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: Understanding the Concept:
The question asks for the variance of the stratified sample mean, \(\bar{y}_{st}\), under proportional allocation. Proportional allocation means the sample size in each stratum, \(n_i\), is proportional to the stratum size, \(N_i\).

Step 2: Key Formula or Approach:
The general formula for the variance of 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 \(W_i = N_i/N\) is the stratum weight (given as \(p_i\)). Under proportional allocation, \( n_i = n \frac{N_i}{N} = n W_i \). We substitute this into the general formula.

Step 3: Detailed Explanation:
Starting with the general formula for \( \text{Var}(\bar{y}_{st}) \): \[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k W_i^2 \left( \frac{1}{n_i} - \frac{1}{N_i} \right) S_i^2 \] 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 the \(W_i^2\): \[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k \left( \frac{W_i^2}{nW_i} - \frac{W_i^2}{N_i} \right) S_i^2 = \sum_{i=1}^k \left( \frac{W_i}{n} - \frac{W_i^2}{N_i} \right) S_i^2 \] 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 becomes \(W_i/N\): \[ \text{Var}(\bar{y}_{st}) = \sum_{i=1}^k \left( \frac{W_i}{n} - \frac{W_i}{N} \right) S_i^2 = \sum_{i=1}^k W_i \left( \frac{1}{n} - \frac{1}{N} \right) S_i^2 \] 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 the notation \(p_i = W_i\), we get: \[ \text{Var}(\bar{y}_{st}) = \left(\frac{1}{n} - \frac{1}{N}\right) \sum_{i=1}^k p_i S_i^2 \]
Step 4: Final Answer:
The variance of the stratified mean under proportional allocation is \( \left(\frac{1}{n} - \frac{1}{N}\right) \sum_{i=1}^k p_i S_i^2 \).