Question 1043430: In a survey of 3095 adults, 1415 say they have started paying bills online in the last year.
Construct a 99% confidence interval for the population proportion. Interpret the results. Round to three decimal places as needed
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 99% CI is z*sqrt{ p*(1-p)/n}
z=2.576
p=1415/3095=0.4572
1-p=0.5428
sqrt of {0.4572*0.5428/3095}=0.00895
multiply that by 2.576 to get the interval width on each side of the point estimate, which is 0.4572.
0.023067
(0.434,0.480)
We don't know the true proportion who stated they have started paying bills online in the last year, but we are extremely confident, 99% so, that the true number lies in the above interval.
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