Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 29 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. Click the icon to view Agresti and Coull's 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.) and A. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is between OB. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between O C. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between D. The proportion of students who eat cauliflower on Jane's campus is between and 90% of the time. and and

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