Question

# In a survey of 2808 ​adults, 1413 say they have started paying bills online in the...

In a survey of 2808 ​adults, 1413 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. 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. 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. The endpoints of the given confidence interval show that adults pay bills online​ 99% of the time

sample proportion, pcap = 1413/ 2808 = 0.5032
sample size, n = 2808
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.5032 * (1 - 0.5032)/2808) = 0.0094

For 99% Confidence level, the z-value = 2.58

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.5032 - 2.58 * 0.0094 , 0.5032 + 2.58 * 0.0094)

CI = (0.479 , 0.527)

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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