Question

The tobacco industry closely monitors all surveys that involve smoking. One survey showed that among 785 randomly selected subjects who completed 4 years of college, 18.3 % smoke. The 99% confidence interval for the true percentage of smokers of all people who completed 4 years of college is closest to

Answer 1

Answer:

The 99% confidence interval for the true percentage of smokers of all people who completed 4 years of college is closest to (0.1475, 0.2185).

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:

There are 785 students, so $n = 785$

18.3% of them smoke, so $p = 0.183$.

99% confidence interval

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.183 - 2.575\sqrt{\frac{0.183*0.817}{785}} = 0.1475$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.183 + 2.575\sqrt{\frac{0.183*0.817}{785}}{119}} = 0.2185$

The 99% confidence interval for the true percentage of smokers of all people who completed 4 years of college is closest to (0.1475, 0.2185).