document.write( "Question 1203043: In a recent poll, 380 people were asked if the like dogs, and 68% said they did. Find the margin of error for this poll, at the 99% confidence level. Give answer to four decimal places if possible.
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Algebra.Com's Answer #838287 by Theo(13342) $\"\"$ $\"About$

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n = sample size = 380
\n" ); document.write( "p = percent of people who said they liked dogs = 68% = .68.
\n" ); document.write( "q = percent of people who did not say they liked dogs = 32% = .32.
\n" ); document.write( "s = standard error = sqrt(p * q / n) = .0012275688.
\n" ); document.write( "critical z-score at 99% two tail confidence interval = plus or minus 2.5758.
\n" ); document.write( "margin of error = (x-m).
\n" ); document.write( "when z = 2.5758, z-score formula becomes 2.5758 = (x-m) / .0012275688.
\n" ); document.write( "solve for (x-m) to get (x-m) = .0031619718.
\n" ); document.write( "with the mean of .68, that would lead to a 99% confidence interval of from .67684 to .68316.
\n" ); document.write( "here's what it looks like on a normal distribution calculator.\r
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