Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 25

25 students, she finds 5

5 who eat cauliflower. Obtain and interpret a 99

99% confidence interval for the proportion of students who eat cauliflower onJane's campus using Agresti andCoull's method.

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Construct and interpret the 99

99% confidence interval. Select the correct choice below and fill in the answer boxes within your choice.

(Round to three decimal places asneeded.)

A.

There is a 99

99% chance that the proportion of students who eat cauliflower onJane's campus is between

nothing

and

nothing

.

B.

There is a 99

99% chance that the proportion of students who eat cauliflower inJane's sample is between

nothing

and

nothing

.

C.

The proportion of students who eat cauliflower onJane's campus is between

nothing

and

nothing

99

99% of the time.

D.

One is 99

99% confident that the proportion of students who eat cauliflower onJane's campus is between

nothing

and

nothing

.