A researcher wants to construct a 90% 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 0.50. Using this estimate,
What sample size is needed so that the confidence interval will have a margin of error of
0)1?
A) 5
B) 7
C) 68
D) 42

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 desired confidence level (for 90% confidence, $Z = 1.645$), $p$ is the estimated proportion (0.50 in this case), and $E$ is the desired margin of error (0.1). Plugging in the values: $n = \left(\frac{1.645^2 \cdot 0.50 \cdot (1-0.50)}{0.1^2}\right) = 67.6025$, which rounds up to 68 to ensure the sample is large enough.

A Researcher Wants to Construct a 90% Confidence Interval for the Proportion