document.write( "Question 440970: . In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Construct a 95% confidence interval for our estimate of the percentage of T.V. viewers who watch network news programs \n" ); document.write( "
Algebra.Com's Answer #304560 by stanbon(75887)\"\" \"About 
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In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Construct a 95% confidence interval for our estimate of the percentage of T.V. viewers who watch network news programs
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\n" ); document.write( "p-hat = 0.4
\n" ); document.write( "ME = z*sqrt(pq/n)
\n" ); document.write( "ME = 1.96*sqrt(0.4*0.6/5100) = 0.0134
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\n" ); document.write( "95% CI: 0.4-0.0134 < p < 0.4+0.0134
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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