SBE_SM13.doc

Chapter 13Analysis of Variance andExperimental Design

Learning Objectives

1. Understand how the analysis of variance procedure can be used to determine if the means of more than two populations are equal.

2. Know the assumptions necessary to use the analysis of variance procedure.

3. Understand the use of the F distribution in performing the analysis of variance procedure.

4. Know how to set up an ANOVA table and interpret the entries in the table.

5. Be able to use output from computer software packages to solve analysis of variance problems.

6. Know how to use Fisher’s least significant difference (LSD) procedure and Fisher’s LSD with the Bonferroni adjustment to conduct statistical comparisons between pairs of populations means.

7. Understand the difference between a completely randomized design, a randomized block design, and factorial experiments.

8. Know the definition of the following terms:

comparisonwise Type I error rate partitioningexperimentwise Type I error rate blockingfactor main effectlevel interactiontreatmentreplication

13 - 1

Chapter 13

Solutions:

1. a. = (30 + 45 + 36)/3 = 37

= 5(30 - 37)2 + 5(45 - 37)2 + 5(36 - 37)2 = 570

MSTR = SSTR /(k - 1) = 570/2 = 285

b. = 4(6) + 4(4) + 4(6.5) = 66

MSE = SSE /(nT - k) = 66/(15 - 3) = 5.5

c. F = MSTR /MSE = 285/5.5 = 51.82

Using F table (2 degrees of freedom numerator and 12 denominator), p-value is less than .01

Actual p-value = .0000

Because p-value = .05, we reject the null hypothesis that the means of the three populations are equal.

d.Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 570 2 285 51.82Error 66 12 5.5Total 636 14

2. a. = (153 + 169 + 158)/3 = 160

= 4(153 - 160)2 + 4(169 - 160) 2 + 4(158 - 160) 2 = 536

MSTR = SSTR /(k - 1) = 536/2 = 268

b. = 3(96.67) + 3(97.33) +3(82.00) = 828.00

MSE = SSE /(nT - k) = 828.00 /(12 - 3) = 92.00

c. F = MSTR /MSE = 268/92 = 2.91

Using F table (2 degrees of freedom numerator and 9 denominator), p-value is greater than .10


Because p-value > = .05, we cannot reject the null hypothesis.

13 - 2

Analysis of Variance and Experimental Design

d. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 536 2 268 2.91Error 828 9 92Total 1364 11

3. a.

= 4(100 - 87) 2 + 6(85 - 87) 2 + 5(79 - 87) 2 = 1,020

MSTR = SSB /(k - 1) = 1,020/2 = 510

b. = 3(35.33) + 5(35.60) + 4(43.50) = 458

MSE = SSE /(nT - k) = 458/(15 - 3) = 38.17

c. F = MSTR /MSE = 510/38.17 = 13.36




d. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 1020 2 510 13.36Error 458 12 38.17Total 1478 14

4. a.Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 1200 3 400 80Error 300 60 5Total 1500 63

b. Using F table (3 degrees of freedom numerator and 60 denominator), p-value is less than .01


Because p-value = .05, we reject the null hypothesis that the means of the 4 populations are equal.

5. a. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 120 2 60 20Error 216 72 3Total 336 74

13 - 3

Chapter 13

b. Using F table (2 numerator degrees of freedom and 72 denominator), p-value is less than .01


Because p-value = .05, we reject the null hypothesis that the 3 population means are equal.

6. Manufacturer 1 Manufacturer 2 Manufacturer 3

Sample Mean 23 28 21Sample Variance 6.67 4.67 3.33

= (23 + 28 + 21)/3 = 24

= 4(23 - 24) 2 + 4(28 - 24) 2 + 4(21 - 24) 2 = 104

MSTR = SSTR /(k - 1) = 104/2 = 52

= 3(6.67) + 3(4.67) + 3(3.33) = 44.01

MSE = SSE /(nT - k) = 44.01/(12 - 3) = 4.89

F = MSTR /MSE = 52/4.89 = 10.63



Because p-value = .05, we reject the null hypothesis that the mean time needed to mix a batch of material is the same for each manufacturer.

7. Superior Peer Subordinate

Sample Mean 5.75 5.5 5.25Sample Variance 1.64 2.00 1.93

= (5.75 + 5.5 + 5.25)/3 = 5.5

= 8(5.75 - 5.5) 2 + 8(5.5 - 5.5) 2 + 8(5.25 - 5.5) 2 = 1

MSTR = SSTR /(k - 1) = 1/2 = .5

= 7(1.64) + 7(2.00) + 7(1.93) = 38.99

MSE = SSE /(nT - k) = 38.99/21 = 1.86

F = MSTR /MSE = 0.5/1.86 = 0.27

13 - 4




Because p-value > = .05, we cannot reject the null hypothesis that the means of the three populations are equal; thus, the source of information does not significantly affect the dissemination of the information.

8. Marketing Managers

Marketing Research Advertising

Sample Mean 5 4.5 6Sample Variance .8 .3 .4

= (5 + 4.5 + 6)/3 = 5.17

= 6(5 - 5.17)2 + 6(4.5 - 5.17) 2 + 6(6 - 5.17) 2 = 7.00

MSTR = SSTR /(k - 1) = 7.00/2 = 3.5

= 5(.8) + 5(.3) + 5(.4) = 7.50

MSE = SSE /(nT - k) = 7.50/(18 - 3) = .5

F = MSTR /MSE = 3.5/.50 = 7.00



Because p-value = .05, we reject the null hypothesis that the mean perception score is the same for the three groups of specialists.

