Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. use a 98% confidence interval to estimate the true proportion of students on financial aid. express the answer in the form ± e and round to the nearest thousandth.

Answer 1

Answer: 0.59 ± 0.081 = (0.671, 0.509)  
Explanation: 
The sample proportion, p = 118/200 = 0.59 
The test statistic at 98% confidence interval is given by : 
p + z*Sqrt(p(1-p)/n) ; p - z*Sqrt(p(1-p)/n) 
z at 98% C.I. is 2.33 
Therefore, 0.59 + 2.33 * sqrt(0.59*(1-0.59)/200) and 0.59 + 2.33 * sqrt(0.59*
(1-0.59)/200)
 =0.59 + 0.081 and 0.59 - 0.081
 =0.671, 0.509