document.write( "Question 1045268: In a survey of 3107 ​adults, 1499 say they have started paying bills online in the last year.\r
\n" ); document.write( "\n" ); document.write( "Construct a 99% confidence interval for the population proportion. ​(Round to three decimal places as​ needed.)\r
\n" ); document.write( "\n" ); document.write( "Interpret your results. Choose the correct answer below.
\n" ); 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.
\n" ); 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.
\n" ); document.write( "C.The endpoints of the given confidence interval show that adults pay bills online​ 99% of the time.
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Algebra.Com's Answer #660772 by Boreal(15235)\"\" \"About 
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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.
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