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Chapter 13

Analysis of Variance and

Experimental 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 Fdistribution 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 Fishers least significant difference (LSD) procedure and Fishers 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 partitioning

experimentwise Type I error rate blocking

factor main effect

level interaction

treatment

replication

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Chapter 13

Solutions:

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

( )2

1

SSTRk

j j

j

n x x=

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

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

b.2

1

SSE ( 1)k

j j

j

n s=

= = 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 Ftable (2 degrees of freedom numerator and 12 denominator),p-value is less than .01

Actualp-value = .0000

Becausep-value = .05, we reject the null hypothesis that the means of the three populations areequal.

d.

Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Treatments 570 2 285 51.82

Error 66 12 5.5

Total 636 14

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

( )2

1

SSTRk

j j

j

n x x=

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

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

b.2

1

SSE ( 1)k

j j

j

n s=

= = 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 Ftable (2 degrees of freedom numerator and 9 denominator),p-value is greater than .10


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

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Analysis of Variance and Experimental Design

d.


Treatments 536 2 268 2.91

Error 828 9 92

Total 1364 11

3. a. 4(100) 6(85) 5(79) 8715

x + += =

( )2

1

SSTRk

j j

j

n x x=

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

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

b.2

1

SSE ( 1)k

j j

j

n s=

= = 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.


Treatments 1020 2 510 13.36Error 458 12 38.17

Total 1478 14

4. a.


Treatments 1200 3 400 80

Error 300 60 5

Total 1500 63

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


Becausep-value = .05, we reject the null hypothesis that the means of the 4 populations areequal.

5. a.


Treatments 120 2 60 20

Error 216 72 3

Total 336 74

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Chapter 13

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


Becausep-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 21

Sample Variance 6.67 4.67 3.33

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

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-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 5.25


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

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

= = 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

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Becausep-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 6

Sample Variance .8 .3 .4

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

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-value = .05, we reject the null hypothesis that the mean perception score is the samefor the three groups of specialists.

9.

Real Estate

Agent Architect Stockbroker

Sample Mean 67.73 61.13 65.80

Sample Variance 117.72 180.10 137.12

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

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

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

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Chapter 13

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

F= MSTR /MSE = 172.74/144.98 = 1.19



Becausep-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 Variance

Source DF SS MS F P

Factor 2 226.1 113.0 3.70 0.042

Error 21 640.8 30.5

Total 23 866.9

Individual 95% CIs For Mean

Based on Pooled StDev

Level 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

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

11. a / 2 .0251 1 1 1

LSD MSE 5.5 2.179 2.2 3.235 5i j

t tn n

= + = + = =

1 2 30 45 15x x = = > LSD; significant difference



b. 1 2 / 21 2

1 1MSE x x t

n n

+

1 1(30 45) 2.179 5.5

5 5

+

-15 3.23 = -18.23 to -11.77

12. a.

Sample 1 Sample 2 Sample 3



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

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1

SSE ( 1)k

j j

j

n s=

= = 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 Ftable (2 degrees of freedom numerator and 9 denominator),p-value is between .01 and .025



b. / 2 .0251 1 1 1

LSD MSE 92 2.262 46 15.344 4i j

t tn n

= + = + = =


1 3 51 58 7x x = = < LSD; no significant difference


13. / 2 .0251 3

1 1 1 1LSD MSE 4.89 2.262 2.45 3.54

4 4t t

n n

= + = + = =

Since 1 3 23 21 2 3.54,x x = = < there does not appear to be any significant difference betweenthe means of population 1 and population 3.

14. 1 2 LSDx x

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

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Chapter 13

1 1 1 1BSD 2.602 MSE 2.602 .5 1.06

6 6i jn n

= + = + =

1 2 5 4.5 .5x x = = < 1.06; no significant difference

1 3 5 6 1x x = = < 1.06; no significant difference

2 3 4.5 6 1.5x x = = > 1.06; significant difference

16. a.

