In: Statistics and Probability
Breakport University wishes to determine the percentage of students who will be taking courses during the summer. How many students must they ask to estimate the percentage with 99% confidence of being within 5 percentage points of the true value?
Solution :
Given that,
= 0.5 ( assume)
1 - = 1 - 0.5 = 0.5
margin of error = E =5 % = 0.05
At 99% confidence level the z is,
= 1 - 99%
= 1 - 0.99 = 0.01
/2 = 0.005
Z/2 = 2.576 ( Using z table )
Sample size = n = (Z/2 / E)2 * * (1 - )
= (2.576 / 0.05)2 * 0.5 * 0.5
=663.5776
Sample size = 664