#### Question Details

[answered] Last name: Enter your LETTER answers HERE 1 2 3 4 5 6 7 8 9

Complete all 30 questions correctly and fill in the missing highlighted parts of the table! Thank you?

Last name:

Enter your LETTER answers HERE

?

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

1 First name: e270Lastname Firstname TEST8 a

b

c

d You run a gizmo factory. Over the past year the average output of the factory is 480 gizmos and the

standard deviation of that output is 10 units per day. In repeated random samples of size n = 40 days, the

expected value of sample mean is:

12

480

76

48 a

b

c

d In the previous question the variance of the sample mean is.

10

2.5

15.8

1.58 a

b

c Using the population mean and standard deviation in question 1, what is the probability that the mean of a

random sample of n = 40 days exceeds 482 gizmos:

0.1020

0.0571

0.0384 2 3 d

4 0.0287 a

b

c

d Using the population mean and standard deviation in question 1, in repeated samples of size n = 40, 95%

of sample means fall within that fall within ?_____ gizmos from the population means.

3.9

3.5

3.1

2.7 a

b

c

d In the population of employees in your company, 35 percent contribute to the annual United Way

campaign. You plan to take a sample of size n = 200 and calculate the sample proportion who contribute to

United Way.

The expected value of the sample proportion is:

0.035

0.25

0.35

0.40 a

b

c

d In the previous question, the standard error of the sample proportion is:

0.0473

0.0415

0.0398

0.0337 a

b

c

d In the previous question, 90 percent of sample proportions from samples of size n = 200 deviate from the

population proportion of 0.35 by no more that ?____ (or ____ percentage points).

0.036 (3.6 percentage points.)

0.042 (4.2 percentage points.)

0.048 (4.8 percentage points.)

0.055 (5.5 percentage points.) a

b

c

d There is a population of 10 families in a small neighborhood. You plan to take a random sample of 4

families (without replacement). How many samples of size n = 4 are possible.

40

80

210

400 5 6 7 8 9 The expression

a

b

c

d 10 a

b

c

d means: In repeated sampling the probability that population mean is within ?1.96?/?n from xx is 0.95.

In repeated sampling the probability that xx is within ?1.96?/?n from the population mean is 0.95.

95% of sample means deviate from the population mean by no more than 1.96?/?n in either direction.

Both b and c are correct.

The population proportion of Americans with diabetes is 9 percent (? = 0.09). In repeated random

samples of n = 800 Americans, 90% of sample proportions of people with diabetes would fall within ?______

from ?.

0.01 (1.0 percentage point)

0.017 (1.7 percentage points)

0.029 (2.9 percentage points)

0.036 (3.6 percentage points) 11 a

b

c

d

12

a

b

c

d

13 a

b

c

d

14 a

b

c

d

15

a

b

c

d

16 As part of a statistics assignment in October to estimate the percentage of voters who would vote for a

mayoral candidate, each of 500 students collects his or her own random sample of likely voters. There are

400 voters in each student?s random sample.

Each student then constructs a 95 percent confidence interval for the population proportion who will vote

for the candidate using his or her own random sample. Considering the 500 intervals constructed by the

students, the expected number of intervals that will contain the population proportion who will vote for

that candidate will be approximately:

380

400

475

500

Suppose the sample proportion of one of the students in the previous question, Beth's sample, is px = 0.46.

Beth's 95% interval estimate of proportion of the population of voters voting for the candidate is:

[0.41 , 0.51]

[0.40 , 0.52]

[0.39 , 0.53]

[0.38 , 0.54]

It is estimated that 80% of Americans go out to eat at least once per week, with a margin of error of 0.04

(for 95% confidence). A 95% confidence interval for the population proportion of Americans who go out

to eat once per week or more is:

[0.722, 0.878]

[0.760, 0.840]

[0.771, 0.829]

[0.798, 0.802]

You are the manager of a political campaign. You think that the population proportion of voters who will

vote for your candidate is 0.50 (use this for a planning value). Your candidate wants to know what

proportion of the population will vote for her. Your candidate wants to know this with a margin-of-error of

? 0.01 (at 95% confidence). How big of a sample of voters should you take?

