Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 387 drivers and find that 298 claim to always buckle up. Construct a 84% confidence interval for the population proportion that claim to always buckle up.

Answer 1

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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)`.

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that
`n = 387, \pi = (298)/(387) = 0.77`

84% confidence level

So
`\alpha = 0.16`, z is the value of Z that has a p-value of
`1 - (0.16)/(2) = 0.92`, so
`Z = 1.405`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.77 - 1.405\sqrt{(0.77*0.23)/(387)} = 0.74`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.77 + 1.405\sqrt{(0.77*0.23)/(387)} = 0.8`

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).