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
, and a confidence interval
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
Z is the zscore that has a pvalue of ![1 - \frac{\alpha}{2}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B%5Calpha%7D%7B2%7D)
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 ![n = 1186, \pi = \frac{248}{1186} = 0.2091](https://tex.z-dn.net/?f=n%20%3D%201186%2C%20%5Cpi%20%3D%20%5Cfrac%7B248%7D%7B1186%7D%20%3D%200.2091)
Construct a 90% confidence interval to estimate the population proportion of Americans age 18-29 who vaped in the past week.
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2091 - 1.645\sqrt{\frac{0.2091*0.7909}{1186}} = 0.1897](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.2091%20-%201.645%5Csqrt%7B%5Cfrac%7B0.2091%2A0.7909%7D%7B1186%7D%7D%20%3D%200.1897)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2091 + 1.645\sqrt{\frac{0.2091*0.7909}= 0.2285](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.2091%20%2B%201.645%5Csqrt%7B%5Cfrac%7B0.2091%2A0.7909%7D%3D%200.2285)
The 90% confidence interval is (0.1897, 0.2285).