Machine 1 Machine 2 Machine 3 Machine 4

Sample Mean 7.1 9.1 9.9 11.4

Sample Variance 1.21 .93 .70 1.02

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

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



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

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

/ 2 .025

1 1 1 1LSD MSE 0.97 2.086 .3233 1.19

6 6i jt t

n n

= + = + = =

2 4 9.1 11.4 2.3x x = = > LSD; significant difference

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 ttable is .005, we will use t.005 = 2.845 as an approximation

for t.004 (20 degrees of freedom)

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1 1BSD 2.845 0.97 1.62

6 6

= + =

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 1.5

Significant ? Yes Yes Yes No Yes No

18. n1 = 8 n2 = 8 n3 = 8

t /2 is based upon 21 degrees of freedom

.025

1 1LSD 30.5 2.080 7.6250 5.74

8 8t

= + = =

Comparing Small and Medium

92.20 89.65 2.55 = < LSD; no significant difference

Comparing Small and Large

92.20 84.80 7.40 = > LSD; significant difference

Comparing Medium and Large

89.65 84.80 4.85 = < LSD; no significant difference

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

( )2

1

SSTRk

j j

j

n x x=

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

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

c.2

1s = 164.42

2s = 131.22

3s = 110.4

2

1

SSE ( 1)k

j j

j

n s=

=

= 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

Using Ftable (2 degrees of freedom numerator and 15 denominator),p-value is between .01 and .

025

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Chapter 13


Becausep-value = .05, we reject the hypothesis that the means for the three treatments areequal.

20. a.

Source of Variation Sum of Squares Degrees of Freedom Mean Square FTreatments 1488 2 744 5.50

Error 2030 15 135.3

Total 3518 17

b. / 21 1 1 1

LSD MSE 2.131 135.3 14.316 6i j

tn n

= + = + =

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

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

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

21.



Error 160 30 5.33

Total 460 34

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

Ha: Not all the population means are equal



Becausep-value = .05, we reject H0

23.



Error 250 16 15.63

Total 400 18


025


Becausep-value = .05, we reject the null hypothesis that the means of the three treatments areequal.

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24.


Treatments 1200 2 600 43.99

Error 600 44 13.64

Total 1800 46



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

25.

A B C



8(119) 10(107) 10(100)

107.9328x

+ += =

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

= = 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.


Treatments 4560 2 2280 9.87

Error 6240 27 231.11Total 10800 29



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Chapter 13

Becausep-value = .05, we reject the null hypothesis that the means of the three assemblymethods are equal.

27.


Between 61.64 3 20.55 17.56

Error 23.41 20 1.17Total 85.05 23



Becausep-value = .05, we reject the null hypothesis that the mean breaking strength of the fourcables is the same.

28.

50 60 70Sample Mean 33 29 28

Sample Variance 32 17.5 9.5

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

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-value > = .05, we cannot reject the null hypothesis that the mean yields for the three

temperatures are equal.

29.

Direct

Experience

Indirect

Experience Combination

Sample Mean 17.0 20.4 25.0


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

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( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1SSE ( 1)

k

j j

jn s

== = 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



Becausep-value = .05, we reject the null hypothesis that the means for the three groups areequal.

30.

Paint 1 Paint 2 Paint 3 Paint 4

Sample Mean 13.3 139 136 144

Sample Variance 47.5 .50 21 54.5

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

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



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

31.

A B C


Sample Variance 1 25 2.5

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

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Chapter 13

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1SSE ( 1)

k

j j

jn s

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

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

F= MSTR /MSE = 35 /2 = 17.5



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

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

/ 2 .025

1 1 1 1LSD MSE 5.09 2.101 1.4543 2.53

7 7i jt t

n n

= + = + = =

1 2 17.0 20.4 3.4x x = = > 2.53; significant difference

1 3 17.0 25.0 8x x = = > 2.53; significant difference

2 3 20.4 25 4.6x x = = > 2.53; significant difference

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

/ 2 .025

1 1 1 1LSD MSE 2 2.179 .8 1.95

5 5i jt t

n n

= + = + = =

1 2 20 21 1x x = = < 1.95; no significant difference

1 3 20 25 5x x = = > 1.95; significant difference

2 3 21 25 4x x = = > 1.95; significant difference

34. Treatment Means:

1xg = 13.6 2xg = 11.0 3xg = 10.6

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Block Means:

1x g= 9 2x g= 7.67 3x g= 15.67 4x g= 18.67 5x g= 7.67

Overall Mean:

x = 176/15 = 11.73

Step 1

( )2

SST iji j

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

Step 2

( )2

SSTR jj

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

Step 3

( )2

SSBL ii

k x x= g = 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


Treatments 26.53 2 13.27 6.60

Blocks 312.32 4 78.08

Error 16.08 8 2.01

Total 354.93 14




35.


