Question

In a hypothetical group, it is given that \( d = 0.05 \), \( p=0.5\alpha \) and \( t = 2 \). If N is large, then the sample size \( n_0 \), is

• A. 250
• B. 325
• C. 400
• D. 550

Hint:

When a question asks for sample size for a proportion but doesn't provide a preliminary estimate for \(p\), always use \(p=0.5\). This is the "worst-case scenario" as it yields the maximum possible sample size needed to achieve the desired margin of error and confidence level.

Answer 1

The Correct Option is C

Step 1: Concept Overview:
This problem involves determining the necessary sample size for estimating a population proportion. While the notation is slightly different, it aligns with the standard sample size formula for large populations (N). The expression \( p=0.5\alpha \) is likely an OCR error and should read \(p=0.5\). The value \(p = 0.5\) maximizes the sample size when there's no prior knowledge of the true proportion.

Step 2: Core Formula:
The formula to calculate the required sample size (\(n_0\)) for estimating a population proportion is: \[ n_0 = \frac{Z^2 p(1-p)}{d^2} \] Where: - \( Z \) is the Z-score for the desired confidence level (represented here as \(t\)).
- \( p \) is the estimated population proportion.
- \( d \) is the margin of error.

Step 3: Step-by-Step Solution:
Based on the standard formula, we have: - Margin of error: \( d = 0.05 \).
- Z-score: \( Z = t = 2 \). A Z-score of 2 corresponds to approximately a 95.45% confidence level.
- Proportion estimate: \( p = 0.5 \). Using \(p=0.5\) maximizes \(p(1-p)\), ensuring a sufficient sample size regardless of the actual proportion.
Substitute these values into the formula: \[ n_0 = \frac{(2)^2 \times 0.5 \times (1-0.5)}{(0.05)^2} \] \[ n_0 = \frac{4 \times 0.5 \times 0.5}{0.0025} \] \[ n_0 = \frac{4 \times 0.25}{0.0025} \] \[ n_0 = \frac{1}{0.0025} \] Since \( 0.0025 = \frac{1}{400} \), \[ n_0 = 1 \div \frac{1}{400} = 400 \]
Step 4: Answer:
The required sample size \(n_0\) is 400.