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. A state-wide survey indicates that the proportion is 0.45. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.08?
A)210
B)8
C)17
D)149

Question

Quizplus · Accepted Answer

The sample size needed for a proportion can be calculated using the formula $n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right)$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion, and $E$ is the desired margin of error. For a 98% confidence interval, the Z-score is approximately 2.33. Plugging in the values: $n = \left(\frac{2.33^2 \cdot 0.45 \cdot (1-0.45)}{0.08^2}\right)$, which calculates to approximately 210.

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