Question

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

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 ____.

Solution :

Given that,

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

1 - = 1 - 0.364 = 0.636

Z/2 = Z0.05 = 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

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