Question

In a survey of 2051 adults in a recent​ year, 724 made a New​ Year's resolution to eat healthier. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

Answer 1

Given, n = 2051, x = 724

P^ = x/n = 724/2051 = 0.3529

90% CI

alpha = 0.1

Zc = 1.645

CI = P^ +/- Zc*SQRT(P^(1-P^)/n)

CI = 0.3529 +/- 1.645*SQRT(0.3529*(1-0.3529)/2051)

CI = (0.3355, 0.3703)

We are 90% confident that the population mean lies between the above interval

width = 0.3703-0.3355 = 0.0348

95% CI

alpha = 0.05

Zc = 1.96

CI = P^ +/- Zc*SQRT(P^(1-P^)/n)

CI = 0.3529 +/- 1.96*SQRT(0.3529*(1-0.3529)/2051)

CI = (0.3322, 0.3736)

We are 95% confident that the population mean lies between the above interval

width = 0.3736-0.3322 = 0.0414

95% CI has large width than 90% CI (not much)