Experimental Design And

Chapter 13 Experimental Design and Analysis of Variance

Learning Objectives

1. Understand the basic principles of an experimental study.

2. Understand the difference between a completely randomized design, a randomized block design, and a factorial experiment.

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

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

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

6. Know how to use the analysis of variance procedure to determine if the means of more than two populations are equal for a completely randomized design, a randomized block design, and a factorial experiment.

7. Know how to use the analysis of variance procedure to determine if the means of more than two populations are equal for an observational study.

8. Be able to use output from computer software packages to solve experimental design problems.

9. 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 population means.

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Solutions:

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

k 2 2 2 2 SSTR  nj x j  x  = 6(156 - 144) + 6(142 - 144) + 6(134 - 144) = 1,488 j1

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

2 2 2 c. s1 = 164.4 s2 = 131.2 s3 = 110.4

k 2 SSE (nj  1) s j = 5(164.4) + 5(131.2) +5(110.4) = 2030 j 1

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

e. Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 1488 2 744 5.50 .0162 Error 2030 15 135.3 Total 3518 17

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

Using F table (2 degrees of freedom numerator and 15 denominator), p-value is between .01 and . 025

Using Excel or Minitab, the p-value corresponding to F = 5.50 is .0162.

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

2. Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 300 4 75 14.07 .0000 Error 160 30 5.33 Total 460 34

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

Ha: Not all the population means are equal

b. Using F table (4 degrees of freedom numerator and 30 denominator), p-value is less than .01

Because p-value   = .05, we reject H0

13 - 2 Experimental Design and Analysis of Variance

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

Using F table (2 degrees of freedom numerator and 44 denominator), p-value is less than .01

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

6. A B C Sample Mean 119 107 100 Sample Variance 146.89 96.43 173.78

8(119) 10(107)  10(100) x   107.93 28

k 2 2 2 2 SSTR  nj x j  x  = 8(119 - 107.93) + 10(107 - 107.93) + 10(100 - 107.93) = 1617.9 j1

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

k 2 SSE (nj  1) s j = 7(146.86) + 9(96.44) + 9(173.78) = 3,460 j 1

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

F = MSTR /MSE = 809.95 /138.4 = 5.85

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7. a. Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 4560 2 2280 9.87 .0006 Error 6240 27 231.11 Total 10800 29

b. Using F table (2 degrees of freedom numerator and 27 denominator), p-value is less than .01

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

8. x = (79 + 74 + 66)/3 = 73

k 2 2 2 2 SSTR  nj x j  x  = 6(79 - 73) + 6(74 - 73) + 6(66 - 73) = 516 j1

MSTR = SSTR /(k - 1) = 516/2 = 258

2 2 2 s1 = 34 s2 = 20 s3 = 32

k 2 SSE (nj  1) s j = 5(34) + 5(20) +5(32) = 430 j 1

MSE = SSE /(nT - k) = 430/(18 - 3) = 28.67

F = MSTR /MSE = 258/28.67 = 9.00

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 516 2 258 9.00 .003 Error 430 15 28.67 Total 946 17

Using Excel or Minitab the p-value corresponding to F = 9.00 is .003.

Because p-value   = .05, we reject the null hypothesis that the means for the three plants are equal. In other words, analysis of variance supports the conclusion that the population mean examination score at the three NCP plants are not equal.

50 60 70 Sample Mean 33 29 28 Sample Variance 32 17.5 9.5

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

k 2 2 2 2 SSTR  nj x j  x  = 5(33 - 30) + 5(29 - 30) + 5(28 - 30) = 70 j1

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

k 2 SSE (nj  1) s j = 4(32) + 4(17.5) + 4(9.5) = 236 j 1

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

F = MSTR /MSE = 35 /19.67 = 1.78

Using F table (2 degrees of freedom numerator and 12 denominator), p-value is greater than .10

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

10. Direct Indirect Experience Experience Combination Sample Mean 17.0 20.4 25.0 Sample Variance 5.01 6.26 4.01

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

k 2 2 2 2 SSTR  nj x j  x  = 7(17 - 20.8) + 7(20.4 - 20.8) + 7(25 - 20.8) = 225.68 j1

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

k 2 SSE (nj  1) s j = 6(5.01) + 6(6.26) + 6(4.01) = 91.68 j 1

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

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equal.

