MWATU

Ms. Lisa Monnin is the budget director for Nexus Media Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information.

  Sales ($)

129

137

142

162

137

145

  Audit  ($)

128

98

128

140

148

110

132

At the 0.1 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff?

(a)

State the decision rule. (Round your answer to 3 decimal places.)

  Reject H0 if t >

(b)

Compute the pooled estimate of the population variance. (Round your answer to 2 decimal places.)

  Pooled variance

(c)

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

  Value of the test statistic

(d)

State your decision about the null hypothesis.

 H0 : μs ≤ μa

(e)

Estimate the p-value. (Round your answers to 3 decimal places.)

  p-value

Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne attire with those of Calvin Klein. Assume the population standard deviations are not the same. The following is the amount ($000) earned per month by a sample of Claiborne models:

$5.4

$4.3

$3.7

$6.7

$4.9

$5.9

$3.1

$5.2

$4.7

$3.5

5.8

4

3.1

5.6

6.9

The following is the amount ($000) earned by a sample of Klein models.

$2.5

$2.6

$3.5

$3.4

$2.8

$3.1

$4

$2.5

$2

$2.9

2.7

2.3

(1)

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

  Degrees of freedom

(2)

State the decision rule for 0.01 significance level: H0: μLC ≤ μCK; H1: μLC > μCK. (Round your answer to 3 decimal places.)

  Reject H0 if t>

(3)

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

  Value of the test statistic

(4)

Is it reasonable to conclude that Claiborne models earn more? Use the 0.01 significance level.

   H0. It is  to conclude that Claiborne models earn more.

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. The information is summarized below.

  Statistic

Men

Women

  Sample mean

23.82   

21.38   

  Population standard deviation

5.91   

4.87   

  Sample size

34   

36   

At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month?

(a)

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

  Value of the test statistic

(b)

What is your decision regarding on null hypothesis?

  The decision is  the null hypothesis that the means are the same.

(c)

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

  p-value

Suppose the manufacturer of Advil, a common headache remedy, recently developed a new formulation of the drug that is claimed to be more effective. To evaluate the new drug, a sample of 210 current users is asked to try it. After a one-month trial, 189 indicated the new drug was more effective in relieving a headache. At the same time a sample of 300 current Advil users is given the current drug but told it is the new formulation. From this group, 255 said it was an improvement.

(1)

State the decision rule for .02 significance level: H0: πn ≤ πc; H1: πn > πc. (Round your answer to 2 decimal places.)

  Reject H0 if z > 

(2)

Compute the value of the test statistic. (Do not round the intermediate value. Round your answer to 2 decimal places.)

  Value of the test statistic

(3)

Can we conclude that the new drug is more effective? Use the .02 significance level.

   H0. We  conclude that the new drug is more effective.

Lester Hollar is vice president for human resources for a large manufacturing company. In recent years he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the last six months. Below are the results.

Employee

Before

After

1

5      

5      

2

5      

3      

3

6      

3      

4

5      

2      

5

4      

2      

6

7      

6      

7

5      

4      

8

5      

7      

At the 0.05 significance level, can he conclude that the number of absences has declined? Hint: For the calculations, assume the "Before" data as the first sample.

Reject H0 if t > . (Round your answer to 3 decimal places.)

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

The p-value is .

Decision:  H0.

What is the critical F value for a sample of four observations in the numerator and seven in the denominator? Use a one-tailed test and the .01 significance level. (Round your answer to 2 decimal places.)

The following hypotheses are given.

Ho : σ1² = σ2²

H1 : σ1² ≠ σ2²

A random sample of eight observations from the first population resulted in a standard deviation of 10. A random sample of six observations from the second population resulted in a standard deviation of 7.

(1)

State the decision rule for .02 significance level: (Round your answer to 1 decimal place.)

  Reject Ho if F >

(2)

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

  Value of the test statistic

(3)

At the .02 significance level, is there a difference in the variation of the two populations?

  Ho. There is in the variations of the two populations.

A study of the effect of television commercials on 12-year-old children measured their attention span, in seconds. The commercials were for clothes, food, and toys.

