In a survey of 3151 adults, 14811481 say they have started paying bills online i

Question

In a survey of 3151 adults, 14811481 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 _____

Interpret your results. Choose the correct answer below.

A. 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.

B. The endpoints of the given confidence interval show that adults pay bills online 99% of the time.

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.

Thanks!

Explanation / Answer

number of people who have paid>

total population=3151

Let p= % paid=1481/3151=47%

Z for 99%=2.575

standard error=sqrt(p(1-p)/n) =sqrt(0.47*0.53/3151) =8.8912x10^-3

Hence 99% conf interval is p+-Z*se

Hence CI=[0.47-2.575*se,0.47+2.575*se]=[0.447,0.493]

Hence final answer is

A)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.

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