Sample Size From Within-Subject Variability

Why Plan Sample Size Around Precision?

In a crossover drug-drug interaction (DDI) study, each participant receives both treatment conditions, so the exposure ratio is estimated from within-participant comparisons. The number of participants needed depends strongly on how variable those comparisons are. A study planned around a confidence-interval width makes that design choice explicit: it asks how many evaluable participants are needed for the estimate to be sufficiently precise, given an expected exposure ratio and a plausible within-subject variability.

This is a precision calculation, not a guarantee that a study will produce a particular conclusion. The expected ratio locates the anticipated confidence interval; within-subject variability and sample size determine its expected width. For planning, use the estimand defined for the study and a variability estimate appropriate to its endpoint, such as log AUC or log Cmax.

Model for a Crossover Exposure Ratio

Definition: Let \(Y_{iT}\) and \(Y_{iR}\) be the positive exposure endpoints for participant \(i\) under test and reference conditions. Define the within-participant log contrast \(D_i=\log(Y_{iT})-\log(Y_{iR})\). The estimated log exposure ratio is \(\bar{D}\), and the estimated geometric mean ratio is \(\exp(\bar{D})\). For a two-sided confidence interval with confidence level \(1-\alpha\), a paired t interval is \(\bar{D}\pm t_{1-\alpha/2,n-1}s_D/\sqrt{n}\), where \(s_D\) is the sample standard deviation of the log contrasts and \(n\) is the number of evaluable participants.

The planning model assumes that participants are independent of one another; the log contrasts are approximately normally distributed, or the sample is large enough for the t-based interval to be adequate; and the crossover design has no material carryover or period effects that invalidate the intended contrast. It also assumes that the variability estimate represents the study population, endpoint, dosing conditions, and conduct planned. Appropriate sequence allocation, washout, and endpoint measurement remain essential: a sample-size formula cannot repair a biased or poorly executed design.

A common planning input is the within-subject coefficient of variation, \(CV_w\), expressed as a proportion rather than a percentage. If the log-scale within-subject residual standard deviation is \(\sigma_w\), then, under a lognormal model, \(CV_w=\sqrt{\exp(\sigma_w^2)-1}\). Thus \(\sigma_w=\sqrt{\log(1+CV_w^2)}\). When the two period-specific residuals are independent with variance \(\sigma_w^2\), the standard deviation of their difference is \(\sigma_D=\sqrt{2}\sigma_w\). This gives the planning standard error \(\sigma_D/\sqrt{n}\).

From Desired Width to Sample Size

Confidence-interval width is most straightforward to plan on the log scale. Let \(R\) be the expected exposure ratio. If the desired interval is to extend by a multiplicative factor \(F\) above and below \(R\), its log-scale half-width is \(h=\log(F)\). For example, a 20% multiplicative margin corresponds to \(F=1.20\); because the interval is symmetric on the log scale, the lower factor is \(1/1.20\), not \(0.80\).

More generally, if the desired ratio-scale limits are \(L\) and \(U\), and the interval is centered on the expected ratio \(R\) on the log scale, the largest half-width that keeps both limits within the target is \(h=\min\{\log(R/L),\log(U/R)\}\). This requires \(L<R<U\). If a total log-scale width \(W\) is specified symmetrically, then \(h=W/2\).

\text{Required condition:}\qquad t_{1-\alpha/2,n-1}\frac{\sqrt{2}\,\sigma_w}{\sqrt{n}}\leq h

Because the t critical value depends on \(n\), the finite-sample calculation is iterative. A useful initial approximation replaces the t critical value with the standard normal critical value \(z_{1-\alpha/2}\):

n_{\text{approx}}= \left(\frac{z_{1-\alpha/2}\sqrt{2}\,\sigma_w}{h}\right)^2

This approximation is a starting point, not the final answer when the sample is small. Substitute candidate integer values of \(n\) into the t-based condition and select the smallest \(n\) that meets the target. Round up, never down. The result is the number of evaluable participants; enrollment may need to be higher to allow for withdrawals or unusable data.

