A sample of 36 observations is selected

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A sample of 36 observations is selected from a normal population. The sample mean is 12, and the population standard deviation is 3. Conduct the following test of hypothesis using the 0.01 significance level.

H0: μ ≤ 10

H1: μ > 10

1.

Value: 10.00 points

Required information

a.

Is this a one- or two-tailed test?

One-tailed test

Two-tailed test

References

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Multiple Choice Difficulty: 2 Intermediate Learning Objective: 10-05 Conduct a test of a hypothesis about a population mean.

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2.

Value: 10.00 points

Required information

b.

What is the decision rule?

Reject H0 when z ≤ 2.326

Reject H0 when z > 2.326

References

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Multiple Choice Difficulty: 2 Intermediate Learning Objective: 10-05 Conduct a test of a hypothesis about a population mean.

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3.

Value: 10.00 points

Required information

c.

What is the value of the test statistic?

Value of the test statistic

References

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Worksheet Difficulty: 2 Intermediate Learning Objective: 10-05 Conduct a test of a hypothesis about a population mean.

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4.

Value: 10.00 points

Required information

d.

What is your decision regarding H0?

Fail to reject H0

Reject H0

References

EBook & Resources

Multiple Choice Difficulty: 2 Intermediate Learning Objective: 10-05 Conduct a test of a hypothesis about a population mean.

eBook: Conduct a test of a hypothesis about a population mean.

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5.

Value: 10.00 points

Required information

e.

What is the p-value?

p-value

References

Given the following hypotheses:

H0 : μ = 400

H1 : μ ≠ 400

A random sample of 12 observations is selected from a normal population. The sample mean was 407 and the sample standard deviation 6. Using the .01 significance level:

a.

State the decision rule. (Negative amount should be indicated by a minus sign. Round your answers to 3 decimal places.)

Reject H0 when the test statistic is the interval (,).

b.

Compute the value of the test statistic. (Round your answer to 3 decimal places.)

Value of the test statistic

c.

What is your decision regarding the null hypothesis?

Do not reject

Reject

The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?

a.

What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Reject H0 : μ ≥ 42.3 when the test statistic is .

b.

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Value of the test statistic

c.

What is your decision regarding H0?

Fail to reject

Reject

Given the following hypotheses:

H0 : μ = 100

H1 : μ ≠ 100

A random sample of six resulted in the following values: 118, 105, 112, 119, 105, and 111. Assume a normal population.

a.

Using the .05 significance level, determine the decision rule? (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

Reject H0 : μ = 100 when the test statistic is (, ).

b.

Compute the value of the test statistic. (Round your answer to 2 decimal places.)

Value of the test statistic

C-1.

What is your decision regarding the H0?

Reject

Do not reject

C-2.

Can we conclude the mean is different from 100?

No

Yes

d.

Estimate the p-value.

The p-value is

A sample of 65 observations is selected from one population with a population standard deviation of 0.75. The sample mean is 2.67. A sample of 50 observations is selected from a second population with a population standard deviation of 0.66. The sample mean is 2.59. Conduct the following test of hypothesis using the .08 significance level.

H0 : μ1 ≤ μ2

H1 : μ1 > μ2

a.

This a -tailed test.

b.

State the decision rule. (Negative values should be indicated by a minus sign. Round your answer to 2 decimal places.)

The decision rule is to reject H0 if z is .

c.

Compute the value of the test statistic. (Round your answer to 2 decimal places.)

Value of the test statistic

d.

What is your decision regarding H0?

H0.

e.

What is the p-value? (Round your answer to 4 decimal places.)

p-value

The null and alternate hypotheses are:

formula77.mml

formula78.mml

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15.

At the .10 significance level, is there a difference in the population means?

a.

This is a -tailed test.

b.

The decision rule is to reject formula79.mmlif formula80.mmlor formula81.mml. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

c.

The test statistic is t =. (Round your answer to 3 decimal places.)

d.

What is your decision regardingformula79.mml?

e.

The p-value is.

The null and alternate hypotheses are:

H0: μ1 ≤ μ2

H1: μ1 > μ2

A random sample of 20 items from the first population showed a mean of 100 and a standard deviation of 15. A sample of 16 items for the second population showed a mean of 94 and a standard deviation of 8. Use the .05 significant level.

a.

Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.)

Degrees of freedom

b.

State the decision rule for .05 significance level. (Round your answer to 3 decimal places.)

Reject H0 if t>.

c.

Compute the value of the test statistic. (Round your answer to 3 decimal places.)

Value of the test statistic

d.

What is your decision regarding the null hypothesis? Use the .05 significance level.

The null hypothesis is.

The following are six observations collected from treatment 1, four observations collected from treatment 2, and five observations collected from treatment 3. Test the hypothesis at the 0.05 significance level that the treatment means are equal.

Treatment 1

Treatment 2

Treatment 3

9

13

10

7

20

9

11

14

15

9

13

14

12

15

10

a.

State the null and the alternate hypothesis.

Ho : μ1 μ2 μ3

H1 : Treatment means all the same.

b.

What is the decision rule? (Round your answer to 2 decimal places.)

Reject Ho if F >

c.

Compute SST, SSE, and SS total. (Round your answers to 2 decimal places.)

SST =

SSE =

SS total =

d.

Complete the ANOVA table. (Round SS, MS and F values to 2 decimal places.)

Source

SS

df

MS

F

Treatments

Error

6.88

Total

152.93

e.

State your decision regarding the null hypothesis.

