Jane wants to estimate the proportion of students on her campus who eat cauliflo

Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 35 students, she finds 2 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.

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. There is a 95% chance that the proportion of students who eat cauliflower on Jane's campus is between __ and__.

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

Explanation / Answer

p = 2 / 35 = 0.057

alpha / 2 = 0.025 Z= 1.96

I: 0.057 +/- 1.96 * SQRT ( 0.057*0.943 / 35 )

0.057 +/- 0.0768

-0.0198< P < 0.1338

C. There is a 95% chance that the proportion of students who eat cauliflower on Jane's campus is between _-0.0198_ and_0.1338_.

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