Worked Calculations

Worked Example: A 20% Multiplicative Precision Target

Suppose a hypothetical two-period crossover study targets a 90% confidence interval and an expected AUC ratio of \(R=1.50\). Prior information suggests \(CV_w=30\%\), and the desired interval should extend no more than 20% multiplicatively above and below the expected ratio. Find the number of evaluable participants.

Convert the variability to the log scale:

\sigma_w=\sqrt{\log(1+0.30^2)} =\sqrt{\log(1.09)} =0.2936,\qquad \sigma_D=\sqrt{2}(0.2936)=0.4152

The target half-width is \(h=\log(1.20)=0.1823\). The normal approximation gives

n_{\text{approx}}= \left(\frac{1.645(0.4152)}{0.1823}\right)^2 =14.04

Check nearby integer values using the 90% t critical values. At \(n=15\), \(df=14\), \(t_{0.95,14}=1.761\), and the margin is \(1.761(0.4152/\sqrt{15})=0.1888\), which exceeds 0.1823. At \(n=16\), \(df=15\), \(t_{0.95,15}=1.753\), and the margin is \(1.753(0.4152/\sqrt{16})=0.1820\), which meets the target. Therefore, plan for 16 evaluable participants. The anticipated interval is approximately \(1.50\exp(\pm0.1820)\), or 1.250 to 1.800, on the ratio scale.

Worked Example: A Target Interval Given as Ratio Limits

For a hypothetical DDI study, suppose the expected exposure ratio is \(R=1.10\), the desired confidence level is 95%, and the target interval limits are \(0.80R\) and \(1.25R\). Assume \(CV_w=20\%\). Find the minimum number of evaluable participants.

The log distances from the expected ratio to each limit are both \(\log(1.25)=0.2231\), so \(h=0.2231\). Convert the coefficient of variation and calculate the paired-contrast standard deviation:

\sigma_w=\sqrt{\log(1+0.20^2)} =\sqrt{\log(1.04)} =0.1980,\qquad \sigma_D=\sqrt{2}(0.1980)=0.2801

The normal approximation is

n_{\text{approx}}= \left(\frac{1.960(0.2801)}{0.2231}\right)^2 =6.05

Check the t condition. For \(n=8\), \(df=7\), \(t_{0.975,7}=2.365\), giving a margin of \(2.365(0.2801/\sqrt{8})=0.2342\), which is too wide. For \(n=9\), \(df=8\), \(t_{0.975,8}=2.306\), giving \(2.306(0.2801/\sqrt{9})=0.2153\), which is within the target. Thus, 9 evaluable participants are required under these assumptions. The anticipated ratio-scale interval is approximately \(1.10\exp(\pm0.2153)\), or 0.887 to 1.364; the target limits are 0.880 and 1.375.

Worked Example: Unequal Log Distances to the Target Limits

A hypothetical study expects a Cmax ratio of \(R=0.70\). The target 90% confidence interval is to fit within \([0.60,0.82]\), and the estimated within-subject coefficient of variation is \(35\%\). Determine the required number of evaluable participants.

The two distances on the log scale are \(\log(0.70/0.60)=0.1542\) and \(\log(0.82/0.70)=0.1582\). The tighter limit determines the allowable half-width, so \(h=0.1542\). The log-scale variability is

\sigma_w=\sqrt{\log(1+0.35^2)} =\sqrt{\log(1.1225)} =0.3399,\qquad \sigma_D=\sqrt{2}(0.3399)=0.4807

The normal approximation gives

n_{\text{approx}}= \left(\frac{1.645(0.4807)}{0.1542}\right)^2 =26.3

At \(n=28\), \(df=27\), and \(t_{0.95,27}=1.703\), so the margin is \(1.703(0.4807/\sqrt{28})=0.1547\), just greater than the allowable 0.1542. At \(n=29\), \(df=28\), and \(t_{0.95,28}=1.701\), so the margin is \(1.701(0.4807/\sqrt{29})=0.1519\), which meets the requirement. Plan for 29 evaluable participants. The anticipated interval is approximately \(0.70\exp(\pm0.1519)\), or 0.601 to 0.815, within the specified limits.

