Question

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 online in the last year is between the endpoints of the given confidence interval.

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

3. 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 = 3316

x = 1433

Point estimate = sample proportion = = x / n = 1433 / 3316 = 0.432

1 - = 1 - 0.432 = 0.568

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 (((0.432 * 0.568) / 3316)

= 0.022

A 99% confidence interval for population proportion p is ,

± E  

= 0.432  ± 0.022

= ( 0.410, 0.454 )

b) 1. 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