Question

In a simple random sample of 352 students at a college, 92 reported that they have at least $1000 of credit card debt. Which interval is the 99% confidence interval for the percent of all the students at that college who have at least $1000 in credit card debt

Answer 1

The 99% confidence interval for the percent of all the students at that college who have at least $1000 in credit card debt is (20.11%, 32.17%).

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

In a simple random sample of 352 students at a college, 92 reported that they have at least $1000 of credit card debt.

This means that
`n = 352, \pi = (92)/(352) = 0.2614`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a p-value of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.2614 - 2.575\sqrt{(0.2614*0.7386)/(352)} = 0.2011`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2614 + 2.575\sqrt{(0.2614*0.7386)/(352)} = 0.3217`

As percent:

0.2011*100% = 20.11%

0.3217*100% = 32.17%.

The 99% confidence interval for the percent of all the students at that college who have at least $1000 in credit card debt is (20.11%, 32.17%).