Question

In a survey of 3125 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.

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

solution:-

given that n = 3125

x = 1415

=> p^ = x/n = 1415/3125 = 0.453

then q = 1 - p = 1 - 0.453 = 0.547

99% confidence for z is 2.58

confidence interval formula

=> p^ +/- z * sqrt(p*q/n)

=> 0.453 +/- 2.58 * sqrt( 0.453*0.547/3125)

**=> (0.430 , 0.476)**

Interpret your results.

**option 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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