Clothes

Food

Toys

26

45

60

21

48

51

43

43

43

35

53

54

28

47

63

31

42

53

17

34

48

31

43

58

20

57

47

47

51

44

51

54

Click here for the Excel Data File

(a)

Complete the ANOVA table. Use .05 significance level. (Round the SS and MS values to 1 decimal place and F value to 2 decimal places. Leave no cells blank - be certain to enter "0" wherever required. Round the DF values to nearest whole number.)

  Source

DF

SS

MS

F

P

  Factors

  Error

  Total

(b)

Find the values of mean and standard deviation. (Round the mean and standard deviation values to 3 decimal places.)

  Level

N

Mean

StDev

  Clothes

  Food

  Toys

(c)

Is there a difference in the mean attention span of the children for the various commercials?

  The hypothesis of identical means can definitely be .

  There is  in the mean attention span.

(d)

Are there significant differences between pairs of means?

Clothes have a mean attention span of at least ten minutes  the other groups.

Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level.

Treatment 1

Treatment 2

Treatment 3

8          

3

3

11          

2

4

10          

1

5

3

4

2

 Click here for the Excel Data File

(a-1)

State the null hypothesis and the alternate hypothesis.

      Null hypothesis

Ho: μ1 = μ2

Ho: μ1 = μ2 = μ3

(a-2)

Alternative hypothesis

H1: Treatment means are all the same

H1: Treatment means are not 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 an ANOVA table. (Round F, SS to 2 decimal places and MS to 3 decimal places.)

  Source

SS

df

MS

F

  Treatments

  Error

  Total

(e)

State your decision regarding the null hypothesis.

Reject H0.

Do not reject H0.

(f)

If H0 is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the 95 percent level of confidence.

  , we  conclude that the treatments 1 and 2 have different means.

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. A sample of 56 day-shift workers showed that the mean number of units produced was 336, with a population standard deviation of 19. A sample of 61 night-shift workers showed that the mean number of units produced was 341, with a population standard deviation of 25 units.

At the .02 significance level, is the number of units produced on the night shift larger?

(1)

This is a -tailed test.

(2)

The decision rule is to reject  if Z < . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

(3)

The test statistic is Z = . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

(4)

What is your decision regarding ?

Each month the National Association of Purchasing Managers publishes the NAPM index. One of the questions asked on the survey to purchasing agents is: Do you think the economy is contracting? Last month, of the 260 responses, 161 answered yes to the question. This month, 172 of the 246 responses indicated they felt the economy was contracting.

At the .02 significance level, can we conclude that a larger proportion of the agents believe the economy is contracting this month?

pc =  . (Do not round the intermediate value. Round your answer to 2 decimal places.)

The test statistic is  . (Negative amount should be indicated by a minus sign. Do not round the intermediate value. Round your answer to 2 decimal places.)

Decision:  the null. H0 : π1 ≥ π2

The manufacturer of an MP3 player wanted to know whether a 10 percent reduction in price is enough to increase the sales of its product. To investigate, the owner randomly selected eight outlets and sold the MP3 player at the reduced price. At seven randomly selected outlets, the MP3 player was sold at the regular price. Reported below is the number of units sold last month at the sampled outlets.

  Regular price

131  

127  

88  

116  

143  

121  

96  

  Reduced price

 124  

135  

152  

131  

112  

101  

117  

115  

At the .100 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume the Reduced price as the first sample.

The pooled variance is . (Round your answer to 2 decimal places.)

The test statistic is . (Round your answer to 2 decimal places.)

 H0.

The null and alternate hypotheses are:

The following paired observations show the number of traffic citations given for speeding by Officer Dhondt and Officer Meredith of the South Carolina Highway Patrol for the last five months.

Day

May

June

July

August

September

  Officer Dhondt

30

22

25

19

26

  Officer Meredith

26

19

20

15

19

At the .05 significance level, is there a difference in the mean number of citations given by the two officers?

(a)

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

Reject H0 if t <  or t > .

(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 H0 ?

 H0

(d)

The p-value is .