11. 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

k 2 2 2 2 2 SSTR  nj x j  x  = 5(133 - 138) + 5(139 - 138) + 5(136 - 138) + 5(144 - 138) = 330 j1

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

k 2 SSE (nj  1) s j = 4(47.5) + 4(50) + 4(21) + 4(54.5) = 692 j 1

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

F = MSTR /MSE = 110 /43.25 = 2.54

Using F table (3 degrees of freedom numerator and 16 denominator), p-value is between .05 and .10

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

12. A B C Sample Mean 20 21 25 Sample Variance 1 25 2.5

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

k 2 2 2 2 SSTR  nj x j  x  = 5(20 - 22) + 5(21 - 22) + 5(25 - 22) = 70 j1

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

k 2 SSE (nj  1) s j = 4(1) + 4(2.5) + 4(2.5) = 24 j 1

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.

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

k 2 2 2 2 SSTR  nj x j  x  = 5(30 - 37) + 5(45 - 37) + 5(36 - 37) = 570 j1

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

k 2 SSE (nj  1) s j = 4(6) + 4(4) + 4(6.5) = 66 j 1

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

F = MSTR /MSE = 285/5.5 = 51.82

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

1 1   1 1  b. LSDt MSE    t 5.5   2.179 2.2  3.23  / 2  .025   ni n j  5 5 

x1 x 2 30  45  15  LSD; significant difference

1 1  c. x1 x 2  t / 2 MSE   n1 n 2 

1 1  (30 45)  2.179 5.5   5 5 

-15  3.23 = -18.23 to -11.77

14. a. Sample 1 Sample 2 Sample 3 Sample Mean 51 77 58 Sample Variance 96.67 97.34 81.99

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

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k 2 2 2 2 SSTR  nj x j  x  = 4(51 - 62) +4(77 - 62) + 4(58 - 62) = 1,448 j1

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

k 2 SSE (nj  1) s j = 3(96.67) + 3(97.34) + 3(81.99) = 828 j 1

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

F = MSTR /MSE = 724/92 = 7.87

Actual p-value = .0106

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

1 1   1 1  b. LSDt MSE    t 92   2.262 46  15.34  / 2  .025   ni n j  4 4 

x1 x 3 51  58  7  LSD; no significant difference

15. a. 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

k 2 2 2 2 SSTR  nj x j  x  = 4(23 - 24) + 4(28 - 24) + 4(21 - 24) = 104 j1

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

k 2 SSE (nj  1) s j = 3(6.67) + 3(4.67) + 3(3.33) = 44.01 j 1

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

F = MSTR /MSE = 52/4.89 = 10.63

Using Excel or Minitab, the p-value corresponding to F = 10.63 is .0043

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.

1 1   1 1  b. LSDt / 2 MSE    t .025 4.89    2.262 2.45  3.54 n1 n 3  4 4 

Since x1 x 3 23  21  2  3.54, there does not appear to be any significant difference between the means for manufacturer 1 and manufacturer 3.

16. x1 x 2  LSD

23 - 28  3.54

-5  3.54 = -8.54 to -1.46

17. a. Marketing Marketing Managers Research Advertising Sample Mean 5 4.5 6 Sample Variance .8 .3 .4

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

k 2 2 2 2 SSTR  nj x j  x  = 6(5 - 5.17) + 6(4.5 - 5.17) + 6(6 - 5.17) = 7.00 j1

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

k 2 SSE (nj  1) s j = 5(.8) + 5(.3) + 5(.4) = 7.50 j 1

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

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

Using Excel or Minitab, the p-value corresponding to F = 7.00 is .0071

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

b. 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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1 1   1 1  BSD 2.602 MSE   2.602 .5   1.06     ni n j  6 6 

x1 x 2 5  4.5  .5  1.06; no significant difference

x1 x 3 5  6  1  1.06; no significant difference

x2 x 3 4.5  6  1.5  1.06; significant difference

18. 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

k 2 2 2 2 2 SSTR  nj x j  x  = 6(7.1 - 9.38) + 6(9.1 - 9.38) + 6(9.9 - 9.38) + 6(11.4 - 9.38) = 57.77 j1

