document.write( "Question 1058521: Assume all samples are simple random samples.
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document.write( "In a poll, 1025 randomly selected adults were surveyed and 29% said they use internet for shopping at least a few times a year.
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document.write( "Find a point estimate of the percentage of adults who use the internet for shopping.
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document.write( "Find a 99% confidence interval estimate of the percentage of adults who use internet for shopping
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Algebra.Com's Answer #673593 by Boreal(15235) You can put this solution on YOUR website! The CI interval width is z(0.995) sqrt{ (p*(1-p))/n} \n" ); document.write( "=2.576*sqrt{(.29*.71)/1025} \n" ); document.write( "=2.576*0.0142 \n" ); document.write( "=0.0365 (rounded at end) \n" ); document.write( "The interval is (0.2535,0.3265) \n" ); document.write( " |