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: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 [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 whichZ is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].For this problem, we have that:There are 785 students, so [tex]n = 785[/tex]18.3% of them smoke, so [tex]p = 0.183[/tex].99% confidence intervalSo [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].The lower limit of this interval is:[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.183 - 2.575\sqrt{\frac{0.183*0.817}{785}} = 0.1475[/tex]The upper limit of this interval is:[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.183 + 2.575\sqrt{\frac{0.183*0.817}{785}}{119}} = 0.2185[/tex]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).