Adjusting Recruitment for Attrition

Why Recruitment Must Exceed the Analyzable Sample Size

A sample-size calculation gives the number of observations needed for the planned analysis under its assumptions. In practice, not everyone recruited will contribute usable data: some participants may withdraw, miss a follow-up visit, or have an outcome that cannot be measured. If recruitment stops as soon as the planned number has enrolled, the final analyzable sample may be too small, and the study may have less power than intended.

The adjustment is a conversion from a required number of analyzable observations to a recruitment target. It depends on an anticipated attrition or unusable-data rate. It does not change the original power calculation; it aims to leave enough observations after expected losses.

Definition: Let \(n\) be the number of analyzable observations required by the planned study design, and let \(r\) be the anticipated fraction of recruited participants who will not provide an analyzable observation. The recruitment target \(N\) is the number to recruit before those losses occur. Under a constant expected retention fraction \(1-r\), the basic adjustment is \(N=n/(1-r)\), rounded up to the next whole participant.

The rate \(r\) is a proportion, not a percentage number: an anticipated 15% loss is \(r=0.15\). The denominator \(1-r\) is the anticipated retention fraction. Dividing by it makes the recruitment target larger than the required analyzable sample whenever \(r>0\).

Deriving the Recruitment Adjustment

Suppose each recruited participant has probability \(1-r\) of contributing analyzable data. For \(N\) recruited participants, the expected number of analyzable observations is the sum of \(N\) individual retention indicators. Each indicator has expectation \(1-r\), so linearity of expectation gives:

$$ E(A)=N(1-r), $$

where \(A\) is the eventual number of analyzable observations. To make this expected count at least the required \(n\), set \(N(1-r)\ge n\). Since \(1-r>0\) for an attrition rate below 100%, division gives \(N\ge n/(1-r)\). Because recruitment must be a whole number, the target is:

N=\left\lceil \frac{n}{1-r}\right\rceil.

The ceiling symbol means round upward, not to the nearest integer. Rounding down could leave the expected analyzable count below the target. The check is straightforward: multiply the recruitment target by the retention fraction and verify that the result is at least \(n\).

Worked Example: A Single-Group Study With 15% Attrition

A hypothetical study requires 120 analyzable participants. Planners expect 15% of those recruited to withdraw or provide unusable outcome data. How many should they recruit?

Here \(n=120\) and \(r=0.15\), so the anticipated retention fraction is \(1-0.15=0.85\). Apply the formula:

N=\left\lceil \frac{120}{0.85}\right\rceil =\left\lceil 141.176\ldots\right\rceil =142.

The expected analyzable count from recruiting 142 is \(142(0.85)=120.7\). Recruiting 141 would give an expected count of \(141(0.85)=119.85\), below the required 120. The recruitment target is therefore 142 participants.

This calculation targets an expected analyzable count of at least 120. It does not mean that exactly 120, or at least 120, will certainly be observed. Actual losses vary from study to study.

Applying the Adjustment to Study Groups

When a study has separate groups, the required analyzable number and the expected attrition rate should be considered for each group. If the planned analysis requires \(n_1\) analyzable observations in one group and \(n_2\) in another, and anticipated attrition rates are \(r_1\) and \(r_2\), calculate the recruitment targets separately:

N_1=\left\lceil\frac{n_1}{1-r_1}\right\rceil, \qquad N_2=\left\lceil\frac{n_2}{1-r_2}\right\rceil.

This matters when rates may differ between groups. Using a single average rate can leave one group short even if the combined expected number looks adequate. The appropriate planned analyzable counts still come from the design and power calculation; attrition adjustment is applied afterward to those counts.

Worked Example: Different Attrition Rates in Two Groups

A hypothetical two-group study requires 80 analyzable participants in each group. Planners anticipate 10% attrition in Group A and 25% in Group B. Find the separate and total recruitment targets.

For Group A, the expected retention fraction is \(0.90\):

N_A=\left\lceil\frac{80}{0.90}\right\rceil =\left\lceil 88.889\ldots\right\rceil =89.

For Group B, the expected retention fraction is \(0.75\):

N_B=\left\lceil\frac{80}{0.75}\right\rceil =\left\lceil 106.667\ldots\right\rceil =107.

The total target is \(89+107=196\) participants: 89 in Group A and 107 in Group B. Their expected analyzable counts are \(89(0.90)=80.1\) and \(107(0.75)=80.25\), respectively. Both meet the planned target in expectation.

Recruiting 98 in each group would also total 196, but would not respect the different expected retention rates: the expected analyzable counts would be 88.2 in Group A and 73.5 in Group B. A total alone is not enough; the allocation between groups matters.

What the Formula Does—and Does Not—Guarantee

The formula is based on an expected count. It is appropriate as a basic inflation when \(r\) is a reasonable estimate of the proportion who will be lost or unusable, and when the same rate is applied to the intended recruitment pool. If the loss of each participant is modeled as independent with the same probability \(r\), then \(A\), the analyzable count, has a binomial distribution with mean \(N(1-r)\). Its value varies around that mean.

