Question

A survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating (either an online dating site or a dating app on their cell phone). 55- to 64-year-olds The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating. If we use this sample to estimate the proportion of all American adults ages 55 to 64 to use online dating, the standard error is 0.016. Find a 99% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating. Round your answers to three decimal places. The 99% confidence interval is ______?

Answer 1

Answer:

The 99% confidence interval is (0.0787, 0.1613).

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:

The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating. This means that $n = 411, \pi = \frac{50}{411} = 0.12$

The 99% confidence interval is

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{411}} = 0.12 - 2.575\sqrt{\frac{0.12*0.88}{411}} = 0.0787$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{411}} = 0.12 + 2.575\sqrt{\frac{0.12*0.88}{411}} = 0.1613$

The 99% confidence interval is (0.0787, 0.1613).