35 Sample Size Estimation

35.1 Solving Group Sample Sizes Using Weights

Let \(n\) be the total sample size obtained by adding two groups such that \(n = n_1 + n_2\). Let \(w\) represent the proportion of \(n\) allocated to \(n_1\), referred to as the weight of \(n_1\). Then \(n_2 = \frac{n_1 \cdot (1 - w)}{w}\).

Furthermore, \(n = n_1 + \frac{n_1 \cdot (1-w)}{w}\).

Proof:

By the assumptions, we know

\[\begin{aligned} n &= n_1 + n_2 \\ &= w \cdot n + (1 - w) \cdot n \end{aligned}\]

This implies

\[\begin{aligned} n_1 &= w \cdot n\\ n_2 &= (1-w) \cdot n \end{aligned}\]

We observe the following:

\[\begin{aligned} \frac{n_2}{n_1} &= \frac{(1-w) \cdot n}{w \cdot n} \\ &= \frac{(1-w)}{w} \\ \Rightarrow n_2 &= \frac{n_1 \cdot (1-w)}{w} \end{aligned}\]

This further implies

\[ n = n_1 + \frac{n_1 \cdot (1-w)}{w}\]

Notice now that both \(n\) and \(n_2\) are defined as functions of \(n_1\) and \(w\). Thus, we may estimate the sample size required in each of two groups by estimating only \(n_1\), provided we know the weight \(w\).

35.1.1 Corollary

For \(k \in \mathbb{N}\), let \(n\) be the total sample size of \(k\) subgroups such that

\[n = n_1 + n_2 + n_3 + ... + n_k\]

Suppose, further, that there exists a vector of weights \(W\) that satisfy the following conditions:

  1. For each \(w_i \in W\), \(0 \leq w_i \leq 1\)
  2. \(\sum\limits_{i=1}^{k} w_i = 1\)
  3. \(n = w_1 \cdot n + w_2 \cdot n + ... + w_k \cdot n = \sum\limits_{i=1}^{k} w_i \cdot n\)

Let us assign the values \(n_1 = w_1 \cdot n\), \(n_2 = w_2 \cdot n\), …, \(n_k = w_k \cdot n\). Then for all \(i | i \leq k\),

\[\begin{aligned} \frac{n_i}{n_1} &= \frac{w_i \cdot n}{w_1 \cdot n} \\ &= \frac{w_i}{w_1} \\ \Rightarrow n_i &= \frac{n_1 \cdot w_i}{w_1} \end{aligned}\]

Thus, each \(n_i\) may be estimated by estimating the value of \(n_1\) provided \(W\) is a fully specified vector of weights for each of the \(k\) groups.