Decision: Ho

A stock analyst wants to determine whether there is a difference in the mean rate of return for three types of stock: utility, retail, and banking stocks. The following output is obtained:

Picture

a.

Using the .05 level of significance, is there a difference in the mean rate of return among the three types of stock?

since the test statistic is the critical value .

b.

Can the analyst conclude there is a difference between the mean rates of return for utility and retail stocks? For utility and banking stocks? For banking and retail stocks? Explain.

Since 0 between and the mean rates of return from utility stocks and retail stocks .

Since 0 between and , the mean rates of return from utility and banking stocks .

Since 0 between and , the mean rates of return from retail and banking stocks .

There are three hospitals in the Tulsa, Oklahoma, area. The following data show the number of outpatient surgeries performed on Monday, Tuesday, Wednesday, Thursday, and Friday at each hospital last week. At the 0.05 significance level, can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week?

Number of Surgeries Performed

Day

St. Luke's

St. Vincent

Mercy

Monday

14

18

24

Tuesday

20

24

14

Wednesday

16

22

14

Thursday

18

20

22

Friday

20

28

24

Picture Click here for the Excel Data File

1.

Set up the null hypothesis and the alternative hypothesis.

For Treatment:

Null hypothesis

H0: µSt. Luke's = µSt. Vincent = µMercy

H0: µSt. Luke's ≠ µSt. Vincent ≠ µMercy

2.

Alternative hypothesis

H1: Not all means are equal.

H1: All means are equal.

3.

For blocks:

Null hypothesis

H0: µMon = µTue = µWed = µThu = µFri

H0: µMon ≠ µTue ≠ µWed ≠ µThu ≠ µFri

4.

Alternative hypothesis

H1: All means are equal.

H1: Not all means are equal.

5.

State the decision rule for .05 significance level. (Round your answers to 2 decimal places.)

For Treatment:

Reject H0 if F>

For blocks:

Reject H0 if F>

6.

Complete the ANOVA table. (Round SS, MS and F to 2 decimal places.)

Source

SS

df

MS

F

Treatments

Blocks

Error

Total

7.

What is your decision regarding the null hypothesis?

The decision for the F value (Treatment) at 0.05 significance is:

Reject H0

Do not Reject H0

8.

The decision for the F value (Block) at 0.05 significance is:

Reject H0

Do not Reject H0

9.

Can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week?

There is in the mean number of surgeries performed by hospital or by day of the week.

Mr. James McWhinney, president of Daniel-James Financial Services, believes there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information. The X column indicates the number of client contacts last month, and the Y column shows the value of sales ($ thousands) last month for each client sampled.

Number of Contacts, X

Sales ($ thousands), Y

Number of Contacts, X

Sales ($ thousands), Y

14

24

23

30

12

14

48

90

20

28

50

85

16

30

55

120

46

80

50

110

PictureClick here for the Excel Data File

a.

Determine the regression equation. (Negative amounts should be indicated by a minus sign. Do not round intermediate calculations. Round final answers to 2 decimal places.)

X

Y

Picture

Picture

(Picture ) 2

(Picture )2

(Picture )(Picture )

14

376.36

1376.41

719.74

12

14

−21.4

−47.1

20

−13.4

179.56

443.54

16

30

−31.1

967.21

46

12.6

357.21

23

−10.4

967.21

48

90

28.9

213.16

421.94

50

85

23.9

275.56

396.74

55

466.56

3469.21

1,272.24

50

110.0

16.6

48.9

Picture

=

Picture

=

sx

=

sy

=

r

=

b =

a =

Y' = + X

b.

Determine the estimated sales if 40 contacts are made.(Do not round intermediate calculations. Round final answers to 2 decimal places.)

We are studying mutual bond funds for the purpose of investing in several funds. For this particular study, we want to focus on the assets of a fund and its five-year performance. The question is: Can the five-year rate of return be estimated based on the assets of the fund? Nine mutual funds were selected at random, and their assets and rates of return are shown below.

Assets

Return

Assets

Return

Fund

($ millions)

(%)

Fund

($ millions)

(%)

AARP High Quality Bond

$622.2

10.8

MFS Bond A

$494.5

11.6

Babson Bond L

160.4

11.3

Nichols Income

158.3

9.5

Compass Capital Fixed Income

275.7

11.4

T. Rowe Price Short-term

681.0

8.2

Galaxy Bond Retail

433.2

9.1

Thompson Income B

241.3

6.8

Keystone Custodian B-1

437.9

9.2

PictureClick here for the Excel Data File

b-1.

Compute the coefficient of correlation. (Round your answer to 3 decimal places. Negative amount should be indicated by a minus sign.)

r =

b-2.

Compute the coefficient of determination. (Round your answer to 3 decimal places.)

r2 =

c.

Give a description of the degree of association between the variables.

There is association between the variables.

d.

Determine the regression equation. Use assets as the independent variable. (Round your answers to 4 decimal places. Negative amounts should be indicated by a minus sign.)

b =

a =

e.

For a fund with $400.0 million in sales, determine the five-year rate of return (in percent). (Round your answer to 4 decimal places.)

formula300.mml= The equation should be used with caution. Assets do not account for much of the variation in the rate of return.

On the first statistics exam, the coefficient of determination between the hours studied and the grade earned was 80%. The standard error of estimate was 10. There were 20 students in the class. Develop an ANOVA table for the regression analysis of hours studied as a predictor of the grade earned on the first statistics exam.

Source

DF

SS

MS

Regression

Error

Total

-2-2

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