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 #324264 by edjones(8007)![]() ![]() You can put this solution on YOUR website! n=750, p=260/750=.347, q=.653 \n" ); document.write( "(a,b)=p+z[alpha/2]*sqrt((p*q)/n) \n" ); document.write( "=.347+-(1.96)*sqrt((.347*.653)/750) Z for the 95% confidence level is 1.96 \n" ); document.write( "=.347+-(1.96)(.0174) \n" ); document.write( "=.347+-.034 \n" ); document.write( "=(.381, .313) \n" ); document.write( ". \n" ); document.write( "Ed \n" ); document.write( " \n" ); document.write( " |