Question 1205760: A survey of an urban university showed that 865 of 1,360 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase?
Select one:
a. [0.559, 0.713]
b. [0.604, 0.668]
c. [0.610, 0.662]
d. [0.616, 0.656]
Answer by math_tutor2020(3817)   (Show Source):
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Answer: [0.610, 0.662] which is choice C
Explanation
At 95% confidence, the z critical value is roughly z = 1.96
This is something to memorize or have on a reference sheet somewhere.
p = population proportion of those who support a fee increase to improve student recreation center
The goal is to estimate p using a confidence interval.
phat = sample proportion
phat's job is to estimate p.
The sample proportion is at the center of the confidence interval.
phat = (number who support the increase)/(number surveyed)
phat = x/n
phat = 865/1360
phat = 0.636029 approximately
Around 63.6% of this sample supports the increase.
E = margin of error of a proportion
E = z*sqrt(phat*(1-phat)/n)
E = 1.96*sqrt(0.636029*(1-0.636029)/1360)
E = 0.025572 approximately
The margin of error determines how wide the confidence interval is.
Then we can compute the lower and upper boundaries (L and U)
L = phat - E
L = 0.636029 - 0.025572
L = 0.610457
L = 0.610
and
U = phat + E
U = 0.636029 + 0.025572
U = 0.661601
U = 0.662
The 95% confidence interval is [0.610, 0.662]
It is of the format [L, U]
Some stats textbooks will use the format (L, U) which basically means the same thing.
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