12_ch08-isbe11

464

Chapter 8 Design of Experiments and Analysis of Variance

8.1 Since only one factor is utilized, the treatments are the four levels (A, B, C, D) of the qualitative factor. 8.2 The treatments are the combinations of levels of each of the two factors. There are 2 × 5 = 10 treatments. They are: (A, 50), (A, 60), (A, 70), (A, 80), (A, 90) (B, 50), (B, 60), (B, 70), (B, 80), (B, 90) 8.3 One has no control over the levels of the factors in an observational experiment. One does have control of

the levels of the factors in a designed experiment. 8.4 a. College GPA's are measured on college students. The experimental units are college students. b. Household income is measured on households. The experimental units are households. c. Gasoline mileage is measured on automobiles. The experimental units are the automobiles of a

particular model. d. The experimental units are the sectors on a computer diskette. e. The experimental units are the states. 8.5 a. This is an observational experiment. The economist has no control over the factor levels or

unemployment rates. b. This is a designed experiment. The manager chooses only three different incentive programs to

compare, and randomly assigns an incentive program to each of nine plants. c. This is an observational experiment. Even though the marketer chooses the publication, he has no

control over who responds to the ads. d. This is an observational experiment. The load on the facility's generators is only observed, not

controlled. e. This is an observational experiment. One has no control over the distance of the haul, the goods

hauled, or the price of diesel fuel. 8.6 a. The response variable is the amount of the purchase.

b. There is one factor in this problem: type of credit card.

c. There are 4 treatments, corresponding to the 4 levels of the factor. The treatments are VISA, MasterCard, American Express, and Discover.

d. The experimental units are the credit card holders.

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

Design of Experiments and Analysis of Variance 465

8.7 a. The response variable is the age when the tumor was first detected.

b. The experimental units are the smokers.

c. There is one factor in this problem: screening method.

d. There are 2 treatments in this problem, corresponding to the 2 levels of the factor. The treatments are CT and chest X-ray.

8.8 a. The experimental units are the accounting alumni. b. The response variable is income. c. There are 2 factors in the problem: Mach score classification and Gender.

d. Mach score classification has 3 levels – high, moderate, and low. Gender has 2 levels male and female.

e. There are a total of 2 x 3 = 6 treatments in his experiment. The treatments are all of the Mach score rating-gender combinations.

8.9 a. The experimental units for this study are the students in the introductory psychology class.

b. The study is a designed experiment because the students are randomly assigned to a particular study group.

c. There are 2 factors in this problem: Class standing and study group.

d. Class standing has 3 levels: Low, Medium, and High. Study group has 2 levels: practice test and

review.

e. There are a total of 3 × 2 = 6 treatments. They are: (Low, Review), (Low, Practice exam), (Medium, Review), (Medium, Practice exam), (High, Review), and (High, Practice exam).

f. The response variable is the final exam score.

8.10 a. The response variable in this problem is the consumer’s opinion on the value of the discount offer.

b. There are two treatments in this problem: Within-store price promotion and between-store price promotion.

c. The experimental units are the consumers.

8.11 a. There are 2 factors in this problem, each with 2 levels. Thus, there are a total of 2 × 2 = 4 treatments.

b. The 4 treatments are: (Within-store, home), (Within-store, in store), (Between-store,home), and (Between-store, in store).

8.12 a. There are 2 factors in the problem: Type of yeast and Temperature. Type of yeast has 2 levels – Brewer’s yeast and baker’s yeast. Temperature has 4 levels – 45o, 48o, 51o and 54oC.

b. The response variable is the autolysis yield.

c. There are a total of 2 × 4 = 8 treatments in this experiment. The treatments are all the type of yeast-temperature combinations.

d. This is a designed experiment.


466 Chapter 8

8.13 a. The dependent variable is the dissolution time.

b. There are 3 factors in this experiment: Binding agent, binding concentration, and relative density. Binding agent has 2 levels – khaya gum and PVP. Binding concentration has 2 levels − .5% and 4.0%. Relative density has 2 levels – high and low.

c. There could be a total of 2 x 2 x 2 = 8 treatments for this experiment. They are:

khaya gum, .5%, high PVP, .5%, high khaya gum, .5%, low PVP, .5%, low khaya gum, 4.0%, high PVP, 4.0%, high khaya gum, 4.0%, low PVP, 4.0%, low 8.14 a. The response is the evaluation by the undergraduate student of the ethical behavior of the

salesperson. b. There are two factors—type of sales job at two levels (high tech. vs. low tech.) and sales task at two

levels (new account development vs. account maintenance). c. The treatments are the 2 × 2 = 4 combinations type of sales job and sales task. d. The experimental units are the college students. 8.15 a. From Table VIII with ν1 = 4 and ν2 = 4, F.05 = 6.39. b. From Table X with ν1 = 4 and ν2 = 4, F.01 = 15.98. c. From Table VII with ν1 = 30 and ν2 = 40, F.10 = 1.54. d. From Table IX with ν1 = 15, and ν2 = 12, F.025 = 3.18. 8.16 a. P(F ≤ 3.48) = 1 - .05 = .95 using Table VIII, Appendix B, with ν1 = 5 and ν2 = 9 b. P(F > 3.09) = .01 using Table X, Appendix B, with ν1 = 15 and ν2 = 20 c. P(F > 2.40) = .05 using Table VIII, Appendix B, with ν1 = 15 and ν2 = 15 d. P(F ≤ 1.83) = 1 − .10 = .90 using Table VII, Appendix B, with ν1 = 8 and ν2 = 40 8.17 a. In the second dot diagram #2, the difference between the sample means is small relative to the

variability within the sample observations. In the first dot diagram #1, the values in each of the samples are grouped together with a range of 4, while in the second diagram #2, the range of values is 8.



b. For diagram #1,

1

17 8 9 9 10 11 54

6 6

xx

n

= 9

2

212 13 14 14 15 16 84

6 6

xx

n

= 14

For diagram #2,

1

15 5 7 11 13 13 54

6 6

xx

n

= 9

2

210 10 12 16 18 18 84

6 6

xx

n

= 14

c. For diagram #1,

SST = 2

2

1

( )i ii

n x x

1 = 6(9 − 11.5)2 + 6(14 − 11.5)2 = 75

54 84

11.512

xx

n

For diagram #2,

SST = 2

2

1

( )i ii

n x x

= 6(9 - 11.5)2 + 6(14 - 11.5)2 = 75

d. For diagram #1,

21s =

2 2121

1

1

54496

61 6 1

xx

n

n

=

2544966

6 1

= 2

22s =

2 2222

2

2

841186

61 6 1

xx

n

n

= 2

SSE = (n1 − 1) 21s + (n2 − 1) 2

2s = (6 − 1)2 + (6 − 1)2 = 20


468 Chapter 8

For diagram #2,

21s =

2 2121

1

1

54558

61 6 1

xx

n

n

= 14.4

22s =

2 2222

2

2

841248

61 6 1

xx

n

n

= 14.4

SSE = (n1 − 1) 21s + (n2 − 1) 2

2s = (6 − 1)14.4 + (6 − 1)14.4 = 144

e. For diagram #1, SS(Total) = SST + SSE = 75 + 20 = 95

SST is SST 75

100% = 100%SS(Total) 95

= 78.95% of SS(Total)

For diagram #2, SS(Total) = SST + SSE = 75 + 144 = 219

SST is SST 75

100% = 100%SS(Total) 219

= 34.25% of SS(Total)

f. For diagram #1, MST = SST 75

1 2 1k

= 75

MSE = SSE 20

12 2n k

= 2 F = MST 75

= MSE 2

= 37.5

For diagram #2, MST = SST 75

1 2 1k

= 75

MSE = SSE 144

12 2n k

= 14.4 F = MST 75

= MSE 14.4

= 5.21

g. The rejection region for both diagrams requires α = .05 in the upper tail of the F-distribution with ν1

= p − 1 = 2 − 1 = 1 and ν2 = n − p = 12 − 2 = 10. From Table VIII, Appendix B, F.05 = 4.96. The rejection region is F > 4.96.

For diagram #1, the observed value of the test statistic falls in the rejection region (F = 37.5 > 4.96).

Thus, H0 is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means at α = .05.

For diagram #2, the observed value of the test statistic falls in the rejection region (F = 5.21 > 4.96).

Thus, H0 is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means at α = .05.

h. We must assume both populations are normally distributed with common variances.



8.18 For each dot diagram, we want to test: H0: μ1 = μ2 Ha: μ1 ≠ μ2 From Exercise 8.17,

Diagram #1 Diagram #2 1x = 9 1x = 9

2x = 14 2x = 14

21s = 2 2

1s = 14.4

22s = 2 2

2s = 14.4

a.

Error! Bookmark not defined.Diagram #1

Diagram #2

2 21 22

p

1 2

22 2

2 ( )2

s ss

n n

2 21 22

p

1 2

214.4 14.4

14.4 ( )2

s ss

n n

In Exercise 8.17, MSE = 2 In Exercise 8.17, MSE = 14.4 The pooled variance for the two-sample t-test is the same as the MSE for the F-test.

b.

Diagram #1 Diagram #2

1 2

2p

1 2

9 14 =

1 11 1 2 + 6 6

= 6.12

x xt =

sn n

1 2

2p

1 2

9 14 =

1 11 1 14.4 + 6 6

= 2.28

x xt =

sn n

In Exercise 8.17, F = 37.5 In Exercise 8.17, F = 5.21 The test statistic for the F-test is the square of the test statistic for the t-test.

c.

Diagram #1 Diagram #2 For the t-test, the rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n1 + n2 − 2 = 6 + 6 − 2 = 10. From Table V, Appendix B, t.025 = 2.228.

For the t-test, the rejection region is the same as Diagram #1 since we are using the same α, n1, and n2 for both tests.

The rejection region is t < −2.228 or t > 2.228. In Exercise 8.17, the rejection region for both diagrams using the F-test is F > 4.96. The tabled F value equals the square of the tabled t value.


470 Chapter 8

d.

Diagram #1 Diagram #2

For the t-test, since the test statistic falls in the rejection region (t = −6.12 < −2.228), we would reject H0. In Exercise 8.17, using the F-test, we rejected H0.

For the t-test, since the test statistic falls in the rejection region (t = −2.28 < −2.228), we would reject H0. In Exercise 8.17, using the F-test, we rejected H0.

e. Assumptions for the t-test:

1. Both populations have relative frequency distributions that are approximately normal. 2. The two population variances are equal. 3. Samples are selected randomly and independently from the populations. Assumptions for the F-test:

1. Both population probability distributions are normal. 2. The two population variances are equal. 3. Samples are selected randomly and independently from the respective populations. The assumptions are the same for both tests. 8.19 Refer to Exercise 8.17, the ANOVA table is:

For diagram #1:

Source Df SS MS F Treatment 1 75 75 37.5Error 10 20 2 Total 11 95

For diagram #2:

Source Df SS MS F Treatment 1 75 75 5.21Error 10 144 14.4 Total 11 219

8.20 a. df for Error is 41 − 6 = 35 SSE = SS(Total) − SST = 46.5 − 17.5 = 29.0

MST = SST 17.5

1 6k

= 2.9167 MSE = SSE 29.0

35n k

= .8286

F = MST 2.9167

= MSE .8286

= 3.52

The ANOVA table is:

Source df SS MS F Treatment 6 17.5 2.9167 3.52Error 35 29.0 .8286Total 41 46.5



b. The number of treatments is k. We know k − 1 = 6 k = 7. c. The total sample size is n = 41 + 1 = 42, where 41 = df Total. d. First, one would number the 42 experimental units from 1 to 42. Then generate over 100 uniform

random numbers from 1 to 42. The first 6 different random numbers will correspond to treatment 1. The next 6 different random numbers will correspond to treatment 2. Repeat the process for treatments, 3, 4, 5, 6, and 7.

e. To determine if there is a difference among the population means, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ7 Ha: At least one of the population means differs from the rest The test statistic is F = 3.52. The rejection region requires α = .10 in the upper tail of the F-distribution with numerator df = k − 1 = 6 and denominator df = n − k = 35. From Table VII, Appendix B, F.10 ≈ 1.98. The

rejection region is F > 1.98. Since the observed value of the test statistic falls in the rejection region (F = 3.52 > 1.98), H0 is

rejected. There is sufficient evidence to indicate a difference among the population means at α = .10. f. The observed significance level is P(F ≥ 3.52). With numerator df = 6 and denominator df = 35, and

Table X, P(F ≥ 3.52) < .01. g. H0: μ1 = μ2 Ha: μ1 ≠ μ2

The test statistic is t = 1 2

1 2

3.7 4.1

1 11 1 .8286MSE6 6

x x

n n

= −.76

The rejection region requires α/2 = .10/2 = .05 in each tail of the t-distribution with df = n − p = 35.

From Table V, Appendix B, t.05 ≈ 1.697. The rejection region is t < −1.697 or t > 1.697. Since the observed value of the test statistic does not fall in the rejection region (t = −.76 −1.697),

H0 is not rejected. There is insufficient evidence to indicate that μ1 and μ2 differ at α = .10. h. For confidence coefficient .90, α = .10 and α/2 = .05. From Table V, Appendix B, with df = 35, t.05 ≈ 1.697. The confidence interval is:

1 2( )x x ± t.05

1 2

1 1MSE

n n

(3.7 − 4.1) ± 1.6971 1

.82866 6

−.4 ± .892 (1.292, .492) i. The confidence interval is:

1x ± t.05 MSE/6 3.7 ± 1.697 .8286 / 6 3.7 ± .631 (3.069, 4.331)


472 Chapter 8

8.21 a. Some preliminary calculations are:

CM = 2 237.1

12

iy =

n

= 114.701

SS(Total) = 2iy − CM = 145.89 − 114.701 = 31.189

SST = 2

CMi

i

T

n =

2 2 216.9 16.0 4.2

5 4 3 − 114.701

= 127.002 − 114.701 = 12.301

SSE = SS(Total) − SST = 31.189 − 12.301 = 18.888

MST = SST 12.301

= 1 3 1k

= 6.1505 MSE = SSE 18.888

= 12 3n k

= 2.0987

F = MST 6.1505

= MSE 2.0987

= 2.931

Source df SS MS F Treatments 2 12.30 6.15 2.93Error 9 18.89 2.10Total 11 31.19

b. H0: μ1 = μ2 = μ3 Ha: At least two treatment means differ The test statistic is F = 2.931. The rejection region requires α = .01 in the upper tail of the F-distribution with ν1 = k − 1 = 3 − 1 = 2

and ν2 = n − k = 12 − 3 = 9. From Table X, Appendix B, F.01 = 8.02. The rejection region is F > 8.02. Since the observed value of the test statistic does not fall in the rejection region (F = 2.93 8.02),

H0 is not rejected. There is insufficient evidence to indicate a difference in the treatment means at α = .01. 8.22 a. This is a completely randomized design because the subjects were randomly assigned to one of three

groups. b. The response variable was the total WTP (willing to pay) value and the treatments were the 3 types

of instructions given. c. To determine if the mean total WTP values differed among the three groups, we test: H0: μ1 = μ2 = μ3 Ha: At least two treatment means differ d. One would number the subjects from 1 to 252. Then, use a random number generator to generate

350 to 400 random numbers from 1 to 252 (We need to generate more than 252 random numbers to account for duplicates.) The first 84 different random numbers will be assigned to group 1, the next 84 different random numbers will be assigned to groups 2, and the rest will be assigned to group 3.



8.23 a. To determine if the mean LUST discount percentages across the seven states differ, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 = μ7 Ha: At least two treatment means differ b. From the ANOVA table, the test statistic is F = 1.60 and the p-value = 0.174.

Since the observed p-value (p = 0.174) is not less than α = .10, H0 is not rejected. There is insufficient evidence to indicate a difference in the mean LUST discount percentages among the seven states at α = .10.

8.24 a. The experimental unit in the study is the college tennis coach. The dependent variable is the

response to the statement “the Prospective Student-Athlete Form on the web site contributes very little to the recruiting process” on a scale from 1 to 7. There is one factor in the study and it is the NCAA division of the college tennis coach. There are 3 levels of this factor, and thus, there are 3 treatments – Division I, Division II, and Division III.

b. To determine if the mean responses of tennis coaches from the different divisions differ, we test:

H0: μ1 = μ2 = μ3 Ha: At least 1 μi differs

c. Since the observed p-value of the test (p < .003) is less than α = .05, H0 is rejected. There is sufficient evidence to indicate differences in mean response among coaches of the 3 divisions.