Consequently, setting the expected count equal to the minimum required count does not ensure a high probability of reaching that count. For example, a recruitment target whose expected analyzable count is 120 may sometimes yield fewer than 120. If the study requires a specified probability of reaching or exceeding the target, planners need to choose \(N\) using the distribution of the retained count, not only its expectation. The simple inflation formula is not that probability calculation.

There is also uncertainty in the attrition rate itself. A rate estimated from a small or different study may not transfer well to the planned setting. When that uncertainty is important, it is useful to calculate targets under several plausible rates and examine the consequences. A small additional operational buffer may also be justified, but it should be stated and explained rather than hidden in an unexplained increase.

Worked Example: Sensitivity to the Anticipated Loss Rate

A hypothetical study needs 240 analyzable observations. Planners are unsure whether the unusable-data rate will be 10%, 18%, or 25%. Calculate the recruitment target under each scenario.

At 10% attrition, \(r=0.10\):

N_{10\%}=\left\lceil\frac{240}{0.90}\right\rceil =\left\lceil 266.667\ldots\right\rceil=267.

At 18% attrition, \(r=0.18\):

N_{18\%}=\left\lceil\frac{240}{0.82}\right\rceil =\left\lceil 292.683\ldots\right\rceil=293.

At 25% attrition, \(r=0.25\):

N_{25\%}=\left\lceil\frac{240}{0.75}\right\rceil =\left\lceil 320\right\rceil=320.

The plausible rates imply targets from 267 to 320. The study team should choose an anticipated rate that is justified for the planned procedures and population, and should recognize how much the target changes across plausible values. Selecting the highest rate simply because it is the most conservative may not be feasible; selecting the lowest without justification risks ending with too few analyzable observations.

Define Attrition at the Right Stage

The rate must use a clearly defined denominator. If \(r\) is the fraction of people recruited who fail to provide an analyzable outcome, then \(N\) must count people at that same recruitment stage. Do not mix losses before enrollment with losses after enrollment, or count the same person twice.

For example, suppose 10% of people approached decline enrollment, and among those enrolled, 15% later have no usable outcome. If the required number of analyzable observations is \(n\), then the expected fraction of approached people who become analyzable is \(0.90\times0.85=0.765\). The target number to approach would be based on \(n/0.765\), rounded up. If instead the recruitment target is defined as the number already enrolled, only the post-enrollment 15% rate belongs in the adjustment. Clearly naming the stages prevents an incorrect denominator.

Likewise, “attrition” may include withdrawal, missed visits, protocol deviations, failed measurements, or other reasons for unusable outcomes. Specify which losses are included and apply the rate to the relevant analysis population. If different causes overlap, estimate the total fraction lost rather than adding overlapping percentages as if they were separate.

Inflation Does Not Fix Missing-Data Bias

Recruitment inflation addresses the number of observations, not the process that causes observations to be missing. It can help preserve the planned analyzable sample size, but it cannot guarantee that the remaining participants represent the original target population or that a treatment comparison is unbiased. If withdrawal is related to health outcomes, treatment response, or group membership, the analyzed participants may differ systematically from those lost.

For instance, recruiting extra participants may restore the intended count after withdrawals, but it does not by itself correct an estimate distorted because participants with worsening symptoms are more likely to leave one group. The study needs an appropriate plan for preventing, documenting, and analyzing missing data as well as a recruitment target. Nor should planners assume that excess recruitment can always compensate for very high attrition: costs, recruitment capacity, and the validity of the analysis remain relevant.

Common Mistakes

Adding the attrition percentage to the required sample size. For 20% attrition, recruiting \(1.20n\) is not enough: 20% of that total would be lost, leaving only \(0.80(1.20n)=0.96n\). Divide by the retention fraction instead: \(n/0.80=1.25n\).
Rounding to the nearest whole number. Round upward so the expected analyzable count is not reduced by rounding.
Treating the target as a guarantee. The formula matches expected retained count to the requirement; random variation can still leave fewer analyzable observations.
Using one rate for groups without checking. When expected losses differ, calculate group targets separately and add them.
Assuming extra recruitment removes bias. It can address anticipated sample-size loss, but not systematic differences between people with and without observed outcomes.

Key takeaway: Start with the required analyzable count, define the loss rate and its denominator, divide by the expected retention fraction, and round up. Interpret the result as a recruitment target based on expected retention—not as a guarantee of sample size or a remedy for missing-data bias.

Check Your Understanding

Try these questions before reviewing the worked examples.

A study needs 150 analyzable observations and anticipates 20% attrition. What is the recruitment target, and what is the expected analyzable count at that target?
Why is dividing by \(1-r\) more appropriate than multiplying the required sample size by \(1+r\)?
A two-group study needs 60 analyzable observations per group. Attrition is expected to be 5% in one group and 20% in the other. Should the same recruitment target be used for both groups? Explain.
What does the basic attrition formula guarantee about the eventual analyzable sample size? What does it not guarantee?
If 12% decline before enrollment and 10% of enrolled participants later have unusable outcomes, what fraction of those approached is expected to provide analyzable data?