Question

A random sample of 80 college students showed that 44 had driven a car during the day before the survey was conducted. Suppose that we are interested in forming a 80 percent confidence interval for the proportion of all college students who drove a car the day before the survey was conducted.

Where appropriate, express your answer as a proportion (not a percentage). Round answers to three decimal places.

Answer 1

The 80% confidence interval for the proportion of all college students who drove a car the day before the survey was conducted is (0.479, 0.621).

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)`.

A random sample of 80 college students showed that 44 had driven a car during the day before the survey was conducted.

This means that
`n = 80, \pi = (44)/(80) = 0.55`

80% confidence level

So
`\alpha = 0.2`, z is the value of Z that has a pvalue of
`1 - (0.2)/(2) = 0.9`, so
`Z = 1.28`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.55 - 1.28\sqrt{(0.55*0.45)/(80)} = 0.479`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.55 + 1.28\sqrt{(0.55*0.45)/(80)} = 0.621`

The 80% confidence interval for the proportion of all college students who drove a car the day before the survey was conducted is (0.479, 0.621).