8.25 a. A completely randomized design was used.

b. There are 4 treatments: 3 robots/colony, 6 robots/colony, 9 robots/colony, and 12 robots/colony.

c. To determine if there was a difference in the mean energy expended (per robot) among the 4 colony sizes, we test:

H0: μ1 = μ2 = μ3 = μ4 Ha: At least two means differ

d. Since the p-value (<.001) is less than α (.05), H0 is rejected. There is sufficient evidence to indicate a

difference in mean energy expended per robot among the 4 colony sizes at α = .05. 8.26 a. To determine if differences exist in the mean rates of return among the three types of fund groups, we test: H0: μ1 = μ2 = μ3

Ha: At least two means differ

b. The rejection region requires α = .01 in the upper tail of the F-distribution with ν1 = k – 1 = 3 – 1 = 2 and ν2 = N – k = 90 – 3 = 87. From Table X, Appendix B, F.01 ≈ 4.98. The rejection region is

F > 4.98.

c. Since the observed value of the test statistic falls in the rejection region (F = 6.965 > 4.98), H0 is rejected. There is sufficient evidence to indicate differences exist in the mean rates of return among the three types of fund groups at α = .01.


474 Chapter 8

8.27 To determine if a driver’s propensity to engage in road rage is related to his/her income, we test: H0: μ1 = μ2 = μ3

Ha: At least two means differ The test statistic is F = 3.90 and the p-value is p < .01. Since the p-value is less than α = .05, H0 is rejected. There is sufficient evidence to indicate a driver’s propensity to engage in road rage is related to his/her income for α > .01. Since the sample means increase as the income increases, it appears that road rage increases as income increases.

8.28 a. To determine if the mean knowledge gain differs among the three groups, we test: H0: μ1 = μ2 = μ3 Ha: At least two population means differ b. Using MINITAB, the results are:

One-way ANOVA: NO, CHECK, FULL

Source DF SS MS F P Factor 2 6.64 3.32 0.45 0.637 Error 72 527.36 7.32 Total 74 534.00

S = 2.706 R-Sq = 1.24% R-Sq(adj) = 0.00%

Individual 95% CIs For Mean Based on

Pooled StDev Level N Mean StDev -+---------+---------+---------+-------- NO 30 2.433 3.431 (-----------*------------) CHECK 25 2.720 2.264 (------------*------------) FULL 20 1.950 1.820 (--------------*--------------) -+---------+---------+---------+-------- 0.80 1.60 2.40 3.20

The ANOVA table is: c. The test statistic is F = 0.45 and the p-value = 0.637.

Since the p-value = 0.637 is larger than any reasonable significance level, H0 is not rejected. There is

insufficient evidence to indicate a difference in the mean knowledge gained among the three levels of assistance for any reasonable value of α.

Practically speaking, there is not one type of assistance that helps students more than another.

Source df SS MS F-value p-value

Assist 2 6.64 3.32 0.45 0.637

Error 72 527.36 7.32

Total 74 534



8.29 a. There is one factor in this problem which is Group. There are 5 treatments in this problem, corresponding to the 5 levels of Group: Casualties, Survivors, Implementers/casualties, Implementers/survivors, and Formulators. The response variable is the ethics score. The experimental units are the employees enrolled in an Executive MBA program.

b. To determine if there are any differences among the mean ethics scores for the five groups, we test:

H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least one of the population means differs from the rest

c. The test statistic is F = 9.85 and the p-value is p = 0.000. Since the p-value (0.000) is less than any

reasonable significance level α, H0 is rejected. There is sufficient evidence to indicate a difference in the mean ethics scores among the five groups of employees for any reasonable value of α.

d. We will check the assumptions of normality and equal variances. Using MINITAB, the histograms are:

Freq

uenc

y

543210

543210

30

20

10

0

543210

30

20

10

0

CASUAL SURVIVE IMPCAS

IMPSUR FORMUL

CASUAL

1.845StDev 1.023N 71

IMPCASMean 1.593StDev 0.6360

Mean

N 27

IMPSURMean 2.545StDev 1.301N 33

FORMUL

1.787

Mean 2.871StDev 1.176N 31

StDev 0.8324N 47

SURVIVEMean

Histogram of CASUAL, SURVIVE, IMPCAS, IMPSUR, FORMULNormal

The data for some of the 5 groups do not look particularly mound-shaped, so the assumption of normality is probably not valid.


476 Chapter 8

Using MINITAB, the boxplots are:

FORMULIMPSURIMPCASSURVIVECASUAL

5

4

3

2

1

Dat

a

Boxplot of CASUAL, SURVIVE, IMPCAS, IMPSUR, FORMUL

The spreads of responses do not appear to be about the same. The groups Implementers/survivors and Formulators have more variability than the other three groups. Thus, the assumption of constant variance is probably not valid.

The assumptions required for the ANOVA F-test do not appear to be reasonably satisfied.

8.30 a. We will select size as the quantitative variable and color as the qualitative variable. To determine if the mean size of diamonds differ among the 6 colors, we test:

H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 Ha: At least two means differ

b. Using MINITAB, the ANOVA table is:

One-way ANOVA: Carats versus Color

Analysis of Variance for Carats Source DF SS MS F P Color 5 0.7963 0.1593 2.11 0.064 Error 302 22.7907 0.0755 Total 307 23.5869 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ----------+---------+---------+------ D 16 0.6381 0.3195 (-------------*------------) E 44 0.6232 0.2677 (-------*-------) F 82 0.5929 0.2648 (-----*-----) G 65 0.5808 0.2792 (------*------) H 61 0.6734 0.2643 (------*------) I 40 0.7310 0.2918 (-------*--------) ----------+---------+---------+------ Pooled StDev = 0.2747 0.60 0.70 0.80

The test statistic is F = 2.11 and the p-value is p = 0.064.

Since the p-value (0.064) is less than α = .10, H0 is rejected. There is sufficient evidence to indicate the mean size of diamonds differ among the 6 colors at α = .10.



c. We will check the assumptions of normality and equal variances. Using MINITAB, the stem-and-leaf plots are: Stem-and-Leaf Display: Carats Stem-and-leaf of Carats Color= 1 N = 16 Leaf Unit = 0.010 1 1 9 3 2 01 5 3 01 5 4 7 5 23 7 6 (4) 7 1156 5 8 5 9 5 10 00011 Stem-and-leaf of Carats Color = 2 N = 44 Leaf Unit = 0.010 1 1 9 4 2 123 11 3 0011345 12 4 6 (11) 5 00012245668 21 6 23 19 7 000123 13 8 113 10 9 10 10 0000011113 Stem-and-leaf of Carats Color = 3 N = 82 Leaf Unit = 0.010 5 1 88999 12 2 1356667 23 3 01124445567 27 4 0178 (21) 5 000111122333345566678 34 6 0 33 7 00001112367 22 8 0012555 15 9 0 14 10 00000011112224 Stem-and-leaf of Carats Color = 4 N = 65 Leaf Unit = 0.010 5 1 88899 12 2 0001359 21 3 000124455 23 4 08 (12) 5 000013556789 30 6 0034 26 7 0000001348 16 8 0125 12 9 12 10 000000011126


478 Chapter 8

Stem-and-leaf of Carats Color = 5 N = 61 Leaf Unit = 0.010 2 1 89 5 2 457 14 3 012344567 16 4 03 21 5 25778 27 6 001466 (13) 7 0001112233448 21 8 0014669 14 9 14 10 0000011111266 1 11 0

Stem-and-leaf of Carats Color = 6 N = 40 Leaf Unit = 0.010

4 2 5689 8 3 0113 11 4 115 13 5 26 15 6 25 20 7 03355 20 8 002 17 9 0 16 10 0000001111114579

The data for the 6 colors do not look particularly mound-shaped, so the assumption of normality is probably not valid. However, departures from this assumption often do not invalidate the ANOVA results. Using MINITAB, the box plots are:

IHGFED

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

Color

Ca

rats

The spreads of all the colors appear to be about the same, so the assumption of constant variance is probably valid.



8.31 a. To determine if the mean level of trust differs among the six treatments, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 Ha: At least one μi differs b. The test statistic is F = 2.21. The rejection region requires α in the upper tail of the F-distribution with ν1 = k − 1 = 6 − 1 = 5 and

ν2 = n − k = 237 − 6 = 231. From Table VII, Appendix B, F.05 ≈ 2.21. The rejection region is F > 2.21. Since the observed value of the test statistic does not fall in the rejection region (F = 2.21 2.21),

H0 is not rejected. There is insufficient evidence to indicate that at least two mean trusts differ at α = .05. c. We must assume that all six samples are drawn from normal populations, the six population

variances are the same, and that the samples are independent. d. I would classify this experiment as designed. Each subject was randomly assigned to receive one of

the six scenarios.

8.32 a. I would classify this experiment as designed. Each subject was randomly assigned to receive one of the three dosages (DM, honey, nothing). There are 3 treatments in the study corresponding to the 3 dosages: DM, honey, nothing.

b. Using MINITAB, the output is:

One-way ANOVA: TotalScore versus Treatment Source DF SS MS F P Treatment 2 318.51 159.25 17.51 0.000 Error 102 927.72 9.10 Total 104 1246.23 S = 3.016 R-Sq = 25.56% R-Sq(adj) = 24.10% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -----+---------+---------+---------+---- C 37 6.514 2.940 (-----*-----) DM 33 8.333 3.256 (-----*------) H 35 10.714 2.855 (-----*-----) -----+---------+---------+---------+---- 6.4 8.0 9.6 11.2 Pooled StDev = 3.016

To determine if differences exist in the mean improvement scores among the 3 treatment groups, we test: H0: μ1 = μ2 = μ3

Ha: At least two treatment means differ The test statistic is F = 17.51 and the p-value = 0.000.

Since the observed p-value (0.000) is less than any reasonable significance level, H0 is rejected. There is sufficient evidence to indicate a difference in the mean improvement scores among the three levels of dosage for any reasonable value of α.


480 Chapter 8

8.33 To determine if the mean THICKNESS differs among the 4 types of housing, we test:

H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ

The test statistic is F = 11.74 and the p-value = 0.000. Since the observed p-value (0.000) is less than any reasonable significance level, Ho is rejected. There is sufficient evidence to indicate a difference in the mean thickness among the four levels of housing for any reasonable value of α.

To determine if the mean WHIPPING CAPACITY differs among the 4 types of housing, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ

The test statistic is F = 31.36 and the p-value = 0.000. Since the observed p-value (0.000) is less than any reasonable significance level, Ho is rejected. There is sufficient evidence to indicate a difference in the mean whipping capacity among the four levels of housing for any reasonable value of α.

To determine if the mean STRENGTH differs among the 4 types of housing, we test:

H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ

The test statistic is F = 1.70 and the p-value = 0.193. Since the observed p-value (0.193) is higher than any reasonable significance level, H0 is not rejected. There is insufficient evidence to indicate a difference in the mean strength among the four levels of housing for any reasonable value of α.

Thus, the mean thickness and the mean percent overrun differ among the 4 housing systems.

8.34 a.

3

1 26(10.5) 25(3.9) 25(1.4) 405.55.3355

76 76 76

i ii

n x

x

3

2 2 2 2

1

SST ( ) 26(10.5 5.3355) 25(3.9 5.3355) 25(1.4 5.3355) 1132.1941i ii

n x x

b. SSE = 2 2 21 1 2 2 3 3( 1) ( 1) ( 1)n s n s n s = (26 – 1) (7.6)2+ (25 – 1) (7.5)2 + (25 – 1) (7.5)2

= 1444 + 1350 + 1350 = 4144 c. SS(Total) = SST + SSE = 1132.1941 + 4144 = 5276.1941

SST 1132.1942

MST 566.0971k 1 3 1

SSE 4144

MSE 56.7671n k 76 3

566.0971

9.9756.7671

MSTF

MSE



The ANOVA table is:

d. To determine if the mean anxiety levels differ among the 3 groups, we test:

H0: μ1 = μ2 = μ3

Ha: At least two treatment means differ The test statistic is F = 9.97.

The rejection region requires α = .01 in the upper tail of the F-distribution with υ1 = k – 1 = 3 – 1 = 2 and υ2 = n – k = 76 – 3 = 73. From Table X, Appendix B, F.01 = 4.92. The rejection region is F > 4.92. Since the observed value of the test statistic does fall in the rejection region (F = 9.97 > 4.92), H0 is rejected. There is sufficient evidence to indicate a difference in the mean anxiety levels among the three groups at α = .01.

e. The assumption of constant variance is satisfied since the three sample variances are all very similar (7.62 = 57.76, 7.52 = 56.25, and 7.52 = 56.25).

We are unable to check the normality assumption since we need the individual anxiety levels to create a histogram or stem-and-leaf plot.

8.35 The number of pairwise comparisons is equal to k(k − 1)/2. a. For k = 3, the number of comparisons is 3(3 − 1)/2 = 3. b. For k = 5, the number of comparisons is 5(5 − 1)/2 = 10. c. For k = 4, the number of comparisons is 4(4 − 1)/2 = 6. d. For k = 10, the number of comparisons is 10(10 − 1)/2 = 45. 8.36 The experimentwise error rate is the probability of making a Type I error for at least one of all of the

comparisons made. If the experimentwise error rate is α = .05, then each individual comparison is made at a value of α which is less than .05.

8.37 A comparisonwise error rate is the error rate (or the probability of declaring the means different when, in

fact, they are not different, which is also the probability of a Type I error) for each individual comparison. That is, if each comparison is run using α = .05, then the comparisonwise error rate is .05.

Source df SS MS F-value

Groups 2 1132.1941 566.0971 9.97

Error 73 4144.00 56.77

Total 75 5276.1941


482 Chapter 8

8.38 a. From the diagram, the following pairs of treatments are significantly different because they are not connected by a line: A and E, A and B, A and D, C and E, C and B, C and D, and E and D. All other pairs of means are not significantly different because they are connected by lines.

b. From the diagram, the following pairs of treatments are significantly different because they are not

connected by a line: A and B, A and D, C and B, C and D, E and B, E and D, and B and D. All other pairs of means are not significantly different because they are connected by lines.

c. From the diagram, the following pairs of treatments are significantly different because they are not

connected by a line: A and E, A and B, and A and D. All other pairs of means are not significantly different because they are connected by lines.

d. From the diagram, the following pairs of treatments are significantly different because they are not

connected by a line: A and E, A and B, A and D, C and E, C and B, C and D, E and D, and B and D. All other pairs of means are not significantly different because they are connected by lines.

8.39 (μ1 − μ2): (2, 15) Since all values in the interval are positive, μ1 is significantly greater than μ2. (μ1 − μ3): (4, 7) Since all values in the interval are positive, μ1 is significantly greater than μ3. (μ1 − μ4): (-10, 3) Since 0 is in the interval, μ1 is not significantly different from μ4. However, since the center of the interval is less than 0, μ4 is larger than μ1. (μ2 − μ3): (-5, 11) Since 0 is in the interval, μ2 is not significantly different from μ3. However, since the center of the interval is greater than 0, μ2 is larger than μ3. (μ2 − μ4): (−12, −6) Since all values in the interval are negative, μ4 is significantly greater than μ2. (μ3 − μ4): (−8, −5) Since all values in the interval are negative, μ4 is significantly greater than μ3. Thus, the largest mean is μ4 followed by μ1, μ2,and μ3. 8.40 a. The test statistic is F = 22.68 and the p-value is p = 0.001. Since the observed p-value (0.001) is less

than any reasonable α-level we select (.01, .05, or .10), we reject H0. There is sufficient evidence to indicate a difference in the mean number of alternatives listed among the three emotional states for any α > .001.

b. The probability of declaring at least one pair of means different when they are not is .05. c. The mean number of alternatives listed under the guilty state is significantly higher than mean

number of alternatives listed under the angry and neutral states. There is no difference in the mean number of alternatives listed under the angry and neutral states.

8.41 The mean response for Division I coaches is significantly higher than the mean responses for the Division II and Division III coaches. There is no difference in the mean responses between Division II and Division III coaches.



8.42 a. The total number of comparisons conducted is k(k – 1)/2 = 4(4 – 1)/2 = 6.

b. The mean energy expended by robots in the 12 robot colony is significantly smaller than the mean energy expended by robots in any of the other size colonies. There is no difference in the mean energy expended by robots in the 3 robot colony, the 6 robot colony, and the 9 robot colony.

8.43 a. Tukey’s multiple comparison method is preferred over other methods because it controls experimental error at the chosen α level. It is more powerful than the other methods.

b. From the confidence interval comparing large-cap and medium-cap mutual funds, we find that 0 is in

the interval. Thus, 0 is not an unusual value for the difference in the mean rates of return between large-cap and medium-cap mutual funds. This means we would not reject H0. There is insufficient evidence of a difference in mean rates of return between large-cap and medium-cap mutual funds at α = .05.

c. From the confidence interval comparing large-cap and small-cap mutual funds, we find that 0 is not

in the interval. Thus, 0 is an unusual value for the difference in the mean rates of return between large-cap and small-cap mutual funds. This means we would reject H0. There is sufficient evidence of a difference in mean rates of return between large-cap and small-cap mutual funds at α = .05.

d. From the confidence interval comparing medium-cap and small-cap mutual funds, we find that 0 is in

the interval. Thus, 0 is not an unusual value for the difference in the mean rates of return between medium-cap and small-cap mutual funds. This means we would not reject H0. There is insufficient evidence of a difference in mean rates of return between medium-cap and small-cap mutual funds at α = .05.

e. From the above, the mean rate of return for large-cap mutual funds is the largest, followed by

medium-cap, followed by small-cap mutual funds. The mean rate of return for large-cap funds is significantly larger than that for small-cap funds. No other differences exist.

f. We are 95% confident of this decision.

