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 * (
= 0.126
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.
orchestra answered 1 year agoRelated Solutions
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