Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. after surveying 35 ​students, she finds 4 who eat cauliflower. obtain and interpret a 95 ​% confidence interval for the proportion of students who eat cauliflower on​ jane's campus using agresti and​ coull's method.

Answer 1

The mean proportion is p = 4/35 = 0.1143, q = 1 - p = 0.8857, and sample of n = 35.
For a 95% CI, z = +/-1.96
The Agresti-Coull method is as follows:
Limits = [p + (z^2 / 2n) + z*sqrt(pq/n + z^2 / 4n^2)] / (1 + z^2 / n)
Using this formula, the lower limit is 0.0454, while the upper limit is 0.2595.