8.44 a. The Bonferroni method is preferred over other multiple comparisons methods because it does not

require equal sample sizes. The five groups of employees do not have the same sample sizes. In addition, it is more powerful than Scheffe’s method.

b. The number of pairwise comparisons for this analysis is:

( 1) 5(5 1) 20

102 2 2

k kc

c. The mean ethics scores for both Groups 4 and 5 are significantly higher than the mean ethics scores

for Groups 1, 2, and 3. There is no difference in the mean ethics scores between Group 4 and Group 5. There is no difference in the mean ethics scores among Groups 1, 2 and 3.

8.45 a. The probability of declaring at least one pair of means different when they are not is .01.

b. There are a total of ( 1) 3(3 1)

32 2

k k pair-wise comparisons. They are:

‘Under $30 thousand’ to ‘Between $30 and $60 thousand’ ‘Under $30 thousand’ to ‘Over $60 thousand’ ‘Between $30 and $60 thousand’ to ‘Over $60 thousand’


484 Chapter 8

c. Means for groups in homogeneous subsets are displayed in the table:

Subsets Income Group N 1 2

Under $30,000 379 4.60

$30,000-$60,000 392 5.08

Over $60,000 267 5.15

d. Two of the comparisons in part b will yield confidence intervals that do not contain 0. They are:

‘Under $30 thousand’ to ‘Between $30 and $60 thousand’ ‘Under $30 thousand’ to ‘Over $60 thousand’

8.46 From Exercise 8.30, we found that there were differences in the mean carats among the 6 levels of color

From Exercise 8.30, the mean carats for the 6 colors are: G 0.5808 F 0.5929 E 0.6232 D 0.6381 H 0.6734 I 0.7310 Using MINITAB, the Tukey confidence intervals are: Tukey's pairwise comparisons

Family error rate = 0.100 Individual error rate = 0.0101

Critical value = 3.66

Intervals for (column level mean) - (row level mean)

D E F G H E -0.1926 0.2225 F -0.1491 -0.1026 0.2395 0.1631 G -0.1411 -0.0964 -0.1059 0.2558 0.1812 0.1302 H -0.2350 -0.1909 -0.2007 -0.2194 0.1644 0.0904 0.0397 0.0341 I -0.3032 -0.2631 -0.2752 -0.2931 -0.2022 0.1174 0.0475 -0.0010 -0.0074 0.0871

There are only 2 intervals that do not contain 0.

The confidence interval for the difference in mean carats between colors G and I is (−0.2931, −0.0074). The confidence interval for the difference in mean carats between colors F and I is (−0.2752, −0.0010). Since 0 is not contained in these confidence intervals, there is sufficient evidence of a difference in the mean number of carats between colors G and I and between colors F and I. No other differences exist.



8.47 The mean level of trust for the "no close" technique is significantly higher than that for the "assumed close" and the "either-or" techniques. The mean level of trust for the "impending event" technique is significantly higher than that for the "either-or" technique. No other significant differences exist.

8.48 Using MINITAB, the multiple comparisons of the means is shown below:

Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons

Individual confidence level = 98.06%

Honey subtracted from:

Lower Center Upper ----+---------+---------+---------+----- DM -4.120 -2.381 -0.642 (-----*------) Control -5.890 -4.201 -2.511 (------*------) ----+---------+---------+---------+----- -5.0 -2.5 0.0 2.5

DM subtracted from:

Lower Center Upper ----+---------+---------+---------+----- Control -3.535 -1.820 -0.104 (------*------) ----+---------+---------+---------+----- -5.0 -2.5 0.0 2.5

None of the three confidence intervals contain 0:

The confidence interval for the difference in mean improvement scores between DM and Honey is (−4.120 and -0.642). Since this confidence interval is strictly below zero, this implies that the improvement scores for Honey are significantly higher than those of DM. The confidence interval for the difference in mean improvement scores between the Control group and Honey is (−5.890 and −2.511). Since this confidence interval is strictly below zero, this implies that the improvement scores for Honey are significantly higher than those of the Control Group. Compared to the Control group (giving no treatment) and DM, honey is a preferable treatment since it has significantly higher improvement scores. The state is appropriate.

8.49 a. The confidence interval for (μCAGE − μBARN) is (−.1250, −.0323). Since 0 is not contained in this interval, there is sufficient evidence of a difference in the mean thickness between cage and barn egg housing systems. Since this interval is negative, this implies that the thickness is larger for the barn egg housing system.

b. The confidence interval for (μCAGE − μFREE) is (−.1233, −.0307). Since 0 is not contained in this

interval, there is sufficient evidence of a difference in the mean thickness between cage and free range egg housing systems. Since this interval is negative, this implies that the thickness is larger for the free range egg housing system.

c. The confidence interval for (μCAGE − μORGANIC) is (−.1050, −.0123). Since 0 is not contained in this

interval, there is sufficient evidence of a difference in the mean thickness between cage and organic egg housing systems. Since this interval is negative, this implies that the thickness is larger for the organic egg housing system.

d. The confidence interval for (μBARN − μFREE) is (−.0501, .0535). Since 0 is contained in this interval,

there is insufficient evidence of a difference in the mean thickness between barn and free range egg housing systems. Since the center of the interval is greater than 0, the sample mean for barn is greater than that for free range.


486 Chapter 8

e. The confidence interval for (μBARN − μORGANIC) is (−.0318, .0718). Since 0 is contained in this interval, there is insufficient evidence of a difference in the mean thickness between barn and organic egg housing systems. Since the center of the interval is greater than 0, the sample mean for barn is greater than that for organic.

f. The confidence interval for (μFREE − μORGANIC) is (−.0335, .0701). Since 0 is contained in this

interval, there is insufficient evidence of a difference in the mean thickness between free range and organic egg housing systems. Since the center of the interval is greater than 0, the sample mean for free range is greater than that for organic.

g. We rank the housing system means as follows:

Housing System: Cage < Organic < Free < Barn We are 95% confident that the mean shell thickness for the cage housing system is significantly less than the mean thickness for the other three housing systems. There is no significant difference in the mean shell thicknesses among the barn, free range and organic housing systems.

8.50 a. There are 3 blocks used since Block df = b − 1 = 2 and 5 treatments since the treatment df = k − 1 = 4. b. There were 15 observations since the Total df = n − 1 = 14. c. H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ

d. The test statistic is F = MST

MSE = 9.109

e. The rejection region requires α = .01 in the upper tail of the F distribution with ν1 = k − 1 = 5 − 1 = 4

and ν2 = n − k − b + 1 = 15 − 5 − 3 + 1 = 8. From Table X, Appendix B, F.01 = 7.01. The rejection region is F > 7.01.

f. Since the observed value of the test statistic falls in the rejection region (F = 9.109 > 7.01), H0 is

rejected. There is sufficient evidence to indicate that at least two treatment means differ at α = .01. g. The assumptions necessary to assure the validity of the test are as follows: 1. The probability distributions of observations corresponding to all the block-treatment

combinations are normal. 2. The variances of all the probability distributions are equal. 8.51 a. Treatment df = k − 1 = 3 − 1 = 2 Block df = b − 1 = 3 − 1 = 2 Error df = n − k − b + 1 = 9 3 − 3 + 1 = 4 Total df = n − 1 = 9 − 1 = 8

SSB = 2

1

bi

i

B

k − CM from Appendix B

where CM = 2 249

9

ix

n

= 266.7778

SSB = 2 2 217 15 17

3 3 3 − 266.7778 = .8889

SSE = SS(Total) − SST − SSB = 30.2222 − 21.5555 − .8889 = 7.7778



MST = SST 21.5555

1 2k

= 10.7778 MSB =

SSB .8889

1 2b

= .4445

MSE = SSE 7.7778

1 4n k b

= 1.9445

FT = MST 10.7778

MSE 1.9445 = 5.54 FB =

MSB .4445

MSE 1.9445 = .23

The ANOVA table is:

Source df SS MS F Treatment 2 21.5555 10.7778 5.54 Block 2 .8889 .4445 .23 Error 4 7.7778 1.9445 Total 8 30.2222

b. H0: μ1 = μ2 = μ3 vs Ha: At least two treatment means differ

c. The test statistic is F = MST

MSE = 5.54

d. A Type I error would be concluding at least two treatment means differ when they do not. A Type II error would be concluding all the treatment means are the same when at least two differ. e. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 3 − 1 = 2

and ν2 = n − k − b + 1 = 9 − 3 − 3 + 1 = 4. From Table VIII, Appendix A, F.05 = 6.94. The rejection region is F > 6.94. Since the observed value of the test statistic does not fall in the rejection region (F = 5.54 6.94),

H0 is not rejected. There is insufficient evidence to indicate at least two of the treatment means differ at α = .05.

8.52 a. The ANOVA Table is as follows:

Source df SS MS F Treatment 2 12.032 6.016 50.958 Block 3 71.749 23.916 202.586 Error 6 .708 .118 Total 11 84.489

b. To determine if the treatment means differ, we test: H0: μA = μB = μC Ha: At least two treatment means differ

The test statistic is F = MST

MSE = 50.958

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 3 − 1 = 2 and ν2 = n − k − b + 1 = 12 − 3 − 4 + 1 = 6. From Table VIII, Appendix B, F.05 = 5.14. The rejection region is F > 5.14.

Since the observed value of the test statistic falls in the rejection region (F = 50.958 > 5.14), H0 is

rejected. There is sufficient evidence to indicate that the treatment means differ at α = .05.


488 Chapter 8

c. To see if the blocking was effective, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two block means differ

The test statistic is F = MSB

MSE = 202.586

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 4 − 1 = 3 and ν2 = n − k − b + 1 = 12 − 3 − 4 + 1 = 6. From Table VIII, Appendix B, F.05 = 4.76. The rejection region is F > 4.76. Since the observed value of the test statistic falls in the rejection region (F = 202.586 > 4.76), H0 is

rejected. There is sufficient evidence to indicate that blocking was effective in reducing the experimental error at α = .05.

d. From the printouts, we are given the differences in the sample means. The difference between

Treatment B and both Treatments A and C are positive (1.125 and 2.450), so Treatment B has the largest sample mean. The difference between Treatment A and C is positive (1.325), so Treatment A has a larger sample mean than Treatment C. So Treatment B has the largest sample mean, Treatment A has the next largest sample mean and Treatment C has the smallest sample mean.

From the printout, all the means are significantly different from each other. e. The assumptions necessary to assure the validity of the inferences above are: 1. The probability distributions of observations corresponding to all the block-treatment

combinations are normal. 2. The variances of all the probability distributions are equal. 8.53 a. SST = .2(500) = 100 SSB = .3(500) = 150 SSE = SS(Total) − SST − SSB = 500 − 100 − 150 = 250

MST =SST 100

1 4 1k=

− − = 33.3333 MSB =

19

150

1

SSB

−=

−b = 18.75

MSE = SSE 250 250

1 36 4 9 1 4n k b= =

− − + − − + = 10.4167

FT = 4167.10

3333.33

MSE

MST = = 3.20 FB = 4167.10

75.18

MSE

MSB = = 1.80

To determine if differences exist among the treatment means, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ The test statistic is F = 3.20. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 4 − 1 = 3

and ν2 = n − k − b + 1 = 36 − 4 − 9 + 1 = 24. From Table VIII, Appendix B, F.05 = 3.01. The rejection region is F > 3.01.


rejected. There is sufficient evidence to indicate differences among the treatment means at α = .05.



To determine if differences exist among the block means, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two block means differ The test statistic is F = 1.80. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = b − 1 = 9 − 1 = 8

and ν2 = n − b − k + 1 = 36 − 9 − 4 + 1 = 24. From Table VIII, Appendix B, F.05 = 2.36. The rejection region is F > 2.36.

Since the observed value of the test statistic does not fall in the rejection region (F = 1.80 >/ 2.36), H0 is not rejected. There is insufficient evidence to indicate differences among

the block means at α = .05. b. SST = .5(500) = 250 SSB = .2(500) = 100

SSE = SS(Total) − SST − SSB = 500 − 250 − 100 = 150

MST =SST 250

1 4 1k=

− − = 83.3333 MSB =

SS 100

1 9 1

B

b=

− − = 12.5

MSE = SSE 150

1 36 4 9 1n k b=

− − + − − + = 6.25

FT = 25.6

3333.83

MSE

MST = = 13.33 FB = MS 12.5

MSE 6.25

B = = 2

To determine if differences exist among the treatment means, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ The test statistic is F = 13.33. The rejection region is F > 3.01 (same as above). Since the observed value of the test statistic falls in the rejection region (F = 13.33 > 3.01), H0 is

rejected. There is sufficient evidence to indicate differences exist among the treatment means at α = .05. To determine if differences exist among the block means, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two block means differ The test statistic is F = 2.00. The rejection region is F > 2.36 (same as above). Since the observed value of the test statistic does not fall in the rejection region (F = 2.00 >/ 2.36), H0 is not rejected. There is insufficient evidence to indicate differences exist

among the block means at α = .05.


490 Chapter 8

c. SST = .2(500) = 100 SSB = .5(500) = 250


MST = SST 100

1 4 1k=

− − = 33.3333 MSB =

SS 250

1 9 1

B

b=

− − = 31.25

MSE = SSE 150

1 36 4 9 1n k b=

− − + − − + = 6.25

FT = 25.6

3333.33

MSE

MST = = 5.33 FB = MS 31.25

MSE 6.25

B = = 5.00

To determine if differences exist among the treatment means, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ The test statistic is F = 5.33. The rejection region is F > 3.01 (same as above). Since the observed value of the test statistic falls in the rejection region (F = 5.33 > 3.01), H0 is

rejected. There is sufficient evidence to indicate differences exist among the treatment means at α = .05. To determine if differences exist among the block means, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two block means differ The test statistic is F = 5.00. The rejection region is F > 2.36 (same as above). Since the observed value of the test statistic falls in the rejection region (F = 5.00 > 2.36), H0 is

rejected. There is sufficient evidence to indicate differences exist among the block means at α = .05. d. SST = .4(500) = 200 SSB = .4(500) = 200


MST = SST 200

1 4 1k=

− − = 66.6667 MSB =

19

200

1

SSB

−=

−b = 25

MSE = SSE 100

1 36 4 9 1n k b=

− − + − − + = 4.1667

FT = 1667.4

6667.66

MSE

MST = = 16.0 FB = 1667.4

25

MSE

MSB = = 6.00

To determine if differences exist among the treatment means, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ The test statistic is F = 16.0. The rejection region is F > 3.01 (same as above).



Since the observed value of the test statistic falls in the rejection region (F = 16.0 > 3.01), H0 is rejected. There is sufficient evidence to indicate differences among the treatment means at α = .05.

To determine if differences exist among the block means, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two block means differ The test statistic is F = 6.00. The rejection region is F > 2.36 (same as above). Since the observed value of the test statistic falls in the rejection region (F = 6.00 > 2.36), H0 is

rejected. There is sufficient evidence to indicate differences exist among the block means at α = .05. e. SST = .2(500) = 100 SSB = .2(500) = 100


MST = SST 100

1 4 1k=

− − = 33.3333 MSB =

19

100

1

SSB

−=

−b = 12.5

MSE = SSE 300

1 36 4 9 1n k b=

− − + − − + = 12.5

FT = 5.12

3333.33

MSE

MST = = 2.67 FB = 5.12

5.12

MSE

MSB = = 1.00

To determine if differences exist among the treatment means, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ The test statistic is F = 2.67. The rejection region is F > 3.01 (same as above). Since the observed value of the test statistic does not fall in the rejection region (F = 2.67 >/ 3.01), H0 is not rejected. There is insufficient evidence to indicate differences exist

among the treatment means at α = .05. To determine if differences exist among the block means, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two block means differ The test statistic is F = 1.00. The rejection region is F > 2.36 (same as above). Since the observed value of the test statistic does not fall in the rejection region (F = 1.00 >/ 2.36), H0 is not rejected. There is insufficient evidence to indicate differences among

the block means at α = .05.


492 Chapter 8

8.54 a. This experimental design is a randomized block design because in part B, the same subjects provided WTP amounts for insuring both a sculpture and a painting. Each subject had 2 responses.

b. The dependent (response) variable is the WTP amount. The treatments are the two scenarios

(sculpture and painting). The blocks are the 84 subjects. c. To determine if there is a difference in the mean WTM amounts between sculptures and paintings,

we test: H0: μ1 = μ2

Ha: μ1 ≠ μ2 8.55 a. A randomized block design should be used to analyze the data because the same employees were

measured at all three time periods. Thus, the blocks are the employees and the treatments are the three time periods.

b. There is still enough information in the table to make a conclusion because the p-values are given. c. To determine if there are differences in the mean competence levels among the three time periods,

we test:

H0: μ1 = μ2 = μ3 Ha: At least two treatment means differ d. The p-value is p = 0.001. At a significance level > .001, we reject H0. There is sufficient evidence to

conclude that there is a difference in the mean competence levels among the three time periods for any value of α > 0.001.

e. With 90% confidence, the mean competence before the training is significantly less than the mean

competence 2-days after and 2-months after. There is no significant difference in the mean competence between 2-days after and 2-months after.

