document.write( "Question 75554: SMOKING AND COLLEGE EDUCATION \r
\n" ); document.write( "\n" ); document.write( "Q1. The tobacco industry closely monitors all surveys that involve smoking. One survey showed that among 785 randomly selected sunjects who completed four years of college, 18.3% smoke(based on data from American Medical Association).\r
\n" ); document.write( "\n" ); document.write( "a. Construct the 98% confidence interval for the true percentage of smokers among all people who completed four years of college.
\n" ); document.write( "b. Based on the result from part (a),does the smoking rate for those with four year of college appear to be substantially different than the 27% rate for the general population?
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Algebra.Com's Answer #54244 by stanbon(75887)\"\" \"About 
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One survey showed that among 785 randomly selected sunjects who completed four years of college, 18.3% smoke(based on data from American Medical Association).
\n" ); document.write( "a. Construct the 98% confidence interval for the true percentage of smokers among all people who completed four years of college.
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\n" ); document.write( "E= z*sqrt(pq/n)
\n" ); document.write( "You don't have \"p\" so use p-hat=0.183
\n" ); document.write( "E= 2.326sqrt(0.183*0.817/785)
\n" ); document.write( "E= 0.032
\n" ); document.write( "Then CI is: (0.183-0.032,0.183+0.032) or (0.151,0.215)
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\n" ); document.write( "\n" ); document.write( "b. Based on the result from part (a),does the smoking rate for those with four year of college appear to be substantially different than the 27% rate for the general population?
\n" ); document.write( "Yes, 27% is significantly outside the range of the 98% confidence interval
\n" ); document.write( "for the smoking percent of four year college students.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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