Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 17 ​students, she finds 5 who eat cauliflower. Obtain and interpret a 90​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus using Agresti and​ Coull's method.

A. There is a 90​% chance that the proportion of students who eat cauliflower on​ Jane's campus is between __ and ____.

B. The proportion of students who eat cauliflower on​ Jane's campus is between ___and ___ 90​% of the time.

C. There is a 90​% chance that the proportion of students who eat cauliflower in​ Jane's sample is between ___ and ___.

D. One is 90​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between __and ___.

Answer 1

Sample proportion = 5 / 17 = 0.294

90% confidence interval for p is

- Z * sqrt( ( 1 - ) / n) < p < + Z * sqrt( ( 1 - ) / n)

0.294 - 1.645 * sqrt( 0.294 * 0.706 / 17) < p < 0.294 + 1.645 * sqrt( 0.294 * 0.706 / 17)

0.112 < p < 0.476

One is 90​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between 0.112

and 0.476