Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 391 drivers and find that 319 claim to always buckle up. Construct a 85% confidence interval for the population proportion that claim to always buckle up.

Answer 1

The 85% confidence interval for the population proportion that claim to always buckle up is (0.7877, 0.8441).

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)`.

They randomly survey 391 drivers and find that 319 claim to always buckle up.

This means that
`n = 391, \pi = (319)/(391) = 0.8159`

85% confidence level

So
`\alpha = 0.15`, z is the value of Z that has a pvalue of
`1 - (0.15)/(2) = 0.925`, so
`Z = 1.44`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.8159 - 1.44\sqrt{(0.8159*0.1841)/(391)} = 0.7877`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.8159 + 1.44\sqrt{(0.8159*0.1841)/(391)} = 0.8441`

The 85% confidence interval for the population proportion that claim to always buckle up is (0.7877, 0.8441).