1499

5037

8888

9604

If your candidate changes her mind and now wants a margin-of-error of ? 0.03 (but still 95% confidence),

you will:

have to select a smaller sample

be able to use the same sample

have to select a larger sample

the margin-of-error does not have anything to do with the sample size

You run a bank and want to estimate the bank?s average number of customers per day (the population is all

the days you are open for business in a year). You take a random sample of 10 days and record the

numbers of customers on those days. The sample data is shown below. What is a 95% confidence interval

for the bank?s average number of customers per day?

450

440 a

b [433, 467]

[430, 470] 470

460 430

420 420

500 490

420 c

d

17 a

b

c

d

18 a

b

c

d

19

a

b

c

d

20

a

b

c

d 21 a

b

c

d 22 [429, 471]

[424, 476]

As the manager of the bank in the previous question, you want to your 95% interval estimate to capture

the population mean customers per day within ?10 customers. Using a planning value of ?? = 35, how

many days should you include in the sample?

40

48

59

68

You work for a charitable organization and you want to estimate the average age of the people who donate

to your organization. You get a random sample of n = 119 donors and the value of the sample mean is 42

years. The value of the sample standard deviation is 19 years.

The lower and upper end of the 95% confidence interval for the average age of people who donate to your

organization are:

[36.4, 47.6]

[37.5, 46.5]

[38.6, 45.4]

[39.8, 44.2]

In the previous question you want to test the hypothesis, at the 5 percent level of significance, that the

average age is at least 45 years. Compute the probability value for the test.

0.0427 Reject the null hypothesis and conclude the mean age is less than 45.

0.0427 Do not reject the null hypothesis. Conclude the mean age is not less than 45.

0.0043 Reject the null hypothesis and conclude the mean age is less than 45.

0.0043 Do not reject the null hypothesis. Conclude the mean age is not less than 45.

Regardless how you answered the previous question, which of the following statements is correct?

If the mean age is in fact greater than 45 and the hypothesis test leads you to conclude that it is less than

45, the you have committed a Type II error.

If the mean age is in fact greater than 45 and the hypothesis test leads you to conclude that it is less than

45, the you have committed a Type I error.

If the mean age is less than 45 and the hypothesis test leads you to conclude that it is greater than 45, the

you have committed a Type I error.

If the mean age is in at least than 45 and the hypothesis test leads you to conclude that it is greater than 45,

the you have committed a Type II error.

You are reading a report that contains a hypothesis test you are interested in. The writer of the report

writes that the p-value for the test you are interested in is 0.091, but does not tell you the value of the test

statistic. From this information you can:

You cannot decide based on this limited information. You need to know the value of the test statistic.

Not reject the hypothesis at a Probability of Type I error = 0.10, and not reject at a Probability of Type I

error = 0.05

Reject the hypothesis at a Probability of Type I error = 0.10, and reject at a Probability of Type I error =

0.05

Reject the hypothesis at a Probability of Type I error = 0.10, but not reject at a Probability of Type I error =

0.05

According to the Census Bureau the sample proportion of American children without health insurance rose

from 0.109 in 2005 to 0.117 in 2006. Using the Census data you test the hypothesis that the change in the

population proportion of American children without health insurance is zero with a Probability of Type I

error = 0.05. You reject the zero change hypothesis. a b c

d

23 a

b

c

d The correct interpretation is:

The odds are 5 percent that if the sample of American children in 2006 had been expanded to include the

entire population of American children, the change in the proportion of children without health insurance

would still not be zero.

Even if the change in the population of American children without health insurance really is zero, 5% of all

repeated samples would have produced test statistics that cause you to reject the hypothesis that the

change is zero.

The test statistic for the null hypothesis that H?: ? = 0.109 is less than the critical value.

95 percent of repeated samples would produce confidence intervals within ? 0.05 points of 0.117.

