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) 167
C) 7
D) 16

Question

Quizplus · Accepted Answer

We use the formula for margin of error: 
margin of error = z*sqrt((p-hat*(1-p-hat))/n)
Since we don't have any prior information, we use 0.5 as the estimate for p-hat.
We are given that the margin of error is 0.09 and we want a 98% confidence interval, so the z-score is 2.33.
0.09 = 2.33*sqrt((0.5*(1-0.5))/n)
Solving for n:
n = (2.33/0.09)^2 * (0.5*(1-0.5))
n = 167.53
Round up to the nearest whole number: n = 168.
Therefore, the researcher needs to sample at least 168 elementary school students.

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