SOLUTION: In a poll of 200 randomly selected U.S. adults, 104 said they favored a new proposition. Based on this poll, compute a 99% confidence interval for the proportion of all U.S. adults
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Question 449362: In a poll of 200 randomly selected U.S. adults, 104 said they favored a new proposition. Based on this poll, compute a 99% confidence interval for the proportion of all U.S. adults in favor of the proposition ( at the time of the poll). Then compute the table below.
Carry your intermediate computations to at least three decimal places. Round your answer to two decimal places.
What is the LOWER limit of the 99% confidence interval?
What is the UPPER limit of the 99% confidence interval? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! In a poll of 200 randomly selected U.S. adults, 104 said they favored a new proposition.
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p-hat = 104/200 = 0.52
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Based on this poll, compute a 99% confidence interval for the proportion of all U.S. adults in favor of the proposition ( at the time of the poll). Then compute the table below.
Carry your intermediate computations to at least three decimal places. Round your answer to two decimal places.
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ME = z*sqrt[pq/n] = 2.5758*sqrt[0.52*0.48/200] = 0.091
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What is the LOWER limit of the 99% confidence interval?
p-hat-ME = 0.52-0.091 = 0.429
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What is the UPPER limit of the 99% confidence interval?
p-hat+ME = 0.52+0.091 = 0.611
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Cheers,
Stan H.
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