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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 _.

Answers

Answer 1

Answer:

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

$\hat p=(535)/(1016)=0.527$ estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement

$\alpha=0.01$ 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

$\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_(\alpha/2)=2.58$

And replacing into the confidence interval formula we got:

$0.527 - 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.487$

$0.527 + 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.567$

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

Answer 2

Answer:

Final answer:

To build a 99% confidence interval, we first calculate our sample proportion by dividing the number of such instances by the total sample size. Next, we determine the standard error of the proportion, then our margin of error by multiplying the standard error by the Z value of the selected confidence level. Lastly, we determine the confidence interval by adding and subtracting the margin of error from the sample proportion.

Explanation:

To construct a 99% confidence interval for the proportion of adults worried about having enough money to live comfortably in retirement, we will utilize statistical methods and proportions. First, we must calculate the sample proportion. The sample proportion (p) is equal to 535 (the number who are worried) divided by 1016 (the total number of adults surveyed).

Then, we find the standard error of the proportion which we get by multiplying the square root of ((p*(1-p))/n) where n is the number of adults sampled. The margin of error is found using the Z value corresponding to the desired confidence level, in this case, 99%. Multiply the standard error by this Z value. Lastly, we construct the confidence interval by taking the sample proportion (p) ± the margin of error.

The result will give you the 99% confidence interval - meaning we are 99% confident that the true proportion of adults who are worried about having enough money to live comfortably in retirement lies within this interval.

Learn more about Confidence Interval here:

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Answers

Answer:

The simplified expression for the area of rectangle ABCD is $( 15(x + 1))/(2(x - 2))$ , and the restriction on x is x≠2 .

Step-by-step explanation:

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Solving,

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So, the answer is:

The simplified expression for the area of rectangle ABCD is $( 15(x + 1))/(2(x - 2))$ , and the restriction on x is x≠2 .

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Answers

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Step-by-step explanation:

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Answers

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Answers

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