Question 982026
In a survey of 3238adults, 1442 say they have started paying bills online in the last year.
Construct a​ 99% confidence interval for the population proportion. Interpret the results.
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p-hat = 1442/3238 = 0.445
ME = 2.5758*sqrt[(1/2)(1/2)/3238) = 0.023
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99%CI:: (0.445-0.023,0.445+0.023) = (0.422,0.468)
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A​ 99% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis(,).
​(Round to three decimal places as​ needed.)
A) The endpoints of the given confidence interval show that adults pay bills online​ 99% of the time. NO
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B) 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. NO
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C) 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. YES
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Cheers,
Stan H.
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