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

k 2 SSE (nj  1) s j = 5(1.21) + 5(.93) + 5(.70) + 5(1.02) = 19.30 j 1

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

1 1   1 1  LSDt MSE    t 0.97   2.086 .3233  1.19  / 2  .025   ni n j  6 6 

x2 x 4 9.1  11.4  2.3  LSD; significant difference

19. 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

for t.004 (20 degrees of freedom)

1 1  BSD 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

20. a. 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

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

b. n1 = 8 n2 = 8 n3 = 8

t/2 is based upon 21 degrees of freedom

1 1  LSDt.025 30.5    2.080 7.6250  5.74 8 8 

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

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21. Treatment Means:

xg1 = 13.6 xg2 = 11.0 xg3 = 10.6

Block Means:

x1g= 9 x2g= 7.67 x3g = 15.67 x4g= 18.67 x5g = 7.67

Overall Mean:

x = 176/15 = 11.73

Step 1

2 2 2 2 SST  xij  x  = (10 - 11.73) + (9 - 11.73) + · · · + (8 - 11.73) = 354.93 i j

Step 2

2 SSTR b x  x 2 2 2  gj  = 5 [ (13.6 - 11.73) + (11.0 - 11.73) + (10.6 - 11.73) ] = 26.53 j

Step 3

2 SSBL k x  x 2 2 2  ig  = 3 [ (9 - 11.73) + (7.67 - 11.73) + (15.67 - 11.73) + i (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 Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 26.53 2 13.27 6.60 .0203 Blocks 312.32 4 78.08 Error 16.08 8 2.01 Total 354.93 14

22. Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 310 4 77.5 17.69 .0005 Blocks 85 2 42.5 Error 35 8 4.38 Total 430 14

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

23. Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 900 3 300 12.60 .0001 Blocks 400 7 57.14 Error 500 21 23.81 Total 1800 31

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

xg1 = 56 xg2 = 44

Block Means:

x1g= 46 x2g= 49.5 x3g = 54.5

Overall Mean:

x = 300/6 = 50

Step 1

2 2 2 2 SST  xij  x  = (50 - 50) + (42 - 50) + · · · + (46 - 50) = 310 i j

Step 2

2 SSTR b x  x 2 2  gj  = 3 [ (56 - 50) + (44 - 50) ] = 216 j

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Step 3

2 SSBL k x  x 2 2 2  ig  = 2 [ (46 - 50) + (49.5 - 50) + (54.5 - 50) ] = 73 i

Step 4

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

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 216 1 216 20.57 .0453 Blocks 73 2 36.5 Error 21 2 10.5 Total 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.

25. The prices were entered into column 1 of the Minitab worksheet. Coding the treatments as 1 for CVS, 2 for Kmart, 3 for Rite-Aid, and 4 for Wegmans, the coded values were entered into column 2. Finally, the corresponding number of each item was entered into column 3. The Minitab output is shown below:

Two-way ANOVA: Price versus Block, Treatment

Source DF SS MS F P Block 12 323.790 26.9825 63.17 0.000 Treatment 3 9.802 3.2672 7.65 0.000 Error 36 15.376 0.4271 Total 51 348.968

S = 0.6535 R-Sq = 95.59% R-Sq(adj) = 93.76%

The p-value corresponding to Treatment is 0.000; because the p-value < a = .05, there is a significant difference in the mean price for the four retail outlets.

xg1 = 16 xg2 = 15 xg3 = 21

Block Means:

x1g= 18.67 x2g= 19.33 x3g = 15.33 x4g= 14.33 x5g = 19

Overall Mean:

x = 260/15 = 17.33

Step 1

2 2 2 2 SST  xij  x  = (16 - 17.33) + (16 - 17.33) + · · · + (22 - 17.33) = 175.33 i j Step 2

