Answer:
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
[tex]\hat p=\frac{535}{1016}=0.527[/tex] estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement
[tex]\alpha=0.01[/tex] 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
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
And replacing into the confidence interval formula we got:
[tex]0.527 - 2.58 \sqrt{\frac{0.527(1-0.527)}{1016}}=0.487[/tex]
[tex]0.527 + 2.58 \sqrt{\frac{0.527(1-0.527)}{1016}}=0.567[/tex]
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