Respuesta :
Answer:
The 90% confidence interval is (0.1897, 0.2285).
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:
From a July 2019 survey of 1186 randomly selected Americans ages 18-29, it was discovered that 248 of them vaped (used an e-cigarette) in the past week. This means that [tex]n = 1186, \pi = \frac{248}{1186} = 0.2091[/tex]
Construct a 90% confidence interval to estimate the population proportion of Americans age 18-29 who vaped in the past week.
So [tex]\alpha = 0.10[/tex] , z is the value of Z that has a pvalue of [tex]1 - \frac{0.10}{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}} = 0.2091 - 1.645\sqrt{\frac{0.2091*0.7909}{1186}} = 0.1897[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2091 + 1.645\sqrt{\frac{0.2091*0.7909}= 0.2285[/tex]
The 90% confidence interval is (0.1897, 0.2285).