2 SSTR b x  x 2 2 2  gj  = 5 [ (16 - 17.33) + (15 - 17.33) + (21 - 17.33) ] = 103.33 j

Step 3

2 SSBL k x  x 2 2 2  ig  = 3 [ (18.67 - 17.33) + (19.33 - 17.33) + · · · + (19 - 17.33) ] = 64.75 i

Step 4

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

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 100.33 2 51.67 56.78 .0000 Blocks 64.75 4 16.19 Error 7.25 8 .91 Total 175.33 14

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

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

Analysis of Variance Source DF SS MS F P Factor 3 19805 6602 22.12 0.000 Error 36 10744 298 Total 39 30550 Individual 95% CIs For Mean Based on Pooled StDev 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

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

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28. Factor B Factor A Level 1 Level 2 Level 3 Means

Level 1 = 150 = 78 = 84 = 104 x11 x12 x13 x1g Factor A

Level 2 = 110 = 116 = 128 = 118 x21 x22 x23 x2g

Factor B Means = 130 = 97 = 106 xg1 xg2 xg3 x = 111

Step 1

2 2 2 2 SST  xijk  x  = (135 - 111) + (165 - 111) + · · · + (136 - 111) = 9,028 i j k

Step 2

2 SSA br x  x 2 2  jg  = 3 (2) [ (104 - 111) + (118 - 111) ] = 588 i

Step 3

2 SSB ar x  x 2 2 2  gj  = 2 (2) [ (130 - 111) + (97 - 111) + (106 - 111) ] = 2,328 j

Step 4

2 SSAB r x  x  x  x 2 2  ij ig g j  = 2 [ (150 - 104 - 130 + 111) + (78 - 104 - 97 + 111) + i j · · · + (128 - 118 - 106 + 111) 2 ] = 4,392

Step 5

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

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 588 1 588 2.05 .2022 Factor B 2328 2 1164 4.06 .0767 Interaction 4392 2 2196 7.66 .0223 Error 1720 6 286.67 Total 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

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

29. Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 26 3 8.67 3.72 .0250 Factor B 23 2 11.50 4.94 .0160 Interaction 175 6 29.17 12.52 .0000 Error 56 24 2.33 Total 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

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30. Factor A is advertising design; Factor B is size of advertisement.

F a c t o r B F a c t o r A

S m a l l L a r g e M e a n s

A = 1 0 = 1 0 = 1 0 x11 x12 x1g

F a c t o r A B x21 = 1 8 x22 = 2 8 x2g= 2 3

C x31 = 1 4 x32 = 1 6 x3g = 1 5

F a c t o r B M e a n s xg1 = 1 4 xg2 = 1 8 x = 1 6

Step 1

2 2 2 2 2 SST  xijk  x  = (8 - 16) + (12 - 16) + (12 - 16) + · · · + (14 - 16) = 544 i j k

Step 2

2 SSA br x  x 2 2 2  ig  = 2 (2) [ (10- 16) + (23 - 16) + (15 - 16) ] = 344 i

Step 3

2 SSB ar x  x 2 2  gj  = 3 (2) [ (14 - 16) + (18 - 16) ] = 48 j

Step 4

2 SSAB r x  x  x  x 2 2  ij ig g j  = 2 [ (10 - 10 - 14 + 16) + · · · + (16 - 15 - 18 +16) ] = 56 i j

Step 5

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

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 344 2 172 172/16 = 10.75 .0104 Factor B 48 1 48 48/16 = 3.00 .1340 Interaction 56 2 28 28/16 = 1.75 .2519 Error 96 6 16 Total 544 11

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

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

Using Excel or Minitab, the p-value corresponding to F =3.00 is .1340.

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.

31. Factor A is method of loading and unloading; Factor B is type of ride.

Factor B Factor A Roller Screaming Log Coaster Demon Flume Means

Method 1 = 42 = 48 = 48 = 46 x11 x12 x13 x1g Factor A

Method 2 = 50 = 48 = 46 = 48 x21 x22 x23 x2g

Step 1

2 2 2 2 SST  xijk  x  = (41 - 47) + (43 - 47) + · · · + (44 - 47) = 136 i j k

Step 2

2 SSA br x  x 2 2  ig  = 3 (2) [ (46 - 47) + (48 - 47) ] = 12 i

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Step 3

2 SSB ar x  x 2 2 2  gj  = 2 (2) [ (46 - 47) + (48 - 47) + (47 - 47) ] = 8 j

Step 4

2 SSAB r x  x  x  x 2 2  ij ig g j  = 2 [ (41 - 46 - 46 + 47) + · · · + (44 - 48 - 47 + 47) ] = 56 i j

Step 5

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 12 1 12 12/10 = 1.2 .3153 Factor B 8 2 4 4/10 = .4 .6870 Interaction 56 2 28 28/10 = 2.8 .1384 Error 60 6 10 Total 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

Using Excel or Minitab, the p-value corresponding to F = .4 is .6870.

