document.write( "Question 1076088: A sample of 469 teenage births at a large city hospital is observed for the number of births resulting in twins. the 95% confidence interval for the proportion for all births among teenage girls in twins was estimated to be (0.006, 0.026). \r
\n" ); document.write( "\n" ); document.write( "What is the sample proportion of teenage births resulting in twins?
\n" ); document.write( "What is the margin of error for this interval?
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Algebra.Com's Answer #690836 by Boreal(15235)\"\" \"About 
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The 95% interval came from p hat +/- sqrt (p*(1-p))/469
\n" ); document.write( "Half the interval width is 0.01
\n" ); document.write( "The sample proportion will be in the middle or 0.016 or 1.6%
\n" ); document.write( "The margin of error is the interval width or +/-0.01.
\n" ); document.write( "Check by using 0.016+/- sqrt (p*(1-p)/469)
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