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Sample Size, Power and Assurance · Tutorial 256 of 696

Adjusting Assurance for Attrition and Nonadherence

Learn to represent attrition and nonadherence in assurance calculations, quantify their effects, and translate an evaluable-sample target into randomized enrollment.

Advanced 12 min read

What You'll Learn

  • Distinguish missing outcomes from treatment nonadherence in trial simulations.
  • State assumptions that justify a simple nonadherence dilution model.
  • Adjust Bayesian success calculations for the number of retained participants.
  • Calculate assurance under a design distribution for the treatment effect.
  • Convert an evaluable-sample target to expected randomized enrollment.
  • Diagnose when simple attrition adjustments are inadequate.

Why Retention and Adherence Belong in Assurance

An assurance calculation describes the probability that a planned trial will meet its success rule across the design scenarios and trial outcomes considered plausible. Its simulation should therefore represent not only the randomized sample size, but also what may happen after randomization. Dropout can reduce the information in the data; nonadherence can change the treatment effect that the randomized comparison is intended to estimate. These are different mechanisms, and a single “inflation factor” does not generally capture both.

The essential planning question is: what data would the proposed trial actually produce, and what effect would its success rule assess? In a simulation, the answer depends on the assumptions about retention, adherence, outcomes after nonadherence, and the analysis of missing outcomes. Those assumptions should be explicit before estimating assurance.

Definition: Attrition is the loss of participants from the planned outcome assessment, so that their outcomes are unavailable or not used in the analysis. Nonadherence is a departure from the assigned treatment condition. Attrition affects which outcomes are observed; nonadherence can affect the outcomes themselves. A trial’s assurance is the probability, averaged over its specified design distribution and trial-data generation process, that its decision rule declares success.

A Planning Model and Its Assumptions

Consider a two-arm trial with \(n\) participants randomized to each arm. Let \(r_T\) and \(r_C\) be the probabilities that a randomized participant in the treatment and control arms, respectively, has an analyzable endpoint. Let \(\delta\) denote the treatment effect under full adherence, measured in endpoint units. In a simple dilution model, a fraction \(q\) of treatment-assigned participants does not adhere, and their mean outcome is assumed to be the same as it would have been under control. The mean contrast by randomized assignment is then \(\theta=(1-q)\delta\).

This model is useful for illustration, but it is not a universal description of nonadherence. It assumes a common adherence probability, no effect of nonadherence other than foregoing the treatment effect, and no additional effect of treatment discontinuation on outcomes. For the examples below, endpoint outcomes are modeled as independent normal observations with a common known standard deviation \(\sigma\); the likelihood uses the diluted mean contrast and treats the residual variance as unchanged. This working approximation does not include extra outcome variability from individual-level mixtures of adherers and nonadherers. A realistic design may need to simulate adherence for each participant and include the resulting variability and analysis model.

For attrition, the simplest model draws observed counts independently as \(n_T\sim\operatorname{Binomial}(n,r_T)\) and \(n_C\sim\operatorname{Binomial}(n,r_C)\). This assumes endpoint availability is independent across participants and does not depend on their unobserved outcomes, conditional on the modeled variables. If dropout depends on prognosis, response, or treatment experience, complete-case analysis can be biased; reducing the sample size alone does not repair that bias.

Modeling principle: Simulate the process that the analysis will face. Draw retention and adherence according to the design assumptions, generate outcomes under those conditions, apply the planned analysis and success rule, and record whether the trial succeeds. Average the success indicators over simulated trials to estimate assurance.

How the Bayesian Success Rule Changes With Observed Sample Size

For a transparent calculation, suppose the analysis prior for the treatment-policy mean contrast \(\theta\) is \(N(0,\tau^2)\), and the observed difference in sample means is \(D=\bar{Y}_T-\bar{Y}_C\). With independent normal outcomes and known standard deviation \(\sigma\), the likelihood is \(D\mid\theta\sim N(\theta,v)\), where \(v=\sigma^2(1/n_T+1/n_C)\). The posterior is normal, with mean \(\tau^2D/(\tau^2+v)\) and variance \(\tau^2v/(\tau^2+v)\).

