Expected Sample Size And Enrollment Cost

Why Average Enrollment Is Not the Whole Design

Two sample-size re-estimation rules can both meet a prespecified power target and control the type I error rate, yet require different amounts of enrollment, staff time, and funding on average. A rule that usually stops at a modest sample size but occasionally reaches a high cap may have a favorable mean and still create substantial capacity risk. Comparing rules therefore requires both statistical operating characteristics and operational consequences.

The expected final sample size is not a fixed number that belongs to a rule independently of the trial conditions. It is calculated under a specified data-generating scenario: for example, a particular treatment effect and variance. The interim data, and hence the sample size selected by the rule, vary across hypothetical repetitions of the trial. Expected enrollment averages over that variation under the stated scenario.

Definition: Let $N$ denote the final total number of participants enrolled under a sample-size re-estimation rule. Under a specified scenario $\theta$, the expected sample size is $E_{\theta}[N]$, the average final enrollment over repeated trials generated under that scenario. If the rule can produce sample sizes $n_1,\ldots,n_J$, with probabilities $q_1,\ldots,q_J$ under $\theta$, then $E_{\theta}[N]=\sum_{j=1}^{J}q_j n_j$, where $\sum_{j=1}^{J}q_j=1$.

Here, a “scenario” specifies the assumptions needed to generate interim data, including the treatment effect and relevant nuisance parameters. Different scenarios can give different probabilities of selecting each sample size, so they can give different expected enrollment. If results are estimated by simulation, the simulated average estimates the expectation for the scenario used in that simulation; it is not automatically an average across every possible clinical setting.

Calculating Expected Enrollment

For an adaptation rule, the final $N$ is a random variable because it depends on the interim data. Once the possible final sample sizes and their probabilities under a chosen scenario are known, expected enrollment is a probability-weighted average. The probabilities must describe the actual rule, including its minimum, maximum, and any rounding or allocation requirements. In particular, probabilities for all possible outcomes must sum to one.

A short derivation clarifies why the weighted average is the appropriate summary. If $N$ takes the value $n_j$ when outcome $j$ occurs, write $I_j=1$ for that outcome and $I_j=0$ otherwise. Exactly one outcome occurs, so $N=\sum_j n_j I_j$. Taking expectations and using linearity gives:

E_{\theta}[N] =E_{\theta}\left[\sum_{j=1}^{J}n_j I_j\right] =\sum_{j=1}^{J}n_j E_{\theta}[I_j] =\sum_{j=1}^{J}q_j n_j.

This calculation describes the expected final total, not the number recruited after the interim analysis. If $n_{\mathrm{int}}$ participants have already been enrolled at the interim, then the expected additional enrollment is $E_{\theta}[N-n_{\mathrm{int}}]=E_{\theta}[N]-n_{\mathrm{int}}$, provided every possible final sample size is at least $n_{\mathrm{int}}$. Reporting both quantities can help distinguish the total trial footprint from the remaining recruitment commitment.

Worked Example: Comparing Expected Enrollment

Suppose a hypothetical trial has enrolled 120 participants by the interim analysis. Two candidate rules have already been checked for their statistical operating characteristics under the planning scenario: each has a simulated type I error estimate of 0.050 and power of 0.90. Rule A finishes with 180 participants with probability 0.60 and 300 with probability 0.40. Rule B finishes with 200 participants with probability 0.80 and 320 with probability 0.20. A fixed design would enroll 240 participants.

For Rule A, the probabilities sum to one, and:

E_{\theta}[N_A] =(0.60)(180)+(0.40)(300) =108+120 =228\text{ participants}.

The expected additional enrollment after the interim is $228-120=108$ participants. Relative to the fixed design, Rule A reduces expected total enrollment by $240-228=12$ participants, or $12/240=0.0500$, which is 5.00%.

For Rule B:

E_{\theta}[N_B] =(0.80)(200)+(0.20)(320) =160+64 =224\text{ participants}.

Rule B therefore has an expected additional enrollment of $224-120=104$ participants and an expected reduction of $240-224=16$ participants, or $16/240=0.0667$, which is 6.67%. Under this scenario, Rule B has the lower expected enrollment by $228-224=4$ participants. However, its maximum is 320 participants, compared with 300 for Rule A. The average alone does not tell a sponsor whether that higher possible enrollment is feasible.

From Enrollment to Recruitment Time

Expected enrollment can be translated into expected recruitment duration only after stating how enrollment proceeds. Suppose participants accrue at a constant rate $r$ participants per month, the same rate applies across candidate rules, and the adaptation creates a fixed pause of $d$ months. If the pause occurs once and enrollment continues at rate $r$ otherwise, then the duration is $T=N/r+d$. Taking expectations gives $E_{\theta}[T]=E_{\theta}[N]/r+d$.

This relationship is not a general law of adaptive trials. It assumes a constant accrual rate and a fixed pause. If rates vary over time, differ by site, or depend on the selected sample size, calendar duration should instead be calculated from the relevant accrual model. A follow-up period may also affect the time until the primary analysis without changing recruitment duration; those are different timelines and should not be combined without explanation.

Worked Example: Converting Expected Enrollment to Duration

Use the expected enrollments from the previous example. Suppose the hypothetical study accrues 24 participants per month and requires a fixed 0.50-month pause to implement either adaptation. Assume accrual is constant and uninterrupted apart from that pause.

For Rule A, expected recruitment time is:

E_{\theta}[T_A] =\frac{E_{\theta}[N_A]}{r}+d =\frac{228\text{ participants}}{24\text{ participants/month}}+0.50\text{ months} =9.50+0.50 =10.00\text{ months}.

For Rule B:

E_{\theta}[T_B] =\frac{224\text{ participants}}{24\text{ participants/month}}+0.50\text{ months} =9.333+0.50 =9.833\text{ months}.

