Question

A survey of 1295 student loan borrowers found that 460 had loans totaling more than $20,000 for their undergraduate education. Give a 98% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education. (Give answers accurate to 3 decimal places.)

Answer 1

The 98% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education is (0.324, 0.386)

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 1295, \pi = (460)/(1295) = 0.355`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.355 - 2.327\sqrt{(0.355*0.645)/(1295)} = 0.324`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.355 + 2.327\sqrt{(0.355*0.645)/(1295)} = 0.386`

The 98% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education is (0.324, 0.386)