Suppose success requires posterior probability greater than 0.975 that \(\theta>0\). Since the posterior is normal, this is equivalent to the posterior mean divided by posterior standard deviation exceeding the standard-normal quantile at 0.975. For a positive \(D\), the resulting data threshold is:

$$ P(\theta>0\mid D)>0.975 \quad\Longleftrightarrow\quad D>\Phi^{-1}(0.975)\frac{\sqrt{v(\tau^2+v)}}{\tau}. $$

The exact cutoff is \(\Phi^{-1}(0.975)\); its familiar value 1.96 is a rounded approximation. The threshold increases as \(v\) increases, so fewer observed participants can make the success rule harder to meet. In an assurance simulation with random observed counts, the threshold should be recalculated using each replicate’s \(n_T\) and \(n_C\), rather than using the planned randomized count as if everyone supplied an outcome.

Worked Examples

Worked Example: Dilution From Nonadherence

Suppose full adherence would produce a mean improvement of 5 units, and 20% of participants assigned to treatment are expected not to adhere. Under the simple dilution assumptions, the randomized-group treatment contrast is:

$$ \theta=(1-q)\delta=(1-0.20)(5)=4.00\text{ units}. $$

Thus the planning model generates a 4-unit mean contrast by randomized assignment, not a 5-unit contrast. An assurance calculation that generates outcomes around 5 units while claiming to represent this nonadherence assumption would overstate the expected treatment-arm mean in this model. This arithmetic is not appropriate if nonadherers receive an alternative treatment, discontinue for outcome-related reasons, or otherwise have outcomes that differ from the control condition; those possibilities require a more detailed outcome model.

Worked Example: Converting Evaluable Participants to Enrollment

Suppose a design calculation requires 100 endpoint observations per arm and the expected retention probability is 0.80 in each arm. The expected number observed from \(n\) randomized participants in one arm is \(0.80n\). Matching the target in expectation gives:

$$ 0.80n=100 \quad\Longrightarrow\quad n=\frac{100}{0.80}=125. $$

The expected-count adjustment is therefore 125 randomized participants per arm. It does not guarantee that 100 will be observed in each arm. Under the stated independent-retention model, the observed count in one arm is \(\operatorname{Binomial}(125,0.80)\), with standard deviation \(\sqrt{125(0.80)(0.20)}=\sqrt{20}=4.472\) participants. A design needing a high probability of reaching at least 100 observations should evaluate \(P\{\operatorname{Binomial}(n,0.80)\ge100\}\) and select \(n\) accordingly, or incorporate the random count directly in assurance simulations.

Worked Example: Assurance With Dilution and an Evaluable Sample Target

Assume the full-adherence effect has design distribution \(\delta\sim N(5,3^2)\) units, nonadherence is represented by a fixed \(q=0.20\), \(\sigma=10\) units, and the analysis prior is \(N(0,10^2)\). For this calculation, take the observed count to be fixed and equal in both arms. The diluted effect has distribution \(\theta=0.80\delta\sim N(4,2.4^2)\). We seek the smallest observed count per arm giving assurance at least 0.65 under the posterior success rule above.

At \(n_T=n_C=93\), the sampling variance of \(D\) is \(v=100(1/93+1/93)=200/93=2.15054\). The data threshold for posterior success is:

$$ c=\Phi^{-1}(0.975)\frac{\sqrt{2.15054(100+2.15054)}}{10} \approx 2.905\text{ units}. $$

Averaging over the design distribution for \(\theta\) and sampling variation gives \(D\sim N(4,2.4^2+2.15054)\), with standard deviation \(\sqrt{7.91054}=2.813\) units. Therefore:

$$ \text{Assurance} =P(D>c) =\Phi\left(\frac{4-2.905}{2.813}\right) =\Phi(0.389) \approx 0.651. $$

For 92 observed participants per arm, \(v=200/92=2.17391\), the threshold is approximately 2.921 units, and the assurance is \(\Phi((4-2.921)/\sqrt{5.76+2.17391})\approx\Phi(0.383)=0.649\). Thus 93 is the first of these adjacent integer counts to meet the 0.65 target. If retention is 0.80 in both arms, enrolling \(93/0.80=116.25\), or 117 per arm, gives an expected observed count of 93.6 per arm. That conversion alone does not preserve the 0.651 assurance, because actual observed counts vary. To plan assurance for randomized enrollment, simulate the binomial retention counts and recompute the decision rule for each replicate.

