Jane wants to estimate the proportion of students on her campus who eat caulifiowèr. 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. A 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.) O A. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between and Agresti and coull's method for constructing confidence intervals O B. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between and O C. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is between and To deal with issues such as the distribution of p not following a normal distribution, A. Agresti and B. Coull proposed a modified approach to constructing confidence intervals for a proportion. A (1 - a)• 100% confidence interval for p is given by the O D. The proportion of students who eat cauliflower on Jane's campus is between and 90% of the time. x+2 bounds shown below, where p = and x is the number of successes in n trials. n+4 P(1-P) Lower bound: p-z, n+4 P(1-p) Upper bound: p +zg: n+4