document.write( "Question 460767: A doctor says that less than 25% of US adults chew tobacco. In a random sample of 170 US adults, 18.5% say they chew tobacco. At alpha = .05, is there enough evidence to reject the doctor's claim? \r
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document.write( "Since I don't have a standard deviation, do I use t-distribution? Or do I use binomial distribution and find it by square root of npq? \n" );
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Algebra.Com's Answer #316053 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A doctor says that less than 25% of US adults chew tobacco. \n" ); document.write( "In a random sample of 170 US adults, 18.5% say they chew tobacco. \n" ); document.write( "At alpha = .05, is there enough evidence to reject the doctor's claim? \n" ); document.write( "------------------------------- \n" ); document.write( "Since I don't have a standard deviation, do I use t-distribution? Or do I use --binomial distribution and find it by square root of npq? \n" ); document.write( "---- \n" ); document.write( "It is a proportion test so you use the z-distribution. \n" ); document.write( "The standard deviation is sqrt[pq/n]. You have all \n" ); document.write( "the information you need. \n" ); document.write( "============================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |