Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 401 drivers and find that 294 claim to always buckle up. Construct a 90% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5].

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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

They randomly survey 401 drivers and find that 294 claim to always buckle up.

This means that [tex]n = 401, \pi = \frac{294}{401} = 0.7332.

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7332 - 1.645\sqrt{\frac{0.7332*0.2668}{401}} = 0.6969[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7332 + 1.645\sqrt{\frac{0.7332*0.2668}{401}} = 0.7695[/tex]

The 90% confidence interval for the population proportion that claim to always buckle up is [0.6969, 0.7695]