Question

9.1.45-1 Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 39 students, she finds 4 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. - Click the icon to view 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. (Round to three decimal places as needed.) O A. One is 90% confident that the proportion of students who eat cauliflower on Jane's campus is between and O B. There is a 90% chance that the proportion of students who eat cauliflower on Jane's campus is between and O C. The proportion of students who eat cauliflower on Jane's campus is between and 90% of the time. OD. There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between and

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 4 / 39 = 0.103

1 - $\hat p$ = 1 - 0.897

Z$\alpha$/2 = 1.645

Margin of error = E = Z$\alpha$ / 2 * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.645 * ($\sqrt$((0.103 * 0.897) / 39)

Margin of error = E = 0.080

A 90% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.103 - 0.080 < p < 0.103 + 0.080

0.023 < p < 0.183

option A. is correct

One is 90% confidence that the proportion of students who eat cauliflower on jane's campus is between 0.023 and 0.183