Question

In a survey of 2088 adults in a recent​ year, 1440 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.

Compare the widths of the confidence intervals. Choose the correct answer below.

Answer 1

Solution :

Given that,

n = 2088

x = 1440

Point estimate = sample proportion = = x / n = 1440 / 2088 = 0.690

1 - = 1 - 0.690 = 0.310

At 90% confidence level

= 1 - 90%

=1 - 0.90 = 0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 (((0.690 * 0.310)) / 2088)

= 0.017

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.690 - 0.017 < p < 0.690 + 0.017

0.673 < p < 0.707

( 0.673 , 0.707 )

The 90% confidence interval for the population proportion p is : ( 0.673 , 0.707 )

b.

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

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

= 1.96 ((((0.690 * 0.310)) / 2088)

= 0.020

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.690 - 0.020 < p < 0.690 + 0.020

0.670 < p < 0.710

( 0.670 , 0.710 )

The 95% confidence interval for the population proportion p is : ( 0.670 , 0.710 )

Answer : The widths of the confidence intervals is 95%