Question

A survey of 1283 student loan borrowers found that 442 had loans totaling more than $20,000 for their undergraduate education. Give a 90% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education. (Round your answers to three decimal places.)

Answer 1

Given in a survey of n = 1283 student loan borrowers found that X = 442 had loans totaling more than $20,000 for their undergraduate education.

Thus the sample proportion is calculated as:

Now the confidence interval for population proportion is calculated as:

Here Zc is critical score at the given confidence level which is here is 90% and is calculated using the excel formula for normal distribution which is =NORM.S.INV(0.95), thus theZc is computed as 1.645

Now the confidence interval is calcuated as:

Thus the confidence interval is calculated as { 0.323, 0.366).