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Blinded Re-Estimation · Tutorial 362 of 699

Nuisance Parameters In Fixed-Effect Sample Size Formulas

You will learn to identify the nuisance inputs in common fixed-effect sample size equations and assess how changes in those inputs alter the required enrollment.

Intermediate 11 min read

What You'll Learn

  • Distinguish the prespecified treatment effect and operating characteristics from nuisance parameters used in the sample size calculation.
  • Derive the equal-allocation continuous-outcome formula from the standard error of a difference in means.
  • See how a pooled event probability enters an approximate risk-difference sample size equation.
  • Apply dispersion and intraclass-correlation design effects without confusing them with the target effect.
  • Inflate a requirement for expected attrition and account for whole-cluster enrollment.
  • Diagnose when a plug-in calculation needs sensitivity analysis or a design-specific calculation.

What the Sample Size Depends On

A fixed-effect sample size calculation starts with a treatment effect that the study is designed to detect, a type I error rate, and a target power. These are design choices: for example, the investigators might prespecify a five-point mean difference, a two-sided significance level of 0.05, and 90% power. The required sample size also depends on quantities that describe outcome variability or data collection. These are nuisance parameters: they affect precision, but are not the treatment effect being targeted.

The nuisance parameter depends on the outcome and design. For a continuous outcome it may be the standard deviation \(\sigma\); for a binary outcome it may be an event probability; for clustered data it may be the intraclass correlation \(\rho\); and for overdispersed data it may be a variance inflation factor. Expected attrition affects how many people must be enrolled to obtain the required number of analyzable observations.

Definition: In a fixed-effect sample size calculation, the target treatment effect, type I error rate, and power are held at their prespecified values while nuisance parameters are supplied or estimated to calculate the required sample size. “Fixed-effect” here means that the calculation targets a specified effect, not that all nuisance parameters are known without uncertainty.

The equations below are planning approximations for standard fixed-effect comparisons. In a group-sequential design, the overall type I error and power remain design targets, but the required information must be calculated using that design’s stopping boundaries or an appropriate design-specific adjustment. A single-final-analysis normal approximation is not a substitute for that calculation.

Continuous Outcomes: Variance Sets the Scale

Suppose two independent groups have normally distributed outcomes with a common standard deviation \(\sigma\), equal group sizes \(n\), and a target difference \(\delta\) in population means. Assume observations are independent within and between groups, the variance is the same in both groups, and the analysis compares the two group means. For a large-sample, two-sided test with type I error \(\alpha\) and power \(1-\beta\), a common approximation is:

$$ n \approx \frac{2\sigma^2\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2}{\delta^2} $$

Here \(n\) is the number of analyzable participants per group, and \(z_q\) is the \(q\)-quantile of the standard normal distribution. The formula assumes a common variance and equal allocation. Its use also requires that the normal approximation is reasonable for the planned test; small samples, substantial nonnormality, unequal variances, or covariate adjustment may require another calculation.

Key interpretation: With the target effect and operating characteristics fixed, the required sample size is proportional to \(\sigma^2\) and inversely proportional to \(\delta^2\). Doubling the assumed standard deviation quadruples the sample size; doubling the target difference reduces it to one quarter.

Worked Example: Updating the Assumed Standard Deviation

A hypothetical two-arm study targets a mean difference of 5 points. It uses a two-sided \(\alpha=0.05\), 90% power, equal allocation, and assumes a common standard deviation of 14 points. Find the approximate requirement per group and compare it with the requirement if the standard deviation were 12 points instead.

For these design targets, \(z_{1-\alpha/2}=1.96\) and \(z_{1-\beta}=z_{0.90}=1.2816\). Thus their sum is \(3.2416\). The model assumptions for this calculation are independent observations, a common outcome variance in the two groups, and the stated mean difference as the target effect.

With \(\sigma=14\) points, substitute into the equation:

$$ n \approx \frac{2(14^2)(1.96+1.2816)^2}{5^2} = \frac{392(3.2416)^2}{25} \approx 164.8 $$

The requirement must be rounded up to a whole participant, so the plan calls for 165 analyzable participants per group. If instead \(\sigma=12\) points, with all other inputs unchanged:

$$ n \approx \frac{2(12^2)(3.2416)^2}{25} \approx 121.1 $$

That requirement rounds up to 122 per group. The difference between the two calculations is due to the nuisance variance assumption, not a change in the target effect, \(\alpha\), or power. These are analyzable sample sizes; expected losses before analysis must be accounted for separately.