Worked Example: Applying the Decision Rule to One Simulated Trial

Suppose a hypothetical simulated trial randomizes 100 participants per arm, with 78 treatment-arm and 81 control-arm outcomes available. The simulated difference is \(D=3.80\) units. With \(\sigma=10\) and analysis prior \(N(0,10^2)\), the likelihood variance is:

$$ v=100\left(\frac{1}{78}+\frac{1}{81}\right)=2.5166. $$

The posterior mean is \(100D/(100+v)=100(3.80)/102.5166=3.707\), and the posterior standard deviation is \(\sqrt{100v/(100+v)}=\sqrt{2.4548}=1.567\). Hence \(P(\theta>0\mid D)=\Phi(3.707/1.567)=\Phi(2.365)\approx0.991\), which exceeds 0.975. This simulated trial meets the decision rule. One successful replicate is not an assurance estimate: assurance is the proportion of successes across the full set of simulated trials.

Simulation Workflow, Diagnostics, and Alternatives

A practical simulation keeps the sources of uncertainty visible. In each replicate, draw the true effect from the chosen design distribution, generate adherence and endpoint availability, generate the outcomes conditional on those events, fit the planned Bayesian analysis, and apply the prespecified success rule. If the design distribution itself includes uncertainty about retention or adherence, draw those rates as well. The final estimate is the number of simulated successes divided by the number of replicates; Monte Carlo uncertainty should be assessed using the methods for simulation error and rare-event assurance.

1
Specify the target contrast.
State whether the simulated effect represents full adherence or the effect under randomized assignment, and make the outcome assumptions explicit.
2
Generate post-randomization events.
Draw retention and adherence according to the assumed arm-specific models. Record unavailable outcomes rather than silently replacing them.
3
Generate and analyze outcomes.
Use the trial’s planned outcome and missing-data model; calculate the posterior and evaluate the actual success rule.
4
Summarize and challenge the result.
Estimate assurance, quantify simulation error, and repeat under plausible less favorable retention, adherence, and outcome assumptions.

Useful diagnostics include checking simulated retention and adherence rates against their specified probabilities, verifying that observed counts vary as expected, and confirming that the analysis uses those counts correctly. Plotting success rates across effect sizes or retention scenarios can reveal unexpected patterns. If lowering the sample size appears to improve assurance, inspect the success boundary and outcome generation: added noise can sometimes increase threshold-crossing probability in a particular high-effect scenario, so the pattern should be understood rather than assumed impossible. Compare simulated results with a simpler fixed-count calculation when its assumptions apply.

Expected-count inflation is a convenient alternative for a quick enrollment approximation, but it ignores the distribution of retained counts and any interaction between counts and the success rule. A fixed dilution factor is similarly convenient, but can miss participant-level heterogeneity and differential adherence. Full simulation is more flexible and can represent random retention and adherence, but it is only as credible as its assumptions and implementation. It does not correct a misspecified missing-data model or justify an inappropriate analysis.

Common Pitfall and Limits

A common mistake is to divide the required sample size by the retention rate and then divide again by the adherence rate, treating both as if they were the same kind of loss. Attrition reduces the number of observed outcomes; nonadherence need not remove observations at all and may instead change the treatment contrast. The correction is to model each mechanism in its proper place: retention in the data-availability process, and nonadherence in the outcome-generating process and the effect being analyzed.

The examples use a simple normal model and a particular success rule. In practice, dropout may be informative, adherence may vary over time, and outcomes after discontinuation may differ from either assigned treatment condition. The model should reflect the trial’s intended interpretation and analysis; the detailed choice of estimand and handling of intercurrent events requires careful specification. Sensitivity analyses are especially important when retention or adherence assumptions are uncertain.

Check Your Understanding

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

  1. If full-adherence effect is 8 units and the fixed dilution fraction is 0.25, what mean contrast does the simple model generate?
  2. With retention probability 0.75, how many participants per arm are needed to obtain 90 observed outcomes per arm in expectation?
  3. Why is that expected-count enrollment not a guarantee of 90 observed participants?
  4. In the normal known-variance model, what happens to \(v=\sigma^2(1/n_T+1/n_C)\) when either observed count falls?
  5. Why should an assurance simulation not treat nonadherence as missing outcomes unless the endpoint is actually unavailable?

Solutions. (1) \((1-0.25)(8)=6\) units. (2) \(90/0.75=120\) randomized per arm. (3) The observed count is random; under independent retention it follows a binomial distribution, so it can fall below its expectation. (4) \(v\) increases, reflecting less precise information. (5) Nonadherence can alter an observed participant’s outcome without making that outcome missing; representing it as dropout changes both the data process and potentially the effect being assessed.

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