Chandler, Jamal, Allison, and Qi are given the following problem. In 1982, Wayne Gretsky set a record for the greatest number of
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Answer:
A 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus is [0.012, 0.270].
Step-by-step 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. = ~ N(0,1)
where, = sample proportion of students who eat cauliflower
n = sample of students
p = population proportion of students who eat cauliflower
<em>Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.</em>
<u>So, 95% confidence interval for the population proportion, p is ;</u>
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 < < 1.96) = 0.95
P( <
<
) = 0.95
P( < p <
) = 0.95
Now, in Agresti and Coull's method; the sample size and the sample proportion is calculated as;
n = = 27.842
=
= 0.141
<u>95% confidence interval for p</u> = [ ,
]
= [ ,
]
= [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.