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
\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 shhoppers who shop on impulse?
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Algebra.Com's Answer #324264 by edjones(8007)\"\" \"About 
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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)
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