Respuesta :
Answer:
The 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623, 0.7369).
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].
In a survey of 1002 people, 701 said that they voted in a recent presidential election.
This means that [tex]n = 1002, \pi = \frac{701}{1002} = 0.6996[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6996 - 2.575\sqrt{\frac{0.6996*0.3004}{1002}} = 0.6623[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6996 + 2.575\sqrt{\frac{0.6996*0.3004}{1002}} = 0.7369[/tex]
The 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623, 0.7369).
