Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 ​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. Select the correct choice below and fill in the answer boxes within your choice.
​(Round to three decimal places as​ needed.)
A. The proportion of students who eat cauliflower on​ Jane's campus is between___ and __ 95​% of the time.
B.There is a 95​% chance that the proportion of students who eat cauliflower in​ Jane's sample is between __ and __.
C. There is a 95​% chance that the proportion of students who eat cauliflower on​ Jane's campus is between __ and__.
D. One is 95​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between __ and __.

Answer 1

A 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus is [0.012, 0.270].

Explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 ​students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of students who eat cauliflower

n = sample of students

p = population proportion of students who eat cauliflower

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

P(
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

Now, in Agresti and​ Coull's method; the sample size and the sample proportion is calculated as;

`n = n + Z^(2)__((_\alpha)/(2))`

n =
`24 + 1.96^(2)` = 27.842

`\hat p = (x+(Z^(2)__((\alpha)/(2)_) )/(2) )/(n)` =
`\hat p = (2+(1.96^(2) )/(2) )/(27.842)` = 0.141

95% confidence interval for p = [
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ]

= [
`0.141 -1.96 * {\sqrt{(0.141(1-0.141))/(27.842) } }` ,
`0.141 +1.96 * {\sqrt{(0.141(1-0.141))/(27.842) } }` ]

= [0.012, 0.270]

Therefore, a 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between 0.012 and 0.270.