A university planner is interested in determining the percen

A university planner is interested in determining the percentage of spring semester students who will attend summer school. She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school.

a.

Construct a 95% confidence interval estimate for the percentage of spring semester students who will return to summer school.

b.

Using the results of the pilot study with a 0.95 probability, how large of a sample would have to be taken to provide a margin of error of 3% or less?

Part A I am clear about to reach the answer 0.276 to 0.424.
I need help with how to achieve the answer of \"972\" for part B. I know that is the answer but do not know how to obtain it.

Solution

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 56/160 = 0.35
ME = 0.03
n = ( 1.96 / 0.03 )^2 * 0.35*0.65
= 971.07111 ~ 972