Treatments 310 4 77.5 17.69Blocks 85 2 42.5

Error 35 8 4.38

Total 430 14


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Chapter 13


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

36.


Treatments 900 3 300 12.60Blocks 400 7 57.14

Error 500 21 23.81

Total 1800 31



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


1xg = 56 2xg = 44

Block Means:

1x g= 46 2x g= 49.5 3x g= 54.5

Overall Mean:

x = 300/6 = 50

Step 1

( )2

SST iji j

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

Step 2

( )2

SSTR jj

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

Step 3

( )2

SSBL ii

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

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Step 4

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


Treatments 216 1 216 20.57

Blocks 73 2 36.5Error 21 2 10.5

Total 310 5

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


Becausep-value = .05, we reject the null hypothesis that the mean tune-up times are the samefor both analyzers.

38.


Treatments 45 4 11.25 7.12Blocks 36 3 12

Error 19 12 1.58

Total 100 19



Becausep-value = .05, we reject the null hypothesis that the mean total audit times for the fiveauditing procedures are equal.


1xg = 16 2xg = 15 3xg = 21

Block Means:

1x g= 18.67 2x g= 19.33 3x g= 15.33 4x g= 14.33 5x g= 19

Overall Mean:

x = 260/15 = 17.33

Step 1

( )2

SST iji j

x x= = (16 - 17.33) 2 + (16 - 17.33) 2 + + (22 - 17.33) 2 = 175.33Step 2

( )2

SSTR jj

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

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Chapter 13

Step 3

( )2

SSBL ii

k x x= g = 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


Treatments 100.33 2 51.67 56.78

Blocks 64.75 4 16.19

Error 7.25 8 .91

Total 175.33 14


Actualp

-value = .0000

Becausep-value = .05, we reject the null hypothesis that the mean times for the three systemsare equal.

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


Source DF SS MS F P

Factor 3 19805 6602 22.12 0.000

Error 36 10744 298

Total 39 30550



Level 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

Becausep-value =.05, we reject the null hypothesis that the mean heart rate for the four methodsare equal.

41.

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Factor A

Level 2

Factor B Means

11x = 150

Level 1 Level 2 Level 3 Means

Factor B Factor A

Level 112x = 78 13x = 84 1x g= 104

21x = 110

1xg = 130

22x= 116

2xg = 97

23x = 128

3xg = 106

2x g= 118

x = 111

Step 1

( )2

SST ijki j k

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

Step 2

( )2

SSA ji

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

Step 3

( )2

SSB jj

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

Step 4

( )2

SSAB ij i ji j

r x x x x= + g g = 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


Factor A 588 1 588 2.05

Factor B 2328 2 1164 4.06

Interaction 4392 2 2196 7.66

Error 1720 6 286.67Total 9028 11

Factor A: F= 2.05

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

Actual p-value = .2022

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

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Chapter 13

Factor B: F= 4.06

UsingFtable (2 degrees of freedom numerator and 6 denominator),p-value is between .05 and .10


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

Interaction:F= 7.66

UsingFtable (2 degrees of freedom numerator and 6 denominator),p-value is between .01 and .025


Becausep-value = .05, Interaction is significant

42.


Factor A 26 3 8.67 3.72

Factor B 23 2 11.50 4.94

Interaction 175 6 29.17 12.52

Error 56 24 2.33

Total 280 35

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

Becausep-value = .05, Factor A is significant.

Using Ftable for Factor B (2 degrees of freedom numerator and 24 denominator),p-value is

between .01 and .025


Becausep-value = .05, Factor B is significant.

Using Ftable for Interaction (6 degrees of freedom numerator and 24 denominator),p-value is less

than .01


Becausep-value = .05, Interaction is significant

43.

