Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 34 ​students, she finds 2 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 method. LOADING... Click the icon to view Agresti and​ Coull method. Construct and interpret the 90​% confidence interval. Select the correct choice below and fill in the answer boxes within your choice. ​(Round to three decimal places as​ needed.) A. There is a 90​% chance that the proportion of students who eat cauliflower on​ Jane's campus is between nothing and nothing. B. There is a 90​% chance that the proportion of students who eat cauliflower in​ Jane's sample is between nothing and nothing. C. One is 90​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between nothing and nothing. D. The proportion of students who eat cauliflower on​ Jane's campus is between nothing and nothing 90​% of the time.