continues is 0.53; the arm-specific continuation probabilities are 0.459 and 0.173. Assume these probabilities are consistent with the scenario being evaluated and that stage-2 increments follow the stated rules.

The expected stage-2 shared-control enrollment is \(30(0.53)=15.9\). The expected increments in the two treatment arms are \(30(0.459)=13.77\) and \(30(0.173)=5.19\). Therefore:

$$ E(N)=120+30(0.53)+30(0.459+0.173) =120+15.9+13.77+5.19 =154.86\approx154.9\text{ participants}. $$

The value is below the maximum \(120+30+30+30=210\) participants because some stage-2 increments are not recruited in some scenarios. The expected value is not obtained by multiplying the maximum enrollment by one overall continuation probability: each increment has its own recruitment rule and probability.

Efficiency Assumptions, Diagnostics, and Alternatives

The expected-enrollment formula makes clear why adjustment can affect enrollment through two channels. First, a credible efficiency gain can reduce the stage increments required for a chosen design target. Second, more precise interim statistics can change the continuation probabilities \(q_{j,k}\) and \(q_{0,k}\). The second channel has no universal direction: it depends on the treatment-effect scenario, boundaries, and which arms are active. For a fair scenario comparison, use continuation probabilities appropriate to each planned design rather than assuming they remain unchanged after the variance model changes.

A useful diagnostic is to vary the prespecified \(R^2\) over a plausible range and recompute both the stage-size inputs and scenario-specific continuation probabilities. Check that the assumed \(R^2\) refers to the analysis covariates and outcome in the relevant population, and verify that rounding has not caused a stage increment to become zero or changed the intended ordering of analyses. The joint design calculation should also retain the shared-control correlation and the planned decision boundaries.

Worked Example: Diagnosing an Optimistic Efficiency Assumption

Return to the hypothetical outcome with \(\sigma^2=100\) points squared. The planned \(R^2=0.36\) gives residual variance 64 and an unrounded sample size of 41.34 per group. Suppose the true variance explained in the trial population is instead \(R^2=0.20\). The residual variance is then \(100(1-0.20)=80\) points squared.

Holding the other inputs fixed, the new sample-size requirement is \(41.34(80/64)=51.68\), or 52 participants per group when rounded up. The standard error at a fixed sample size would be larger than planned by \(\sqrt{80/64}=\sqrt{1.25}=1.118\), about 11.8%. Thus, relying on the \(R^2=0.36\) assumption when the actual value is 0.20 can lead to smaller stages and less precision than intended.

An unadjusted analysis is a straightforward alternative: plan using the full outcome variance and do not claim a sample-size gain from covariates. This may be a prudent sensitivity case when the efficiency assumption is uncertain, although it can require more participants. A prespecified adjusted analysis uses the anticipated variance reduction but relies on model and population assumptions. Neither approach makes the treatment effect larger; adjustment changes precision, not the effect measured. For binary outcomes, the continuous-outcome shortcut \(\sigma^2(1-R^2)\) should not be applied mechanically: the variance and efficiency calculation must match the planned binary-outcome model and effect scale.

Common Mistakes and Practical Limits

  • Using \(R^2\) as a percentage reduction in sample size without stating assumptions. The factor \(1-R^2\) applies to the unrounded variance-based calculation under the specified model and fixed design inputs. Stagewise rounding and boundary calibration can change the final design.
  • Assuming adjustment changes the treatment effect. The planning gain described here is a reduction in variance under the stated assumptions, not a change in the target contrast.
  • Scaling expected enrollment by the variance factor. Expected enrollment depends on stage increments and the probabilities that each arm-specific or shared-control increment is recruited. Those probabilities can change when the interim statistics change.
  • Counting shared control once per treatment arm. In a multi-arm design, control recruitment follows the design's shared-control rule; use that rule and the probability that activates it.
  • Treating the assumed efficiency as guaranteed. If the covariates explain less variation than expected, the realized precision and power can be below plan. Include plausible alternatives in sensitivity checks.

The main limitation is that a single \(R^2\) is a summary of a relationship that can vary across populations and outcomes. Moreover, the simple formula does not account by itself for model misspecification, treatment-by-covariate interactions, missing outcomes, or non-normal small-sample behavior. Those features require assumptions or calculations beyond the basic continuous-outcome efficiency adjustment.

Key takeaway: For a suitable continuous-outcome model, a prespecified \(R^2\) changes the planning variance by the factor \(1-R^2\), and therefore scales a fixed-input, unrounded sample-size calculation by the same factor. To obtain MAMS stage sizes and expected enrollment, then account for rounding, shared-control rules, decision boundaries, and scenario-specific continuation probabilities.

Check Your Understanding

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

  1. If \(R^2=0.25\), by what factor does the planning residual variance change, and by what factor does the unrounded sample size change when all other inputs are fixed?
  2. A three-arm MAMS design has two treatment arms. At a later stage, each continuing arm recruits 20 participants and shared control recruits 20 if at least one arm continues. The probabilities are 0.40 for arm 1, 0.30 for arm 2, and 0.55 for at least one arm. What is the expected later-stage enrollment?
  3. Why is it generally incorrect to multiply a design's maximum enrollment by one overall continuation probability?
  4. What should be reconsidered if the actual covariates explain substantially less outcome variance than the planning assumption?

Solutions. (1) The factor is \(1-0.25=0.75\) for both residual variance and the unrounded fixed-input sample size. (2) Expected shared-control enrollment is \(20(0.55)=11\); treatment increments contribute \(20(0.40)+20(0.30)=14\). The expected later-stage total is \(11+14=25\) participants. (3) Different increments have different recruitment rules and probabilities, and the shared-control probability is not the sum of treatment-arm probabilities. (4) Recalculate the residual variance and sample-size inputs under a more plausible efficiency value, and reassess stage sizes, precision, and scenario-specific continuation probabilities.

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