Question

In a survey of 3287 ​adults, 1459 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. 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 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. 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.

Answer 1

SolutionA:

n=3287

x=1459

p^=sample proportion=x/n=1459/3287=0.4438698

Z crit for 99%=2.576

99% confidence interval for true proportion ,p is

p^-z sqrt(p^(1-p^)/n,p^+z sqrt(p^(1-p^)/n

0.4438698-2.576*sqrt(0.4438698((1-0.4438698)/3287),0.4438698+2.576*sqrt(0.4438698((1-0.4438698)/3287)

0.422,0.466

lower limit=0.422

upper limit=0.466

0.422<p<0.466

We are 99% confident that the true population proportion of f adults who say they have started paying bills online in the last year lies in between 0.422 and 0.486

Solutionb:

confidence interavl is given for populaton proprtion

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.