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).