Assumptions, Alternatives, and Failure Modes

The sample-size calculation is useful only to the extent that its inputs and design assumptions are credible. The within-subject variability must correspond to the endpoint and conditions being studied. A variability estimate from AUC does not automatically represent Cmax, and an estimate from a different dose, population, or measurement procedure may not transport well. Sensitivity calculations across plausible values of \(CV_w\) can show whether the proposed sample size is robust to this uncertainty.

Precision-based sizing is not the same as power-based sizing. Precision sizing chooses \(n\) to make the expected confidence interval no wider than a target. Power sizing chooses \(n\) to achieve a specified probability of rejecting a null hypothesis or meeting a decision criterion under an assumed true effect. A study can have a narrow interval that is still inconclusive for a decision, or high power for a particular test while delivering an interval wider than a desired precision target. The objective should be stated before selecting the formula.

A paired crossover calculation is usually more efficient than a parallel-group calculation when within-participant comparisons are valid, because stable differences between participants are removed from the contrast. A parallel design instead estimates a treatment difference across independent groups and must account for between-subject variability. The paired formula here should not be applied to a parallel study. For more complex designs, unequal period variances, missing periods, or nonstandard estimands, use the variance and degrees of freedom for the actual analysis model rather than forcing the simple paired formula.

The main diagnostics are practical as well as statistical. Check that the chosen \(CV_w\) is reported on the intended scale, that the target interval is correctly transformed to the log scale, and that \(n\) satisfies the t-based condition after rounding. Review expected withdrawals separately: if the target is 16 evaluable participants and anticipated attrition is 10%, enrollment should be at least \(\lceil16/(1-0.10)\rceil=18\). This adjustment does not change the required number of completers.

Using between-subject variability. This can substantially overstate or misstate the variability of a within-participant contrast. Use a within-subject estimate appropriate to the crossover endpoint.
Entering a percentage as if it were a log standard deviation. A \(CV_w\) of 30% is entered as 0.30 and converted using \(\sqrt{\log(1+CV_w^2)}\); it is not itself \(\sigma_w\).
Using a normal critical value as the final small-sample calculation. The t critical value depends on \(n-1\), so verify the candidate with the t-based condition.
Confusing ratio-scale symmetry with log-scale symmetry. A symmetric log interval around \(R\) has reciprocal multiplicative factors, not equal additive distances from \(R\).
Assuming sample size fixes design bias. Carryover, period effects, poor endpoint measurement, or a mismatch between the estimand and analysis can compromise interpretation regardless of \(n\).

Key takeaway: For a paired log exposure ratio, plan from the standard deviation of the within-participant log contrast and the log-scale half-width of the desired interval. Use the finite-sample t critical value to confirm the smallest integer number of evaluable participants that meets the precision target.

Check Your Understanding

Work through each question, then compare your reasoning with the solutions.

Why is the standard deviation of the paired log contrast \(\sqrt{2}\sigma_w\) under the independent, equal-variance period-error assumption?
A target confidence interval is centered at an expected ratio of 0.80 and must remain within 0.70 to 0.92. What log-scale half-width should be used?
For \(CV_w=25\%\), write the expression for \(\sigma_w\). Why is 0.25 not the log-scale standard deviation?
State one difference between precision-based and power-based sample-size planning.

Solutions. (1) If the two period-specific log errors each have variance \(\sigma_w^2\) and are independent, the variance of their difference is \(\sigma_w^2+\sigma_w^2=2\sigma_w^2\), so its standard deviation is \(\sqrt{2}\sigma_w\). (2) The distances are \(\log(0.80/0.70)=0.1335\) and \(\log(0.92/0.80)=0.1398\); use the smaller value, \(h=0.1335\). (3) \(\sigma_w=\sqrt{\log(1+0.25^2)}=0.2462\), approximately. The coefficient of variation is on the original scale and requires the lognormal conversion. (4) Precision planning targets a confidence-interval width; power planning targets the probability of a specified test or decision outcome under an assumed effect.