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Factor A B

Factor B Means

11x = 10

Small Large Means

Factor B Factor B

12x = 10 1x g= 10

21x = 18

1xg = 14

22x = 28

2xg = 18

2x g= 23

x = 16

A

C31x = 14 32x = 16 3x g= 15

Step 1

( )2

SST ijki j k

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

Step 2

( )2

SSA ii

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

Step 3

( )2

SSB jj

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

Step 4

( )2

SSAB ij i ji j

r x x x x= + g g = 2 [ (10 - 10 - 14 + 16) 2 + + (16 - 15 - 18 +16) 2 ] = 56

Step 5

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


Factor A 344 2 172 172/16 = 10.75

Factor B 48 1 48 48/16 = 3.00Interaction 56 2 28 28/16 = 1.75

Error 96 6 16

Total 544 11

Using Ftable for Factor A (2 degrees of freedom numerator and 6 denominator),p-value is

between .01 and .025


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Chapter 13

Becausep-value = .05, Factor A is significant; there is a difference due to the type ofadvertisement design.

Using Ftable for Factor B (1 degree of freedom numerator and 6 denominator),p-value is greater

than .01


Becausep-value > = .05, Factor B is not significant; there is not a significant difference due to

size of advertisement.

Using Ftable for Interaction (2 degrees of freedom numerator and 6 denominator),p-value is greater

than .10


Becausep-value > = .05, Interaction is not significant.

44.

Factor A

Method 2

Factor B Means

11x = 42

Roller

Coaster

Screaming

Demon

LogFlume Means

Factor B Factor A

Method 112x = 48 13x = 48 1x g= 46

21x = 50

1xg = 46

22x = 48

2xg = 48

23x = 46

3xg = 47

2x g= 48

x = 47

Step 1

( )2

SST ijki j k

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

Step 2

( )2

SSA ii

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

Step 3

( )2

SSB jj

ar x x= g = 2 (2) [ (46 - 47) 2 + (48 - 47) 2 + (47 - 47) 2 ] = 8Step 4

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( )2

SSAB ij i ji j

r x x x x= + g g = 2 [ (41 - 46 - 46 + 47) 2 + + (44 - 48 - 47 + 47) 2 ] = 56

Step 5



Factor A 12 1 12 12/10 = 1.2

Factor B 8 2 4 4/10 = .4

Interaction 56 2 28 28/10 = 2.8

Error 60 6 10

Total 136 11

UsingFtable for Factor A (1 degree of freedom numerator and 6 denominator),p-value is greater

than .10


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

UsingFtable for Factor B (2 degrees of freedom numerator and 6 denominator), p-value is greater

than .10


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

UsingFtable for Interaction (2 degrees of freedom numerator and 6 denominator),p-value is greaterthan .10


Becausep-value >= .05, Interaction is not significant

45.

Factor B Factor A

Under junior

high school

Senior high

schoolabove college Means

Factor A

Male11

x

=29.5114

12x

=35.2943 13x

=50.3092

1x

=38.3716

Female 21x =20.6957 22x

=27.1644 23x

= 42.0532x

=29.9709

Factor B Means1

x

=25.1036

2x

=31.2294 3x

=46.1809 x =34.1713

Step 1

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Chapter 13

( )2

SST ijki j k

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

= 3,270.7354

Step 2

( )2

SSA ii

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

Step 3

( )2

SSB jj

ar x x= g = 2(2) [(25.1036 - 34.1713) 2 + (31.2294 - 34.1713) 2 + (46.1809 - 34.1713) 2]= 940.4381

Step 4

( )2

SSAB ij i ji j

r x x x x= + g g = 2[(29.5114 - 38.3716 - 25.1036 + 34.1713 ) 2 + + (42.0526- 29.9709 - 46.1809 + 34.1713) 2] = 0.2663

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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 2.3987Factor 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

UsingFtable for Factor A (1 degree of freedom numerator and 24 denominator), p-value is greater

than .1


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

UsingFtable for Factor B (2 degrees of freedom numerator and 24 denominator), p-value is lessthan .05