Binary Outcomes: The Event Probability Affects Precision

For a binary outcome, let \(p_0\) and \(p_1\) be the event probabilities in the control and treatment groups, and let the target risk difference be \(\delta=p_1-p_0\). With equal group sizes, a commonly used normal approximation for a two-sided risk-difference test is:

$$ n \approx \frac{ \left[ z_{1-\alpha/2}\sqrt{2\bar{p}(1-\bar{p})} + z_{1-\beta}\sqrt{p_0(1-p_0)+p_1(1-p_1)} \right]^2 }{(p_1-p_0)^2}, \qquad \bar{p}=\frac{p_0+p_1}{2} $$

The first variance term represents the pooled event probability under the null hypothesis of equal rates; the second represents the variance under the specified alternative rates. This is a planning approximation for independent observations with equal allocation and a normal approximation to the risk-difference test. Sparse events or small expected cell counts can make the approximation unreliable.

The target difference alone does not determine the sample size. For example, two pairs of event rates can have the same difference but different variances. The baseline event rate, or another justified specification that determines \(p_0\) and \(p_1\), is therefore needed. A pooled probability is a nuisance input to the variance calculation; it is not itself the treatment effect.

Worked Example: Planning for a Risk Difference

In a hypothetical trial, the control event probability is \(p_0=0.30\) and the target treatment event probability is \(p_1=0.20\). The target risk difference is \(-0.10\). Use a two-sided \(\alpha=0.05\), 80% power, equal allocation, and the approximation above.

The pooled probability is \(\bar{p}=(0.30+0.20)/2=0.25\). The null-variance contribution is \(1.96\sqrt{2(0.25)(0.75)}\), and the alternative-variance contribution is \(0.8416\sqrt{(0.30)(0.70)+(0.20)(0.80)}\). Retaining precision through the calculation gives:

$$ \begin{aligned} n &\approx \frac{ \left[ 1.96\sqrt{0.375} + 0.8416\sqrt{0.37} \right]^2 }{(-0.10)^2}\\ &\approx \frac{(1.20024997+0.51192529)^2}{0.01}\\ &\approx 293.1544 \end{aligned} $$

Rounding up gives 294 analyzable participants per group. The negative sign indicates the direction of the specified risk difference; the squared denominator uses its magnitude. The calculation assumes the stated event probabilities are appropriate planning values and that participants’ binary outcomes are independent.

Dispersion, Clustering, and Attrition

For some outcomes, the variance is larger than the variance assumed by a basic model. If the planned estimator’s variance is multiplied by a dispersion factor \(\phi\), while the effect and other assumptions stay fixed, the required sample size is approximately multiplied by \(\phi\) as well. This scaling is appropriate only when the factor describes the variance of the estimator in the planned analysis; it is not a universal replacement for an outcome-specific sample size equation.

Worked Example: Allowing for Overdispersion

Suppose a hypothetical independent-observation calculation requires 120 analyzable participants per group under its basic variance model. An appropriate model for the planned analysis indicates that the variance is 1.5 times the basic variance, so \(\phi=1.5\). Assuming this factor scales the estimator variance as specified, the adjusted requirement is:

$$ n_{\text{adjusted}} \approx \phi n_{\text{basic}} =1.5(120)=180 $$

The adjusted target is 180 analyzable participants per group. The calculation does not change the target effect or operating characteristics; it allows for greater variability. The factor must be justified for the outcome and analysis rather than selected merely to obtain a preferred sample size.

In a cluster-randomized design, outcomes from people in the same cluster are generally correlated. For equal cluster size \(m\) and intraclass correlation \(\rho\), a common design effect is:

$$ DE=1+(m-1)\rho $$

Under this approximation, multiplying the independent-observation requirement by \(DE\) accounts for the loss of information due to within-cluster correlation. It assumes clusters have the stated size and a common intraclass correlation; unequal cluster sizes, few clusters, or more complex correlation patterns may require a more careful design calculation.

Worked Example: Combining Intraclass Correlation and Attrition

A hypothetical cluster-randomized study would require 200 analyzable individuals per group if observations were independent. The planned average cluster size is 20, the intraclass correlation is \(\rho=0.02\), and 10% of enrolled individuals are expected to lack analyzable outcomes. Assume equal cluster sizes and that the design-effect approximation is suitable.

