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.

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Answers

I thinks that’s the answer

1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260. Use this sample information to construct a 95% confidence interval for the average daily revenue. The store manager believes that the coffee and pastry strategy would lead to an average daily revenue of $1,200. Is the manager correct based on the 95% confidence interval?

Answers

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                (Xbar - \mu)/((s)/(√(n) ) ) ~ t_n_-_1

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, \mu is given by;

P(-2.032 < t_3_4 < 2.032) = 0.95

P(-2.032 < (Xbar - \mu)/((s)/(√(n) ) ) < 2.032) = 0.95

P(-2.032 * {(s)/(√(n) ) < {Xbar - \mu} < 2.032 * {(s)/(√(n) ) ) = 0.95

P(X bar - 2.032 * {(s)/(√(n) ) < \mu < X bar + 2.032 * {(s)/(√(n) ) ) = 0.95

95% confidence interval for \mu = [ X bar - 2.032 * {(s)/(√(n) ) , X bar + 2.032 * {(s)/(√(n) ) ]

                                            = [ 1080 - 2.032 * {(260)/(√(35) ) , 1080 + 2.032 * {(260)/(√(35) ) ]

                                             = [ 990.70 , 1169.30 ]

No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.

Therefore, the store manager believe is not correct.

Final answer:

The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.

Explanation:

In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:

  • Calculate the standard error by dividing the standard deviation by the square root of the sample size. Here, the standard deviation is $260, and the sample size is 5 weeks * 7 days/week = 35 days. So, the standard error is $260 / sqrt(35) = $43.89.
  • Multiply the standard error by the z-score to get the margin of error. So, $43.89 * 1.96 = $86.03.
  • Calculate the lower and upper bounds of the 95% confidence interval by subtracting and adding the margin of error from/to the sample mean. So, ($1080 - $86.03, $1080 + $86.03) = ($993.97, $1166.03).

The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.

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What is the probability of a die coming up with the number 10?

Answers

Answer:

0

Step-by-step explanation:

0 is the probability of a die coming up with the number 10

because a die contains numbers from 1 to 6 so probability of numbers coming between are possible but 10 is not possible

What is 0.035 as a simplified reduced fraction

Answers

Answer:

7/200

Step-by-step explanation:

0.035= 35/1000= 7*5/200*5=7/200

7/200 is the correct answer

Evaluate the given integral by changing to polar coordinates. sin(x2 y2) dA R , where R is the region in the first quadrant between the circles with center the origin and radii 2 and 4

Answers

Answer:

I = 1.47001

Step-by-step explanation:

we have the function

f(x,y)=sin(x^2y^2)\n

In polar coordinates we have

x=rcos\theta\ny=rsin\theta

and dA is given by

dA=rdrd\theta

Hence, the integral that we have to solve is

I=\int \limt_2^4 \int \limit_0^(\pi /2)sin(r^4cos^2\theta sin^2\theta)rdrd\theta

This integral can be solved in a convenient program of your choice (it is very difficult to solve in an analytical way, I use Wolfram Alpha on line)

I = 1.47001

Hope this helps!!!

Allison buys a spool of thread for sewing. There are 10 yards of thread on the spool. She uses 9 meters. How much thread is left on thespool in meters? Round your answer to the nearest thousandth, if necessary.

Answers

The amount of meters left is  0.144 meters

First and foremost, it should be noted that: 1 yard = 0.9144 meter

Therefore, 10 yards to meters will be:

= 0.9144 × 10

= 9.144 meters

Since Allison uses 9 meters, the amount of thread that is left on the

spool will be:

= 9.144 meters - 9 meters

= 0.144 meters

Read related link on:

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there are 9.144 meters in 10 yards, therefore, there would be .144 meters left of thread.