Answer to Question #116383 in Statistics and Probability for Mary-Grace Antonio

Question

According to the National Institute on Alcohol Abuse and Alcoholism (NIAAA), and the National Institutes of Health (NIH), 41%
of college students nationwide engage in binge drinking behavior, having 5 or more drinks in one occasion during the past two
weeks. In a commissioned study, 462 students are selected randomly from a list of all students enrolled at the college. Of these,
162 admitted to having engaged in binge drinking. Based on the results of the study, find a 98% confidence interval for the
proportion of all students at this college that engage in binge drinking

Answer 1

$\hat p=\frac{162}{462}=0.3506.$

$98\%CI=(\hat p-z_{0.01}\sqrt{\frac{\hat p(1-\hat p)}{n}},\hat p+z_{0.01}\sqrt{\frac{\hat p(1-\hat p)}{n}})=$

$=(0.3506-2.326\sqrt{\frac{0.3506(1-0.3506)}{426}},0.3506+2.326\sqrt{\frac{0.3506(1-0.3506)}{426}})=$

$=(0.299,0.402).$