Question

In a survey of 2075 adults in a recent​ year, 1432 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.

A. The​ 90% confidence interval for the population proportion p is..

B.The​ 95% confidence interval for the population proportion p is..

C.Determine the widths of the intervals..

Answer 1

Solution :

Given that,

n = 2075

x = 1432

Point estimate = sample proportion = = x / n = 1432 / 2075 = 0.690

1 - = 1 - 0.690 = 0.310

A)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 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) / 2075)

= 0.017

Width = 2 * 0.017 = 0.034

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

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

B)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.690 * 0.310) / 2075)

= 0.020

Width = 2 * 0.020 = 0.040

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

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