Under these assumptions, Rule B reduces expected recruitment time by $10.00-9.833=0.167$ months, about five days using a 30-day month. This is an expectation, not a promise that an individual trial will finish in that time. Actual duration depends on realized enrollment and operational variation.

Expected Cost and Other Resource Measures

Resource use can include direct trial costs, site and staff capacity, investigational product, laboratory tests, and the administrative work of implementing an adaptation. A transparent calculation separates costs that vary with enrollment from costs incurred regardless of final sample size. For a simple budget model with per-participant cost $c$, fixed cost $F$, and adaptation cost $A$, total cost is $C=F+A+cN$. If $F$, $A$, and $c$ are fixed for a rule, then $E_{\theta}[C]=F+A+cE_{\theta}[N]$. If any costs depend on which sample size is selected, calculate cost within each outcome and then weight those costs by their probabilities.

For resource planning, expected use is only one useful summary. A sponsor may also need the maximum sample size, the probability of exceeding a budget or site-capacity threshold, and the expected use of scarce resources. These quantities answer different questions: the mean describes average use across repetitions, while a hard limit asks whether a particular outcome is feasible.

Worked Example: Comparing Expected Costs and a Budget Ceiling

Suppose the hypothetical trial costs $18,000 per enrolled participant. Rule A has $80,000 in fixed costs and $25,000 in adaptation-related costs. Rule B has $110,000 in fixed costs and $10,000 in adaptation-related costs. Assume these costs do not vary by which sample size the rule selects. Both rules retain the type I error and power performance stated in the first example.

For Rule A, the expected participant-related cost is:

(228\text{ participants})(\$18{,}000/\text{participant})=\$4{,}104{,}000.

Adding fixed and adaptation costs gives $E_{\theta}[C_A]=\$4{,}104{,}000+\$80{,}000+\$25{,}000=\$4{,}209{,}000$.

For Rule B, expected participant-related cost is:

(224\text{ participants})(\$18{,}000/\text{participant})=\$4{,}032{,}000.

Thus $E_{\theta}[C_B]=\$4{,}032{,}000+\$110{,}000+\$10{,}000=\$4{,}152{,}000$. Rule B has expected cost lower by $\$4{,}209{,}000-\$4{,}152{,}000=\$57{,}000$.

Now suppose the sponsor has a hard total-cost ceiling of $5,700,000. Rule A’s maximum cost is $(300)(\$18{,}000)+\$105{,}000=\$5{,}505{,}000$, below the ceiling. Rule B’s maximum is $(320)(\$18{,}000)+\$120{,}000=\$5{,}880{,}000$, above it. Rule B has the lower expected cost but does not satisfy the ceiling in every possible outcome. If exceeding the ceiling is unacceptable, the maximum-cost constraint changes the decision; expected cost alone is not enough.

How to Compare Rules Responsibly

A fair comparison starts by evaluating each candidate under the same scenarios and cost assumptions. Statistical performance remains a constraint: a rule with low expected enrollment is not attractive if it fails the required type I error control or power. Among rules that meet those requirements, expected enrollment, expected duration, and expected resource use help describe operational consequences. It can also be useful to report the maximum enrollment and the probability of crossing a meaningful capacity or budget threshold.

Simulation estimates should be checked for stability. If a rule’s outcomes are generated by simulation, increase the number of repetitions or examine Monte Carlo uncertainty when the reported means or tail probabilities are close to a decision threshold. Repeat the comparison under plausible planning scenarios: because the probability of each final sample size depends on interim data, expected enrollment and cost can change as the assumed treatment effect or nuisance parameters change. Report the scenario alongside each expectation.

These summaries have limitations. Expected cost depends on the completeness of the cost model, and a constant per-participant cost may be unrealistic if later recruitment requires additional sites or overtime. Expected duration based on a single accrual rate may hide site-opening delays or pauses. A single expectation also does not describe the distribution: two rules with the same mean can have very different upper tails. Where capacity matters, examine the distribution or relevant exceedance probabilities in addition to the mean.

Common mistake: “Rule B has the smaller expected sample size, so it is operationally better.” That conclusion ignores the assumptions behind the expectation, the rule’s upper range, and the trial’s hard resource limits. The correction is to compare statistical requirements first, then report expected use alongside relevant maximums, exceedance probabilities, and scenario-specific operational assumptions.

The central comparison is therefore not a single ranking by expected enrollment. It is a structured assessment: verify statistical performance under the intended scenarios, calculate expected enrollment and any derived duration or cost with explicit assumptions, and check whether the full range of possible outcomes can be supported.

Check Your Understanding

Answer each question, then compare with the solutions below.

A rule selects 150 participants with probability 0.70 and 250 with probability 0.30. What is its expected final enrollment?
At a constant rate of 20 participants per month, with a fixed 0.25-month adaptation pause, what is expected recruitment time for a rule with expected enrollment 210?
A rule has expected enrollment 200, a fixed cost of $50,000, an adaptation cost of $15,000, and a per-participant cost of $12,000. What is its expected total cost if those costs are constant across outcomes?
Why can a rule with lower expected enrollment still be infeasible for a trial with a hard enrollment cap?

Solutions: (1) $E[N]=(0.70)(150)+(0.30)(250)=105+75=180$ participants. (2) $E[T]=210/20+0.25=10.5+0.25=10.75$ months, under the stated constant-rate and fixed-pause assumptions. (3) Expected participant cost is $(200)(\$12{,}000)=\$2{,}400{,}000$; adding $50,000 and $15,000 gives expected total cost of $2,465,000. (4) The expectation is an average, not the maximum. A rule may have a low mean but sometimes select an enrollment above the cap, so feasibility also requires checking its possible outcomes or the probability of exceeding the limit.