Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 27 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 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.) O A. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between and. O B. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between and OC. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is between and O D. The proportion of students who eat cauliflower on Jane's campus is between and 90% of the time.

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neip
Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 27 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 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.)
O A. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between
and
O B. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between
and
O C. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus
is
between
and
O D. The proportion of students who eat cauliflower on Jane's campus is between
and 90% of the time.
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Transcribed Image Text:neip Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 27 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 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.) O A. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between and O B. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between and O C. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is between and O D. The proportion of students who eat cauliflower on Jane's campus is between and 90% of the time. Question Viewer Click to select and enter your answer(s) and then click Check Answer. All parts showing Clear All Check Answer MacBook Air 888 DII DD esc F2 F3 F4 F5 F6 F7 FB F9 F11 F12 # $ へ & ( 1 2 3 4 5 6 7 8 9 delete { [ } Q W E Y tab %3D A S D G J K caps lock return く C V M .. .- つ エ く (C B
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