Question

According to the National Institute on Alcohol Abuse and Alcoholism (NIAAA), 41% of USA college students nationwide engage in “binge drinking” behavior, having five or more drinks in one occasion during the past two weeks. A college president wonders if the proportion of students enrolled at her college that binge drink is actually lower than the national proportion. In a commissioned study, 348 students are selected randomly from a list of all students enrolled at the college. Of these, 132 admitted to having engaged in binge drinking.
Based on the results of the test and using the "plus four" method, a 95% confidence interval for the proportion of all students at this college that engage in binge drinking is

Select one:

a. 0.338 to 0.423.

b. 0.330 to 0.432.

c. 0.341 to 0.420.

d. 0.343 to 0.418.

Answer 1

Here sample proportion by "plus four"method = (132 + 2)/ (348 + 4) = 0.381

standard error of proportion = sqrt [0.381 * (1 - 0.381)/352] = 0.02588

critical test statistic for 95% confidence interval = 1.96

margin of error = 1.96 * 0.02588 = 0.0507

95% confidence interval = 0.381 +- 0.051 = (0.330, 0.432)

Option b is correct here.