Question

6.3.11-T In a survey of 2480 adults in a recent​ year, 1352 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. The​ 90% confidence interval for the population proportion p is left parenthesis nothing comma nothing right parenthesis . ​(Round to three decimal places as​ needed.) The​ 95% confidence interval for the population proportion p is left parenthesis nothing comma nothing right parenthesis . ​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 2480

x = 1352

Point estimate = sample proportion = = x / n = 1352/2480=0.545

1 - = 1 - 0.545 = 0.455

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 (((0.545 * 0.455) / 2480)

= 0.016

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.545 - 0.016 < p < 0.545 + 0.016

(0.529 < p < 0.561)

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 (((0.545 * 0.455) / 2480)

= 0.020

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.545 - 0.020 < p < 0.545 + 0.020

(0.525 < p < 0.565)

The confidence level increases, the width is increases