document.write( "Question 1057942: A recent poll of 380 randomly selected Americans showed that 24% (\r
\n" ); document.write( "\n" ); document.write( " = 0.24) are happy with their cell phone carriers.
\n" ); document.write( "The country’s largest cell phone carrier would like to know, within a 90% confidence level, the margin of error for this poll. (90% confidence level z*-score of 1.645)
\n" ); document.write( "Remember, the margin of error, E, can be determined using the formula E = z*\r
\n" ); document.write( "\n" ); document.write( ".
\n" ); document.write( "To the nearest tenth of a percent, the margin of error for the poll is %.
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Algebra.Com's Answer #672942 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
A recent poll of 380 randomly selected Americans showed that 24% (
\n" ); document.write( "= 0.24) are happy with their cell phone carriers.
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\n" ); document.write( "The country’s largest cell phone carrier would like to know, within a 90% confidence level, the margin of error for this poll. (90% confidence level z*-score of 1.645)
\n" ); document.write( "Remember, the margin of error, E, can be determined
\n" ); document.write( "using the formula E = z*sqrt(pq/n)
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\n" ); document.write( "E = 1.645*sqrt[0.24*0.76/380] = 0.0360..\r
\n" ); document.write( "\n" ); document.write( "To the nearest tenth of a percent,
\n" ); document.write( "the margin of error for the poll is 3.6%
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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