9. Real Estate

Agent Architect StockbrokerSample Mean 67.73 61.13 65.80Sample Variance 117.72 180.10 137.12

= (67.73 + 61.13 + 65.80)/3 = 64.89

= 15(67.73 - 64.89) 2 + 15(61.13 - 64.89) 2 + 15(65.80 - 64.89) 2 = 345.47

MSTR = SSTR /(k - 1) = 345.47/2 = 172.74

= 14(117.72) + 14(180.10) + 14(137.12) = 6089.16

13 - 5

Chapter 13

MSE = SSE /(nT - k) = 6089.16/(45-3) = 144.98

F = MSTR /MSE = 172.74/144.98 = 1.19



Because p-value > = .05, we cannot reject the null hypothesis that the job stress ratings are the same for the three occupations.

10. The Minitab output is shown below:

Analysis of VarianceSource DF SS MS F PFactor 2 226.1 113.0 3.70 0.042Error 21 640.8 30.5Total 23 866.9 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ---------+---------+---------+-------Small 8 92.200 6.365 (-------*--------) Medium 8 89.650 4.973 (-------*-------) Large 8 84.800 5.129 (--------*-------) ---------+---------+---------+-------Pooled StDev = 5.524 85.0 90.0 95.0

Because p-value = .05, we reject the null hypothesis that the mean service ratings are equal.

11. a

LSD; significant difference



b.

-15 3.23 = -18.23 to -11.77

12. a. Sample 1 Sample 2 Sample 3


13 - 6


= (51 + 77 + 58)/3 = 62

= 4(51 - 62) 2 +4(77 - 62) 2 + 4(58 - 62) 2 = 1,448

MSTR = SSTR /(k - 1) = 1,448/2 = 724

= 3(96.67) + 3(97.34) + 3(81.99) = 828

MSE = SSE /(nT - k) = 828/(12 - 3) = 92

F = MSTR /MSE = 724/92 = 7.87

Using F table (2 degrees of freedom numerator and 9 denominator), p-value is between .01 and .025



b.


LSD; no significant difference


13.

Since there does not appear to be any significant difference between

the means of population 1 and population 3.

14.

23 - 28 3.54

-5 3.54 = -8.54 to -1.46

15. Since there are only 3 possible pairwise comparisons we will use the Bonferroni adjustment.

= .05/3 = .017

t.017/2 = t.0085 which is approximately t.01 = 2.602

13 - 7

Chapter 13

1.06; no significant difference


1.06; significant difference

16. a.Machine 1 Machine 2 Machine 3 Machine 4

Sample Mean 7.1 9.1 9.9 11.4Sample Variance 1.21 .93 .70 1.02

= (7.1 + 9.1 + 9.9 + 11.4)/4 = 9.38

= 6(7.1 - 9.38) 2 + 6(9.1 - 9.38) 2 + 6(9.9 - 9.38) 2 + 6(11.4 - 9.38) 2 = 57.77

MSTR = SSTR /(k - 1) = 57.77/3 = 19.26

= 5(1.21) + 5(.93) + 5(.70) + 5(1.02) = 19.30

MSE = SSE /(nT - k) = 19.30/(24 - 4) = .97

F = MSTR /MSE = 19.26/.97 = 19.86



Because p-value = .05, we reject the null hypothesis that the mean time between breakdowns is the same for the four machines.

b. Note: t/2 is based upon 20 degrees of freedom


17. C = 6 [(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)]

= .05/6 = .008 and /2 = .004

Since the smallest value for /2 in the t table is .005, we will use t.005 = 2.845 as an approximation

13 - 8


for t.004 (20 degrees of freedom)

Thus, if the absolute value of the difference between any two sample means exceeds 1.62, there is sufficient evidence to reject the hypothesis that the corresponding population means are equal.

Means (1,2) (1,3) (1,4) (2,3) (2,4) (3,4)| Difference | 2 2.8 4.3 0.8 2.3 1.5Significant ? Yes Yes Yes No Yes No

18. n1 = 8 n2 = 8 n3 = 8

t/2 is based upon 21 degrees of freedom

Comparing Small and Medium


Comparing Small and Large


Comparing Medium and Large


19. a. = (156 + 142 + 134)/3 = 144

= 6(156 - 144) 2 + 6(142 - 144) 2 + 6(134 - 144) 2 = 1,488

b. MSTR = SSTR /(k - 1) = 1488/2 = 744

c. = 164.4 = 131.2 = 110.4

= 5(164.4) + 5(131.2) +5(110.4) = 2030

d. MSE = SSE /(nT - k) = 2030/(12 - 3) = 135.3

e. F = MSTR /MSE = 744/135.3 = 5.50


13 - 9

Chapter 13


Because p-value = .05, we reject the hypothesis that the means for the three treatments are equal.

