Question

In a survey of 3398 adults, 1487 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. 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.

Answer 1

p= 1487/3398=0.4376

q=1-p= 0.5624

n= 3398

alpha=0.01 then Z(alpha/2)= 2.576

Margin of error E= Z(alpha/2)*sqrt(p*q/n)

=2.576*sqrt(0.4376*0.5624/3398)

=0.02192

99% Confidence interval for population proportion =(p-E,p+E)

rounded lower bound= 0.4157

rounded upper bound= 0.4595

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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