document.write( "Question 1113821: In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percentage of T.V. viewers who watch network news programs.\r
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document.write( "A) 2.00%\r
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document.write( "B) 1.34%\r
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document.write( "C) 1.54%\r
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document.write( "D) 1.76%\r
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document.write( "E) 1.12%\r
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Algebra.Com's Answer #728901 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percentage of T.V. viewers who watch network news programs. \n" ); document.write( "ME = z*sqrt(p*q/n) \n" ); document.write( "ME = 1.96*sqrt(0.4*0.6/5100) = 1.34% \r \n" ); document.write( "\n" ); document.write( "A) 2.00% \n" ); document.write( "B) 1.34% \n" ); document.write( "C) 1.54% \n" ); document.write( "D) 1.76% \n" ); document.write( "E) 1.12 \n" ); document.write( "-------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "--------- \n" ); document.write( " \n" ); document.write( " |