Question

you randomly selected 138 students from CSU and find that 57 are on scholarship or financial aid. You would like to contract a 90% confidence interval to estimate the true proportion of students on scholarship or financial aid. Based on this intial sample what is the minimum sample size you must select in order to have a margin of error of 3%?

A. 443

B. 444

C. 728

D. 729

E. None of the above

Answer 1

solution:

Given that,

= 57/138=0.413

1 - = 1 - 0.413 = 0.587

margin of error = E = 3% = 0.03

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.03)2 * 0.413 * 0.587

=728.9

Sample size = 729 rounded