20. a. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 1488 2 744 5.50Error 2030 15 135.3Total 3518 17

b.

|156-142| = 14 < 14.31; no significant difference

|156-134| = 22 > 14.31; significant difference

|142-134| = 8 < 14.31; no significant difference

21. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 300 4 75 14.07Error 160 30 5.33Total 460 34

22. a. H0: u1 = u2 = u3 = u4 = u5

Ha: Not all the population means are equal



Because p-value = .05, we reject H0




Because p-value = .05, we reject the null hypothesis that the means of the three treatments are equal.

13 - 10





Because p-value = .05, we reject the hypothesis that the treatment means are equal.

25. A B C


= 8(119 - 107.93) 2 + 10(107 - 107.93) 2 + 10(100 - 107.93) 2 = 1617.9

MSTR = SSTR /(k - 1) = 1617.9 /2 = 809.95

= 7(146.86) + 9(96.44) + 9(173.78) = 3,460

MSE = SSE /(nT - k) = 3,460 /(28 - 3) = 138.4

F = MSTR /MSE = 809.95 /138.4 = 5.85




26. a. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 4560 2 2280 9.87Error 6240 27 231.11Total 10800 29



13 - 11

Chapter 13

Because p-value = .05, we reject the null hypothesis that the means of the three assembly methods are equal.

27. Source of Variation Sum of Squares Degrees of Freedom Mean Square FBetween 61.64 3 20.55 17.56Error 23.41 20 1.17Total 85.05 23



Because p-value = .05, we reject the null hypothesis that the mean breaking strength of the four cables is the same.

28.50 60 70

Sample Mean 33 29 28Sample Variance 32 17.5 9.5

= (33 + 29 + 28)/3 = 30

= 5(33 - 30) 2 + 5(29 - 30) 2 + 5(28 - 30) 2 = 70

MSTR = SSTR /(k - 1) = 70 /2 = 35

= 4(32) + 4(17.5) + 4(9.5) = 236

MSE = SSE /(nT - k) = 236 /(15 - 3) = 19.67

F = MSTR /MSE = 35 /19.67 = 1.78



Because p-value > = .05, we cannot reject the null hypothesis that the mean yields for the three temperatures are equal.

29. Direct

ExperienceIndirect

Experience CombinationSample Mean 17.0 20.4 25.0Sample Variance 5.01 6.26 4.01

= (17 + 20.4 + 25)/3 = 20.8

13 - 12


= 7(17 - 20.8) 2 + 7(20.4 - 20.8) 2 + 7(25 - 20.8) 2 = 225.68

MSTR = SSTR /(k - 1) = 225.68 /2 = 112.84

= 6(5.01) + 6(6.26) + 6(4.01) = 91.68

MSE = SSE /(nT - k) = 91.68 /(21 - 3) = 5.09

F = MSTR /MSE = 112.84 /5.09 = 22.17



Because p-value = .05, we reject the null hypothesis that the means for the three groups are equal.

30. Paint 1 Paint 2 Paint 3 Paint 4

Sample Mean 13.3 139 136 144Sample Variance 47.5 .50 21 54.5

= (133 + 139 + 136 + 144)/3 = 138

= 5(133 - 138) 2 + 5(139 - 138) 2 + 5(136 - 138) 2 + 5(144 - 138) 2 = 330

MSTR = SSTR /(k - 1) = 330 /3 = 110

= 4(47.5) + 4(50) + 4(21) + 4(54.5) = 692

MSE = SSE /(nT - k) = 692 /(20 - 4) = 43.25

F = MSTR /MSE = 110 /43.25 = 2.54



Because p-value > = .05, we cannot reject the null hypothesis that the mean drying times for the four paints are equal.

31. A B C

Sample Mean 20 21 25Sample Variance 1 25 2.5

13 - 13

Chapter 13

= (20 + 21 + 25)/3 = 22

= 5(20 - 22) 2 + 5(21 - 22) 2 + 5(25 - 22) 2 = 70

MSTR = SSTR /(k - 1) = 70 /2 = 35

= 4(1) + 4(2.5) + 4(2.5) = 24

MSE = SSE /(nT - k) = 24 /(15 - 3) = 2

F = MSTR /MSE = 35 /2 = 17.5



Because p-value = .05, we reject the null hypothesis that the mean miles per gallon ratings are the same for the three automobiles.

32. Note: degrees of freedom for t/2 are 18




33. Note: degrees of freedom for t/2 are 12




34. Treatment Means:

= 13.6 = 11.0 = 10.6

13 - 14


Block Means:

= 9 = 7.67 = 15.67 = 18.67 = 7.67

Overall Mean:

= 176/15 = 11.73

Step 1

= (10 - 11.73) 2 + (9 - 11.73) 2 + · · · + (8 - 11.73) 2 = 354.93

Step 2

= 5 [ (13.6 - 11.73) 2 + (11.0 - 11.73) 2 + (10.6 - 11.73) 2 ] = 26.53

Step 3

= 3 [ (9 - 11.73) 2 + (7.67 - 11.73) 2 + (15.67 - 11.73) 2 +

(18.67 - 11.73) 2 + (7.67 - 11.73) 2 ] = 312.32

Step 4

SSE = SST - SSTR - SSBL = 354.93 - 26.53 - 312.32 = 16.08

Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 26.53 2 13.27 6.60Blocks 312.32 4 78.08Error 16.08 8 2.01Total 354.93 14




35. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 310 4 77.5 17.69Blocks 85 2 42.5Error 35 8 4.38Total 430 14


13 - 15

Chapter 13


Because p-value = .05, we reject the null hypothesis that the means of the treatments are equal.

36. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 900 3 300 12.60Blocks 400 7 57.14Error 500 21 23.81Total 1800 31



Because p-value = .05, we reject the null hypothesis that the means of the treatments are equal.


= 56 = 44

Block Means:

= 46 = 49.5 = 54.5

Overall Mean:

= 300/6 = 50

Step 1

= (50 - 50) 2 + (42 - 50) 2 + · · · + (46 - 50) 2 = 310

Step 2

= 3 [ (56 - 50) 2 + (44 - 50) 2 ] = 216

Step 3

= 2 [ (46 - 50) 2 + (49.5 - 50) 2 + (54.5 - 50) 2 ] = 73

13 - 16


Step 4

SSE = SST - SSTR - SSBL = 310 - 216 - 73 = 21

Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 216 1 216 20.57Blocks 73 2 36.5Error 21 2 10.5Total 310 5

Using F table (1 degree of freedom numerator and 2 denominator), p-value is between .025 and .05


Because p-value = .05, we reject the null hypothesis that the mean tune-up times are the same for both analyzers.

38. Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 45 4 11.25 7.12Blocks 36 3 12Error 19 12 1.58Total 100 19



Because p-value = .05, we reject the null hypothesis that the mean total audit times for the five auditing procedures are equal.


= 16 = 15 = 21

Block Means:

= 18.67 = 19.33 = 15.33 = 14.33 = 19

Overall Mean:

= 260/15 = 17.33

Step 1

= (16 - 17.33) 2 + (16 - 17.33) 2 + · · · + (22 - 17.33) 2 = 175.33

13 - 17

Chapter 13

Step 2

= 5 [ (16 - 17.33) 2 + (15 - 17.33) 2 + (21 - 17.33) 2 ] = 103.33

Step 3

= 3 [ (18.67 - 17.33) 2 + (19.33 - 17.33) 2 + · · · + (19 - 17.33) 2 ] = 64.75

Step 4

SSE = SST - SSTR - SSBL = 175.33 - 103.33 - 64.75 = 7.25

Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 100.33 2 51.67 56.78Blocks 64.75 4 16.19Error 7.25 8 .91Total 175.33 14



Because p-value = .05, we reject the null hypothesis that the mean times for the three systems are equal.

40. The Minitab output for these data is shown below:

Analysis of VarianceSource DF SS MS F PFactor 3 19805 6602 22.12 0.000Error 36 10744 298Total 39 30550 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -----+---------+---------+---------+-Tread 10 178.00 16.77 (---*----) ArmCrank 10 171.00 18.89 (---*----) Shovel 10 175.00 14.80 (---*---) SnowThro 10 123.60 18.36 (---*----) -----+---------+---------+---------+-Pooled StDev = 17.28 125 150 175 200

Because p-value =.05, we reject the null hypothesis that the mean heart rate for the four methods are equal.

41.

13 - 18


Step 1

= (135 - 111) 2 + (165 - 111) 2 + · · · + (136 - 111) 2 = 9,028

Step 2

= 3 (2) [ (104 - 111) 2 + (118 - 111) 2 ] = 588

Step 3

= 2 (2) [ (130 - 111) 2 + (97 - 111) 2 + (106 - 111) 2 ] = 2,328

Step 4

= 2 [ (150 - 104 - 130 + 111) 2 + (78 - 104 - 97 + 111) 2 +

· · · + (128 - 118 - 106 + 111) 2 ] = 4,392

Step 5

SSE = SST - SSA - SSB - SSAB = 9,028 - 588 - 2,328 - 4,392 = 1,720

Source of Variation Sum of Squares Degrees of Freedom Mean Square FFactor A 588 1 588 2.05Factor B 2328 2 1164 4.06Interaction 4392 2 2196 7.66Error 1720 6 286.67Total 9028 11

Factor A: F = 2.05

Using F table (1 degree of freedom numerator and 6 denominator), p-value is greater than .10


Because p-value > = .05, Factor A is not significant

13 - 19

Chapter 13

Factor B: F = 4.06



Because p-value > = .05, Factor B is not significant

Interaction: F = 7.66



Because p-value = .05, Interaction is significant

42. Source of Variation Sum of Squares Degrees of Freedom Mean Square FFactor A 26 3 8.67 3.72Factor B 23 2 11.50 4.94Interaction 175 6 29.17 12.52Error 56 24 2.33Total 280 35

Using F table for Factor A (3 degrees of freedom numerator and 24 denominator), p-value is .025

Because p-value = .05, Factor A is significant.

Using F table for Factor B (2 degrees of freedom numerator and 24 denominator), p-value is between .01 and .025


Because p-value = .05, Factor B is significant.

Using F table for Interaction (6 degrees of freedom numerator and 24 denominator), p-value is less than .01


Because p-value = .05, Interaction is significant

43.

