Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 32 ​students, she finds 4 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 __.

Answer 1

please rate if it is helpfulll

Answer 2

We can use Agresti and Coull's method to obtain a confidence interval for the proportion of students who eat cauliflower on Jane's campus.

The formula for the adjusted sample size is:

n' = n + z^2

where n is the original sample size, z is the critical value for a normal distribution at the desired confidence level (95% in this case), and ^2 represents "squared."

Using the formula, we have:

n' = 32 + (1.96)^2 = 71.9616

We can now calculate the adjusted sample proportion, which is:

p' = (4 + (1.96^2)/2) / 71.9616 = 0.1423

Next, we can calculate the margin of error using the formula:

ME = z*sqrt(p'(1-p')/n')

where p' is the adjusted sample proportion, n is the adjusted sample size, and z is the critical value for a normal distribution at the desired confidence level.

Using the formula, we have:

ME = 1.96sqrt(0.1423(1-0.1423)/71.9616) = 0.116

Finally, we can construct the confidence interval by adding and subtracting the margin of error from the adjusted sample proportion:

CI = p' ± ME = 0.1423 ± 0.116

Rounding to three decimal places, we get:

Lower endpoint: 0.026 Upper endpoint: 0.258

Therefore, we are 95% confident that the true proportion of students who eat cauliflower on Jane's campus is between 0.026 and 0.258.

The correct choice is D: One is 95% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.026 and 0.258.