To test the hypothesis, at a 5% level of significance, that the proportion of American children without

health insurance has increased from 0.109, a random sample of 1020 children revealed a sample

proportion of 0.121. Compute the p-value.

0.1112 Conclude that proportion of children without health insurance has increased.

0.1112 Conclude that proportion of children without health insurance has not increased.

0.0556 Conclude that proportion of children without health insurance has increased.

0.0056 Conclude that proportion of children without health insurance has not increased. Next SEVEN questions are based on the following regression model

To determine the impact of variations in price on sales the management of Big Bob's Burger Barn sets different

prices in its burger joints in 75 stores located in different cities.

Using the sales and price data, a simple regression is run with sales (in thousands of dollars) as the dependent

variable and price (in dollars) as the independent variable.

Use the following calculations and the accompanying regression summary output to answer the next seven

questions.

?xy = 32847.677

xx =

5.6872

?x? = 2445.7074

yx =

77.3747

SUMMARY OUTPUT

Regression Statistics

Multiple R

R Square

Adjusted R Square

Standard Error

Observations 0.38296261

75 ANOVA

df SS MS F

Significance F

46.927903

1.97E-09 Regression

Residual

Total 3115.48187 Intercept

PRICE CoefficientsStandard Error t Stat

P-value

Lower 95% Upper 95%

6.5262907 18.6783242 1.588E-029 108.893295 134.907052

1.97E-09 24

a

b

c

d The model predicts that when raising the price by $1, sales would change by $_______ thousand.

-$7.83

-$8.26

-$9.04

-$9.65 25 The predicted sales for a price of $6.00 per burger is $ _______ thousand. a

b

c

d

26

a

b

c

d

27

a

b

c

d $71.4

$74.9

$78.6

$82.5

Given that ?(y ? yx )? = 1219.091, the sample data show that _______ fraction of variations is sales

is explained by price.

0.68

0.48

0.39

0.25

The regression result shows the observed sales deviate from the predicted sales, on average, by

$ _______ thousand.

$4.2

$5.1

$6.2

$7.3 28

a

b

c

d The standard error of the slope coefficient b? is ________.

2.846

2.341

1.143

0.985 29

a

b

c

d The test statistic for the null hypothesis that a change in price has no impact on sales is:

-6.850

-5.798

-4.746

-3.694 30

a

b

c

d The margin of error for a 95% interval estimate for the population slope parameter is:

4.02

3.34

2.95

2.28

**Solution details:**

Answered

QUALITY

Approved

ANSWER RATING

This question was answered on: * Sep 18, 2020 *

* * Solution~0001013074.zip (25.37 KB)

This attachment is locked

We have a ready expert answer for this paper which you can use for in-depth understanding, research editing or paraphrasing. You can buy it or order for a fresh, original and plagiarism-free copy from our tutoring website www.aceyourhomework.com (Deadline assured. Flexible pricing. TurnItIn Report provided)

##### Pay using PayPal (No PayPal account Required) or your credit card . All your purchases are securely protected by .

#### About this Question

STATUSAnswered

QUALITYApproved

DATE ANSWEREDSep 18, 2020

EXPERTTutor

ANSWER RATING

#### GET INSTANT HELP/h4>

We have top-notch tutors who can do your essay/homework for you at a reasonable cost and then you can simply use that essay as a template to build your own arguments.

You can also use these solutions:

- As a reference for in-depth understanding of the subject.
- As a source of ideas / reasoning for your own research (if properly referenced)
- For editing and paraphrasing (check your institution's definition of plagiarism and recommended paraphrase).

#### NEW ASSIGNMENT HELP?

### Order New Solution. Quick Turnaround

Click on the button below in order to Order for a New, Original and High-Quality Essay Solutions.
New orders are original solutions *and precise to your writing instruction requirements. Place a New Order using the button below.*

WE GUARANTEE, THAT YOUR PAPER WILL BE WRITTEN FROM SCRATCH AND WITHIN YOUR SET DEADLINE.