8.56 a. This is a randomized block design. The blocks are the 12 plots of land. The treatments are the three

methods used on the shrubs: fire, clipping, and control. The response variable is the mean number of flowers produced. The experimental units are the 36 shrubs.



b. Plot c. To determine if there is a difference in the mean number of flowers produced among the three

treatments, we test: H0: μ1 = μ2 = μ3 Ha: The mean number of flowers produced differ for at least two of the methods. The test statistic is F = 5.42 and p = .009. We can reject the null hypothesis at the α > .009 level of significance. At least two of the methods differ with respect to mean number of

flowers produced by pawpaws. d. The means of Clipping and Burning do not differ significantly. The means of treatment Burning and

treatment Clipping exceeds that of the Control. 8.57 a. The time of the year (month) could affect the number of rigs running, so a

randomized complete block design was used to “block” out the month to month variation.

b. There are 3 treatments in this experiment. They are the three states – California, Utah, and Alaska.

c. There are 3 blocks in this experiment – the three months selected: November 2000, October 2001, and November 2001.

d. To determine if there is a difference in the mean number of rigs running among the three states, we

test:

H0: μ1 = μ2 = μ3

e. From the printout, the test statistic is F = 38.07 and the p-value is p = 0.002. Since the p-value is so small, we would reject H0 for any value of α > .002. There is sufficient evidence to indicate a difference in the mean number of oil rigs running among the three states.

f. From the SPSS printout, there is no significant difference in the mean number of oil rigs running in

Alaska and Utah. However, both of these states have a significantly smaller number of rigs running than does California. Thus, California has the largest mean number of oil rigs running.


494 Chapter 8

8.58 Using SAS, the ANOVA Table is: The ANOVA Procedure Dependent Variable: temp Sum of Source DF Squares Mean Square F Value Pr > F Model 11 18.53700000 1.68518182 0.52 0.8634 Error 18 58.03800000 3.22433333 Corrected Total 29 76.57500000 R-Square Coeff Var Root MSE temp Mean 0.242076 1.885189 1.795643 95.25000 Source DF Anova SS Mean Square F Value Pr > F STUDENT 9 18.41500000 2.04611111 0.63 0.7537 PLANT 2 0.12200000 0.06100000 0.02 0.9813 To determine if there are differences among the mean temperatures among the three treatments, we test:

H0: μ1 = μ2 = μ3 Ha: At least two treatment means differ The test statistic is F = 0.02. The associated p-value is p = .9813. Since the p-value is very large, there is

no evidence of a difference in mean temperature among the three treatments. Since there is no difference, we do not need to compare the means. It appears that the presence of plants or pictures of plants does not reduce stress.

8.59 Using MINITAB, the ANOVA table is:

Two-way ANOVA: Rate versus Week, Day

Analysis of Variance for Rate Source DF SS MS F P Week 8 575.2 71.9 6.10 0.000 Day 4 94.2 23.5 2.00 0.118 Error 32 376.9 11.8 Total 44 1046.4

Individual 95% CI Day Mean -+---------+---------+---------+---------+ 1 8.8 (--------*---------) 2 4.6 (--------*---------) 3 5.8 (--------*--------) 4 5.4 (--------*---------) 5 6.4 (---------*--------) -+---------+---------+---------+---------+ 2.5 5.0 7.5 10.0 12.5

To determine if there is a difference in mean rate of absenteeism among the 5 days of the week, we test:

H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ

The test statistic is F = 2.00.

Since no α was given, we will select α = .05. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k – 1 = 5 – 1 = 4 and ν2 = n – k – b + 1 = 45 – 5 – 9 + 1 = 32. From Table VIII, Appendix B, F.05, 4, 32 ≈ 2.69. The rejection region is F > 2.69.



Since the observed value of the test statistic does not fall in the rejection region (F = 2.00 >/ 2.69), H0 is not rejected. There is insufficient evidence to indicate a difference in mean rate of absenteeism among the 5 days of the week at α = .05.

To test for the effectiveness of blocking, we test: H0: All block means are the same Ha: At least two block means differ

The test statistic is F = 6.10.

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = b – 1 = 9 – 1 = 8 and ν2 = n – k – b + 1 = 45 – 5 – 9 + 1 = 32. From Table VIII, Appendix B, F.05, 8, 32 ≈ 2.27. The rejection region is F > 2.27. Since the observed value of the test statistic falls in the rejection region (F = 6.10 > 2.27), H0 is rejected. There is sufficient evidence to indicate blocking was effective at α = .05.

8.60 a. The experimenters expected there to be much variation in the number of participants from week to

week (more participants at the beginning and fewer as time goes on). Thus, by blocking on weeks, this extraneous source of variation can be controlled.

b. df(Week) = b − 1 = 6 − 1 = 5

MS(Prompt) = SST 1185.0

df 4 = 296.25

F(Prompt) = MST 296.25

MSE 7.43 = 39.87

The ANOVA table is:

Source df SS MS F p Prompt 4 1185.0 296.25 39.87 0.0001 Week 5 386.4 77.28 10.40 0.0001 Error 20 148.6 7.43 Total 29 1720.0

c. To determine if a difference exists in the mean number of walkers per week among the five walker

groups, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ where μi represents the mean number of walkers in group i. The test statistic is F = 39.87. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 5 − 1 = 4


Since the observed value of the test statistic falls in the rejection region (F = 39.87 > 2.69), H0 is rejected. There is sufficient evidence to indicate differences exist among the mean number of walkers per week among the 5 walker groups at α = .05.


496 Chapter 8

d. The following conclusions are drawn: The mean number of walkers per week in the "Frequent/High" group is significantly higher than the

mean number of walkers per week in the "Infrequent/Low" group, the "Infrequent/High" group, and the "Control" group. The mean number of walkers per week in the "Frequent/Low" group is significantly higher than the mean number of walkers per week in the "Infrequent/Low" group, the "Infrequent/High" group, and the "Control" group. The mean number of walkers per week in the "Infrequent/Low" group is significantly higher than the mean number of walkers per week in the "Control" group. The mean number of walkers per week in the "Infrequent/High" group is significantly higher than the mean number of walkers per week in the "Control group.

e. In order for the above inferences to be valid, the following assumptions must hold: 1) The probability distributions of observations corresponding to all block-treatment conditions

are normal. 2) The variances of all the probability distributions are equal. 8.61 Using MINITAB, the ANOVA table is: Two-way ANOVA: Corrosion versus Time, System Source DF SS MS F P Time 2 63.1050 31.5525 337.06 0.000 System 3 9.5833 3.1944 34.12 0.000 Error 6 0.5617 0.0936 Total 11 73.2500 S = 0.3060 R-Sq = 99.23% R-Sq(adj) = 98.59% Individual 95% CIs For Mean Based on Pooled StDev System Mean ------+---------+---------+---------+--- 1 9.0667 (----*-----) 2 9.7333 (-----*----) 3 11.0667 (----*-----) 4 8.7333 (----*-----) ------+---------+---------+---------+--- 8.80 9.60 10.40 11.20

To determine if there is a difference in mean corrosion rates among the 4 systems, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ The test statistic is F = 34.12. Since no α level was given, we will select α = .05. The rejection region requires α = .05 in the upper tail

of the F distribution with ν1 = k – 1 = 4 – 1 = 3 and ν2 = n – k – b + 1 = 12 – 4 – 3 + 1 = 6. From Table VIII, Appendix B, F.05, 3, 6 = 4.76. The rejection region is F > 4.76.



Since the observed value of the test statistic falls in the rejection region (F = 34.12 > 4.76), H0 is rejected. There is sufficient evidence to indicate a difference in mean corrosion rates among the 4 systems at α = .05.

Using SAS, Tukey’s multiple comparison results are: Tukey's Studentized Range (HSD) Test for CORROSION NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ. Alpha 0.05 Error Degrees of Freedom 6 Error Mean Square 0.093611 Critical Value of Studentized Range 4.89559 Minimum Significant Difference 0.8648 Means with the same letter are not significantly different. Tukey Grouping Mean N SYSTEM A 11.0667 3 3 B 9.7333 3 2 B C B 9.0667 3 1 C C 8.7333 3 4 The mean corrosion rate for system 3 is significantly larger than all of the other mean corrosion rates. The

mean corrosion rate of system 2 is significantly larger than the mean for system 4. If we want the system (epoxy coating) with the lowest corrosion rate, we would pick either system 1 or system 4. There is no significant difference between these two groups and they are in the lowest corrosion rate group.

8.62 a. There are two factors. b. No, we cannot tell whether the factors are qualitative or quantitative. c. Yes. There are four levels of factor A and three levels of factor B. d. A treatment would consist of a combination of one level of factor A and one level of factor B. There

are a total of 4 × 3 = 12 treatments. e. One problem with only one replicate is there are no degrees of freedom for error. This is overcome

by having at least two replicates.


498 Chapter 8

8.63 a. The ANOVA table is:

Source df SS MS F A 2 .8 .4000 3.69B 3 5.3 1.7667 16.31AB 6 9.6 1.6000 14.77Error 12 1.3 .1083Total 23 17.0

df for A is a − 1 = 3 − 1 = 2

df for B = b − 1 = 4 − 1 = 3 df for AB is (a − 1)(b − 1) = 2(3) = 6

df for Error is n − ab = 24 − 3(4) = 12 df for Total is n − 1 = 24 − 1 = 23

SSE = SS(Total) − SSA − SSB − SSAB = 17.0 − .8 − 5.3 − 9.6 = 1.3

MSA = SS .8

1 3 1

A

a

= .40 MSB =

SS 5.3

1 4 1

B

b

= 1.7667

MSAB = SS 9.6

( 1)( 1) (3 1)(4 1)

AB

a b

= 1.60

MSE = SSE 1.3

24 3(4)n ab

= .1083

FA = MS .4000

= MSE .1083

A = 3.69 FB =

MS 1.7667 =

MSE .1083

B = 16.31

FAB = MS 1.6000

= MSE .1083

AB = 14.77

b. Sum of Squares for Treatment = SSA + SSB + SSAB = .8 = 5.3 + 2.6 = 15.7

MST = SST 15.7

1 3(4) 1ab

= 1.4273 FT = MST 1.4273

= MSE .1083

= 13.18

To determine if the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ12 Ha: At least two treatments means differ The test statistic is F = 13.18. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 3(4) −

1 = 11 and ν2 = n − ab = 24 − 3(4) = 12. From Table VIII, Appendix B, F.05 ≈ 2.75. The rejection region is F > 2.75.


rejected. There is sufficient evidence to indicate the treatment means differ at α = .05. c. Yes. We need to partition the Treatment Sum of Squares into the Main Effects and Interaction Sum

of Squares. Then we test whether factors A and B interact. Depending on the conclusion of the test for interaction, we either test for main effects or compare the treatment means.



d. Two factors are said to interact if the effect of one factor on the dependent variable is not the same at different levels of the second factor. If the factors interact, then tests for main effects are not necessary. We need to compare the treatment means for one factor at each level of the second.

e. To determine if the factors interact, we test: H0: Factors A and B do not interact to affect the response mean Ha: Factors A and B do interact to affect the response mean

The test statistic is F = MS

MSE

AB = 14.77

The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) =

(3 − 1)(4 − 1) = 6 and ν2 = n − ab = 24 − 3(4) = 12. From Table VIII, Appendix B, F.05 = 3.00. The rejection region is F > 3.00.


rejected. There is sufficient evidence to indicate the two factors interact to affect the response mean at α = .05.

f. No. Testing for main effects is not warranted because interaction is present. Instead, we compare the

treatment means of one factor at each level of the second factor. 8.64 a. Factor A has 3 + 1 = 4 levels and factor B has 1 + 1 = 2 levels. b. There are a total of 23 + 1 = 24 observations and 4 × 2 = 8 treatments. Therefore, there were 24/8 =

3 observations for each treatment. c. AB df = (a − 1)(b − 1) = (4 − 1)(2 − 1) = 3 Error df = n − ab = 24 − 4(2) = 16

MSA = SS

1

A

a SSA = (a − 1)MSA = (4 − 1)(.75) = 2.25

MSB = SS .95

1 2 1

B

b

= .95

MSAB = SS

( 1)( 1)

AB

a b SSAB = (a − 1)(b − 1)MSAB = (4 − 1)(2 − 1)(.30) = .9

SSE = SS(Total) − SSA − SSB − SSAB = 6.5 − 2.25 − .95 − .9 = 2.4

MSE = SSE 2.4

24 - 4(2)n ab

= .15

SST = SSA + SSB + SSAB = 2.25 + .95 + .90 = 4.1

Treatment df = ab − 1 = 4(2) − 1 = 7

MST = SST 4.1

1 7ab

= .5857 FT = MST .5857

MSE .15

= 3.90

FA = MSA .75

= = 5.00MSE .15

FB = MSB .95

= = 6.33MSE .15

FAB = MSAB .30

= = 2.00MSE .15


500 Chapter 8

The ANOVA table is:

Source df SS MS FTreatments 7 4.1 .59 3.90A 3 2.25 .75 5.00 B 1 .95 .95 6.33 AB 3 .90 .30 2.00 Error 16 2.40 .15 Total 23 6.50

d. To determine whether the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ8 Ha: At least two treatment means differ


MSE = 3.90

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = ab − 1 = 4(2) −

1 = 7 and ν2 = n − ab = 24 − 4(2) = 16. From Table VII, Appendix B, F.10 = 2.13. The rejection region is F > 2.13.


rejected. There is sufficient evidence to indicate the treatment means differ at α = .10. e. To determine if the factors interact, we test: H0: Factors A and B do not interact to affect the response mean Ha: Factors A and B do interact to affect the response mean The test statistic is F = 2.00. The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) =

(4 − 1)(2 − 1) = 3 and ν2 = n − ab = 24 − 4(2) = 16. From Table VII, Appendix B, F.10 = 2.46. The rejection region is F > 2.46.

Since the observed value of the test statistic does not fall in the rejection region (F = 2.00 2.46),

H0 is not rejected. There is insufficient evidence to indicate factors A and B interact at α = .10. To determine if the four means of factor A differ, we test: H0: There is no difference in the four means of factor A Ha: At least two of the factor A means differ The test statistic is F = 5.00. The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = a − 1 = 4 − 1 =

3 and ν2 = n − ab = 24 - 4(2) = 16. From Table VII, Appendix B, F.10 = 2.46. The rejection region is F > 2.46.


rejected. There is sufficient evidence to indicate at least two of the four means of factor A differ at α = .10.



To determine if the 2 means of factor B differ, we test: H0: There is no difference in the two means of factor B Ha: At least two of the factor B means differ The test statistic is F = 6.33. The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = b − 1 = 2 − 1 = 1

and ν2 = n − ab = 24 − 4(2) = 16. From Table VII, Appendix B, F.10 = 3.05. The rejection region is F > 3.05.


rejected. There is sufficient evidence to indicate the two means of factor B differ at α = .10. All of the tests performed are warranted because interaction was not significant.

8.65 a. The treatments for this experiment consist of a level for factor A and a level for factor B. There are six treatments—(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), and (2, 3) where the first number represents the level of factor A and the second number represents the level of factor B.

The treatment means appear to be different

because the sample means are quite different. The factors appear to interact because the lines are not parallel.

b. SST = SSA + SSB + SSAB = 4.4408 + 4.1267 + 18.0667 = 26.5742

MST = SST

1ab =

26.5742

2(3) 1 = 5.315 FT =

MST

MSE =

5.315

.246 = 21.62

To determine whether the treatment means differ, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 Ha: At least two treatment means differ


MSE = 21.62


1 = 5 and ν2 = n − ab = 12 − 2(3) = 6. From Table VIII, Appendix B, F.05 = 4.39. The rejection region is F > 4.39.


rejected. There is sufficient evidence to indicate that the treatment means differ at α = .05. This supports the plot in a.


