Questions: What sample size is needed to give a margin of error within ± 1.5% in estimating a population proportion with 95% confidence? Round your answer up to the nearest integer.

Solution Steps

We need to determine the sample size \( n \) required to achieve a margin of error of \( \pm 1.5\% \) (or \( 0.015 \) in decimal form) when estimating a population proportion with a confidence level of \( 95\% \).

For a \( 95\% \) confidence level, the corresponding Z-score is approximately \( Z = 1.96 \).

The formula for the margin of error in estimating a population proportion is given by:

\[ \text{Margin of Error} = Z \times \sqrt{\frac{p(1-p)}{n}} \]

Rearranging this formula to solve for \( n \) gives:

\[ n = \frac{Z^2 \times p(1-p)}{\text{Margin of Error}^2} \]

Assuming the worst-case scenario for the population proportion \( p = 0.5 \):

\[ n = \frac{(1.96)^2 \times 0.5 \times (1 - 0.5)}{(0.015)^2} \]

Calculating the components:

• \( Z^2 = (1.96)^2 = 3.8416 \)
• \( p(1-p) = 0.5 \times 0.5 = 0.25 \)
• \( \text{Margin of Error}^2 = (0.015)^2 = 0.000225 \)

Substituting these values into the formula:

\[ n = \frac{3.8416 \times 0.25}{0.000225} \]

Calculating \( n \):

\[ n = \frac{0.9604}{0.000225} \approx 4268.8889 \]

Since the sample size must be a whole number, we round up to the nearest integer:

\[ n = 4269 \]

Final Answer