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

UsingFtable for Interaction (2 degrees of freedom numerator and 24 denominator),p-value is

greater than .9

Actual p-value =.9985

Becausep-value > = .05, Interaction is not significant

46. 1x g= (1.13 + 1.56 + 2.00)/3 = 1.563

2x g= (0.48 + 1.68 + 2.86)/3 = 1.673

1xg = (1.13 + 0.48)/2 = 0.805

2xg = (1.56 + 1.68)/2 = 1.620

3xg = (2.00 + 2.86)/2 = 2.43

x = (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

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Chapter 13

( )2

SSA ii

br x x= g = 3(25)[(1.563 - 1.618)2 + (1.673 - 1.618)2] = 0.4538Step 3

( )2

SSB jj

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

Step 4

( )2

SSAB ij i ji j

r x x x x= + g g = 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


Factor A 0.4538 1 0.4538 0.2648

Factor B 66.1059 2 33.0080 19.2608

Interaction 14.2525 2 7.1263 4.1583

Error 246.7778 144 1.7137

Total 327.5000 149

Factor A: Actualp-value = .6076. Becausep-value > = .05, Factor A is not significant.

Factor B: Actualp-value = .0000. Becausep-value = .05, Factor B is significant.

Interaction: Actualp-value = .0176. Becausep-value = .05, Interaction is significant.

47. a.Area 1 Area 2

Sample Mean 96 94

Sample Variance 50 40

x = (96 + 94)/2 = 95

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1

SSE ( 1)k

j j

j

n s=

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

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

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

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

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

b.Area 1 Area 2 Area 3


Sample Variance 50 40 42

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

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1

SSE ( 1)k

j j

j

n s=

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

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

F= MSTR /MSE = 196 /44 = 4.45



Becausep-value > = .05, we cannot reject the null hypothesis that the mean asking prices for all

three areas are equal.



Source DF SS MS F P

Factor 2 753.3 376.6 18.59 0.000

Error 27 546.9 20.3

Total 29 1300.2



Level 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 themean resale value for sport utility vehicles or full-size pickup trucks.

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Chapter 13



Source DF SS MS F P

Factor 3 2.603 0.868 2.94 0.046

Error 36 10.612 0.295

Total 39 13.215Individual 95% CIs For Mean


Level 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

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

50.

LawyerPhysicalTherapist

CabinetMaker

SystemsAnalyst

Sample Mean 50.0 63.7 69.1 61.2

Sample Variance 124.22 164.68 105.88 136.62

50.0 63.7 69.1 61.261

4x

+ + += =

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-value

= .05, we reject the null hypothesis that the mean job satisfaction rating is the

same for the four professions.

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Source DF SS MS F P

Factor 2 4339 2169 3.66 0.039

Error 27 15991 592

Total 29 20330



Level 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 VARIANCE

SOURCE DF SS MS F p

FACTOR 3 1271.0 423.7 8.74 0.000

ERROR 36 1744.2 48.4

TOTAL 39 3015.2

INDIVIDUAL 95 PCT CI'S FOR MEAN

BASED ON POOLED STDEV

LEVEL 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

Becausep-value = 0.05, we reject the null hypothesis that that the mean base salary for artdirectors is the same for each of the four regions.



Source DF SS MS F P

Factor 2 12.402 6.201 9.33 0.001

Error 37 24.596 0.665

Total 39 36.998Individual 95% CIs For Mean


Level 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

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Chapter 13

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

54.

x y z

Sample Mean 92 97 84Sample Variance 30 6 35.33

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

( )2

1

SSTRk

j j

j

n x x=

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

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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-value = .05, we reject the null hypothesis that the mean absorbency ratings for thethree brands are equal.

55. The EXCEL output is shown below:


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

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Becausep-value = .05, we reject the null hypothesis that the mean consumption is the same forthe three sectors.

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Chapter 13

56.

Method A Method B Method C



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

( )2

1

SSTRk

j j

j

n x x=

= = 10(90 - 85) 2 + 10(84 - 85) 2 + 10(81 - 85) 2 = 420MSTR = SSTR /(k- 1) = 420 /2 = 210

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-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 30,300

Sample Variance 2,102,500 2,325,625 2,722,500 1,960,000

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

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



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

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58.

Design A Design B Design C



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

( )2

1

SSTRk

j j

j

n x x=

= = 4(90 - 102) 2 +4(107 - 102) 2 +(109 - 102) 2 = 872MSTR = SSTR /(k- 1) = 872 /2 = 436

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-value = .05, we reject the null hypothesis that the mean lifetime in hours is the samefor the three designs.

