Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 36 students, she finds 3 who eat cauliflower. Obtain and interpret a 90% confidence interval for...

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Extracted text: Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 36 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 metho 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.) A. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between and B. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between and C. The proportion of students who eat cauliflower on Jane's campus is between and 90% of the time. D. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is between and Agresti and coull's method for constructing confidence intervals 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 X+2 bounds shown below, where p = and x is the number of successes in n trials. n+4 P(1-) Lower bound: p - za n+4 P(1-P) Upper bound: p +za* n+4 Print Done