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

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

32. Factor A is gender; Factor B is occupation

Factor B Factor A Financial Computer Manager Programmer Pharmacist Means

Male = 979 = 797 = 1047 = 941 x11 x12 x13 x1g Factor A

Female = 635 = 741 = 931 = 769 x21 x22 x23 x2g

Step 1

2 2 2 2 SST  xijk  x  = (872 - 979) + (859 - 979) + · · · + (817 - 931) = 864,432 i j k

Step 2

2 SSA br x  x 2 2  ig  = 3(5) [(941 - 855) + (769 - 855) ] = 221,880 i

Step 3

2 SSB ar x  x 2 2 2  gj  = 2(5) [(807 - 855) + (769 - 855) + (989 - 855) ] = 276,560 j

Step 4

2 SSAB r x  x  x  x 2 2  ij ig g j  = 5[(979 - 941 - 807 - 855) + (797 - 941 - 769 + 855) + i j · · · + (989 - 769 - 989 + 855) 2] = 115,440

Step 5

SSE = SST - SSA - SSB - SSAB = 864,432 - 221,880 - 276,560 - 115,440 = 250,552

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 221,880 1 221,880 21.25 .0001 Factor B 276,560 2 138,280 13.25 .0001 Interaction 115,440 2 57,720 5.53 .0106 Error 250,552 24 10,440 Total 864,432 29

Using F table for Factor A (1 degree of freedom numerator and 24 denominator), p-value is less than .01

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Because p-value   = .05, Factor A (gender) is significant

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

Because p-value   = .05, Factor B (occupation) is significant

33. Factor A is time pressure (low and moderate); Factor B is level of knowledge (naïve, declarative and procedural).

x1g= (1.13 + 1.56 + 2.00)/3 = 1.563

x2g= (0.48 + 1.68 + 2.86)/3 = 1.673

xg1 = (1.13 + 0.48)/2 = 0.805

xg2 = (1.56 + 1.68)/2 = 1.620

xg3 = (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

2 SSA br x  x 2 2  ig  = 3(25)[(1.563 - 1.618) + (1.673 - 1.618) ] = 0.4538 i Step 3

2 SSB ar x  x 2 2 2  gj  = 2(25)[(0.805 - 1.618) + (1.62 - 1.618) + (2.43 - 1.618) ] = 66.0159 j

Step 4

2 SSAB r x  x  x  x 2  ij ig g j  = 25[(1.13 - 1.563 - 0.805 + 1.618) + (1.56 - 1.563 - 1.62 i j

+ 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 Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 0.4538 1 0.4538 0.2648 .6076 Factor B 66.1059 2 33.0080 19.2608 .0000 Interaction 14.2525 2 7.1263 4.1583 .0176 Error 246.7778 144 1.7137 Total 327.5000 149

Factor A: Using Excel or Minitab, the p-value corresponding to F = .2648 is .6076. Because p-value > = .05, Factor A is not significant.

Factor B: Using Excel or Minitab, the p-value corresponding to F = 19.2608 is .0000. Because p- value   = .05, Factor B is significant.

Interaction: Using Excel or Minitab, the p-value corresponding to F = 4.1583 is .0176. Because p- value   = .05, Interaction is significant.

34. x y z Sample Mean 92 97 84 Sample Variance 30 6 35.33

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

k 2 2 2 2 SSTR  nj x j  x  = 4(92 - 91) + 4(97 - 91) + 4(84 - 91) = 344 j1

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

k 2 SSE (nj  1) s j = 3(30) + 3(6) + 3(35.33) = 213.99 j 1

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.

