Question

In a survey of 3447 ​adults, 1453 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Interpret the results. A​ 99% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis . ​(Round to three decimal places as​ needed.) Interpret your results. Choose the correct answer below. A. The endpoints of the given confidence interval show that adults pay bills online​ 99% of the time. 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. C. 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.

Answer 1

Sample proportion = 1453 / 3447 = 0.422

99% confidence interval for p is

- Z * sqrt ( ( 1 - ) / n) < p < + Z * sqrt ( ( 1 - ) / n)

0.422 - 2.576 * sqrt( 0.422 * 0.578 / 3447) < p < 0.422 + 2.576 * sqrt( 0.422 * 0.578 / 3447)

0.400 < p < 0.444

99% CI is ( 0.400 , 0.444 )

Interpretation -

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.