Respuesta :
Answer:
The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
They randomly survey 387 drivers and find that 298 claim to always buckle up.
This means that [tex]n = 387, \pi = \frac{298}{387} = 0.77[/tex]
84% confidence level
So [tex]\alpha = 0.16[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.16}{2} = 0.92[/tex], so [tex]Z = 1.405[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8[/tex]
The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).