13 - 20


Step 1

= (8 - 16) 2 + (12 - 16) 2 + (12 - 16) 2 + · · · + (14 - 16) 2 = 544

Step 2

= 2 (2) [ (10- 16) 2 + (23 - 16) 2 + (15 - 16) 2 ] = 344

Step 3

= 3 (2) [ (14 - 16) 2 + (18 - 16) 2 ] = 48

Step 4

= 2 [ (10 - 10 - 14 + 16) 2 + · · · + (16 - 15 - 18 +16) 2 ] = 56

Step 5

SSE = SST - SSA - SSB - SSAB = 544 - 344 - 48 - 56 = 96

Source of Variation

Sum of Squares Degrees of Freedom Mean Square F

Factor A 344 2 172 172/16 = 10.75Factor B 48 1 48 48/16 = 3.00Interaction 56 2 28 28/16 = 1.75Error 96 6 16Total 544 11

Using F table for Factor A (2 degrees of freedom numerator and 6 denominator), p-value is between .01 and .025

13 - 21

Chapter 13


Because p-value = .05, Factor A is significant; there is a difference due to the type of advertisement design.

Using F table for Factor B (1 degree of freedom numerator and 6 denominator), p-value is greater than .01


Because p-value > = .05, Factor B is not significant; there is not a significant difference due to size of advertisement.

Using F table for Interaction (2 degrees of freedom numerator and 6 denominator), p-value is greater than .10


Because p-value > = .05, Interaction is not significant.

44.

Step 1

= (41 - 47) 2 + (43 - 47) 2 + · · · + (44 - 47) 2 = 136

Step 2

= 3 (2) [ (46 - 47) 2 + (48 - 47) 2 ] = 12

Step 3

= 2 (2) [ (46 - 47) 2 + (48 - 47) 2 + (47 - 47) 2 ] = 8

13 - 22


Step 4

= 2 [ (41 - 46 - 46 + 47) 2 + · · · + (44 - 48 - 47 + 47) 2 ] = 56

Step 5


Source of Variation

Sum of Squares Degrees of Freedom Mean Square F

Factor A 12 1 12 12/10 = 1.2Factor B 8 2 4 4/10 = .4Interaction 56 2 28 28/10 = 2.8Error 60 6 10Total 136 11

Using F table for Factor A (1 degree of freedom numerator and 6 denominator), p-value is greater than .10


Because p-value > = .05, Factor A is not significant

Using F table for Factor B (2 degrees of freedom numerator and 6 denominator), p-value is greater than .10


Because p-value > = .05, Factor B is not significant



Because p-value > = .05, Interaction is not significant

45.

Factor B Factor A

Under junior high school

Senior high school

above college Means

Factor AMale =29.5114 =35.2943 =50.3092 =38.3716

Female =20.6957 =27.1644 = 42.053 =29.9709

Factor B Means =25.1036 =31.2294 =46.1809 =34.1713

Step 1

=(28.0559 - 34.1713 ) 2 + (29.0387 - 34.1713) 2 + · · · + (49.0288 -

13 - 23

Chapter 13

34.1713) 2 = 3,270.7354

Step 2

= 3(2) [(38.3716 - 34.1713 ) 2 + (29.9709 - 34.1713 ) 2] = 211.7166

Step 3

= 2(2) [(25.1036 - 34.1713) 2 + (31.2294 - 34.1713) 2 + (46.1809 - 34.1713) 2]

= 940.4381

Step 4

= 2[(29.5114 - 38.3716 - 25.1036 + 34.1713 ) 2 + · · · +

(42.0526 - 29.9709 - 46.1809 + 34.1713) 2] = 0.2663

13 - 24


Step 5

SSE = SST - SSA - SSB - SSAB = 2,118.3143

Source of Variation

Sum of Squares

Degrees of Freedom Mean Square F

Factor A 211.7166 1 211.7166 2.3987 Factor B 940.4381 2 470.2191 5.3275 Interaction 0.2663 2 0.1332 0.0015 Error 2,118.3143 24 88.2631 Total 3,270.7353 29

Using F table for Factor A (1 degree of freedom numerator and 24 denominator), p-value is greater than .1


Because p-value > = .05, Factor A (gender) is not significant

Using F table for Factor B (2 degrees of freedom numerator and 24 denominator), p-value is less than .05


Because p-value = .05, Factor B (educational level) is significant


Actual p-value =.9985

Because p-value > = .05, Interaction is not significant

46. = (1.13 + 1.56 + 2.00)/3 = 1.563

= (0.48 + 1.68 + 2.86)/3 = 1.673

= (1.13 + 0.48)/2 = 0.805

= (1.56 + 1.68)/2 = 1.620

= (2.00 + 2.86)/2 = 2.43

= (1.13 + 1.56 + 2.00 + 0.48 + 1.68 + 2.86)/6 = 1.618

Step 1

SST = 327.50 (given in problem statement)

Step 2

= 3(25)[(1.563 - 1.618)2 + (1.673 - 1.618)2] = 0.4538

13 - 25

Chapter 13

Step 3

= 2(25)[(0.805 - 1.618)2 + (1.62 - 1.618) 2 + (2.43 - 1.618) 2] = 66.0159

Step 4

= 25[(1.13 - 1.563 - 0.805 + 1.618) 2 + (1.56 - 1.563 - 1.62

+ 1.618) 2 + · · · + (2.86 - 1.673 - 2.43 + 1.618) 2] = 14.2525

Step 5

SSE = SST - SSA - SSB - SSAB = 327.50 - 0.4538 - 66.0159 - 14.2525

Source of Variation Sum of Squares Degrees of Freedom Mean Square FFactor A 0.4538 1 0.4538 0.2648Factor B 66.1059 2 33.0080 19.2608Interaction 14.2525 2 7.1263 4.1583Error 246.7778 144 1.7137Total 327.5000 149

Factor A: Actual p-value = .6076. Because p-value > = .05, Factor A is not significant.

