Information With Unequal Cluster Sizes

Why participant totals are not enough

In a study with clustered observations, participants in the same cluster may have more similar outcomes than participants in different clusters. A cluster might be a clinic, school, or site. If outcomes within a clinic are correlated, adding another participant to that clinic generally adds less information than adding a participant in a new, independent clinic. Unequal cluster sizes make the calculation more involved: a total of 100 participants spread evenly across clusters need not provide the same information as 100 participants concentrated in a few large clusters.

The aim here is to calculate the information for a mean, or for a difference between two means, under a simple correlation model. Information is the reciprocal of the variance of the estimator. Thus it has units of inverse outcome-units squared, and greater information means a smaller model-based variance. The calculations help distinguish a participant count from the precision that the study’s clustering supports.

Model and information from one cluster

Suppose there are $K$ independent clusters. Cluster $j$ contains $m_j$ participants, and $N=\sum_{j=1}^K m_j$ is the total participant count. For an outcome $Y_{ij}$ from participant $i$ in cluster $j$, assume the random-intercept model

Y_{ij}=\mu+b_j+\epsilon_{ij}, \qquad b_j\sim N(0,\rho\sigma^2), \qquad \epsilon_{ij}\sim N(0,(1-\rho)\sigma^2).

The cluster effects $b_j$ are independent across clusters; the individual errors $\epsilon_{ij}$ are independent across participants and clusters, and independent of the cluster effects. Here $\mu$ is the common mean, $\sigma^2$ is the marginal outcome variance, and $\rho$ is the intracluster correlation (ICC): the correlation between two distinct participants in the same cluster. Assume $0\leq\rho<1$, and, for the information formula below, treat $\sigma^2$ and $\rho$ as known. The model implies a common marginal variance $\sigma^2$ and covariance $\rho\sigma^2$ for two outcomes in the same cluster.

The cluster mean is $\bar{Y}_j=m_j^{-1}\sum_iY_{ij}$. Its variance is

\operatorname{Var}(\bar{Y}_j) =\rho\sigma^2+\frac{(1-\rho)\sigma^2}{m_j} =\sigma^2\frac{1+(m_j-1)\rho}{m_j}.

Because clusters are independent, their means are independent. With a common mean and known cluster-mean variances, the information about $\mu$ is the sum of the reciprocal variances. This gives the central calculation:

Definition: Under the stated model, the information about the common mean contributed by clusters of sizes $m_1,\ldots,m_K$ is $$ \mathcal I_\mu=\sum_{j=1}^K\frac{1}{\operatorname{Var}(\bar{Y}_j)} =\frac{1}{\sigma^2}\sum_{j=1}^K\frac{m_j}{1+(m_j-1)\rho}. $$ The model-based variance of the inverse-variance weighted mean estimator is $1/\mathcal I_\mu$.

Each term depends on both the cluster size and the ICC. If $\rho=0$, the term for cluster $j$ is $m_j/\sigma^2$, so information adds in proportion to participants. With positive $\rho$, the denominator grows with cluster size: extra participants in an existing cluster still add information, but not as much as independent participants would.

Justification and implications of unequal sizes

For independent normally distributed cluster means with common expectation $\mu$ and variances $v_j=\operatorname{Var}(\bar{Y}_j)$, the log-likelihood for $\mu$, up to terms not involving $\mu$, is $-\tfrac12\sum_j(\bar{Y}_j-\mu)^2/v_j$. Its second derivative with respect to $\mu$ is $-\sum_j1/v_j$. The Fisher information is the negative expected second derivative, so it equals $\sum_j1/v_j$, yielding the formula above. Equivalently, the estimator weights cluster means in proportion to their precision $1/v_j$.

The formula also explains why two studies with the same $N$ can have different information. For a fixed number of clusters and a fixed total number of participants, the function

f(m)=\frac{m}{1+(m-1)\rho}

is concave in $m$ when $0<\rho<1$, since $f''(m)=-2\rho(1-\rho)/[1+(m-1)\rho]^3<0$. Consequently, more balanced cluster sizes yield a larger sum of $f(m_j)$ than more dispersed sizes with the same total and cluster count. This comparison holds under the model and for the specified information calculation; it is not a claim that every possible design feature is otherwise equal in practice.

Worked Example: Information From Two Unequal Clusters

Suppose a hypothetical study estimates one mean from two independent clusters of sizes 4 and 8. Let the marginal outcome variance be $25$ outcome-units squared and the ICC be $0.10$. Under the model, both clusters have the same marginal variance, and the correlation is common within clusters. Calculate the information and the model-based standard error.

The cluster contributions before dividing by $\sigma^2$ are

\frac{4}{1+(4-1)(0.10)}=\frac{4}{1.3}=3.0769, \qquad \frac{8}{1+(8-1)(0.10)}=\frac{8}{1.7}=4.7059.

Therefore,

\mathcal I_\mu =\frac{3.0769+4.7059}{25} =0.3113\ \text{inverse outcome-units squared}.

The estimated mean has model-based variance $1/0.3113=3.212$ outcome-units squared and standard error $\sqrt{3.212}=1.792$ outcome units. As a benchmark, treating all 12 observations as independent would give information $12/25=0.4800$. That benchmark is not appropriate under the specified positive within-cluster correlation; it overstates precision.

Worked Example: Same Participant Total, Different Size Patterns

Compare two hypothetical studies with four independent clusters, 24 participants in total, marginal variance $\sigma^2=16$ outcome-units squared, and ICC $\rho=0.20$. Study A has cluster sizes $6,6,6,6$; Study B has sizes $3,3,9,9$. Assume the same common-mean model in both studies.

For Study A, each cluster contributes $6/[1+5(0.20)]=6/2=3$ before division by the variance. Thus

\mathcal I_A=\frac{4(3)}{16}=0.7500.