502 Chapter 8

c. Yes. Since there are differences among the treatment means, we test for interaction. To determine whether the factors A and B interact, we test:

H0: Factors A and B do not interact to affect the mean response Ha: Factors A and B do interact to affect the mean response

The test statistic is F = MSAB 9.0033

= MSE .24583

= 36.62

The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (2 − 1)(3 − 1) = 2 and ν2 = n − ab = 12 − 2(3) = 6. From Table VIII, Appendix

B, F.05 = 5.14. The rejection region is F > 5.14. Since the observed value of the test statistic falls in the rejection region (F = 36.62 > 5.14), H0 is

rejected. There is sufficient evidence to indicate that factors A and B interact to affect the response mean at α = .05.

d. No. Because interaction is present, the tests for main effects are not warranted. e. The results of the tests in parts b and c support the visual interpretation in part a. 8.66 a. The treatments are the combinations of the levels of factor A and the levels of factor B. There are 2 ×

2 = 4 treatments. The treatment means are:

11

1129.6 35.2

2 2

xx

= 32.4

1212

47.3 42.1

2 2

xx

21

2112.9 17.6

2 2

xx

= 15.25

2222

28.4 22.7

2 2

xx

The factors do not appear to interact—the lines

are almost parallel. The treatment means do appear to differ because the sample means range from 15.25 to 44.7.



b. CM = 2 2235.8

8

ix

n

SS(Total) = 2x − CM = 7922.92 − 6950.205 = 972.715

SSA =

2 2 2154.2 81.6CM=

2(2) 2(2)

iA

br

= 7609.05 − 6950.205 = 658.845

SSB = 2 2 295.3 140.5

CM=2(2) 2(2)

iB

ar

= 7205.585 − 6950.205 = 255.38

SSAB = 2ijAB

r

− SSA − SSB − CM =

2 2 2 264.8 89.4 30.5 51.1

2 4 2 2

− 658.845 − 255.38 − 6950.205 = 7866.43 − 7864.43 = 2 SSE = SS(Total) − SSA − SSB − SSAB = 972.715 − 658.845 − 255.38 − 2 = 56.49 A df = a − 1 = 2 − 1 = 1 B df = b − 1 = 2 − 1 = 1 AB df = (a − 1)(b − 1) = (2 − 1)(2 − 1) = 1 Error df = n − ab = 8 − 2(2) = 4 Total df = n − 1 = 8 − 1 = 7

MSA = SSA 658.845

1 1a

= 658.845 MSB = SSB 255.38

1 1b

= 255.38

MSAB = SSAB 2

( 1)( 1) 1a b

= 2 MSE = SSE 56.49

= n - ab 4

= 14.1225

FA = MSA 658.845

= = 46.65MSE 14.1225

FB = MSB 255.38

18.08MSE 14.1225

FAB = MSAB 2

.14MSE 14.1225

The ANOVA table is:

Source df SS MS F A 1 658.845 658.845 46.65 B 1 255.380 255.380 18.08 AB 1 2.000 2.000 .14 Error 4 56.490 14.1225 Total 7 972.715

c. SST = SSA + SSB + SSAB = 658.845 + 255.380 + 2.000 = 916.225

df = ab − 1 = 2(2) − 1 = 3

MST = SST 916.225

= = 305.408 1 3ab

FT = MST 305.408

= MSE 14.1225

= 21.63


504 Chapter 8

To determine whether the treatment means differ, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two of the treatment means differ The test statistic is F = 21.63. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 2(2) −



rejected. There is sufficient evidence to indicate the treatment means differ at α = .05. This agrees with the conclusion in part a. d. Since there are differences among the treatment means, we test for the presence of interaction: H0: Factors A and B do not interact to affect the response means Ha: Factors A and B do interact to affect the response means The test statistic is F = .14. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) =


Since the observed value of the test statistic does not fall in the rejection region (F = .14 7.71), H0

is not rejected. There is insufficient evidence to indicate the factors interact at α = .05. e. Since the interaction was not significant, we test for main effects. To determine whether the two means of factor A differ, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2 The test statistic is F = 46.65. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = a − 1 = 2 − 1 = 1

and ν2 = n − ab = 8 − 2(2) = 4. From Table VIII, Appendix B, F.05 = 7.71. The rejection region is F > 7.71.


rejected. There is sufficient evidence to indicate the two means of factor A differ at α = .05. To determine whether the two means of factor B differ, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2 The test statistic is F = 18.08.



The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = b − 1 = 2 − 1 = 1 and ν2 = n − ab = 8 − 2(2) = 4. From TableVIII, Appendix B, F.05 = 7.71. The rejection region is

F > 7.71. Since the observed value of the test statistic falls in the rejection region (F = 18.08 > 7.71), H0 is

rejected. There is sufficient evidence to indicate the two means of factor B differ at α = .05. f. The results of all the tests agree with those in part a. g. Since no interaction is present, but the means of both factors A and B differ, we compare the two

means of factor A and compare the two means of factor B. Since there are only two means to compare for each factor, the higher population mean corresponds to the higher sample mean.

Factor A: 1

129.6 35.2 47.3 42.1

2(2)

xx

br

= 38.55

2

212.9 17.6 28.4 22.7

2(2)

xx

br

= 20.4

The mean for level 1 of factor A is significantly higher than the mean for level 2.

Factor B: 1

129.6 35.2 12.9 17.6

2(2)

xx

ar

= 23.825

2

247.3 42.1 28.4 22.7

2(2)

xx

ar

= 35.125

The mean for level 2 of factor B is significantly higher than the mean for level 1. 8.67 a. SSA = .2(1000) = 200, SSB = .1(1000) = 100, SSAB = .1(1000) = 100 SSE = SS(Total) − SSA − SSB − SSAB = 1000 − 200 − 100 − 100 = 600 SST = SSA + SSB + SSAB = 200 + 100 + 100 = 400

MSA = SS 200

100 1 3 1

A

a

MSB =

SS 100 50

1 3 1

B

b

MSAB = SS 100

25( 1)( 1) (3 1)(3 1)

AB

a b

MSE = SSE 600

27 3(3)n ab

= 33.333 MST = SST 400

1 3(3) 1ab

= 50

FA = MS 100

= = 3.00MSE 33.333

A FB =

MS 50= = 1.50

MSE 33.333

B

FAB = MS 25

= = .75MSE 33.333

AB FT =

MST 50= = 1.50

MSE 33.333

Source df SS MS F A 2 200 100 3.00 B 2 100 50 1.50 AB 4 100 25 .75 Error 18 600 33.333 Total 26 1000


506 Chapter 8

To determine whether the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two treatment means differ


MSE = 1.50

Suppose α = .05. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 3(3) − 1 = 8 and ν2 = n − ab = 27 − 3(3) = 18. From Table VIII, Appendix B, F.05 =

2.51. The rejection region is F > 2.51. Since the observed value of the test statistic does not fall in the rejection region (F = 1.50 . 2.51), H0

is not rejected. There is insufficient evidence to indicate the treatment means differ at α = .05. Since there are no treatment mean differences, we have nothing more to do.

b. SSA = .1(1000) = 100, SSB = .1(1000) = 100, SSAB = .5(1000) = 500 SSE = SS(Total) − SSA − SSB − SSAB = 1000 − 100 − 100 − 500 = 300 SST = SSA + SSB + SSAB = 100 + 100 + 500 = 700

MSA = SS 100

50 1 3 1

A

a

MSB =

SS 100 50

1 3 1

B

b

MSAB = SS 500

125( 1)( 1) (3 1)(3 1)

AB

a b

MSE = SSE 300

27 3(3)n ab

= 16.667 MST = SST 700

1 9 1ab

= 87.5

FA = MS 50

= = 3.00MSE 16.667

A FB =

MS 50= = 3.00

MSE 16.667

B

FAB = MS 125

= = 7.50MSE 16.667

AB FT =

MST 87.5= = 5.25

MSE 16.667

Source df SS MS F A 2 100 50 3.00B 2 100 50 3.00AB 4 500 125 7.50Error 18 300 16.667 Total 26 1000

To determine if the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two treatment means differ


MSE = 5.25



The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 3(3) − 1 = 8 and ν2 = n − ab = 27 - 3(3) = 18. From Table VIII, Appendix B, F.05 = 2.51. The

rejection region is F > 2.51. Since the observed value of the test statistic falls in the rejection region (F = 5.25 > 2.51), H0 is rejected.

There is sufficient evidence to indicate the treatment means differ at α = .05. Since the treatment means differ, we next test for interaction between factors A and B. To determine

if factors A and B interact, we test: H0: Factors A and B do not interact to affect the mean response Ha: Factors A and B do interact to affect the mean response


MSE

AB = 7.50




rejected. There is sufficient evidence to indicate the factors A and B interact at α = .05. Since interaction is present, no tests for main effects are necessary.

c. SSA = .4(1000) = 400, SSB = .1(1000) = 100, SSAB = .2(1000) = 200

SSE = SS(Total) − SSA − SSB − SSAB = 1000 − 400 − 100 − 200 = 300

SST = SSA + SSB + SSAB = 400 + 100 + 200 = 700

MSA = SS 400

50 1 3 1

A

a

MSB =

SS 100 50

1 3 1

B

b

MSAB = SS 200

50( 1)( 1) (3 1)(3 1)

AB

a b

MSE = SSE 300

16.667 27 3(3)n ab

MST = SST 700

87.5 1 3(3) 1ab

FA = MS 200

= = 12.00MSE 16.667

A FB =

MS 50= = 3.00

MSE 16.667

B

FAB = MS 50

= = 3.00MSE 16.667

AB FT =

MST 87.5= = 5.25

MSE 16.667




508 Chapter 8


MSE= 5.25


1 = 8 and ν2 = n − ab = 27 - 3(3) = 18. From Table VIII, Appendix B, F.05 = 2.51. The rejection region is F > 2.51.


rejected. There is sufficient evidence to indicate the treatment means differ at α = .05. Since the treatment means differ, we next test for interaction between factors A and B. To determine


The test statistic is F = MSAB

MSE = 3.00





d. SSA = .4(1000) = 400, SSB = .4(1000) = 400, SSAB = .1(1000) = 100

SSE = SS(Total) − SSA − SSB − SSAB = 1000 − 400 − 400 − 100 = 100

SST = SSA + SSB + SSAB = 400 + 400 + 100 = 900

MSA = SS 400

200 1 3 1

A

a

MSB =

SS 400 200

1 3 1

B

b

MSAB = SS 100

25( 1)( 1) (3 1)(3 1)

AB

a b

MSE = SSE 100

5.556 27 3(3)n ab

MST = 900

112.5 1 3(3) 1

SST

ab

FA = MS 200

= = 36.00MSE 5.556

A FB =

MS 200= = 36.00

MSE 5.556

B

FAB = MS 25

= = 4.50MSE 5.556

AB FT =

MST 112.5= = 20.25

MSE 5.556






MSE = 20.25




rejected. There is sufficient evidence to indicate the treatment means differ at α = .05. Since the treatment means differ, we next test for interaction between factors A and B. To determine



MSE

AB = 4.50





8.68 a. There are a total of 2 × 4 = 8 treatments.

b. The interaction between temperature and type was significant. This means that the effect of type of yeast on the mean autolysis yield depends on the level of temperature.

c. To determine if the main effect of type of yeast is significant, we test:

H0: μBa = μBr Ha: μBa ≠ μBr

To determine if the main effect of temperature is significant, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least one mean differs

d. The tests for the main effects should not be run until after the test for interaction is conducted. If interaction is significant, then these interaction effects could cover up the main effects. Thus, the main effect tests would not be informative.

If the test for interaction is not significant, then the main effect tests could be run.


510 Chapter 8

e. Baker’s yeast: The mean yield for temperature 54o is significantly lower than the mean yields for the other 3

temperatures. There is no difference in the mean yields for the temperatures 45o, 48o and 51o. Brewer’s yeast: The mean yield for temperature 54o is significantly lower than the mean yields for theother 3

temperatures. There is no difference in the mean yields for the temperatures 45o, 48o and 51o. 8.69 a. This is a complete 6 × 6 factorial design.

b. There are 2 factors – Coagulant and pH level. There are 6 levels of coagulant: 5, 10, 20, 50, 100, and 200 mg / liter. There are 6 levels of pH: 4.0, 5.0, 6.0, 7.0, 8.0, and 9.0.

There are 6 x 6 = 36 treatments. In the pairs, let the coagulant level be the first number and pH level the second. The 36 treatments are:

(5, 4.0) (5, 5.0) (5, 6.0) (5, 7.0) (5, 8.0) (5, 9.0) (10, 4.0) (10, 5.0) (10, 6.0) (10, 7.0) (10, 8.0) (10, 9.0) (20, 4.0) (20, 5.0) (20, 6.0) (20, 7.0) (20, 8.0) (20, 9.0) (50, 4.0) (50, 5.0) (50, 6.0) (50, 7.0) (50, 8.0) (50, 9.0) (100, 4.0) (100, 5.0) (100, 6.0) (100, 7.0) (100, 8.0) (100, 9.0) (200, 4.0) (200, 5.0) (200, 6.0) (200, 7.0) (200, 8.0) (200, 9.0)

8.70 a. There are a total of 2 x 4 = 8 treatments for this study. They include all combinations of Insomnia

status and Education level. The 8 treatments are: Normal sleeper, College Graduate Normal sleeper, Some college Normal sleeper, High School graduate Normal sleeper, High School Dropout Chronic insomnia, College Graduate Chronic insomnia, Some college Chronic insomnia, High School graduate Chronic insomnia, High School Dropout



b. Since Insomnia and Education did not interact, this means that the effect of Insomnia on the Fatigue Severity Scale does not depend on the level of Education. In a graph, the lines will be parallel. A possible graph of this situation is:

Insomnia

FSS

2.01.81.61.41.21.0

11

10

9

8

7

6

5

4

3

2

Education

34

12

Scatterplot of FSS vs Insomnia

c. This means that the researchers can infer that the population mean FSS for people who had insomnia is higher than the population mean FSS for normal sleepers.

d. This means that at least one level of education had a mean FSS score that differed from the rest. There may be more than one difference, but there is at least one.

e. With 95% confidence, we can conclude that the mean FSS value for high school dropouts is significantly higher than the mean FSS values for the 3 other education levels. There is no significant difference in the mean FSS values for college graduates, those with some college and high school graduates.

8.71 a. This is a complete 2 × 2 factorial design. The 2 factors are Color and Question. There are two levels of color – Blue and Red. There are two levels of question – difficult and simple. The 4 treatments are: blue/difficult, blue/simple, red/difficult, red/simple.

b. There is a significant interaction between color and question. The effect of color on the mean score is different at each level of question.

c. Using MINITAB, the graph is:

Simple Difficult

80

70

60

50

40

Question

Mea

n S

core

Blue

Red

Since the lines are not parallel, it indicates that there is significant interaction between color and question.


512 Chapter 8

8.72 a. The degrees of freedom for “Type of message retrieval system” is a − 1 = 2 − 1 = 1. The degrees of freedom for “Pricing option” is b − 1 = 2 − 1 = 1. The degrees of freedom for the interaction of Type of message retrieval system and Pricing option is (a − 1)(b – 1) = (2 − 1)(2 − 1) = 1. The degrees of freedom for error is n − ab = 120 − 2(2) = 116.

Source Df SS MS F Type of message retrieval system 1 - - 2.001 Pricing Option 1 - - 5.019 Type of system × pricing option 1 - - 4.986 Error 116 - - Total 119

b. To determine if “Type of system” and “Pricing option” interact to affect the mean willingness to buy,

we test: H0: “Type of system” and “Pricing option” do not interact Ha: “Type of system” and “Pricing option” interact

c. The test statistic is F = MSAB

MSE = 4.986

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = (a − 1)(b − 1) = (2 − 1)(2 − 1) = 1 and ν2 = n − ab = 120 − 2(2) = 116. From Table VIII, Appendix

B, F.05 ≈ 3.92. The rejection region is F > 3.92. Since the observed value of the test statistic falls in the rejection region (F = 4.986 > 3.92), H0 is

rejected. There is sufficient evidence to indicate “Type of system” and “Pricing option” interact to affect the mean willingness to buy at α = .05.

d. No. Since the test in part c indicated that interaction between “Type of system” and “Pricing option”

is present, we should not test for the main effects. Instead, we should proceed directly to a multiple comparison procedure to compare selected treatment means. If interaction is present, it can cover up the main effects.

8.73 a. There are two factors for this experiment, housing system and weight class. There are a total of 4 ×

2 = 8 treatments. The treatments are: Cage, M Cage, L Free, M Free, L Barn, M Barn, L Organic, M Organic, L



b. Using SAS, the results are:

The GLM Procedure

Dependent Variable: OVERRUN

Sum of Source DF Squares Mean Square F Value Pr > F

Model 7 11364.52381 1623.50340 14.93 <.0001

Error 20 2175.33333 108.76667

Corrected Total 27 13539.85714

R-Square Coeff Var Root MSE OVERRUN Mean

0.839339 2.061383 10.42913 505.9286

Source DF Type I SS Mean Square F Value Pr > F

HOUSING 3 10787.79048 3595.93016 33.06 <.0001 WTCLASS 1 329.14286 329.14286 3.03 0.0973 HOUSING*WTCLASS 3 247.59048 82.53016 0.76 0.5303

Source DF Type III SS Mean Square F Value Pr > F

HOUSING 3 10787.79048 3595.93016 33.06 <.0001 WTCLASS 1 320.47407 320.47407 2.95 0.1015 HOUSING*WTCLASS 3 247.59048 82.53016 0.76 0.5303

c. To determine if interaction between housing system and weight class exists, we test:

H0: Housing system and weight class do not interact Ha: Housing system and weight class do interact

The test statistic is F = 0.76 and the p-value is p = .5303. Since the p-value is not less than α (p = .5303 .05), H0 is not rejected. There is insufficient evidence to indicate that housing system and weight class interact at α = .05.

d. To determine if there is a difference in mean whipping capacity among the 4 housing systems, we

test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two means differ

The test statistic is F = 33.06 and the p-value is less than .0001. Since the p-value is less than α (p < .0001 < .05), H0 is rejected. There is sufficient evidence to indicate a difference in mean whipping capacity among the 4 housing systems at α = .05.

e. To determine if there is a difference in mean whipping capacity between the 2 weight classes, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2

The test statistic is F = 2.95 and the p-value is .1015. Since the p-value is not less than α (p = .1015 .05), H0 is not rejected. There is insufficient evidence to indicate a difference in mean whipping capacity between the 2 weight classes at α = .05.


