Respuesta :
Answer:
The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).
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 interval [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].
For this problem, we have that:
Of 150 adults selected randomly from one town, 30 of them smoke. This means that [tex]n = 150, p = \frac{30}{150} = 0.2[/tex]
99% confidence interval
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue 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}} = 0.20 - 2.575\sqrt{\frac{0.20*0.80}{150}} = 0.1159[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 + 2.575\sqrt{\frac{0.20*0.80}{150}}{119}} = 0.2841[/tex]
The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).
