SOLUTION: survey of 3161 adults, 1410 say they have started paying bills online in the last year.
Construct a 99% confidence interval for the population proportion. Interpret the res
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Question 1136378: survey of 3161 adults, 1410 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 ___
(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 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 Boreal(15235) (Show Source): You can put this solution on YOUR website!
A 99% CI interval is mean +/- z(0.995)sqrt (p*(1-p)/n); z=2.576
p=1410/3161=0.4461
the half-interval is 2.576*sqrt (.4461*0.5539/3161)=0.0228
the interval is (0.4233, 0.4689)
We don't know the true proportion but we are extremely confident it lies in this interval.This is C. CIs are about populations, not samples when they are interpreted. That is why they are constructed.
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