One of the music industry's most pressing questions is: Can paid download stores contend nose-to-nose with free peer-to-peer download services? Data gathered over the last 12 months show Apple's iTunes was used by an average of 1.61 million households with a sample standard deviation of .49 million family units. Over the same 12 months WinMX (a no-cost P2P download service) was used by an average of 2.20 million families with a sample standard deviation of .28 million. Assume the population standard deviations are not the same.

(a)

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

  Degrees of freedom

(b)

State the decision rule for .02 significance level: H0: A = W; H1: A ≠ W . (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.)

  Reject H0 if t <  or t > 

(c)

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign.Round your answer to 2 decimal places.)

  Value of the test statistic

(d)

Test the hypothesis of no difference in the mean number of households picking either variety of service to download songs. Use the .02 significance level.

 H0. There is  difference in the mean number of households picking either variety of service to download songs.

When only two treatments are involved, ANOVA and the Student t test (Chapter 11) result in the same conclusions. Also, . As an example, suppose that 14 randomly selected students were divided into two groups, one consisting of 6 students and the other of 8. One group was taught using a combination of lecture and programmed instruction, the other using a combination of lecture and television. At the end of the course, each group was given a 50-item test. The following is a list of the number correct for each of the two groups. Using analysis of variance techniques, test the null hypothesis,  that the two mean test scores are equal.

Lecture and Programmed Instruction

Lecture and Television

14

33

13

27

24

36

22

21

14

28

12

25

26

24

(a-1)

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

  Source

SS

df

MS

F

  Factors

  Error

  Total

(a-2)

Use a  level of significance. (Round your answer to 2 decimal places.)

  The test statistic is F

(b)

Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

  t

(c)

There is  in the mean test scores.

The following hypotheses are given.

Ho : σ1² ≤ σ2²

H1 : σ1² > σ2²

A random sample of five observations from the first population resulted in a standard deviation of 12. A random sample of seven observations from the second population showed a standard deviation of 7. At the .01 significance level, is there more variation in the first population?

The test statistic is . (Round your answer to 2 decimal places.)

Decision:   Ho

Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level.

Treatment 1

Treatment 2

Treatment 3

8          

3

3

11          

2

4

10          

1

5

3

4

2

 Click here for the Excel Data File

(a-1)

State the null hypothesis and the alternate hypothesis.

      Null hypothesis

Ho: μ1 = μ2

Ho: μ1 = μ2 = μ3

(a-2)

Alternative hypothesis

H1: Treatment means are all the same

H1: Treatment means are not 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 an ANOVA table. (Round F, SS to 2 decimal places and MS to 3 decimal places.)

  Source

SS

df

MS

F

  Treatments

  Error

  Total

(e)

State your decision regarding the null hypothesis.

Do not reject H0.

Reject H0.

(f)

If H0 is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the 95 percent level of confidence.

  , we  conclude that the treatments 1 and 2 have different means.

There are four radio stations in Midland. The stations have different formats (hard rock, classical, country/western, and easy listening), but each is concerned with the number of minutes of music played per hour. From a sample of 10 hours from each station, the following sample means were offered.

                                                      SS total = 650.75

(a)

SST = . (Round your answer to 3 decimal places.)

(b)

SSE = . (Round your answer to 3 decimal places.)

(c)

Complete an ANOVA table. (Round SS, MS, F to 3 decimal places and df to nearest whole number.)

SS

df

MS

F

Treatments

Error

Total

(d)

At the .05 significance level, is there a difference in the treatment means? 

Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level.

Treatment 1

Treatment 2

Treatment 3

3

9

6

2

6

3

5

5

5

1

6

5

3

8

5

1

5

4

4

1

7

5

6

4

Click here for the Excel Data File

(a)

Ho : μ1  μ2   μ3.

H1 : Treatment means  all the same

(b)

Reject   Ho  if F >  .(Round your answer to 2 decimal places.)

(c)

SST =   SSE =   SS total =  (Round your answers to 2 decimal places.)

(d)

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

  Source

SS

df

MS

F

  Treatments

  Error

  Total

(e)

Decision:  Ho

(f)

Find the 95% confidence interval for the difference between treatment 2 and 3. (Round your answers to 2 decimal places.)

95% confidence interval is:   ±  

We can conclude that the treatments 2 and 3 are