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.

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Answer:

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

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

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

This means that [tex]n = 391, \pi = \frac{319}{391} = 0.8159[/tex]

85% confidence level

So [tex]\alpha = 0.15[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.15}{2} = 0.925[/tex], so [tex]Z = 1.44[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8159 - 1.44\sqrt{\frac{0.8159*0.1841}{391}} = 0.7877[/tex]

The upper limit of this interval is:

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

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