document.write( "Question 371232: In a random sample of 500 people of a ctty, it was
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document.write( "found that 160 preferred seafood. Find a 95%
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document.write( "confidence interval for the actual proportion of
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document.write( "people who preferred seafood. \n" );
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Algebra.Com's Answer #264516 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! In a random sample of 500 people of a city, it was \n" ); document.write( "found that 160 preferred seafood. \n" ); document.write( "Find a 95% confidence interval for the actual proportion of \n" ); document.write( "people who preferred seafood. \n" ); document.write( "----- \n" ); document.write( "p-hat = 160/500 = 0.32 \n" ); document.write( "--- \n" ); document.write( "ME = 1.96*sqrt(0.32*0.68/500) = 0.0410 \n" ); document.write( "---- \n" ); document.write( "95% CI: 0.32-0.0410 < p < 0.32+0.0410 \n" ); document.write( "----- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |