Powering a Cluster-Randomized Grant When the ICC Is Unknown

Symptoms

• You know approximately how many participants each site can recruit, but lack a reliable ICC.
• A proposal uses a single small ICC without a sensitivity table or allowance for lost sites.

What this usually means

The sample size depends on similarity within clusters as well as the effect size. For equal cluster sizes in a simple parallel design, the design effect is 1 + (m − 1) × ICC. An uncertain ICC creates an uncertain sample requirement; a single optimistic value hides that uncertainty.

Common causes

• An ICC from another outcome, population, or adjustment model has been borrowed without checking comparability.
• A large participant target is mistaken for a large number of independent clusters.
• Attrition is represented by inflating participants only, even though entire sites might be lost.

Run these checks

1. Define the outcome scale, target effect, design, and intended analysis.
2. Find ICC evidence from comparable settings and show a plausible range.
3. Specify retained participants per cluster and realistic variation in cluster size.
4. Include separate scenarios for individual dropout and cluster loss.
5. Validate the final plan with a method that handles the actual cluster count, covariates, and allocation.

What not to do

Do not report this design-effect screen as an exact guarantee of 80% power. It uses a normal approximation and omits small-cluster degrees-of-freedom penalties. Do not use it unchanged for binary outcomes, stepped-wedge trials, repeated cross sections, or highly unequal clusters.

Treatment options

Use the range to test feasibility before committing to a design. If a plausible ICC makes the plan infeasible, evaluate additional clusters, a different target effect justified by practical importance, or design improvements. Predefine baseline adjustment and use a dedicated calculation or simulation aligned with the final analysis.

Worked example

For a continuous outcome with standardized difference 0.35, two-sided α = .05, target power .80, and 25 retained participants per cluster, the individually randomized normal approximation needs 128.14 participants per arm. Applying the design effect gives this screening table:

At ICC .05, retaining 20 rather than 25 participants requires 13 analyzable clusters per arm by the same approximation. To retain 12 clusters with 10% cluster loss, the simple recruitment allowance is 14 per arm. These are separate scenarios, not a combined final recommendation.

What to tell the reviewers

We evaluated recruitment requirements across ICCs .01–.10 and separately considered participant attrition and cluster loss. The design-effect calculation was used for feasibility screening. The final sample justification will use the planned analysis and a method that accounts for the number and sizes of clusters.

See it in R and Python

Both languages use the same normal-approximation feasibility screen. This is not a final small-cluster power calculation.

Python dependencies: NumPy and SciPy. Install with python -m pip install numpy scipy.

# Continuous outcome, parallel CRT, equal clusters; base R only.
# Normal approximation SCREEN, not a final small-cluster power calculation.
d <- 0.35; m <- 25; alpha <- 0.05; target <- 0.80
n_ind <- 2*(qnorm(1-alpha/2)+qnorm(target))^2/d^2
icc <- c(0.01,0.03,0.05,0.10)
de <- 1+(m-1)*icc
clusters <- ceiling(n_ind*de/m)
print(data.frame(ICC=icc,design_effect=de,clusters_per_arm=clusters,people_per_arm=clusters*m),row.names=FALSE,quote=FALSE)
cat(sprintf('Individual-randomization normal approximation: %.2f per arm\n',n_ind))
# Participant attrition changes retained cluster size, not just total sample.
retained <- 20
cat(sprintf('At ICC=.05 and 20 retained per cluster: %d clusters per arm\n',ceiling(n_ind*(1+(retained-1)*.05)/retained)))
# Cluster loss must be allowed for separately.
cat(sprintf('For %d analyzable clusters/arm and 10%% cluster loss: recruit %d/arm\n',clusters[3],ceiling(clusters[3]/.90)))
stopifnot(all(diff(clusters)>=0))

# DASS Analysis Clinic Case 005; Python dependencies: numpy, scipy.
from pathlib import Path
import numpy as np
from scipy import stats, optimize
HERE = Path(__file__).resolve().parent

def ols(X, y):
    b = np.linalg.lstsq(X, y, rcond=None)[0]
    residual = y - X @ b
    df = len(y) - X.shape[1]
    cov = (residual @ residual / df) * np.linalg.inv(X.T @ X)
    se = np.sqrt(np.diag(cov))
    ci = np.column_stack((b-stats.t.ppf(.975, df)*se,b+stats.t.ppf(.975,df)*se))
    return b, cov, ci

def power(n, d):
    # Matches R power.t.test(strict=FALSE): rejection tail in effect direction.
    return stats.nct.sf(stats.t.ppf(.975,2*n-2),2*n-2,d*np.sqrt(n/2))

effect=.35; m=25
n=2*(stats.norm.ppf(.975)+stats.norm.ppf(.8))**2/effect**2
print('Normal approximation only; individual n per arm:',n)
for icc in [.01,.03,.05,.10]:
 de=1+(m-1)*icc; k=int(np.ceil(n*de/m)); print(f'ICC={icc:.2f}; DE={de:.2f}; clusters/arm={k}; people/arm={k*m}')
print('20 retained per cluster, ICC .05:',int(np.ceil(n*(1+19*.05)/20)))
print('12 analyzable clusters and 10% cluster loss:',int(np.ceil(12/.9)))
assert [int(np.ceil(n*(1+24*r)/25)) for r in [.01,.03,.05,.10]]==[7,9,12,18]

Download Python script

Sources

• Campbell et al. (2012). CONSORT 2010 statement: extension to cluster randomised trials. BMJ, 345, e5661.
• Hemming et al. (2017). How to design efficient cluster randomised trials. BMJ, 358, j3064.