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35. Physical Cabinet Systems Lawyer Therapist Maker Analyst 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  61.2 x   61 4

k 2 2 2 2 2 SSTR  nj x j  x  = 10(50.0 - 61) + 10(63.7 - 61) + 10(69.1 - 61) + 10(61.2 - 61) = 1939.4 j1

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

k 2 SSE (nj  1) s j = 9(124.22) + 9(164.68) + 9(105.88) + 9(136.62) = 4,782.60 j 1

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.

36. The Minitab output is shown below:

Analysis of Variance Source DF SS MS F P Factor 3 2.603 0.868 2.94 0.046 Error 36 10.612 0.295 Total 39 13.215 Individual 95% CIs For Mean Based on Pooled StDev 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

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

37. The Minitab output is shown below:

Analysis of Variance Source DF SS MS F P Factor 2 3840 1920 15.64 0.000 Error 21 2578 123 Total 23 6418 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ------+------+------+------Computer 8 81.25 11.13 (----*-----) Constr. 8 56.88 10.55 (-----*----) Engineer 8 85.63 11.54 (----*-----) ------+------+------+------Pooled StDev = 11.08 60 75 90

Because p-value   = .05, we reject the null hypothesis that the mean salary is the same for the three job fields.

38. Method A Method B Method C Sample Mean 90 84 81 Sample Variance 98.00 168.44 159.78

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

k 2 2 2 2 SSTR  nj x j  x  = 10(90 - 85) + 10(84 - 85) + 10(81 - 85) = 420 j1

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

k 2 SSE (nj  1) s j = 9(98.00) + 9(168.44) + 9(159.78) = 3,836 j 1

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.

39. a. Nonbrowser Light Browser Heavy Browser Sample Mean 4.25 5.25 5.75 Sample Variance 1.07 1.07 1.36

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

k 2 2 2 2 SSTR  nj x j  x  = 8(4.25 - 5.08) + 8(5.25 - 5.08) + 8(5.75 - 5.08) = 9.33 j1

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

k 2 SSW (nj  1) s j = 7(1.07) + 7(1.07) + 7(1.36) = 24.5 j1

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 for the three groups.

1 1   1 1  b. LSDt MSW    2.080 1.17   1.12  / 2     ni n j  8 8 

Since the absolute value of the difference between the sample means for nonbrowsers and light browsers is 4.25 5.25  1, we cannot reject the null hypothesis that the two population means are equal.

40. a. Treatment Means:

x1 = 22.8 x2 = 24.8 x3 = 25.80

Block Means:

x1 = 19.67 x2 = 25.67 x3 = 31 x4 = 23.67 x5 = 22.33

Overall Mean:

x = 367 /15 = 24.47

Step 1

2 2 2 2 SST  xij  x  = (18 - 24.47) + (21 - 24.47) + · · · + (24 - 24.47) = 253.73 i j

Step 2

2 SSTR b x  x 2 2 2  gj  = 5 [ (22.8 - 24.47) + (24.8 - 24.47) + (25.8 - 24.47) ] = 23.33 j

Step 3

2 SSBL k x  x 2 2 2  ig  = 3 [ (19.67 - 24.47) + (25.67 - 24.47) + · · · + (22.33 - 24.47) ] = 217.02 i

Step 4

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

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatment 23.33 2 11.67 6.99 .0175 Blocks 217.02 4 54.26 32.49 Error 13.38 8 1.67 Total 253.73 14

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 Sample Mean 22.8 24.8 25.8 Sample Variance 21.2 9.2 27.2

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

k 2 2 2 2 SSTR  nj x j  x  = 5(22.8 - 24.47) + 5(24.8 - 24.47) + 5(25.8 - 24.47) = 23.33 j1 MSTR = SSTR /(k - 1) = 23.33 /2 = 11.67

k 2 SSE (nj  1) s j = 4(21.2) + 4(9.2) + 4(27.2) = 230.4 j 1

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

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

Using Excel or Minitab, the p-value corresponding to F = .61 is .4406.

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.

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Randomized Completely Sum of Squares Block Design 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  for the randomized block design is substantially smaller than it is for the completely randomized design.

41. 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 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.

