Why Recruitment Targets Exceed Analyzable Targets
A sample size calculation gives the number of participants whose data must be usable for the planned analysis. It does not, by itself, account for participants who withdraw, fail to complete required study periods, or have pharmacokinetic (PK) profiles that cannot support the endpoint calculation. The number randomized or recruited therefore often needs to be larger than the number required for analysis.
This adjustment is a planning calculation, not a guarantee. Its simplest form assumes that each recruited participant has the same probability of providing analyzable data, and that this probability is known well enough to use as a planning value. The consequences of violating those assumptions can be substantial: fewer usable observations usually reduce power, while losses related to treatment or outcomes may also compromise the interpretation of the comparison.
The calculation assumes that the starting value \(n_A\) is already the appropriate analyzable sample size for the study design and planned analysis. It should not be confused with a new power calculation: it adjusts recruitment to compensate for anticipated losses. In a parallel-group study, the calculation is generally made separately for each group. In a crossover, the relevant retention probability is often the probability of having the required usable data for the subject-level comparison, not merely the probability of completing any one period.
Worked Example: Inflating a Parallel-Group Target
Suppose a hypothetical parallel-group bioequivalence study requires \(n_A=26\) analyzable participants in each group. Based on planning assumptions, 15% of randomized participants in either group may not provide analyzable data. Find the recruitment target per group using expected-count inflation.
The assumed retention probability is \(q=1-0.15=0.85\). Thus, $$ n_R=\left\lceil\frac{26}{0.85}\right\rceil =\left\lceil30.588\right\rceil =31. $$
At 31 randomized participants per group, the expected analyzable count is \(31(0.85)=26.35\). The recruitment target is therefore 31 per group, or 62 in total. The expected count is slightly above 26 because participants must be recruited in whole numbers.
This calculation does not mean that 31 recruits will always yield at least 26 analyzable participants. It only makes the expected count at least 26. If reaching the target with a specified high probability matters, a probability-based calculation is more appropriate.
Expected Counts Versus Probability of Meeting the Target
Under independent, equal-probability retention, the number of analyzable participants among \(n_R\) recruits follows a binomial distribution: \(X\sim\operatorname{Binomial}(n_R,q)\). Expected-count inflation ensures \(E(X)=n_Rq\geq n_A\), but does not set \(P(X\geq n_A)\) to any particular value. Because the count varies around its expectation, a study can fall short even when its expected count equals the target.
If the design requires, for example, at least a 90% chance of reaching an analyzable target, recruitment can instead be selected using the binomial tail probability: $$ P(X\geq n_A) =\sum_{k=n_A}^{n_R}{n_R\choose k}q^k(1-q)^{n_R-k}. $$ The smallest \(n_R\) for which this probability meets the chosen assurance level is the recruitment target under this model. This is a probability-of-sufficient-retention calculation, not the study’s bioequivalence power: it addresses whether enough participants will remain analyzable, conditional on the retention assumptions.
Worked Example: Planning for a High Probability of Sufficient Retention
A hypothetical study needs at least 26 analyzable participants in a group, and the planning retention probability is \(q=0.85\). Illustrate a target with approximately a 90% or greater chance of reaching 26, rather than merely an expected count of 26.
For \(n_R=33\), the expected count is \(33(0.85)=28.05\). The probability of reaching 26 is calculated from $$ P(X\geq26) =\sum_{k=26}^{33}{33\choose k}(0.85)^k(0.15)^{33-k}. $$ A normal approximation with continuity correction has mean \(28.05\) and standard deviation \(\sqrt{33(0.85)(0.15)}\approx2.051\). It gives $$ P(X\geq26)\approx P\left(Z\geq\frac{25.5-28.05}{2.051}\right) =P(Z\geq-1.243)\approx0.893. $$
For \(n_R=34\), the mean is \(34(0.85)=28.90\), and the standard deviation is \(\sqrt{34(0.85)(0.15)}\approx2.082\). The corresponding approximation is $$ P(X\geq26)\approx P\left(Z\geq\frac{25.5-28.90}{2.082}\right) =P(Z\geq-1.633)\approx0.949. $$
Thus 34 recruits per group gives an approximate probability above 90% under the stated model, while the approximation for 33 is just below 90%. A final target should be checked using the exact binomial tail when that threshold is important. The normal calculations are illustrations; they depend on the assumed independent, common retention probability and do not account for uncertainty in that probability.
Dropout and Nonevaluable PK Profiles Are Not the Same
A participant may remain in the study but still lack a usable profile for a particular PK endpoint. Examples include a missed sample near a critical part of the concentration-time profile, a collection or processing problem, or a prespecified reason that a concentration cannot be used. Conversely, a participant may withdraw after providing some data. Whether those data contribute depends on the design and analysis; the participant should not automatically be counted as analyzable just because one period or some samples were completed.
For a crossover comparison requiring usable profiles from both periods, a simple planning model can multiply period-specific probabilities only if the relevant profile-availability events are independent. If availability in period 1 is \(q_1\) and in period 2 is \(q_2\), then under independence the probability of an evaluable pair is \(q_{\text{pair}}=q_1q_2\). If the events are dependent, this product is not justified: the joint probability must be estimated or specified directly. A common participant-level problem could make failures in the two periods strongly associated.
Worked Example: Allowing for Paired Profile Availability
A hypothetical two-period crossover requires 24 participants with usable profiles in both periods. Planning assumptions are a 94% chance of a usable profile in period 1 and a 90% chance in period 2. Assume, for this illustration only, that profile usability across periods is independent and that there are no additional losses.