59. a.

Nonbrowser Light Browser Heavy Browser

Sample Mean 4.25 5.25 5.75


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

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSW ( 1)k

j j

j

n s=

= = 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


05


Becausep-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. / 21 1 1 1

LSD MSW 2.080 1.17 1.128 8i j

tn n

= + = + =

Since the absolute value of the difference between the sample means for nonbrowsers and lightbrowsers is 4.25 5.25 1 = , we cannot reject the null hypothesis that the two population means areequal.

60. a. Treatment Means:

1x = 22.8 2x = 24.8 3x = 25.80

Block Means:

1x = 19.67 2x = 25.67 3x = 31 4x = 23.67 5x = 22.33

Overall Mean:

x = 367 /15 = 24.47

Step 1

( )2

SST iji j

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

Step 2

( )

2

SSTR jj

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

Step 3

( )2

SSBL ii

k x x= g = 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


Treatment 23.33 2 11.67 6.99

Blocks 217.02 4 54.26 32.49

Error 13.38 8 1.67

Total 253.73 14



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

b.

I II III

Sample Mean 22.8 24.8 25.8Sample Variance 21.2 9.2 27.2

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

( )2

1

SSTRk

j j

j

n x x=

= = 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

2

1

SSE ( 1)k

j j

j

n s=

= = 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



Becausep-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 ofgasoline. The following table illustrates the relationship between the randomized block design and

the completely randomized design.

Sum of Squares

Randomized

Block Design

Completely

Randomized Design

SST 253.73 253.73

SSTR 23.33 23.33

SSBL 217.02 does not exist

SSE 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 2 for the randomized block design issubstantially 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

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Chapter 13

Source DF SS MS F P

Product 14 217.236 15.517 64.18 0.000

Store 2 2.728 1.364 5.64 0.009

Error 28 6.769 0.242

Total 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.



Source DF SS MS F P

Factor 2 731.75 365.88 93.16 0.000

Error 63 247.42 3.93

Total 65 979.17



Level 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 thesame 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.

Factor A

System 2

Factor B Means

11x = 10

Spanish French German Means

Factor B Factor A

System 112x = 12 13x = 14 1x g= 12

21x = 8

1xg = 9

22x= 15

2xg = 13.5

23x = 19

3xg = 16.5

2x g= 14

x = 13

Step 1

( )2

SST ijki j k

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

Step 2

( )2

SSA ii

br x x= g = 3 (2) [ (12 - 13) 2 + (14 - 13) 2 ] = 12Step 3

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( )2

SSB jj

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

Step 4

( )2

SSAB ij i ji j

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

Step 5



Factor A 12 1 12 1.38

Factor B 114 2 57 6.57

Interaction 26 2 12 1.50

Error 52 6 8.67

Total 204 11

Factor A: Actualp-value = .2846. Becausep-value > = .05, Factor A (translator) is not significant.

Factor B: Actualp-value = .0308. Becausep-value = .05, Factor B (language translated) issignificant.

Interaction: Actualp-value = .2963. Becausep-value > = .05, Interaction is not significant.

64.

Factor A

Machine 2

Factor B Means

11x = 32

Manual Automatic Means

Factor B Factor B

12x = 28 1x g = 36

21x = 21

1xg = 26.5

22x = 26

2xg = 27

2x g= 23.5

x = 26.75

Machine 1

Step 1

( )2

SST ijki j k

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

Step 2

( )2

SSA ii

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

Step 3

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Chapter 13

( )2

SSB jj

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

Step 4

( )2

SSAB ij i ji j

r x x x x= + g g = 2[(30 - 30 - 26.5 + 26.75) 2 + + (28 - 23.5 - 27 + 26.75) 2]= 40.5

Step 5

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


Factor A 84.5 1 84.5 13

Factor B .5 1 .5 .08

Interaction 40.5 1 40.5 6.23

Error 26 4 6.5

Total 151.5 7

Factor A: Actualp-value = .0226. Becausep-value = .05, Factor A (machine) is significant.

Factor B: Actualp-value = .7913. Becausep-value > = .05, Factor B (loading system) is not

significant.

Interaction: Actualp-value = .0671. Becausep-value > = .05, Interaction is not significant.

sbe_sm13

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