Question

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.
Calculate the sample proportion of Americans ages 18-29 who vaped in the past week. Round this value to four decimal places.
Using the sample proportion obtained in (a), construct a 90% confidence interval to estimate the population proportion of Americans age 18-29 who vaped in the past week. Please do this "by hand" using the formula and showing your work (please type your work, no images accepted here). Round your confidence limits to four decimal places.

Answer 1

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 $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$

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 $n = 1186, \pi = \frac{248}{1186} = 0.2091$

Construct a 90% confidence interval to estimate the population proportion of Americans age 18-29 who vaped in the past week.

So $\alpha = 0.10$ , z is the value of Z that has a pvalue of $1 - \frac{0.10}{2} = 0.95$, so $z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2091 - 1.645\sqrt{\frac{0.2091*0.7909}{1186}} = 0.1897$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2091 + 1.645\sqrt{\frac{0.2091*0.7909}= 0.2285$

The 90% confidence interval is (0.1897, 0.2285).