Factor B: Actual p-value = .0000. Because p-value = .05, Factor B is significant.

Interaction: Actual p-value = .0176. Because p-value = .05, Interaction is significant.

47. a. Area 1 Area 2

Sample Mean 96 94Sample Variance 50 40

= (96 + 94)/2 = 95

= 4(96 - 95) 2 + 4(94 - 95) 2 = 8

MSTR = SSTR /(k - 1) = 8 /1 = 8

= 3(50) + 3(40) = 270

MSE = SSE /(nT - k) = 270 /(8 - 2) = 45

F = MSTR/MSE = 8/45 = .18

Using F table (1 degree of freedom numerator and 6 denominator), p-value is greater than .10

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Because p-value > = .05, the means are not significantly different.

b. Area 1 Area 2 Area 3

Sample Mean 96 94 83Sample Variance 50 40 42

= (96 + 94 + 83)/3 = 91

= 4(96 - 91) 2 + 4(94 - 91) 2 + 4(83 - 91) 2 = 392

MSTR = SSTR /(k - 1) = 392 /2 = 196

= 3(50) + 3(40) + 3(42) = 396

MSTR = SSE /(nT - k) = 396 /(12 - 3) = 44

F = MSTR /MSE = 196 /44 = 4.45



Because p-value > = .05, we cannot reject the null hypothesis that the mean asking prices for all three areas are equal.


Analysis of VarianceSource DF SS MS F PFactor 2 753.3 376.6 18.59 0.000Error 27 546.9 20.3Total 29 1300.2 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ---------+---------+---------+-------SUV 10 58.600 4.575 (-----*-----) Small 10 48.800 4.211 (-----*----) FullSize 10 60.100 4.701 (-----*-----) ---------+---------+---------+-------Pooled StDev = 4.501 50.0 55.0 60.0

Because the p-value = .000 < = .05, we can reject the null hypothesis that the mean resale value is the same. It appears that the mean resale value for small pickup trucks is much smaller than the mean resale value for sport utility vehicles or full-size pickup trucks.


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Analysis of VarianceSource DF SS MS F PFactor 3 2.603 0.868 2.94 0.046Error 36 10.612 0.295Total 39 13.215 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -------+---------+---------+---------Midcap 10 1.2800 0.2394 (--------*--------) Smallcap 10 1.6200 0.3795 (-------*--------) Hybrid 10 1.6000 0.7379 (--------*--------) Specialt 10 2.0000 0.6583 (--------*--------) -------+---------+---------+---------Pooled StDev = 0.5429 1.20 1.60 2.00

Because p-value = .05, we reject the null hypothesis that the mean expense ratios are equal.

50.

LawyerPhysical Therapist

CabinetMaker

Systems Analyst

Sample Mean 50.0 63.7 69.1 61.2Sample Variance 124.22 164.68 105.88 136.62

= 10(50.0 - 61) 2 + 10(63.7 - 61) 2 + 10(69.1 - 61) 2 + 10(61.2 - 61) 2 =

1939.4

MSTR = SSTR /(k - 1) = 1939.4 /3 = 646.47

= 9(124.22) + 9(164.68) + 9(105.88) + 9(136.62) = 4,782.60

MSE = SSE /(nT - k) = 4782.6 /(40 - 4) = 132.85

F = MSTR /MSE = 646.47 /132.85 = 4.87



Because p-value = .05, we reject the null hypothesis that the mean job satisfaction rating is the same for the four professions.

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Analysis of VarianceSource DF SS MS F PFactor 2 4339 2169 3.66 0.039Error 27 15991 592Total 29 20330 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ---+---------+---------+---------+---West 10 108.00 23.78 (-------*-------) South 10 91.70 19.62 (-------*-------) NE 10 121.10 28.75 (-------*------) ---+---------+---------+---------+---Pooled StDev = 24.34 80 100 120 140

Because the p-value = .039 < = .05, we can reject the null hypothesis that the mean rate for the three regions is the same.


ANALYSIS OF VARIANCESOURCE DF SS MS F pFACTOR 3 1271.0 423.7 8.74 0.000ERROR 36 1744.2 48.4TOTAL 39 3015.2 INDIVIDUAL 95 PCT CI'S FOR MEAN BASED ON POOLED STDEVLEVEL N MEAN STDEV --+---------+---------+---------+----West 10 60.000 7.218 (------*-----) South 10 45.400 7.610 (------*-----) N.Cent 10 47.300 6.778 (------*-----) N.East 10 52.100 6.152 (-----*------) --+---------+---------+---------+----POOLED STDEV = 6.961 42.0 49.0 56.0 63.0

Because p-value = 0.05, we reject the null hypothesis that that the mean base salary for art directors is the same for each of the four regions.


