CI98 = p ± 2.33 √(pq/n)
p = 42/200 = .21
q = 1 - p = 1 - .21 = .79
n = 200
With your data:
CI98 = .21 + 2.33 √[(.21)(.79)/200]
I'll let you take it from here to finish.
In a study of 200 students, 42 were found to have a GPA greater than 3.0. Find a
98% confidence interval estimate of the proportion, p, of all students that have a GPA greater than 3.0.
1 answer
1 answer