A survey of 104 students is selected randomly on a large university campus. They
ID: 3057303 • Letter: A
Question
A survey of 104 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. The result of the survey is that 52 of the 104 students responded "yes." An approximate 98% confidence interval is (0.386, 0.614).
Complete parts a and b below.
a) How would the confidence interval change if the confidence level had been 90% instead of 98%?
The new confidence interval would be ??? (either narrower or wider)
The new confidence interval would be??? (Round to three decimal places as needed.)
b) How would the confidence interval change if the sample size had been 156 instead of 104? (Assume the same sample proportion.)
The new confidence interval would be??? (narrrower or wider)
The new confidence interval would be??? (Round to three decimal places as needed.)
Explanation / Answer
Here sample size = 104
sample proportion of people say "yes" = 52/104 = 0.50
Here 98% confidence interval = (0.386, 0.614)
(a) As we decrease the confidence interval to 90% that would make confidence interval narrower.
It would be = p^ + Z90% sqrt [p^ * (1-p^)/N]
= 0.50 +- 1.645 * sqrt [0.5 * 0.5/104]
= 0.50 +- 0.081
= (0.419, 0.581)
(b) Here if the sample size is changed to 156 instead of 104 so
that would make the new confidence interval narrower.
The new confidence interval = p^ + Z90% sqrt [p^ * (1-p^)/N]
= 0.50 +- 2.326 * sqrt [0.5 * 0.5/156]
= 0.50 +- 0.093
= (0.407, 0.593)
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