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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\n" ); document.write( "A) 2.00%\r
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\n" ); document.write( "B) 1.34%\r
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\n" ); document.write( "C) 1.54%\r
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\n" ); document.write( "D) 1.76%\r
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\n" ); document.write( "E) 1.12%\r
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Algebra.Com's Answer #728901 by stanbon(75887)\"\" \"About 
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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
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\n" ); document.write( "Stan H.
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