Question

If \(n_i \propto N_i\) and \(p_i = \frac{N_i}{N}\) and k is the number of strata and \(N_i\) is the number of units in the \(i^{th}\) stratum then, Var(\(\bar{y}_{stratified}\)) is:

• A. \( \left(\frac{1}{n} - \frac{1}{N}\right) \sum_{i=1}^k p_i S_i^2 \)
• B. \( \left(\frac{1}{n} - \frac{1}{N}\right) \sum_{i=1}^k p_i^2 S_i^2 \)
• C. \( \frac{1}{n} \sum_{i=1}^k p_i S_i^2 - \frac{1}{N} \sum_{i=1}^k p_i^2 S_i^2 \)
• D. \( \sum_{i=1}^k \left(\frac{1}{n_i} - \frac{1}{N_i}\right) p_i S_i^2 \)

Hint:

Proportional allocation simplifies the variance formula for stratified sampling significantly. Memorizing this specific result can save considerable time compared to re-deriving it from the general formula during an exam.

Answer 1

The Correct Option is A

Step 1: Concept Explanation:
The problem requires finding the variance of the stratified sample mean, \(\bar{y}_{st}\), using proportional allocation. This means the sample size within each stratum (\(n_i\)) is proportional to the size of that stratum (\(N_i\)).

Step 2: Formula and Approach:
The general variance formula for the stratified sample mean is: \[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k W_i^2 \text{Var}(\bar{y}_i) = \sum_{i=1}^k W_i^2 \left( \frac{N_i - n_i}{N_i} \right) \frac{S_i^2}{n_i} \] where \(W_i = N_i/N\) represents the stratum weight (equivalent to \(p_i\)). With proportional allocation, \( n_i = n \frac{N_i}{N} = n W_i \). We will substitute this into the general formula.

Step 3: Step-by-Step Derivation:
Starting from the general variance formula:\[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k W_i^2 \left( \frac{1}{n_i} - \frac{1}{N_i} \right) S_i^2 \]Substitute \(n_i = nW_i\):\[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k W_i^2 \left( \frac{1}{nW_i} - \frac{1}{N_i} \right) S_i^2 \]Distribute \(W_i^2\):\[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k \left( \frac{W_i^2}{nW_i} - \frac{W_i^2}{N_i} \right) S_i^2 = \sum_{i=1}^k \left( \frac{W_i}{n} - \frac{W_i^2}{N_i} \right) S_i^2 \]Substitute \(W_i = N_i/N\):\[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k \left( \frac{W_i}{n} - \frac{(N_i/N)^2}{N_i} \right) S_i^2 = \sum_{i=1}^k \left( \frac{W_i}{n} - \frac{N_i}{N^2} \right) S_i^2 \]Since \(N_i/N = W_i\), the second term simplifies to \(W_i/N\):\[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k \left( \frac{W_i}{n} - \frac{W_i}{N} \right) S_i^2 = \sum_{i=1}^k W_i \left( \frac{1}{n} - \frac{1}{N} \right) S_i^2 \]Factor out the constant term \((\frac{1}{n} - \frac{1}{N})\):\[ \text{Var}(\bar{y}_{st}) = \left(\frac{1}{n} - \frac{1}{N}\right) \sum_{i=1}^k W_i S_i^2 \]Using \(p_i = W_i\), we have:\[ \text{Var}(\bar{y}_{st}) = \left(\frac{1}{n} - \frac{1}{N}\right) \sum_{i=1}^k p_i S_i^2 \]
Step 4: Solution:
Therefore, the variance of the stratified mean with proportional allocation is \( \left(\frac{1}{n} - \frac{1}{N}\right) \sum_{i=1}^k p_i S_i^2 \).