514 Chapter 8

8.74 a. The treatments are the 3 × 3 = 9 combinations of PES and Trust. The nine treatments are: (BC, Low), (PC, Low), (NA, Low), (BC, Med), (PC, Med), (NA, Med), (BC, High), (PC, High), and (NA, High).

b. df(Trust) = 3 − 1 = 2; SSE = SSTot − SS(PES) − SS(Trust) − SSPT = 161.1162 − 2.1774 − 7.6367 − 1.7380 = 149.5641

MS(PES) = SS(PES)

df(PES) =

2.1774

2 = 1.0887

MS(Trust) = SS(Trust)

df(Trust) =

7.6367

2 = 3.81835

MS(PT) = SS(PT)

df(PT) =

1.7380

4 = 0.4345

MSE = SSE

df(Error) =

149.5641

206 = 0.7260

FPES = MS(PES)

MSE =

1.0887

0.7260 = 1.50 FTrust =

MS(Trust)

MSE =

3.81835

0.7260 = 5.26

FPT = MS(PT)

MSE =

0.4345

0.7260 = 0.60

The ANOVA table is:

Source df SS MS F PES 2 2.1774 1.0887 1.50Trust 2 7.6367 3.81835 5.26PES × Trust 4 1.7380 0.4345 0.60Error 206 149.5641 0.7260Total 214 161.1162

c. To determine if PES and Trust interact, we test: H0: PES and Trust do not interact to affect the mean tension Ha: PES and Trust do interact to affect the mean tension The test statistic is F = 0.60. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) =

(3 − 1)(3 − 1) = 4 and ν2 = n − ab = 215 − 3(3) = 206. From Table VIII, Appendix B, F.05 ≈ 2.37. The rejection region is F > 2.37.


H0 is not rejected. There is insufficient evidence to indicate that PES and Trust interact at α = .05.



d. The plot of the treatment means is: The mean tension scores for Low

Trust are relatively the same for each level of PES. Similarly, the mean tension scores for Medium Trust are relatively the same for each level of PES. However, the mean tension scores for High Trust are not the same for each level of PES. For both PES levels BC and PC, as the level of trust increases, the mean tension scores decrease. However, for PES level NA, as trust goes from low to medium, the mean tension decreases. As the trust goes from medium to high, the mean tension increases. This indicates that interaction is present which was also found in part d.

e. Because the interaction of PES and Trust was found to be significant, the tests for the main effects

are irrelevant. If the factors interact, the interaction effect can cover up any main effect differences. In addition, interaction implies that the effects of one factor on the dependent variable are different at different levels of the second factor. Thus, there is no one "main" effect of the factor.

8.75 Yes. Using MINITAB, a plot of the data is:

Density

Me

an

Tim

e

HighLow

8

7

6

5

4

3

2

1

0 0

AgentGumPVP

Scatterplot of Time vs Density

Since the lines are not parallel, this indicates interaction is present. The increase in mean time when

density is increased from low to high for PVP is not as great as the increase in mean time when density is increased from low to high for GUM.


516 Chapter 8

8.76 Using MINITAB, the ANOVA results are:

General Linear Model: Deviation versus Group, Trail

Factor Type Levels Values Group fixed 4 F G M N Trail fixed 2 C E

Analysis of Variance for Deviatio, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P Group 3 16271.2 13000.6 4333.5 5.91 0.001 Trail 1 46445.5 46445.5 46445.5 63.34 0.000 Group*Trail 3 2245.2 2245.2 748.4 1.02 0.386 Error 112 82131.7 82131.7 733.3 Total 119 147093.6

First, we must test for treatment effects.

SST = SS(Group) + SS(Trail) + SS(GxT) = 16,271.2 + 46,445.5 + 2,245.2 = 64,961.9.

The df = 3 + 1 + 3 = 7.

SST 64, 961.9MST 9, 280.2714

1 4(2) 1ab

MST 9, 280.2714F= 12.66

MSE 733.3

To determine if there are differences in mean ratings among the 8 treatments, we test: H0: All treatment means are the same Ha: At least two treatment means differ The test statistic is F = 12.66.

Since no α was given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = ab – 1 = 4(2) – 1 = 7 and ν2 = n – ab = 120 – 4(2) = 112. From Table VIII, Appendix B, F.05 ≈ 2.09. The rejection region is F > 2.09. Since the observed value of the test statistic falls in the rejection region (F = 12.66 > 2.09), H0 is rejected. There is sufficient evidence that differences exist among the treatment means at α = .05. Since differences exist, we now test for the interaction effect between Trail and Group. To determine if Trail and Group interact, we test: H0: Trail and Group do not interact Ha: Trail and Group do interact The test statistic is F = 1.02 and p = .386 Since the p-value is greater than α (p = .386 > .05), H0 is not rejected. There is insufficient evidence that Trail and Group interact at α = .05. Since the interaction does not exist, we test for the main effects of Trail and Group. To determine if there are differences in the mean rating between the two levels of Trail, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2



The test statistics is F = 63.34 and p = 0.000. Since the p-value is greater than α (p = .000 < .05), H0 is rejected. There is sufficient evidence that the mean trail deviations differ between the fecal extract trail and the control trail α = .05. To determine if there are differences in the mean rating between the four levels of Group, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least 2 means differ The test statistics is F = 5.91 and p = 0.001. Since the p-value is less than α (p = 0.001 < .05), H0 is rejected. There is sufficient evidence that the mean trail deviations differ among the four groups at α = .05.

8.77 a. Low Load, Ambiguous: 1 1 1Total 25(18) 450n x

High Load, Ambiguous: 2 2 2Total 25(6.1) 152.5n x

Low Load, Common: 3 3 3Total 25(7.8) 195n x

High Load, Common: 4 4 4Total 25(6.3) 157.5n x

b. 2 2 2(sum of all observations) (450 152.5 195 157.5) 955

CM 9,120.25n 100 100

c. Low Load total is 450 + 195 = 645. High Load total is 152.5 + 157.5 = 310.

2

2 21 645 310

SS( ) CM 9,120.25 10, 242.5 9,120.25 1,122.252(25) 2(25)

a

ii

A

Loadbr

Ambiguous total is 450 + 152.5 = 602.5. Common total is 195 + 157.5 = 352.5

2

2 21 602.5 352.5

SS( ) CM 7, 700.0625 9, 745.25 9,120.25 6252(25) 2(25)

b

jj

B

Namear

2

1 1

2 2 2 2

SS(Load Name) (Load) SS(Name)

450 152.5 195 157.5 1,122.25 625 9,120.25

25 25 25 25 11 543 5 1,122.25 625 9,120.25

a b

iji j

AB

SS CMr

, .

676

d. Low Load, Ambiguous: 2 21 15 225s 2

1 1( 1) (25 1)225 5, 400n s

High Load, Ambiguous: 2 22 9.5 90.25s 2

2 2( 1) (25 1)90.25 2,166n s

Low Load, Common: 2 23 9.5 90.25s 2

3 3( 1) (25 1)90.25 2,166n s

High Load, Common: 2 24 10 100s 2

4 4( 1) (25 1)100 2, 400n s

e. 2 2 2 21 1 2 2 3 3 4 4SSE ( 1) ( 1) ( 1) ( 1)

5, 400 2,166 2,166 2, 400 12,132

n s n s n s n s


518 Chapter 8

f. SS(Total) = SS(Load) + SS(Name) + SS(Load x Name) + SSE = 1,122.25 + 625 + 676 + 12,132 = 14,555.25 g. The ANOVA table is:

Source df SS MS F Load 1 1,122.25 1,122.25 8.88 Name 1 625.00 625.00 4.95 Load x Name 1 676.00 676.00 5.35 Error 96 12,132.00 126.375 Total 99 14,555.25

h. Yes. We computed 5.35, which is almost the same as 5.34. The difference could be due to round-

off error.

i. To determine if interaction between Load and Name is present, we test:

H0: Load and Name do not interact Ha: Load and Name class do interact

The test statistic is F = 5.35. The rejection region requires α = .05 in the upper tail of the F distribution with υ1 = (a – 1)(b – 1) = (2 – 1)(2 – 1) = 1 and υ2 = n – ab = 100 – 4 = 96. From Table VIII, Appendix B, F.05 3.96. The rejection region is F > 3.96. Since the observed value of the test statistic falls in the rejection region (F = 5.35 > 3.96), H0 is rejected. There is sufficient evidence to indicate that Load and Name interact at α = .05. Using MINITAB, a graph of the results is:

Load

Mea

n

HighLow

17.5

15.0

12.5

10.0

7.5

5.0

Name12

Scatterplot of Mean vs Load

From the graph, the interaction is quite apparent. For Low load, the mean number of jelly beans taken for the ambiguous name is much higher than the mean number taken for the common name. However, for High load, there is essentially no difference in the mean number of jelly beans taken between the two names.



j. We must assume that: 1. The response distributions for each Load-Name combination (treatment) is normal. 2. The response variance is constant for all Load-Name combinations. 3. Random and independent samples of experimental units are associated with each Load-Name

combination. 8.78 A one-way ANOVA has only one factor with 2 or more levels. A two-way ANOVA has 2 factors, each at

2 or more levels. 8.79 In a completely randomized design, independent random selection of treatments to be assigned to

experimental units is required. In a randomized block design, the experimental units are first grouped into blocks such that within the blocks the experimental units are homogeneous and between the blocks the experimental units are heterogeneous.

8.80 There are 3 × 2 = 6 treatments. They are A1B1, A1B2, A2B1, A2B2, A3B1, and A3B2. 8.81 When the overall level of significance of a multiple comparisons procedure is α, the level of significance

for each comparison is less than α. 8.82 a. SSE = SSTot − SST = 62.55 − 36.95 = 25.60 df Treatment = p − 1 = 4 − 1 = 3 df Error = n − p = 20 − 4 = 16 df Total = n − 1 = 20 − 1 = 19

MST = SST/df = 36.95

3 = 12.32

MSE = SSE/df = 25.60

16 = 1.60

F = MST

MSE =

12.32

1.60 = 7.70

The ANOVA table:

Source df SS MS F Treatment 3 36.95 12.32 7.70 Error 16 25.60 1.60 Total 19 62.55


520 Chapter 8

b. To determine if there is a difference in the treatment means, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two of the means differ where the μi represents the mean for the ith treatment.


MSE = 7.70

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = (p − 1) = (4 − 1) = 3 and ν2 = (n − p) = (20 − 4) = 16. From Table VII, Appendix B, F.10 = 2.46. The

rejection region is F > 2.46. Since the observed value of the test statistic falls in the rejection region (F = 7.70 > 2.46), H0 is

rejected. There is sufficient evidence to conclude that at least two of the means differ at α = .10.

c. 4x = 4

4

57

5

x

n

= 11.4

For confidence level .90, α = .10 and α/2 = .10/2 = .05. From Table V, Appendix B, with df = 16,

t.05 = 1.746. The confidence interval is:

4x ± .05 4MSE/t n 11.4 ± 1.746⋅ 1.6 / 5 11.4 ± .99 (10.41, 12.39)

8.83 a. SS(Treatment) = SS(Total) – SS(Block) – SSE = 22.31 – 10.688 - .288 = 11.334

SS(Treatment) 11.334

MS(Treatment) 3.778 - 1 4 1k

, df = k – 1 = 4 – 1 = 3

SS(Block) 10.688

MS(Block) 2.672 1 5 1b

, df = b – 1 = 5 – 1 = 4

SSE .288

MSE .024 1 20 4 5 1n k b

, df = n – k – b + 1 = 20 – 4 – 5 + 1 = 12

MS(Treatment) 3.778

Treatment F 157.42MSE .024

MS(Block) 2.672

Block F 111.33MSE .024



The ANOVA Table is:

Source df SS MS F

Treatment 3 11.334 3.778 157.42

Block 4 10.688 2.672 111.33

Error 12 0.288 0.024

Total 19 22.310

b. To determine if there is a difference among the treatment means, we test:

H0: μA = μB = μC = μD Ha: At least two treatment means differ

The test statistic is MS(Treatment)

157.42MSE

F

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k – 1 = 4 – 1 = 3 and ν2 = n – k – b + 1 = 20 – 4 – 5 + 1 = 12. From Table VIII, Appendix B, F.05 = 3.49. The rejection region is F > 3.49.

Since the observed value of the test statistic falls in the rejection region (F = 157.42 > 3.49), H0 is rejected. There is sufficient evidence to indicate a difference among thetreatment means at α = .05.

c. Since there is evidence of differences among the treatment means, we need to compare the treatment

means. The number of pairwise comparisons is ( 1) 4(4 1)

62 2

k k .

d. To determine if there are difference among the block means, we test:

H0: All block means are the same Ha: At least two block means differ

The test statistic is MS(Block)

111.33MSE

F

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = b – 1 = 5 – 1 = 4 and ν2 = n – k – b + 1 = 20 – 4 – 5 + 1 = 12. From Table VIII, Appendix B, F.05 = 3.26. The rejection region is F > 3.26.

Since the observed value of the test statistic falls in the rejection region (F = 111.33 > 3.26), H0 is rejected. There is sufficient evidence that the block means differ at α = .05.


522 Chapter 8

8.84 a. df(AB) = (a − 1)(b - 1) = 3(5) = 15 df(Error) = n − ab = 48 − 4(6) = 24 SSAB = MSAB(df) = 3.1(15) = 46.5 SS(Total) = SSA + SSB + SSAB + SSE = 2.6 + 9.2 + 46.5 + 18.7 = 77

MSA = SS 2.6

.8667 1 3

A

a

MSB =

SS 9.2 1.84

1 5

B

b

MSE = SSE 18.7

.7792 24n ab

FA = MSA .8667

1.11MSE .7792

FB = MSB 1.84

2.36MSE .7792

FAB = MS 3.1

3.98MSE .7792

AB

Source df SS MS F A 3 2.6 .8667 1.11B 5 9.2 1.84 2.36AB 15 46.5 3.1 3.98Error 24 18.7 .7792 Total 47 77.0

b. Factor A has a = 3 + 1 = 4 levels and factor B has b = 5 + 1 = 6 levels. The number of treatments is

ab = 4(6) = 24. The total number of observations is n = 47 + 1 = 48. Thus, two replicates were performed.

c. SST = SSA + SSB + SSAB = 2.6 + 9.2 + 46.5 = 58.3

MST = SST 58.3

2.5347 1 4(6) 1ab

F = MST 2.5347

3.25MSE .7792

To determine whether the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ24 Ha: At least one treatment mean is different


MSE = 3.25


1 = 23 and ν2 = n − ab = 48 - 4(6) = 24. From Table VIII, Appendix B, F.05 ≈ 2.03. The rejection region is F > 2.03.


rejected. There is sufficient evidence to indicate the treatment means differ at α = .05.



d. Since there are differences among the treatment means, we test for the presence of interaction: H0: Factor A and factor B do not interact to affect the response mean Ha: Factor A and factor B do interact to affect the response mean


MSE

AB = 3.98




rejected. There is sufficient evidence to indicate factors A and B interact to affect the response means at α = .05.

Since the interaction is significant, no further tests are warranted. Multiple comparisons need to be

performed. 8.85 a. The data are collected as a completely randomized design because five boxes of each size were

randomly selected and tested. b. Yes. The confidence intervals surrounding each of the means do not overlap. This would indicate

that there is a difference in the means for the two sizes. c. No. Several of the confidence intervals overlap. This would indicate that the mean compression

strengths of the sizes that have intervals that overlap are not significantly different. 8.86 a.. This is a two-factor factorial design. It is also a completely randomized design. b. The two factors are "involvement in topic" and "question wording." Both are qualitative variables

because neither are measured on numerical scales. c. There are two levels of "involvement in topic": high and low. There are two levels of "question

wording": positive and negative. d. There are 2 × 2 = 4 treatments. The are: (high, positive), (high, negative), (low, positive), and (low, negative) e. The experiment's dependent variable is the level of agreement. 8.87 a. The experimental design used in this example was a randomized block design.

b. The experimental units in this problem are the electronic commerce and internet-based companies. The response variable is the rate of return for the stock of the companies. The treatments are the 4 categories of companies: e-companies, internet software and service, internet hardware, and internet communication. The blocks are the 3 age categories: 1 year-old, 3 year-old, and 5 year-old.