Analysis of VarianceSource DF SS MS F PFactor 2 12.402 6.201 9.33 0.001Error 37 24.596 0.665Total 39 36.998 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ------+---------+---------+---------+Receiver 15 7.4133 0.8855 (-------*------) Guard 13 6.1077 0.7399 (-------*------) Tackle 12 7.0583 0.8005 (-------*-------) ------+---------+---------+---------+Pooled StDev = 0.8153 6.00 6.60 7.20 7.80

Because p-value = .05, we reject the null hypothesis that the mean rating for the three positions is the same. It appears that wide receivers and tackles have a higher mean rating than guards.

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Chapter 13

54. x y z

Sample Mean 92 97 84Sample Variance 30 6 35.33

= (92 + 97 + 44) /3 = 91

= 4(92 - 91) 2 + 4(97 - 91) 2 + 4(84 - 91) 2 = 344

MSTR = SSTR /(k - 1) = 344 /2 = 172

= 3(30) + 3(6) + 3(35.33) = 213.99

MSE = SSE /(nT - k) = 213.99 /(12 - 3) = 23.78

F = MSTR /MSE = 172 /23.78 = 7.23



Because p-value = .05, we reject the null hypothesis that the mean absorbency ratings for the three brands are equal.

55. The EXCEL output is shown below:

Analysis of Variance

Summary

Level N Sum Average Variation

Professionals 8 1241.8 155.225 20.66211

Services 8 1183.9 147.9875 1.64185

Manufacturing 8 1147.78 143.4725 2.296879

ANOVA

Source SS DF MS F P-value

Factor 562.3677 2 281.1839 34.28954 0.000000243

Error 172.2059 21 8.200281

Total 734.5736 23

Because p-value = .05, we reject the null hypothesis that the mean consumption is the same for the three sectors.

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56. Method A Method B Method C


= (90 + 84 + 81) /3 = 85

= 10(90 - 85) 2 + 10(84 - 85) 2 + 10(81 - 85) 2 = 420

MSTR = SSTR /(k - 1) = 420 /2 = 210

= 9(98.00) + 9(168.44) + 9(159.78) = 3,836

MSE = SSE /(nT - k) = 3,836 /(30 - 3) = 142.07

F = MSTR /MSE = 210 /142.07 = 1.48



Because p-value > = .05, we can not reject the null hypothesis that the means are equal.

57. Type A Type B Type C Type D

Sample Mean 32,000 27,500 34,200 30,300Sample Variance 2,102,500 2,325,625 2,722,500 1,960,000

= (32,000 + 27,500 + 34,200 + 30,000) /4 = 31,000

= 30(32,000 - 31,000) 2 + 30(27,500 - 31,000) 2 + 30(34,200 - 31,000) 2 +

30(30,300 - 31,000) 2 = 719,400,000

MSTR = SSTR /(k - 1) = 719,400,000 /3 = 239,800,000

= 29(2,102,500) + 29(2,325,625) + 29(2,722,500) + 29(1,960,000)

= 264,208,125

MSE = SSE /(nT - k) = 264,208,125 /(120 - 4) = 2,277,656.25

F = MSTR /MSE = 239,800,000 /2,277,656.25 = 105.28



Because p-value = .05, we reject the null hypothesis that the population means are equal.

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58. Design A Design B Design C


= (90 + 107 + 109) /3 = 102

= 4(90 - 102) 2 +4(107 - 102) 2 +(109 - 102) 2 = 872

MSTR = SSTR /(k - 1) = 872 /2 = 436

= 3(82.67) + 3(68.67) + 3(100.67) = 756.03

MSE = SSE /(nT - k) = 756.03 /(12 - 3) = 84

F = MSTR /MSE = 436 /84 = 5.19



Because p-value = .05, we reject the null hypothesis that the mean lifetime in hours is the same for the three designs.

59. a. Nonbrowser Light Browser Heavy Browser


= (4.25 + 5.25 + 5.75) /3 = 5.08

= 8(4.25 - 5.08) 2 + 8(5.25 - 5.08) 2 + 8(5.75 - 5.08) 2 = 9.33

MSB = SSB /(k - 1) = 9.33 /2 = 4.67

= 7(1.07) + 7(1.07) + 7(1.36) = 24.5

MSW = SSW /(nT - k) = 24.5 /(24 - 3) = 1.17

F = MSB /MSW = 4.67 /1.17 = 3.99



Because p-value = .05, we reject the null hypothesis that the mean comfort scores are the same

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Chapter 13

for the three groups.

b.

Since the absolute value of the difference between the sample means for nonbrowsers and light

browsers is , we cannot reject the null hypothesis that the two population means are

equal.

