Question

When 293 college students are randomly selected and surveyed, it is found that 114 own a car. Construct a 99% confidence interval for the true proportion of all college students who own a car. Round your answer to 3 decimal places.

Answer 1

The 99% confidence interval for the true proportion of all college students who own a car is (0.315, 0.463).

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)`.

When 293 college students are randomly selected and surveyed, it is found that 114 own a car.

This means that
`n = 293, \pi = (114)/(293) = 0.389`

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)} = 0.389 - 2.575\sqrt{(0.389*0.621)/(293)} = 0.315`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.389 + 2.575\sqrt{(0.389*0.621)/(293)} = 0.463`

The 99% confidence interval for the true proportion of all college students who own a car is (0.315, 0.463).