Question

In: Statistics and Probability

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

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 29 ​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. LOADING... Click the icon to view Agresti and​ Coull's method.

Solutions

Expert Solution

Solution :

Given that,

n = 29

x = 4

Point estimate = sample proportion = = x / n = 0.138

1 - = 0.862

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * (((0.138 * 0.862) / 29)

= 0.126

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.138 - 0.126 < p < 0.138 + 0.126

0.012 < p < 0.264

The 95% confidence interval for the population proportion p is : (0.012 , 0.264)

A 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus using Agresti and​ Coull's method is between 0.012 and 0.264.


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