how to solve? Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 34 students, she finds 3 who eat cauliflower. Obtain and interpreta 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. The proportion of students who eat cauliflower on Jane's campus is betweenand90% of the time.O B. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is betweenandO C. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is betweenandO D. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is betweenand

Given that, x=3,n=34 The sample proportion is, Critical values: By using the standard normal table of Z, the critical value at 90% confidence level for two tailed test is, z0.10/2=±1.645The 90% Confidence interval for the proportion is: Interpretation: One can be 90% confident that the proportion of students who eat cauliflower on janes campus is between 0.008 and 0.168. Hence, the correct answer is option (D) The options (A), (B) and (C) are incorrect because these options represents the interpretation of the probability but not confidence interval.