document.write( "Question 452767: Out of the 200 insomniacs, 98 reported regularly watching The Late Show with David Letterman before they began to count sheep.Calculate the margin of error for a 78% confidence interval of the true proportion of insomniacs who regularly watch David Letterman before counting sheep.\r
\n" ); document.write( "\n" ); document.write( " A. 0.164
\n" ); document.write( " B. 0.056
\n" ); document.write( " C. 0.043
\n" ); document.write( " D. 0.136
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Algebra.Com's Answer #311290 by stanbon(75887)\"\" \"About 
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Out of the 200 insomniacs, 98 reported regularly watching The Late Show with David Letterman before they began to count sheep.Calculate the margin of error for a 78% confidence interval of the true proportion of insomniacs who regularly watch David Letterman before counting sheep.
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\n" ); document.write( "p = 98/200 = 0.49
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\n" ); document.write( "ME = z*sqrt[pq/n]
\n" ); document.write( "---
\n" ); document.write( "ME = 1.226528*sqrt[0.49*0.51/200] = 0.0434
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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\n" ); document.write( "A. 0.164
\n" ); document.write( "B. 0.056
\n" ); document.write( "C. 0.043
\n" ); document.write( "D. 0.136 \r
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