document.write( "Question 472922: 1) A recent study of 750 internet users in Europe found that 260 of internet users were women. What is the 95% confidence interval of the true proportion of women in Europe who use the internet?\r
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document.write( "2) A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98& confidence interval for the true proportion of women shhoppers who shop on impulse? \n" );
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Algebra.Com's Answer #324260 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! 1) A recent study of 750 internet users in Europe found that 260 of internet users were women. What is the 95% confidence interval of the true proportion of women in Europe who use the internet? \n" ); document.write( "---------------------- \n" ); document.write( "sample proportion: 260/750 = 0.3467 \n" ); document.write( "ME = 1.96*sqrt[0.3467*0.6533/750] = 0.0744 \n" ); document.write( "--- \n" ); document.write( "95% CI: 0.35-0.07 < p < 0.35+0.07 \n" ); document.write( "============================================= \n" ); document.write( " \n" ); document.write( "2) A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98& confidence interval for the true proportion of women shoppers who shop on impulse? \n" ); document.write( "--- \n" ); document.write( "Same procedure as above. \n" ); document.write( "================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "================= \n" ); document.write( " \n" ); document.write( " |