For Study B, the size-3 clusters each contribute $3/[1+2(0.20)]=3/1.4=2.1429$, and the size-9 clusters each contribute $9/[1+8(0.20)]=9/2.6=3.4615$. Hence

\mathcal I_B =\frac{2(2.1429)+2(3.4615)}{16} =\frac{11.2088}{16} =0.7006.

Both studies enroll 24 participants and have four clusters, but Study B has less information under the model. The difference is $0.7500-0.7006=0.0494$, or about 6.59% of Study A’s information. The corresponding standard errors are $\sqrt{1/0.7500}=1.155$ and $\sqrt{1/0.7006}=1.195$ outcome units. The calculation checks the cluster sizes, variance, and ICC on the same basis in both studies.

Information for a two-arm mean contrast

For a two-arm study, assume the model applies separately in each arm, clusters are independent both within and between arms, and the two arm means have the same marginal variance and ICC. Let $\mathcal I_1$ and $\mathcal I_0$ be the information calculated from the cluster sizes in the two arms. The estimated treatment contrast is the difference between the arm means. Its variance is the sum of the independent arm-mean variances:

\operatorname{Var}(\hat{\mu}_1-\hat{\mu}_0) =\frac{1}{\mathcal I_1}+\frac{1}{\mathcal I_0}, \qquad \mathcal I_{\text{contrast}} =\frac{1}{1/\mathcal I_1+1/\mathcal I_0}.

Thus, contrast information is not the sum of the two arm informations. Each arm must estimate its own mean, and uncertainty from both estimates contributes to the difference.

Worked Example: Information for a Treatment Contrast

Consider a hypothetical trial with cluster sizes $5,7,6$ in one arm and $4,8,6$ in the other. Assume independent clusters, common marginal outcome variance $36$ outcome-units squared, and a common ICC of $0.15$. The model uses the same variance and ICC in both arms.

For the first arm, the unscaled contributions are $5/1.6=3.125$, $7/1.9=3.6842$, and $6/1.75=3.4286$. Their sum is $10.2378$, so $\mathcal I_1=10.2378/36=0.2844$. For the second arm, the contributions are $4/1.45=2.7586$, $8/2.05=3.9024$, and $6/1.75=3.4286$. Their sum is $10.0896$, so $\mathcal I_0=10.0896/36=0.2803$.

The contrast variance and standard error are

\frac{1}{0.2844}+\frac{1}{0.2803} \approx 3.516+3.568=7.084, \qquad SE=\sqrt{7.084}=2.662\ \text{outcome units}.

Accordingly, the contrast information is approximately $1/7.084=0.1412$ inverse outcome-units squared. This is a precision calculation under the assumed model, not evidence by itself that the treatment contrast is nonzero.

Planning, diagnostics, and limitations

For an interim information target, calculate information using the clusters whose outcomes are available at that look, then compare it with the planned final information. If cluster sizes differ across looks, the information fraction need not equal the fraction of participants enrolled. This calculation is meaningful only if the ICC and variance used at the interim and final analyses are appropriate to the same outcome model.

Check the inputs before relying on the result:

Cluster definition: confirm that the units treated as independent clusters are genuinely independent for the outcome and analysis.
Cluster-size record: use the observed or planned size of each cluster, not just the average size, when sizes vary.
ICC and variance: justify values using relevant prior information or sensitivity analyses. A single assumed ICC may not represent all settings.
Outcome model: examine whether a common marginal variance and common within-cluster correlation are credible. Large differences across clusters or arms can make this calculation misleading.
Number of clusters: information from this known-variance model does not account for extra uncertainty from estimating variance components or for small-sample inferential corrections. A large participant total cannot compensate automatically for having very few independent clusters.

A common mistake is to calculate $N/\sigma^2$ and call it the information, because all $N$ participants have measured outcomes. That reasoning treats correlated observations as independent. The correction is to calculate each cluster’s contribution $m_j/\{\sigma^2[1+(m_j-1)\rho]\}$ and add those contributions. Another mistake is to apply an equal-cluster-size formula using only the average size when cluster sizes are highly variable; the unequal-size formula uses each $m_j$ directly.

The calculation is model-based. An individual-level mixed model or a cluster-level analysis can be used as an analysis approach, but the model, weighting, and variance assumptions must match the planning calculation. If the outcome is non-normal, the ICC varies, the cluster sizes are related to expected outcomes, or the analysis targets a different parameter, the displayed information may only be a rough approximation and should be checked using a more suitable model or simulation. The calculation alone does not address missing outcomes, covariate adjustment, or uncertainty in the estimated ICC.

Check Your Understanding

Answer each question using the model and information formula in this tutorial. Short solutions follow the questions.

A single cluster contains 10 participants, with $\rho=0.10$ and $\sigma^2=25$. What information does it contribute about the common mean?
When $\rho=0$, how does the information formula simplify for clusters of any sizes?
Two independent arms have information $0.20$ and $0.30$, in inverse outcome-units squared. What is the information for their mean difference?
Why can an equal-cluster-size formula be inappropriate when cluster sizes vary substantially?

Solutions. (1) The contribution is $10/[25\{1+9(0.10)\}]=10/47.5=0.2105$ inverse outcome-units squared. (2) Each cluster contributes $m_j/\sigma^2$, so total information is $\sum_jm_j/\sigma^2=N/\sigma^2$. (3) The contrast variance is $1/0.20+1/0.30=5+3.333=8.333$ outcome-units squared, so its information is $1/8.333=0.1200$ inverse outcome-units squared. (4) With positive ICC, each cluster’s contribution depends nonlinearly on its own size. Replacing unequal sizes with their average can therefore give a different sum of contributions than calculating from each cluster separately.