Question

A study was being done to estimate the proportion of students who receive financial aid. If the study would like to construct 95% confidence interval for the proportion of students receiving financial aid within 3%, what sample size would be needed? A previous study stated that 57% of students receive some form of financial aid.

Answer 1

Solution :

Given that,

= 0.57

1 - = 1 - 0.57 = 0.43

margin of error = E = 3% = 0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.03)2 * 0.57 * 0.43

= 1046.1957

Sample size =1047