Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 17 ​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.

Answer 1

Solution :

Given that,

n = 17

x = 2Point estimate = sample proportion

= = x / n = 2/17=0.118

1 -   = 0.882

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2   * ((( * (1 - )) / n)

= 1.96 (((0.118*0.882) / 17)

E = 0.078

A 95% confidence interval is ,

- E < p < + E

0.118- 0.078 < p < 0.118+0.078

0.04< p < 0.196

(0.04 , 0.196)