60. a. Treatment Means:

= 22.8 = 24.8 = 25.80

Block Means:

= 19.67 = 25.67 = 31 = 23.67 = 22.33

Overall Mean:

= 367 /15 = 24.47

Step 1

= (18 - 24.47) 2 + (21 - 24.47) 2 + · · · + (24 - 24.47) 2 = 253.73

Step 2

= 5 [ (22.8 - 24.47) 2 + (24.8 - 24.47) 2 + (25.8 - 24.47) 2 ] = 23.33

Step 3

= 3 [ (19.67 - 24.47) 2 + (25.67 - 24.47) 2 + · · · + (22.33 - 24.47) 2 ] = 217.02

Step 4

SSE = SST - SSTR - SSBL = 253.73 - 23.33 - 217.02 = 13.38

Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatment 23.33 2 11.67 6.99Blocks 217.02 4 54.26 32.49Error 13.38 8 1.67Total 253.73 14



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Because p-value = .05, we reject the null hypothesis that the mean miles per gallon ratings for the three brands of gasoline are equal.

b. I II III


= (22.8 + 24.8 + 25.8) /3 = 24.47

= 5(22.8 - 24.47) 2 + 5(24.8 - 24.47) 2 + 5(25.8 - 24.47) 2 = 23.33

MSTR = SSTR /(k - 1) = 23.33 /2 = 11.67

= 4(21.2) + 4(9.2) + 4(27.2) = 230.4

MSE = SSE /(nT - k) = 230.4 /(15 - 3) = 19.2

F = MSTR /MSE = 11.67 /19.2 = .61



Because p-value > = .05, we cannot reject the null hypothesis that the mean miles per gallon ratings for the three brands of gasoline are equal.

Thus, we must remove the block effect in order to detect a significant difference due to the brand of gasoline. The following table illustrates the relationship between the randomized block design and the completely randomized design.

Sum of SquaresRandomizedBlock Design

CompletelyRandomized Design

SST 253.73 253.73SSTR 23.33 23.33SSBL 217.02 does not existSSE 13.38 230.4

Note that SSE for the completely randomized design is the sum of SSBL (217.02) and SSE (13.38) for the randomized block design. This illustrates that the effect of blocking is to remove the block effect from the error sum of squares; thus, the estimate of for the randomized block design is substantially smaller than it is for the completely randomized design.

61. The blocks correspond to the 15 items in the market basket (Product) and the treatments correspond to the grocery chains (Store).

The Minitab output is shown below:

Analysis of Variance for Price Source DF SS MS F P

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Chapter 13

Product 14 217.236 15.517 64.18 0.000Store 2 2.728 1.364 5.64 0.009Error 28 6.769 0.242Total 44 226.734

Because the p-value for Store = .009 is less than = .05, there is a significant difference in the mean price per item for the three grocery chains.


Analysis of VarianceSource DF SS MS F PFactor 2 731.75 365.88 93.16 0.000Error 63 247.42 3.93Total 65 979.17 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ---+---------+---------+---------+---UK 22 12.052 1.393 (--*--) US 22 14.957 1.847 (--*--) Europe 22 20.105 2.536 (--*--) ---+---------+---------+---------+---Pooled StDev = 1.982 12.0 15.0 18.0 21.0

Because the p-value = .05, we can reject the null hypothesis that the mean download time is the same for Web sites located in the three countries. Note that the mean download time for Web sites located in the United Kingdom (12.052 seconds) is less than the mean download time for Web sites in the United States (14.957) and Web sites located in Europe (20.105).

63.

Step 1

= (8 - 13) 2 + (12 - 13) 2 + · · · + (22 - 13) 2 = 204

Step 2

= 3 (2) [ (12 - 13) 2 + (14 - 13) 2 ] = 12

Step 3

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= 2 (2) [ (9 - 13) 2 + (13.5 - 13) 2 + (16.5 - 13) 2 ] = 114

Step 4

= 2 [(8 - 12 - 9 + 13) 2 + · · · + (22 - 14 - 16.5 +13) 2] = 26

Step 5


Source of Variation Sum of Squares Degrees of Freedom Mean Square FFactor A 12 1 12 1.38Factor B 114 2 57 6.57Interaction 26 2 12 1.50Error 52 6 8.67Total 204 11

Factor A: Actual p-value = .2846. Because p-value > = .05, Factor A (translator) is not significant.

Factor B: Actual p-value = .0308. Because p-value = .05, Factor B (language translated) is significant.

Interaction: Actual p-value = .2963. Because p-value > = .05, Interaction is not significant.

64.

Step 1

= (30 - 26.75) 2 + (34 - 26.75) 2 + · · · + (28 - 26.75) 2 = 151.5

Step 2

= 2 (2) [ (30 - 26.75) 2 + (23.5 - 26.75) 2 ] = 84.5

Step 3

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Chapter 13

= 2 (2) [ (26.5 - 26.75) 2 + (27 - 26.75) 2 ] = 0.5

Step 4

= 2[(30 - 30 - 26.5 + 26.75) 2 + · · · + (28 - 23.5 - 27 + 26.75) 2]

= 40.5Step 5

SSE = SST - SSA - SSB - SSAB = 151.5 - 84.5 - 0.5 - 40.5 = 26

Source of Variation Sum of Squares Degrees of Freedom Mean Square FFactor A 84.5 1 84.5 13Factor B .5 1 .5 .08Interaction 40.5 1 40.5 6.23Error 26 4 6.5Total 151.5 7

Factor A: Actual p-value = .0226. Because p-value = .05, Factor A (machine) is significant.

Factor B: Actual p-value = .7913. Because p-value > = .05, Factor B (loading system) is not significant.

Interaction: Actual p-value = .0671. Because p-value > = .05, Interaction is not significant.

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