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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.
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.
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.
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.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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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