<PRE><B>Question 449362</B>
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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