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 99% confidence interval for the proportion of students who eat cauliflower on Jane's campus using Agresti and Coull's method. Click the icon to view Agresti and Coull's method. Construct and interpret the 99% 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 99% chance that the proportion of students who eat cauliflower on Jane's campus is between Jane's and OB. One is 99% confident that the proportion of students who eat cauliflower on Jane's campus is between OC. There is a 99% chance that the proportion of students who eat cauliflower in Jane's sample is between OD. The proportion of students who eat cauliflower on Jane's campus is between and 99% of the time. and and​