First calculate the design effect and the cluster-adjusted analyzable requirement:

$$ DE=1+(20-1)(0.02)=1.38, \qquad 200(1.38)=276 $$

Thus, approximately 276 analyzable individuals per group are needed after accounting for clustering. To allow for 10% attrition, divide by the expected retention proportion \(1-0.10=0.90\):

$$ n_{\text{enroll}} \approx \frac{276}{0.90}=306.7 $$

If participants are enrolled in clusters of 20, at least \(\lceil306.7/20\rceil=16\) clusters per group are required, or 320 randomized individuals per group. Rounding at the cluster level is necessary because partial clusters cannot be enrolled. This calculation presumes attrition does not selectively remove particular clusters in a way that changes the design.

For individual randomization, the basic attrition adjustment is similarly \(n_{\text{enroll}}=n_{\text{analyzable}}/(1-r)\), where \(r\) is the anticipated proportion without analyzable outcomes. Attrition inflation addresses expected loss; it does not repair bias caused by outcome-dependent missingness or guarantee that the assumed attrition rate will occur.

How to Check a Nuisance-Parameter Calculation

A sample size is only as defensible as its inputs. Before treating a calculated number as a plan, check that each nuisance parameter matches the outcome definition, population, follow-up period, and planned analysis. In particular, check that the standard deviation is in the outcome’s units, event probabilities refer to the correct endpoint and time horizon, and the assumed cluster size and intraclass correlation are plausible together.

  • Check direction and scale. A larger standard deviation or dispersion factor should not reduce the sample size when other inputs are fixed. A smaller absolute target effect should not reduce it either.
  • Check independence. A formula for independent individuals is not appropriate without adjustment when observations are clustered or repeatedly measured.
  • Check the event-rate scenario. For a binary outcome, different plausible baseline rates can change the variance even if the target risk difference is unchanged.
  • Check the analysis design. A group-sequential design’s stopping boundaries affect the information required. A conventional single-analysis formula alone does not account for those boundaries.
  • Check rounding and losses. Round required enrollment upward, and distinguish analyzable sample size from the number randomized or enrolled.

A useful alternative to choosing a single uncertain nuisance value is to show a sensitivity analysis over defensible values. A conservative design may use a value that produces a larger requirement, while a blinded reassessment may use available information to update nuisance assumptions without changing the prespecified target effect. Neither approach removes uncertainty: the resulting requirement remains conditional on the assumptions used. When plausible inputs lead to very different sample sizes, report that dependence rather than presenting one calculated number as certain.

Common Mistake and Practical Limits

A common mistake is to treat every number that enters the equation as if it were a target effect. For example, someone may say, “The event rate changed, so the treatment effect must have changed.” That reasoning is not justified: an event probability can enter the variance term and alter the calculated sample size while the prespecified risk difference remains unchanged. The correction is to identify each input’s role explicitly: the effect specifies the contrast to detect; nuisance parameters describe the variability and information available for estimating that contrast.

Plug-in calculations also give an apparently precise answer even when nuisance inputs are uncertain. They do not automatically account for uncertainty in an estimated variance, event rate, dispersion factor, or intraclass correlation. They can also fail when the normal approximation is poor, the model assumptions do not hold, or the planned analysis differs from the one used to derive the equation. In such cases, use a design-appropriate method and examine sensitivity to plausible nuisance values rather than relying on false precision.

Key takeaway: Prespecifying the effect, type I error, and power does not make the sample size independent of the data-generating conditions. Variance, event probability, dispersion, intraclass correlation, and attrition determine how much information must be planned to meet those targets.

Check Your Understanding

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

  1. If the assumed standard deviation in an equal-allocation continuous-outcome calculation is multiplied by 1.5, by what factor does the approximate sample size change, with other inputs fixed?
  2. Why can two binary-outcome designs with the same target risk difference require different sample sizes?
  3. A cluster design has average cluster size \(m=10\) and intraclass correlation \(\rho=0.03\). Find the design effect under the equal-size approximation.
  4. If 240 analyzable participants are required per group and 20% are expected to have no analyzable outcome, how many should be enrolled per group before rounding?

Solutions. 1. The sample size is proportional to \(\sigma^2\), so it is multiplied by \(1.5^2=2.25\). 2. The event probabilities affect the null and alternative variance terms; equal risk differences do not imply equal variances. 3. \(DE=1+(10-1)(0.03)=1.27\). 4. \(240/(1-0.20)=300\) participants per group should be enrolled before any additional design-specific rounding.

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