Question

If under SRSWOR, \(U = \sum_{i=1}^{n_1} y_i = n_1 \bar{y}_1\) and \(V = \sum_{j=n_1+1}^{n} y_j = (n-n_1)\bar{y}_2\), then the Var(V) is

• A. \( \frac{n_1(N-(n-n_1))}{N}S^2 \)
• B. \( \frac{(n-n_1)(N-(n-n_1))}{N}S^2 \)
• C. \( \frac{n_1(N-(n-n_1))}{N}S^2 \)
• D. \( \frac{(n-n_1)(N-(n-n_1))}{Nn}S^2 \)

Hint:

When dealing with variances of sums or means in finite population sampling, always remember the finite population correction (FPC) factor, \( \frac{N-n}{N} \). The variance of a sum is not simply \(m . S^2\), but is scaled by the FPC.

Answer 1

The Correct Option is B

Step 1: Problem Definition:
The problem requires finding the variance of the sum of a subset of observations from a finite population, sampled using Simple Random Sampling Without Replacement (SRSWOR). A sample of size \(n\) is divided into two groups of sizes \(n_1\) and \(n-n_1\). V represents the sum of observations in the second group.

Step 2: Relevant Formula:
The variance of the sum of \(m\) units randomly selected from a population of size \(N\) using SRSWOR is: \[ \text{Var}\left(\sum_{i=1}^m y_i\right) = m^2 \text{Var}(\bar{y}_m) \] where \(\bar{y}_m\) is the sample mean of size \(m\). The variance of the sample mean is: \[ \text{Var}(\bar{y}_m) = \frac{N-m}{N} \frac{S^2}{m} \] Therefore: \[ \text{Var}\left(\sum_{i=1}^m y_i\right) = m^2 \left( \frac{N-m}{N} \frac{S^2}{m} \right) = m \frac{N-m}{N} S^2 \]
Step 3: Solution:
V is the sum of \(m = n-n_1\) observations, representing a simple random sample of size \(m\) from a population of size \(N\). Applying the formula with \(m = n-n_1\): \[ \text{Var}(V) = \text{Var}\left(\sum_{j=n_1+1}^{n} y_j\right) = (n-n_1) \frac{N-(n-n_1)}{N} S^2 \] This corresponds to option (B), assuming a typo in the option replaces \(n_1\) with \(m\).
Step 4: Answer:
The variance of V is \( \frac{(n-n_1)(N-(n-n_1))}{N}S^2 \).