Question

In: Statistics and Probability

In a survey of 2580 ​adults, 1416 say they have started paying bills online in the...

In a survey of 2580 ​adults, 1416 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.

Solutions

Expert Solution

n = 2580

p = 1416 / 2580 = 0.549

Z for 99% confidence interval = Z0.005 = 2.575

Confidence interval

                               

                               

                                = (0.524 , 0.574)

Option-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

orchestra answered 2 years ago
Next > < Previous

Related Solutions

In a survey of 3203 ​adults, 1462 say they have started paying bills online in the...
In a survey of 3203 ​adults, 1462 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.)
In a survey of 30833083 ​adults, 14201420 say they have started paying bills online in the...
In a survey of 30833083 ​adults, 14201420 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.)
In a survey of 2536 ​adults, 1477 say they have started paying bills online in the...
In a survey of 2536 ​adults, 1477 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.)
In a survey of 2901 adults, 1495 say they have started paying bills online in the...
In a survey of 2901 adults, 1495 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
In a survey of 3316 ​adults, 1433 say they have started paying bills online in the...
In a survey of 3316 ​adults, 1433 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Interpret the results. (a) A​ 99% confidence interval for the population proportion is ( ),( ). ​(Round to three decimal places as​ needed.) (b) Interpret your results. Choose the correct answer below: 1. With​ 99% confidence, it can be said that the population proportion of adults who say they have started paying bills...
In a survey of 2844 adults , 1435 says they have started paying bills online in...
In a survey of 2844 adults , 1435 says they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
In a survey of 2075 adults in a recent​ year, 1432 say they have made a...
In a survey of 2075 adults in a recent​ year, 1432 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. A. The​ 90% confidence interval for the population proportion p is.. B.The​ 95% confidence interval for the population proportion p is.. C.Determine the widths of the intervals..
In a survey of 2231 adults in a recent​ year, 1438 say they have made a...
In a survey of 2231 adults in a recent​ year, 1438 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The​ 90% confidence interval for the population proportion p is left parenthesis nothing comma nothing right parenthesis,. ​(Round to three decimal places as​ needed.)
In a survey of 2433 adults in a recent year, 1369 say they have made a...
In a survey of 2433 adults in a recent year, 1369 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the population proportion. interpret the results and compare the withs of the confidence intervals. 90% (__,__) 95%(__,__) (Round to three decimal places as needed)
In a survey of 2097 adults in a recent​ year, 1419 say they have made a...
In a survey of 2097 adults in a recent​ year, 1419 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. Please breakdown the how you solve the answer. Thank you
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT