Question

In a survey of 1018 ​adults, a polling agency​ asked, "When you​ retire, do you think you will have enough money to live comfortably or not. Of the 1018 surveyed,

537 stated that they were worried about having enough money to live comfortably in retirement. Construct a 95% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.

Select the correct choice below and fill in the answer boxes to complete your choice.

​(Use ascending order. Round to three decimal places as​ needed.)

A.There is a 95​% probability that the true proportion of worried adults is between __ and __

B.There is 95​% confidence that the true proportion of worried adults is between __ and __

C. 95​% of the population lies in the interval between __ and __

Answer 1

Solution :

Given that,

n = 1018

x = 537

Point estimate = sample proportion = = x / n = 537/1018 = 0.528

1 - = 1 -0.527 = 0.472

Z_/2 = 1.960

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

= 1.960 * ((0.528*(0.472) /1018 )

= 0.031

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.528 - 0.031 < p < 0.528+0.031

0.497< p < 0.559

C. There is a 95% confidence that the true proportion of worried adults is between  0.497 and 0.559