524 Chapter 8

8.88 a. The response variable for this study is the safety rating of nuclear power plants. b. There are three treatments in this study. The treatment groups are the scientists, the journalists, and

the federal government policymakers. c. To determine whether there are differences in the attitudes of scientists, journalists, and government

officials regarding the safety of nuclear power plants, we test: H0: μ1 = μ2 = μ3 Ha: At least two means differ

d. For MST = 11.280, F = MST 11.28

= MSE 2.355

= 4.79

e. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k − 1 = 3 − 1 = 2

and ν2 = n − k = 300 − 3 = 297. From Table VIII, Appendix B, F.05 ≈ 3.00. The rejection region is F > 3.00. Since the observed value of the test statistic falls in the rejection region (F = 4.79 > 3.00), H0 is

rejected. There is sufficient evidence to indicate there are differences in the attitudes of scientists, journalists, and government officials regarding the safety of nuclear power plants at α = .05

f. There will be c = ( 1) 3(3 1)

2 2

k k

= 3 pairwise comparisons.

g. Comparing the mean safety scores for government officials and journalists, the difference in mean

safety scores is 4.2 − 3.7 = .5, The critical value for the Tukey comparison is .23. Since .5 > .23, we conclude that the mean safety score for government officials is higher than the mean safety score for journalists.

Comparing the mean safety scores for government officials and scientists, the difference in mean

safety scores is 4.2 − 4.1 = .1. Since .1 < .23, we conclude that there is no difference in mean safety scores between government officials and scientists.

Comparing the mean safety scores for scientists and journalists, the difference in mean safety scores

is 4.1 − 3.7 = .4, The critical value for the Tukey comparison is .23. Since .4 > .23, we conclude that the mean safety score for scientists is higher than the mean safety score for journalists.

A display of these conclusions is: Journalists Scientists Gov. Officials 3.7 4.1 4.2 8.89 a. To determine if leadership style affects behavior of subordinates, we test: H0: All four treatment means are the same Ha: At least two treatment means differ The test statistic is F = 30.4. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 2(2) −

1 = 3 and ν2 = n − ab = 257 − 2(2) = 253. From Table VIII, Appendix B, F.05 ≈ 2.60. The rejection region is F > 2.60.



Since the observed value of the test statistic falls in the rejection region (F = 30.4 > 2.60), H0 is rejected. There is sufficient evidence to indicate that leadership style affects behavior of subordinates at α = .05.

b. From the table, the mean response for High control, low consideration is significantly higher than for

any other treatment. The mean response for Low control, low consideration is significantly higher than that for High control, high consideration and for Low control, high consideration. No other significant differences exist.

c. The assumptions for Bonferroni's method are the same as those for the ANOVA. Thus, we must

assume that: i. The populations sampled from are normal. ii. The population variances are the same. iii. The samples are independent. 8.90 From the printout, the p-value for treatments or Decoy is p = .589. Since the p-value is not small, we

cannot reject H0. There is insufficient evidence to indicate a difference in mean percentage of a goose flock to approach to within 46 meters of the pit blind among the three decoy types. This conclusion is valid for any reasonable value of α.

8.91 a. This is an observational experiment. The researcher recorded the number of users per hour for each of 24 hours per day, 7 days per week, for 7 weeks. The researcher did not manipulate

the weeks or days or hours. b. The two factors are (1) the day of the week with 7 levels and (2) the hour of the day with 24 levels. c. In a factorial experiment, a is the number of levels of factor A and b is the number of levels of factor

B. If we let factor A be the day of the week and factor B be the hour of the day, then a = 7 and b = 24. d. To determine if the a × b = 7 × 24 = 168 treatment means differ, we test: H0: μ1 = μ2 = μ3 = . . . = μ168 Ha: At least two means differ

The test statistic is F = MST 1143.99

= MSE 45.65

= 25.06

The rejection region requires α = .01 in the upper tail of the F distribution with v1 = p − 1 = 168 − 1 =

167 and v2 = n − p = 1172 – 168 = 1004. From Table X, Appendix B, F.01 ≈ 1.00. The rejection region is F > 1.00.


rejected. There is sufficient evidence to indicate a difference in mean usage among the day-hour combinations at α = .01.

e. The hypotheses used to test if an interaction effect exists are: H0: Days and hours do not interact to affect the mean usage Ha: Days and hours interact do affect the mean usage


526 Chapter 8

f. The test statistic is F = MSAB 55.69

= MSE 45.65

= 1.22

The p-value is p = .0527. Since the p-value is not less than α = .01, H0 is not rejected. There is

insufficient evidence to indicate days and hours interact to affect usage at α = .01. g. To determine if the mean usage differs among the days of the week, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 = μ7 Ha: At least two means differ

The test statistic is F = MSA 3122.02

= MSE 45.65

= 68.39

The p-value is p = .0001. Since the p-value is less than α = .01, H0 is rejected. There is sufficient

evidence to indicate the mean usage differs among the days of the week at α = .01. To determine if the mean usage differs among the hours of the day, we test: H0: μ1 = μ2 = μ3 = . . . = μ24 Ha: At least two means differ

The test statistic is F = MSB 7157.82

= MSE 45.65

= 156.80

The p-value is p = .0001. Since the p-value is less than α = .01, H0 is rejected. There is sufficient

evidence to indicate the mean usage differs among the hours of the day at α = .01. 8.92 a. The response is the weight of a brochure. There is one factor and it is carton. The treatments are the

five different cartons, while the experimental units are the brochures.

b. CM = 2 2.75005

40

y

n

= .01406437506

SS(Total) = 2y − CM = .014066537 − .01406437506 = .00000216264

SST = 2

i

i

T

n − CM = 2 2 2 2 2. . . . .14767 15028 14962 15217 15031

8 8 8 8 8 − .01406437506

= .01406568209 - .01406437506 = .00000130703

SSE = SS(Total) − SST = .00000216264 - .00000130703 = .00000085561

MST = SST .00000130703

1 5 1k

= .000000326756

MSE = SSE .00000085561

40 5n k

= .000000024446

F = MST .000000326756

MSE .000000024446

= 13.37



Source df SS MS F Treatments 4 .00000130703 .000000326756 13.37Error 35 .00000085561 .000000024446Total 39 .00000216264

To determine whether there are differences in mean weight per brochure among the five cartons, we

test: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ The test statistic is F = 13.37. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k − 1 = 5 − 1 = 4

and ν2 = n − k = 40 − 5 = 35. From Table VIII, Appendix B, F.05 ≈ 2.53. The rejection region is F > 2.53. Since the observed value of the test statistic falls in the rejection region (F = 13.37 > 2.53), H0 is

rejected. There is sufficient evidence to indicate a difference in mean weight per brochure among the five cartons at α = .05.

c. We must assume that the distributions of weights for the brochures in the five cartons are normal,

that the variances of the weights for the brochures in the five cartons are equal, and that random and independent samples were selected from each of the cartons.

d. Using MINITAB, the results of Tukey’s multiple comparison procedure are: Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---+---------+---------+---------+----- Carton1 8 0.018459 0.000105 (-----*-----) Carton2 8 0.018785 0.000101 (----*-----) Carton3 8 0.018703 0.000109 (----*-----) Carton4 8 0.019021 0.000232 (-----*-----) Carton5 8 0.018789 0.000188 (----*-----) ---+---------+---------+---------+------ 0.01840 0.01860 0.01880 0.01900 Pooled StDev = 0.000156 Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons Individual confidence level = 99.32% Carton1 subtracted from: Lower Center Upper Carton2 0.0001013 0.0003262 0.0005512 Carton3 0.0000188 0.0002437 0.0004687 Carton4 0.0003375 0.0005625 0.0007875 Carton5 0.0001050 0.0003300 0.0005550 ------+---------+---------+---------+--- Carton2 (-----*------) Carton3 (-----*-----) Carton4 (-----*-----) Carton5 (-----*------) ------+---------+---------+---------+--- -0.00035 0.00000 0.00035 0.00070


528 Chapter 8

Carton2 subtracted from: Lower Center Upper Carton3 -0.0003075 -0.0000825 0.0001425 Carton4 0.0000113 0.0002363 0.0004612 Carton5 -0.0002212 0.0000037 0.0002287 ------+---------+---------+---------+--- Carton3 (------*-----) Carton4 (------*-----) Carton5 (-----*------) ------+---------+---------+---------+--- -0.00035 0.00000 0.00035 0.00070 Carton3 subtracted from: Lower Center Upper Carton4 0.0000938 0.0003187 0.0005437 Carton5 -0.0001387 0.0000862 0.0003112 ------+---------+---------+---------+--- Carton4 (-----*------) Carton5 (-----*------) ------+---------+---------+---------+--- -0.00035 0.00000 0.00035 0.00070 Carton4 subtracted from: Lower Center Upper Carton5 -0.0004575 -0.0002325 -0.0000075 ------+---------+---------+---------+--- Carton5 (-----*------) ------+---------+---------+---------+--- -0.00035 0.00000 0.00035 0.00070

The means arranged in order are: Carton 1 Carton 3 Carton 2 Carton 5 Carton 4 .018459 .018703 .018785 .018789 .019021 The interpretation of the Tukey results are: The mean weight for carton 4 is significantly higher than the mean weights of all the other cartons. The mean weights of cartons 5, 4, and 3 are significantly higher than the mean weight of carton 1. e. Since there are differences among the cartons, management should sample from many cartons. 8.93 a. The df for Groups = ν1 = k – 1 = 3 – 1 = 2. The df for Error = ν2 = n – k = 71 – 3 = 68. The completed ANOVA table is:

Source df SS MS F Groups 2 128.70 64.35 0.16 Error 68 27,124.52 398.89



b. To determine if the total number of activities undertaken differed among the three groups of entrepreneurs, we test:

H0: μ1 = μ2 = μ3 Ha: At least one mean differs The test statistic is F = 0.16. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k − 1 = 3 − 1 = 2

and ν2 = n − k = 71 − 3 = 68. From Table VIII, Appendix B, F.05 ≈ 3.15. The rejection region is F > 3.15.


H0 is not rejected. There is insufficient evidence to indicate that the total number of activities differed among the groups of entrepreneurs at α = .05.

c. The p-value of the test is P(F > 0.16). From Table VII, Appendix B, with ν1 = 2 and ν2 = 68, P(F >

0.16) > .10. d. No. Since our conclusion was that there was no evidence of a difference in the total number of

activities among the groups, there would be no evidence to indicate a difference between two specific groups.

8.94 a. This is a randomized block design. Response: the length of time required for a cut to stop bleeding Factor: drug Factor type: qualitative Treatments: drugs A, B, and C Experimental units: subjects

b. Using MINITAB, the results are:

General Linear Model: Y versus Drug, Person

Factor Type Levels Values Drug fixed 3 A B C Person fixed 5 1 2 3 4 5

Analysis of Variance for Y, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P Drug 2 156.4 156.4 78.2 3.91 0.066 Person 4 7645.8 7645.8 1911.5 95.51 0.000 Error 8 160.1 160.1 20.0 Total 14 7962.3

Tukey 90.0% Simultaneous Confidence Intervals Response Variable Y All Pairwise Comparisons among Levels of Drug

Drug = A subtracted from:

Drug Lower Center Upper -----+---------+---------+---------+- B -11.56 -4.820 1.922 (-------*-------) C -3.72 3.020 9.762 (--------*-------) -----+---------+---------+---------+- -8.0 0.0 8.0 16.0


530 Chapter 8

Drug = B subtracted from:

Drug Lower Center Upper -----+---------+---------+---------+- C 1.098 7.840 14.58 (--------*-------) -----+---------+---------+---------+- -8.0 0.0 8.0 16.0

Let μ1, μ2, and μ3 represent the mean clotting time for the three drugs. H0: μ1 = μ2 = μ3 Ha: At least two means differ

The test statistic is F = MS(Drug)

MSE = 3.91

The p-value is p = 0.066. Since the observed level of significance is less than α = .10, H0 is rejected. There is sufficient evidence to indicate differences in the mean clotting times among the three drugs at α = .10.

c. The observed level of significance is given as 0.066. d. To determine if there is a significant difference in the mean response over blocks, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two block means differ

The test statistic is F = MS(Person)

MSE = 95.51

The p-value is p = 0.000. Since the observed level of significance is less than α = .10, H0 is rejected. There is sufficient evidence to indicate differences in the mean clotting times among the five people at α = .10.

e. The confidence interval to compare drugs A and B is (-11.56, 1.922). Since 0 is in the interval, there

is no evidence of a difference in mean clotting times between drugs A and B.

The confidence interval to compare drugs A and C is (-3.72, 9.762). Since 0 is in the interval, there is no evidence of a difference in mean clotting times between drugs A and C. The confidence interval to compare drugs B and C is (1.098, 14.58). Since 0 is not in the interval, there is evidence of a difference in mean clotting times between drugs B and C. Since the numbers are positive, the mean clotting time for drug C is greater than that for drug B. In summary, the mean clotting time for drug C is greater than that for drug B. No other differences exist.

8.95 a. The quality of the steel ingot. b. There are two factors: temperature and pressure. They are quantitative factors since they are

numerical. c. The treatments are the 3 × 5 = 15 factor-level combinations of temperature and pressure. d. The steel ingots are the experimental units.



8.96 a. MSA = SS

A

A

df =

243.2

1 = 243.2 MSB =

SS

B

B

df =

57.8

1 = 57.8

SSAB = SSTot- SSA − SSB − SSE = 976.3 − 243.2 − 57.8 − 670.8 = 4.5

MSAB = SS

AB

AB

df =

4.5

1 = 4.5 MSE =

SSE

Edf =

670.8

77 = 8.712

FA = MS

MSE

A =

243.2

8.712 = 27.92 FB =

MSB

MSE =

57.8

8.712 = 6.63

FAB = MSAB

MSE =

4.5

8.712 = 0.52

The ANOVA table is:

Source df SS MS F Recent Performance (A) 1 243.2 243.2 27.92Risk attitude(B) 1 57.8 57.8 6.63AB 1 4.5 4.5 0.52Error 77 670.8 8.712 Total 80 976.3

b. To determine if factors A and B interact, we test: H0: Factors A and B do not interact to affect the mean decision Ha: Factors A and B do interact to affect the mean decision The test statistic is F = 0.52. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) =

(2 − 1)(2 − 1) = 1 and ν2 = n − ab = 81 − 2(2) = 77. From Table VIII, Appendix B, F.05 ≈ 4.00. The rejection region is F > 4.00.

Since the observed value of the test statistic does not fall in the rejection region (F = .52 4.00), H0

is not rejected. There is insufficient evidence to indicate that factors A and B interact at α = .05. c. Since the interaction is not significant, the main effect tests are meaningful. To determine if an individual's risk attitude affects his or her budgetary decisions, we test: H0: No difference exists between the risk attitude means Ha: The risk attitude means differ The test statistic is F = 6.63. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = b − 1 = 2 − 1 = 1

and ν2 = n − ab = 81 − 2(2) = 77. From Table VIII, Appendix B, F.05 ≈ 4.00. The rejection region is F > 4.00. Since the observed value of the test statistic falls in the rejection region (F = 6.63 > 4.00), H0 is

rejected. There is sufficient evidence to indicate an individual's risk attitude affects his or her budgetary decisions at α = .05.


532 Chapter 8

d. To determine if recent performance affects budgeting decisions, we test: H0: No difference exists between the recent performance means Ha: The recent performance means differ The test statistic is F = 27.92. The rejection region requires α = .01 in the upper tail of the F-distribution with ν1 = a − 1 = 2 − 1 = 1

and ν2 = n − ab = 81 − 2(2) = 77. From Table X, Appendix B, F.01 ≈ 7.08. The rejection region is F > 7.08. Since the observed value of the test statistic falls in the rejection region (F = 27.92 > 7.08), H0 is

rejected. There is sufficient evidence to indicate that recent performance affects his or her budgetary decisions at α = .01.

