Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 15​ 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.

One is 95% confident that the poplulation of students who eat cauliflower on janes campus is between___and ____?

*Use the corrent z store interval for 95% confidence!!

Round to three decimal places as needed

Answer 1

Let to finding 95% confidence interval for proportion of student who eat cauliflower on janis compose using arrest and calls method given n = 15 x = 0.13 x = 2 p = x/n Rightarrow lower limit P + 2^2 H_2/2n + 2H_2 squareroot pq/n + z^2 H_2/4n^2/1 + 2^2 H_2/(n) = 0.13 + 3.8416 - 1.96 squareroot 0.1131/75 + 3.8416/900/1 + 3.8416/15 = 0.13 + 0.1280 - 1.46(0.1086)/1.2516 = 0.13 - 0.848/1.2516 = 0.0360 \r\n upper limit p + 2^2 H_2/2n - 2h_2 squareroot pq/n + 2^2 H_2/4n_2/ 1 + 2^2 H_2/n = 0.13 + 3.816/30 + 1.96 squareroot 0.1131/15 + 3.8416/900/1 + 3.8416/15 = 0.13 + 0.1280 + 1.96(0.1086)/1.2516 = 0.4708/1.2576 = 0.3761 hence required 95% confidence interval is given by (0.0360,0.3761)