document.write( "Question 1045268: In a survey of 3107 adults, 1499 say they have started paying bills online in the last year.\r
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document.write( "Construct a 99% confidence interval for the population proportion. (Round to three decimal places as needed.)\r
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document.write( "Interpret your results. Choose the correct answer below.
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document.write( "A.With 99% confidence, it can be said that the sample proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.
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document.write( "B.With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.
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document.write( "C.The endpoints of the given confidence interval show that adults pay bills online 99% of the time. \n" );
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Algebra.Com's Answer #660772 by Boreal(15235) You can put this solution on YOUR website! CI=phat +/- 2.576 sqrt{(p)(1-p)/n} \n" ); document.write( "1499/3107=0.482 \n" ); document.write( "0.482+/-2.576*sqrt(0.482)(0.518)/3107; the SE is 0.00896, the interval itself (SE*2.576) is 0.023 \n" ); document.write( "(0.459,0.505) \n" ); document.write( "The answer is B. We are completely confident in what the sample says; we use confidence to describe what we think of the population. \n" ); document.write( " |