Need help answering this stats problem

Question

In a college student​ poll, it is of interest to estimate the proportion p of students in favor of changing from a​ quarter-system to a​ semester-system. How many students should be polled so that p can be estimated to within 0.09 using a​ 99% confidence​ interval?

A)261

B)182

C)114

D)205

Answer 1

Your margin of error for the confidence interval is Z * √(p * (1-p)) / n

Where Z is your z-score for a 99% confidence interval, p is the population proportion and n is the sample size.

We know that your margin of error should be .09, so we can set the equation equal to that number.

.09 = Z * √(p * (1-p)) / n

The Z value for a 99% confidence interval is 2.575 and we will use p=0.5 (since we are not given a p) which will give us the most conservative answer. Now solve for n.

0.09 = 2.575 * √(0.5 * (0.5)) / n

0.0349 = √0.25/ n

0.00122 = .25/n

n = 204.6489

Since this is a decimal and we need our answer in number of people to sample, we round up to n=205.