Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 27 students, she finds 3 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.

ANSWER-BETWEEN9.38%and12.9%Step-by-step explanation:p=p’±[Z(√p'(1−p’)/N]9.38%and 12.9%(after converting the answer we got to percentage).