Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 12

​students, she finds 33 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.

    Construct and interpret the 90​% confidence interval. Select the correct choice below and fill in the answer boxes within your choice.

A. The proportion of students who eat cauliflower on​ Jane's campus is between ____and ___ 90​% of the time.

B. There is a 90​% chance that the proportion of students who eat cauliflower in​ Jane's sample is between __ and __.

C. One is 90​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between ___ and ___.

D. There is a 90​% chance that the proportion of students who eat cauliflower on​ Jane's campus is between ___ and ____.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n 12 / 33 = 0.364

1 - = 1 - 0.364 = 0.636

Z/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.364 * 0.636) / 33)

= 0.138

A 90% confidence interval for population proportion p is ,

± E   

= 0.364  ± 0.138

= ( 0.226, 0.502 )

C. One is 90​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between 0.226  and 0.502