For a two-group comparison with equal variances the sample size for each of the two groups is n = 2(zα + zβ)2(σ/δ)2; see our earlier Result 2. This result assumes that the two treatments are assigned to the experimental units (e.g., subjects, mice, and others) at random. However, sometimes the randomization is carried out on clusters that consist of groupings of the experimental units. Clusters may be cages of animals, and experimental units could be mice. Clusters may be patients, and experimental units could be eyes. The randomization is at the cluster level: the treatment groups (experimental and control) are assigned to clusters at random, and each of m experimental units in a cluster is assigned to the same treatment. Although the data of interest comes from the experimental units in the two experimental groups, the randomization is carried out on the clusters. In the example with patient eyes, we may assign n = 10 patients each to one of the two treatments, for a total of 20 patients. For cluster size m = 2, this generates a total of 40 eyes, with 20 eyes for each treatment.
Usually subjects from the same cluster tend to be alike. Because observations from the same cluster are most likely correlated, with intracluster correlation coefficient ρ > 0, the
m observations in a cluster do not carry the same weight as
m independent observations. For the retinal thickness example in Ledolter et al.,
14 the intracluster correlation is approximately 0.8.
Result: The required number of observations
n (number of clusters,
k, times number of observations in each cluster,
m) in each treatment group is
\begin{equation*}n = km = 2{({z_\alpha } + {z_\beta })^2}{(\sigma /\delta )^2}\left[ {1 + (m - 1)\rho } \right]\end{equation*}
Discussion: The intracluster correlation inflates the sample size that we obtain under complete random sampling, 2(zα + zβ)2(σ/δ)2, by the factor [1 + (m − 1)ρ]. For ρ = 0, we are back to our earlier Result 2. For ρ = 1, we must multiply the sample size that we obtain under complete random sampling by the number of experimental units in the cluster (m). Here experimental units in a cluster are carbon-copies of each other. The m experimental units in a cluster basically count as one unit (and not as m).
In the presence of large intracluster correlation it is important to randomize over many clusters so as to maximize the efficiency of the experiment. Taking more and more replicates within the cluster does not increase the power of the experiment by much, but taking more clusters does. With perfect correlation you may as well leave off one eye from each cluster and save yourself the work collecting measurements on the second eye. For ρ = 0.80 and say n = km = 2(zα + zβ)2(σ/δ)2[1 + (m − 1)ρ] = 100 eyes, you take 50 subjects and analyze both of their eyes. It would be wrong to ignore the intracluster correlation and calculate the number of eyes from n = km = 100/[1 + (m − 1)ρ] = 56.55, taking only 28 subjects with their 56 eyes.