Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 22 students, she finds 22 who eat cauliflower. Obtain and interpret a 95% 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 95% confidence interval. Select the correct choice below and fill in the answer boxes within your choice.

(Round to three decimal places as needed.)

A.The proportion of students who eat cauliflower on Jane's campus is between __and _95% of the time.

B.There is a 95% chance that the proportion of students who eat cauliflower in Jane's sample is between __and __.

C.One is 95% confident that the proportion of students who eat cauliflower on Jane's campus is between __and __.

D.There is a 95% 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 ModifyingAbove p with caretp not following a normal distribution, A. Agresti and B. Coull proposed a modified approach to constructing confidence intervals for a proportion. A left parenthesis 1 minus alpha right parenthesis times 100 %(1)100% confidence interval for p is given by the bounds shown below, where ModifyingAbove p with tilde equals StartFraction x plus 2 Over n plus 4 EndFractionp=x+2n+4 and x is the number of successes in n trials.

Lower bound: ModifyingAbove p with tilde minus z Subscript StartFraction alpha Over 2 EndFraction Baseline times StartRoot StartFraction ModifyingAbove p with tilde left parenthesis 1 minus ModifyingAbove p with tilde right parenthesis Over n plus 4 EndFraction EndRootpz2p1pn+4

Upper bound: ModifyingAbove p with tilde plus z Subscript StartFraction alpha Over 2 EndFraction Baseline times StartRoot StartFraction ModifyingAbove p with tilde left parenthesis 1 minus ModifyingAbove p with tilde right parenthesis Over n plus 4 EndFraction EndRoot