Question

In a survey of 2363 adults in a recent​ year, 1366 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 (__,__)

Answer 1

Solution :

Given that n = 2363 , x = 1366

=> Proportion p = x/n

= 1366/2363

= 0.5781

=> q = 1 - p = 0.4219

A.
=> for 90% confidence interval, Z = 1.645

=> The​ 90% confidence interval for the population proportion p is

=> p +/- Z*sqrt(p*q/n)

=> 0.5781 +/- 1.645*sqrt(0.5781*0.4219/2363)

=> (0.5614,0.5948)

B.
=> for 95% confidence interval, Z = 1.96

=> The​ 95% confidence interval for the population proportion p is

=> p +/- Z*sqrt(p*q/n)

=> 0.5781 +/- 1.96*sqrt(0.5781*0.4219/2363)

=> (0.5582,0.5980)

=> With the given confidence,it can be said that the population proportion of adults who say they have made a new years resolution is between the endpoints of the given confidence interval.

=> The 95% confidence interval is wider than 90% confidence interval