10: How do you solve? Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 4 who eat cauliflower. Obtain and interpret a 90% confidence interval forthe 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.).OA. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is betweenOB. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is betweenand 90% of the time.OC. The proportion of students who eat cauliflower on Jane's campus is betweenOD. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is betweenandandand

From the provided information, Sample size (n) = 20 From which 4 eat cauliflower. Sample proportion (p̂) = 4/20 = 0.2 Confidence level = 90%Level of significance (α) = 1 – 0.90 = 0.10 And α/2 = 0.10/2 = 0.05 The critical value of z can be obtained using excel function as: =NORM.S.INV(0.90+0.05) The excel output is given as: The critical z value is 1.645 approximately. The required 90% confidence interval can be obtained as: CI=p^±zα/2p^1-p^n=0.2±1.6450.21-0.220=0.2±0.147=0.053,0.347 One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.053 and 0.347. Thus, the correct option is A).