document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "What is the LOWER limit of the 99% confidence interval?\r
\n" ); document.write( "\n" ); document.write( "What is the UPPER limit of the 99% confidence interval?
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Algebra.Com's Answer #309169 by stanbon(75887)\"\" \"About 
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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\n" ); document.write( "p-hat = 104/200 = 0.52
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\n" ); document.write( "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.
\n" ); document.write( "Carry your intermediate computations to at least three decimal places. Round your answer to two decimal places.
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\n" ); document.write( "ME = z*sqrt[pq/n] = 2.5758*sqrt[0.52*0.48/200] = 0.091
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\n" ); document.write( "\n" ); document.write( "What is the LOWER limit of the 99% confidence interval?
\n" ); document.write( "p-hat-ME = 0.52-0.091 = 0.429
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\n" ); document.write( "What is the UPPER limit of the 99% confidence interval?
\n" ); document.write( "p-hat+ME = 0.52+0.091 = 0.611
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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