A researcher wants to construct a 98% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school lunches. What sample size is needed so that the confidence interval will have a margin of error of 0.09?
A)119
B)7
C)167
D)16

Question

Quizplus · Accepted Answer

To determine the sample size needed for a confidence interval with a specific margin of error, the formula is $n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right)$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion (if unknown, 0.5 is used for maximum sample size), and $E$ is the margin of error. For a 98% confidence interval, the Z-score is approximately 2.33. Assuming $p = 0.5$ for maximum variability, and $E = 0.09$, the calculation is $n = \left(\frac{2.33^2 \cdot 0.5 \cdot (1-0.5)}{0.09^2}\right)$, which results in a value close to 167, making option C the correct choice.

A Researcher Wants to Construct a 98% Confidence Interval for the Proportion