8.97 a. A completely randomized design was used. There are five treatments. They are the five different

educational levels. b. To determine if the mean concern ratings differ for at least two education levels, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ where μi represents the mean concern rating of the ith education level. The test statistic is F = 3.298. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = p − 1 = 5 − 1 = 4

and ν2 = n − p = 315 − 5 = 310. From Table VIII, Appendix B, F.05 ≈ 2.37. The rejection region is F > 2.37. Since the observed value of the test statistic falls in the rejection region (F = 3.298 > 2.37), H0 is

rejected. There is sufficient evidence to indicate a difference in the mean concern ratings among the 5 education levels at α = .05.

c. The mean concern rating for those with post-graduate education is significantly greater than the mean

concern rating for the four other education level groups. There are no other significant differences. 8.98 a. Some preliminary calculations are:

CM = 2 22.95

10

y

n

= .435125

SS(Total) = 2y − CM = .4705 − .435125 = .035375

SST = SS(DRUG) = 2 2

1 2T T

b b − CM =

2 21.62 1.33

10 10 − .435125 = .004205

MST =SST .004205

1 2 1k

= .004205, df = k − 1 = 1



SSB = SS(DOG) = 2 21 2B B

k k + ⋅⋅⋅ +

210B

k − CM

= 2 2 2 2 2 2 2 2 2 2.32 .38 .27 .36 .42 .31 .19 .19 .3 .21

2

− .435125 = .028925

MSB = SSB .028925

1 10 1b

= .003214, df = b − 1 = 9

SSE = SS(Total) − SST − SSB = .035375 − .004205 − .028925 = .002245

MSE = SSE .002245

1 20 2 10 1n k b

= .0002494

F = MST .004205

MSE .0002494 = 16.86 F =

MSB .003214

MSE .0002494 = 12.89

To determine if there is a difference in mean pressure readings for the two treatments, we test: H0: μA = μB Ha: μA ≠ μB


MSE = 16.86

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 2 − 1 = 1



rejected. There is sufficient evidence to indicate a difference in mean pressure readings for the two drugs at α = .05.

b. Since there is expected to be much variation between the dogs, we use the dogs as blocks to eliminate

this identified source of variation. c.

Dog Drug A Drug B (A − B) Differences

1 .17 .15 .02 2 .20 .18 .02 3 .14 .13 .01 4 .18 .18 .00 5 .23 .19 .04 6 .19 .12 .07 7 .12 .07 .05 8 .10 .09 .01 9 .16 .14 .02

10 .13 .08 .05


534 Chapter 8

Some preliminary calculations are:

.29

10

i

d

dd

n

= .029

2 22

d2d

d

(.29).0129

.00449101 10 1 9

i

i

dd

ns

n

= .0004989

sd = 2d .0004989s = .02234

To determine if there is a difference in mean pressure readings for the two treatments, we test: H0: μA = μB Ha: μA ≠ μB

The test statistic is t = d d

0 .029 0

/ .02234 / 10

d

s n

= 4.105

The rejection region requires α/2 = .05/2 = .025 in each tail of the t distribution with df = n − 1 = 10

− 1 = 9. From Table V, Appendix B, t.025 = 2.262. The rejection region is t < −2.262 or t > 2.262. Since the observed value of the test statistic falls in the rejection region (t = 4.105 > 2.262), H0 is

rejected. There is sufficient evidence to indicate a difference in the treatment means at α = .05. d. In part a, F = 16.86; and in part c, t = 4.105. Note that t2 = 4.1052 = 16.85 = F.

In part a, F.05 = 5.12; and in part c, t.025 = 2.262. Note that 2.025t = 2.2622 = 5.12 = F.05.

e. p-value = P(F ≥ 16.86) with ν1 = 1 and ν2 = 9. Using Table X, Appendix B, P(F ≥ 10.56) < .01. Thus, the p-value is < .01. The probability of a test statistic this extreme if the treatment means are the same is less than .01.

This is very significant. We would reject H0 in favor of Ha if α is larger than the p-value. 8.99 a. A 6 × 5 factorial design was used for this experiment. There are 6 cylinders and 5 batches. b. The two factors are cylinders with 6 levels and batches with 5 levels. c. There are a total of a × b = 6 × 5 = 30 treatments.

d. ix = 1 + 1 + 2 + . . . + 2 = 145

CM = 2 2(145)

90

ix

n

= 233.61111



SS(Batch) = 2iA

br

− CM =

2 22 2 257 2812 24 246(3) 6(3) 6(3) 6(3) 6(3)

− 233.61111

= 296.05556 − 233.61111 = 62.44445

SS(Cyl) = 2iB

br

− CM =

2 2 22 2 246 31 2614 14 145(3) 5(3) 5(3) 5(3) 5(3) 5(3)

− 233.61111

= 289.4 − 233.61111 = 55.78889

SS(B × C) = 2ijAB

r

− SS(Batch) − SS(Cyl) − CM

= 2 22 2 3 64 1

3 3 3 3 − 62.44445 − 55.78889 − 233.61111

= 1201

3 − 62.44445 − 55.78889 − 233.61111 = 48.48888

SSTot = 2ijx − CM = 513 – 233.61111 = 279.38889

SSE = SSTot − SS(Batch) − SS(Cyl) - SS(B × C) = 279.38889 − 62.44445 − 55.78889 − 48.48888 = 112.66667 MS(Batch) = = 15.6111

MS(Cyl) = SS(Cyl) 55.78889

1 6 1b

= 11.1578

MS(B × C) = SS( ) 48.48888

( 1)( 1) (5 1)(6 1)

B x C

a b

= 2.4244

MSE = SSE 112.66667

90 5(6)n ab

= 1.8778

FB = MS(Batch) 15.6111

= MSE 1.8778

= 8.31 FC = MS(Cyl) 11.1578

= MSE 1.8778

= 5.94

FB×C = MS( ) 2.4244

= MSE 1.8778

B C = 1.29


536 Chapter 8

The ANOVA Table is: Source df SS MS F__ Batch 4 62.444 15.611 8.31 Cyl 5 55.789 11.158 5.94 B × C 20 48.489 2.424 1.29 Error 60 112.667 1.878_________ Total 89 279.389 SST = SS(Batch) + SS(Cyl) + SS(B × C) = 62.44444 + 55.788889 + 48.48888 = 166.72221

MST = SST 166.72221

1 5(6) 1ab

= 5.749 FT = MST

MSE = 3.06

To determine if differences exist among the treatment means, we test: H0: μ1 = μ2 = μ3 = . . . = μ30 Ha: At least two means differ The test statistic is FT = 3.06. Since no α is given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the

F distribution with ν1 = ab − 1 = 5(6) − 1 = 29 and ν2 = n − ab = 90 − 5(6) = 60. From Table VIII, Appendix B, F.05 ≈ 1.65. The rejection region is F > 1.65.


rejected. There is sufficient evidence to indicate differences exist among the treatment means at α = .05. e. If batch and cylinder interact to affect the mean weight, this means that the effect of batch on mean

weight depends on the level of cylinder. Batch 1 may have the highest mean weight on cylinder 2, but Batch 4 may have the highest mean weight on cylinder 6.

f. To determine if Batch and Cylinder interact to affect mean weight, we test: H0: Batch and Cylinder do not interact Ha: Batch and Cylinder interact

The test statistic is F = MS( )

MSE

B C = 1.29

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = (a − 1)(b − 1) = (5 − 1)(6 − 1) = 20 and ν2 = n − ab = 90 − 5(6) = 60. From Table VIII, Appendix B,

F.05 = 1.75. The rejection region is F > 1.75. Since the observed value of the test statistic does not fall in the rejection region (F = 1.29 1.75),

H0 is not rejected. There is insufficient evidence to indicate Batch and Cylinder interact to affect the mean weight at α = .05.



g. Since we did not find any evidence of interaction in part f, we will test for the main effects.

To determine if the mean weights differ among the batches, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two means differ The test statistic is FB = 8.31. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = a − 1 = 5 − 1 = 4

and ν2 = n − ab = 90 − 5(6) = 60. From Table VIII, Appendix B, F.05 = 2.53. The rejection region is F > 2.53. Since the observed value of the test statistic falls in the rejection region (F = 8.31 > 2.53), H0 is

rejected. There is sufficient evidence to indicate differences exist among the batches at α = .05. To determine if the mean weights differ among the cylinders, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 Ha: At least two means differ The test statistic is FC = 5.94. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = b − 1 = 6 − 1 = 5

and ν2 = n − ab = 90 − 5(6) = 60. From Table VIII, Appendix B, F.05 ≈ 2.37. The rejection region is F > 2.37. Since the observed value of the test statistic falls in the rejection region (F = 5.94 > 2.37), H0 is

rejected. There is sufficient evidence to indicate differences exist among the cylinders at α = .05. 8.100 a. There are a total of a × b = 3 × 3 = 9 treatments in this study.

b. Using MINITAB, the ANOVA results are:

General Linear Model: Y versus Display, Price

Factor Type Levels Values Display fixed 3 1 2 3 Price fixed 3 1 2 3

Analysis of Variance for Y, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P Display 2 1691393 1691393 845696 1709.37 0.000 Price 2 3089054 3089054 1544527 3121.89 0.000 Display*Price 4 510705 510705 127676 258.07 0.000 Error 18 8905 8905 495 Total 26 5300057

To get the SS for Treatments, we must add the SS for Display, SS for Price, and the SS for Interaction. Thus, SST = 1,691,393 + 3,089,054 + 510,705 = 5,291,152. The df = 2 + 2 + 4 = 8.

SST 5, 291,152MST 661,394

1 3(3) 1ab

MST 661, 394F = 1336.15

MSE 495


538 Chapter 8

To determine whether the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two treatment means differ


MSE = 1336.15

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = ab − 1 = 3(3) − 1 = 8 and ν2 = n − ab = 27 − 3(3) = 18. From Table VII, Appendix B, F.10 = 2.04.

The rejection region is F > 2.04. Since the observed value of the test statistic falls in the rejection region (F = 1336.15 > 2.04), H0 is

rejected. There is sufficient evidence to indicate the treatment means differ at α = .10. c. Since there are differences among the treatment means, we next test for the presence of interaction. H0: Factors A and B do not interact to affect the response means Ha: Factors A and B do interact to affect the response means


MSE

AB = 258.07

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (3 − 1)(3 − 1) = 4 and ν2 = n − ab = 17 − 3(3) = 18. From Table VII, Appendix B,

F.10 = 2.29. The rejection region is F > 2.29. Since the observed value of the test statistic falls in the rejection region (F = 258.07 > 2.29), H0 is

rejected. There is sufficient evidence to indicate the two factors interact at α = .10. d. The main effect tests are not warranted since interaction is present in part c. e. The nine treatment means need to be compared. f. From the graph, if the like letters are connected, the lines are not parallel. This implies interaction is

present. This agrees with the results of part c. 8.101 a. This is a 2 × 2 factorial experiment. b. The two factors are the tent type (treated or untreated) and location (inside or outside). There are 2 ×

2 = 4 treatments. The four treatments are (treated, inside), (treated, outside), (untreated, inside), and (untreated, outside).

c. The response variable is the number of mosquito bites received in a 20 minute interval. d. There is sufficient evidence to indicate interaction is present. This indicates that the effect of the tent

type on the number of mosquito bites depends on whether the person is inside or outside.



8.102 a. This is a completely randomized design with a complete four-factor factorial design. b. There are a total of 2 × 2 × 2 × 2 = 16 treatments. c. Using SAS, the output is:

Analysis of Variance Procedure

Dependent Variable: Y

Sum of Mean

Source DF Squares Square F Value Pr > F

Model 15 546745.50 36449.70 5.11 0.0012

Error 16 114062.00 7128.88

Corrected Total 31 660807.50

R-Square C.V. Root MSE Y Mean

0.827390 41.46478 84.433 203.63

Source DF Anova SS Mean Square F Value Pr > F

SPEED 1 56784.50 56784.50 7.97 0.0123

FEED 1 21218.00 21218.00 2.98 0.1037

SPEED*FEED 1 55444.50 55444.50 7.78 0.0131

COLLET 1 165025.13 165025.13 23.15 0.0002

SPEED*COLLET 1 44253.13 44253.13 6.21 0.0241

FEED*COLLET 1 142311.13 142311.13 19.96 0.0004

SPEED*FEED*COLLET 1 54946.13 54946.13 7.71 0.0135

WEAR 1 378.13 378.13 0.05 0.8208

SPEED*WEAR 1 1540.13 1540.13 0.22 0.6483

FEED*WEAR 1 946.13 946.13 0.13 0.7204

SPEED*FEED*WEAR 1 528.13 528.13 0.07 0.7890

COLLET*WEAR 1 1682.00 1682.00 0.24 0.6337

SPEED*COLLET*WEAR 1 512.00 512.00 0.07 0.7921

FEED*COLLET*WEAR 1 72.00 72.00 0.01 0.9212

SPEE*FEED*COLLE*WEAR 1 1104.50 1104.50 0.15 0.6991

d. To determine if the interaction terms are significant, we must add together the sum of squares for all interaction terms as well as the degrees of freedom.

SS(Interaction) = 55,444.50 + 44,253.13 + 142,311.13 + 54,946.13 + 1,540.13 + 946.13 + 528.13 + 1,682.00 + 512.00 + 72.00 + 1,104.50 = 303,339.78 df(Interaction) = 11

MS(Interaction) = SS(Interacton)

df(Interaction) =

303, 339.78

11 = 27,576.34364

F(Interaction) = MS(Interaction)

MSE =

27, 576.34364

7128.88 = 3.87


540 Chapter 8

To determine if interaction effects are present, we test: H0: No interaction effects exist Ha: Interaction effects exist The test statistic is F = 3.87. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = 11 and ν2 = 16.

From Table VIII, Appendix B, F.05 ≈ 2.49. The rejection region is F > 2.49. Since the observed value of the test statistic falls in the rejection region (F = 3.87 > 2.49), H0 is

rejected. There is sufficient evidence to indicate that interaction effects exist at α = .05. Since the sums of squares for a balanced factorial design are independent of each other, we can look

at the SAS output to determine which of the interaction effects are significant. The three-way interaction between speed, feed, and collet is significant (p = .0135). There are three two-way interactions with p-values less than .05. However, all of these two-way interaction terms are imbedded in the significant three-way interaction term.

e. Yes. Since the significant interaction terms do not include wear, it would be necessary to perform

the main effect test for wear. All other main effects are contained in a significant interaction term. To determine if the mean finish measurements differ for the different levels of wear, we test: H0: The mean finish measurements for the two levels of wear are the same Ha: The mean finish measurements for the two levels of wear are different The test statistic is t = 0.05. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = 1 and ν2 = 16.

From Table VIII, Appendix B, F.05 = 4.49. The rejection region is F > 4.49. Since the observed value of the test statistic does not fall in the rejection region (F = .05 4.49), H0

is not rejected. There is insufficient evidence to indicate that the mean finish measurements differ for the different levels of wear at α = .05.

f. We must assume that: i. The populations sampled from are normal. ii. The population variances are the same. iii. The samples are random and independent.



8.103 Using MINITAB, the ANOVA Table is:

General Linear Model: Rating versus Prep, Standing

Factor Type Levels Values Prep fixed 2 PRACTICE REVIEW Standing fixed 3 HI LOW MED

Analysis of Variance for Rating, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P Prep 1 54.735 54.735 54.735 14.40 0.000 Standing 2 16.500 16.500 8.250 2.17 0.118 Prep*Standing 2 13.470 13.470 6.735 1.77 0.174 Error 126 478.955 478.955 3.801 Total 131 563.659

Tukey 95.0% Simultaneous Confidence Intervals Response Variable Rating All Pairwise Comparisons among Levels of Prep

Prep = PRACTICE subtracted from:

Prep Lower Center Upper ---+---------+---------+---------+--- REVIEW -1.960 -1.288 -0.6162 (-----------*----------) ---+---------+---------+---------+--- -1.80 -1.20 -0.60 0.00

First, we must test for treatment effects. SST = SS(Prep) + SS(Stand) + SS(PxS) = 54.735 + 16.500 + 13.470 = 84.705.

The df = 1 + 2 + 2 = 5.

SST 84.705

16.9411 2(3) 1

MSTab

MST 16.941

4.46MSE 3.801

F

To determine if there are differences in mean ratings among the 6 treatments, we test: H0: All treatment means are the same Ha: At least two treatment means differ The test statistic is F = 4.46.

Since no α was given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = ab – 1 = 2(3) – 1 = 5 and ν2 = n – ab = 132 – 2(3) = 126. From Table VIII, Appendix B, F.05 ≈ 2.29. The rejection region is F > 2.29.


rejected. There is sufficient evidence that differences exist among the treatment means at α = .05. Since differences exist, we now test for the interaction effect between Preparation and Class Standing.

To determine if Preparation and Class Standing interact, we test: H0: Preparation and Class Standing do not interact Ha: Preparation and Class Standing do interact The test statistic is F = 1.77 and p = .174


542 Chapter 8

Since the p-value is greater than α (p = .174 > .05), H0 is not rejected. There is insufficient evidence that Preparation and Class Standing interact at α = .05. Since the interaction does not exist, we test for the main effects of Preparation and Class standing.

To determine if there are differences in the mean rating between the three levels of Class standing, we test: H0: μL = μM = μH Ha: At least 2 means differ The test statistics is F = 2.17 and p = 0.118. Since the p-value is greater than α (p = .118 > .05), H0 is not rejected. There is insufficient evidence

that the mean ratings differ among the 3 levels of Class Standing at α = .05.

To determine if there are differences in the mean rating between the two levels of Preparation, we test:

H0: μP = μS Ha: μP ≠ μS The test statistics is F = 14.40 and p = 0.000. Since the p-value is less than α (p = 0.000 < .05), H0 is rejected. There is sufficient evidence that the

mean ratings differ between the two levels of preparation at α = .05. There are only 2 levels of Preparation. The mean rating for Practice is higher than the mean rating

Review.


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