In a survey of 1016 ? adults, a polling agency? asked, "when you? retire, do you think you will have enough money to live comfortably or not. of the 1016 ? surveyed, 535 stated that they were worried about having enough money to live comfortably in retirement. construct a 99? % confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.

a. there is a 99? % probability that the true proportion of worried adults is between and

b. 99? % of the population lies in the interval between and

c. there is 99? % confidence that the proportion of worried adults is between and

C. There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Step-by-step explanation:

1) Data given and notation  

n=1016 represent the random sample taken    

X=535 represent the people stated that they were worried about having enough money to live comfortably in retirement

estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement

represent the significance level

Confidence =0.99 or 99%

z would represent the statistic

p= population proportion of people stated that they were worried about having enough money to live comfortably in retirement

2) Confidence interval

The confidence interval would be given by this formula

For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

And replacing into the confidence interval formula we got:

And the 99% confidence interval would be given (0.487;0.567).

There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

step-by-step explanation:

if x = 2 you have to find x to finish the problem

555.99.

step-by-step explanation:

hope this