The probability that one randomized participant has both profiles usable is $$ q_{\text{pair}}=0.94(0.90)=0.846. $$ The expected-count recruitment target is therefore $$ n_R=\left\lceil\frac{24}{0.846}\right\rceil =\left\lceil28.369\right\rceil =29. $$ At 29 randomized participants, the expected number with both profiles is \(29(0.846)=24.534\), approximately 24.5 participants.
The product \(0.94(0.90)\) is valid only under the stated independence assumption. If participants likely to have a usable first-period profile are also more likely to have a usable second-period profile, or if a common operational issue affects both periods, the joint probability may differ. Planning should then use a defensible estimate of the probability of a usable pair rather than multiplying marginal probabilities automatically.
How Differential Losses Affect Precision and Interpretation
Equal expected retention across groups is a convenient assumption, not a requirement of the inflation formula. If retention differs by group, calculate recruitment targets separately using group-specific probabilities. If the same number is randomized to both groups despite unequal loss rates, the expected analyzable group sizes will differ. For a two-group comparison with independent groups and a common analyzable-outcome variance \(\sigma^2\), the variance of the difference between group means is $$ \operatorname{Var}(\bar{x}_T-\bar{x}_R) =\sigma^2\left(\frac{1}{n_T}+\frac{1}{n_R}\right), $$ where \(n_T\) and \(n_R\) in this equation denote the analyzable counts in the test and reference groups. Unequal counts can increase this variance compared with a balanced allocation having the same total analyzable sample size.
For example, if 40 participants per group are randomized, but expected retention is 75% in the test group and 90% in the reference group, the expected analyzable counts are 30 and 36. The variance multiplier, excluding \(\sigma^2\), is \(1/30+1/36=0.06111\). With 33 analyzable participants in each group, the corresponding multiplier is \(2/33=0.06061\), slightly smaller despite a similar total. More importantly, the study has fewer analyzable participants than a plan requiring 40 per group. The exact effect on power depends on the design, variability, target effect, and analysis; the comparison illustrates why the usable sample size, not just the randomized total, matters.
Differential dropout is not only a precision issue. If loss depends on treatment, response, tolerability, or another factor related to the endpoint, the remaining participants may no longer represent comparable groups. A larger recruitment target can replace numbers but cannot generally remove bias created by outcome-related missingness. The reason for each loss, its timing, and its relationship to treatment and observed data therefore matter alongside the count.
Specify the endpoint, required periods and profiles, and any prespecified usability criteria. Distinguish pre-randomization screening failures from post-randomization losses.
Use a rate appropriate to the design and endpoint. For a crossover requiring both profiles, use the probability of an evaluable pair, not a single-period completion rate.
Use expected-count inflation for a simple expected target, or a binomial probability criterion when the chance of falling short must be controlled.
Evaluate unequal retention, dependence among profile failures, and whether losses may be related to treatment or endpoint values. Revisit the calculation if the assumptions are not credible.
Diagnostics, Alternatives, and Limitations
During conduct, compare the observed recruitment, withdrawal, and endpoint-evaluability rates with the assumptions used for planning. Summaries by group, period, and reason for loss can reveal patterns that an overall attrition percentage hides. A rising rate of unusable profiles in one period, for example, may indicate a collection or operational problem rather than ordinary random loss. Where possible, examine whether the observed loss is associated with recorded participant characteristics, treatment, or available endpoint-related measurements.
The simple alternatives serve different purposes. Expected-count inflation is transparent and easy to communicate, but does not control the probability of falling short. Binomial assurance directly addresses that probability, but relies on a specified retention probability and independence across participants. A fixed recruitment reserve is convenient when supported by relevant experience, but is not a substitute for stating what loss rate it implies. None of these adjustments repairs bias from informative missingness, guarantees that the original power calculation remains appropriate under a changed analysis, or accounts automatically for uncertainty in the assumed loss rate.
A common mistake is to say, “A 15% dropout rate means we should add 15% to the required sample.” If 26 analyzable participants are needed, adding 15% gives \(26(1.15)=29.9\), or 30 recruits. But with 15% loss, those 30 recruits yield an expected \(30(0.85)=25.5\), which is below 26. The correct expected-count calculation divides by the retention probability: \(26/0.85=30.588\), rounded up to 31. Even this corrected calculation targets the expectation, not a guaranteed minimum.
Check Your Understanding
Answer the questions, then compare your reasoning with the solutions.
- A study needs 36 analyzable participants in one group and anticipates 10% non-evaluability. What is the expected-count recruitment target?
- Why does recruiting a number whose expected analyzable count equals the target not ensure the target will be reached?
- A crossover requires usable profiles from both periods. Period-specific usability probabilities are 0.95 and 0.92. What evaluable-pair probability follows if the period events are independent?
- Two groups have different dropout rates. Why might a larger total randomized sample fail to resolve the main concern?
Solutions. (1) Retention is \(q=0.90\), so \(n_R=\lceil36/0.90\rceil=\lceil40\rceil=40\). (2) The analyzable count varies randomly; meeting the target in expectation does not make the probability of reaching it equal to 1. A binomial tail calculation is needed to quantify that probability under an independent, common-retention model. (3) Under independence, \(q_{\text{pair}}=0.95(0.92)=0.874\), or 87.4%. (4) Unequal losses can leave unequal analyzable group sizes and reduce precision; if loss is related to treatment or endpoint values, it can also create bias that additional recruitment alone does not remove.