42. The prices were entered into column 1 of the Minitab worksheet. Coding the treatments as 1 for Boston, 2 for Miami, 3 for San Diego, 4 for San Jose, and 5 for Washington the coded values were entered into column 2. Finally, the corresponding number of bedrooms was entered into column 3. The Minitab output is shown below:

Two-way ANOVA: Rent versus Bedrooms, Area

Source DF SS MS F P Bedrooms 2 874308 437154 48.29 0.000 Area 4 350202 87550 9.67 0.004 Error 8 72425 9053 Total 14 1296935

S = 95.15 R-Sq = 94.42% R-Sq(adj) = 90.23%

The p-value corresponding to Area is 0.004; because the p-value < α = .05, there is a significant difference in the mean fair market monthly rent for the five metropolitan areas.

43. Factor B Factor A Spanish French German Means

System 1 x = 10 x = 12 x = 14 x = 12 11 12 13 1g Factor A

System 2 x = 8 x = 15 x = 19 x = 14 21 22 23 2g

Factor B Means x = 9 x = 13.5 x = 16.5 g1 g2 g3 x = 13

Step 1

2 2 2 2 SST  xijk  x  = (8 - 13) + (12 - 13) + · · · + (22 - 13) = 204 i j k

Step 2

2 SSA br x  x 2 2  ig  = 3 (2) [ (12 - 13) + (14 - 13) ] = 12 i Step 3

2 SSB ar x  x 2 2 2  gj  = 2 (2) [ (9 - 13) + (13.5 - 13) + (16.5 - 13) ] = 114 j

Step 4

2 SSAB r x  x  x  x 2 2  ij ig g j  = 2 [(8 - 12 - 9 + 13) + · · · + (22 - 14 - 16.5 +13) ] = 26 i j Step 5

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 12 1 12 1.38 .2846 Factor B 114 2 57 6.57 .0308 Interaction 26 2 12 1.50 .2963 Error 52 6 8.67 Total 204 11

Factor A: Using Excel or Minitab, the p-value corresponding to F = 1.38 is .2846. Because p-value >  = .05, Factor A (translator) is not significant.

Factor B: Using Excel or Minitab, the p-value corresponding to F = 6.57 .0308. Because p-value   = .05, Factor B (language translated) is significant.

Interaction: Using Excel or Minitab, the p-value corresponding to F = 1.50 is .2963. Because p-value

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> = .05, Interaction is not significant.

Factor B Factor B Manual Automatic Means

Machine 1 = 32 = 28 = 36 x11 x12 x1g Factor A

Machine 2 = 21 = 26 = 23.5 x21 x22 x2g

Factor B Means = 26.5 = 27 xg1 xg2 x = 26.75

Step 1

2 2 2 2 SST  xijk  x  = (30 - 26.75) + (34 - 26.75) + · · · + (28 - 26.75) = 151.5 i j k

Step 2

2 SSA br x  x 2 2  ig  = 2 (2) [ (30 - 26.75) + (23.5 - 26.75) ] = 84.5 i Step 3

2 SSB ar x  x 2 2  gj  = 2 (2) [ (26.5 - 26.75) + (27 - 26.75) ] = 0.5 j

Step 4

2 SSAB r x  x  x  x 2 2  ij ig g j  = 2[(30 - 30 - 26.5 + 26.75) + · · · + (28 - 23.5 - 27 + 26.75) ] i j

= 40.5 Step 5 SSE = SST - SSA - SSB - SSAB = 151.5 - 84.5 - 0.5 - 40.5 = 26

Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Factor A 84.5 1 84.5 13 .0226 Factor B .5 1 .5 .08 .7913 Interaction 40.5 1 40.5 6.23 .0671 Error 26 4 6.5 Total 151.5 7

Factor A: Using Excel or Minitab, the p-value corresponding to F = 13 is .0226. Because p-value   = .05, Factor A (machine) is significant.

Factor B: Using Excel or Minitab, the p-value corresponding to F = .08 is .7913. Because p-value >  = .05, Factor B (loading system) is not significant.

Interaction: Using Excel or Minitab, the p-value corresponding to F = 6.23 is .0671. Because p-value > = .05, Interaction is not significant.

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