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10 EXAMPLE: EFFECTIVENESS OF PAINT ON BRICKS WITH UNEQUAL SLOPES

10 EXAMPLE: EFFECTIVENESS OF PAINT ON BRICKS WITH UNEQUAL SLOPES

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Analysis of Covariance for Nonreplicated Experiments



TABLE 17.14

Estimate Statements for PROC GLM to Provide Estimates of the Adjusted

Means and Comparison of the Adjusted Means

estimate ‘d=1’ intercept 1 d 1 x 21.4 z 1.0875;

estimate ‘d=–1’ intercept 1 d –1 x 21.4 z 1.0875;

estimate ‘d=1 minus d=–1’ d 2;

estimate ‘a=–1,b=–1,c=–1 d=–1’ intercept 1 d –1 a*b 1 b*c 1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=–1,b=–1,c=1 d=–1’ intercept 1 d –1 a*b 1 b*c –1 a*b*c

1 x 21.4 z 1.0875;

estimate ‘a=–1,b=1,c=–1 d=–1’ intercept 1 d –1 a*b –1 b*c –1 a*b*c

1 x 21.4 z 1.0875;

estimate ‘a=–1,b=1,c= 1 d=–1’ intercept 1 d –1 a*b –1 b*c 1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=–1,b=–1,c=–1 d=1’ intercept 1 d 1 a*b 1 b*c 1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=–1,b=–1,c=1 d=1’ intercept 1 d 1 a*b 1 b*c –1 a*b*c 1

x 21.4 z 1.0875;

estimate ‘a=–1,b=1,c=–1 d=1’ intercept 1 d 1 a*b –1 b*c –1 a*b*c

1 x 21.4 z 1.0875;

estimate ‘a=–1,b=1,c= 1 d=1’ intercept 1 d 1 a*b –1 b*c 1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=1,b=–1,c=–1 d=–1’ intercept 1 d –1 a*b –1 b*c 1 a*b*c

1 x 21.4 z 1.0875;

estimate ‘a=1,b=–1,c=1 d=–1’ intercept 1 d –1 a*b –1 b*c –1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=1,b=1,c=–1 d=–1’ intercept 1 d –1 a*b 1 b*c –1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=1,b=1,c= 1 d=–1’ intercept 1 d –1 a*b 1 b*c 1 a*b*c

1 x 21.4 z 1.0875;

estimate ‘a=1,b=–1,c=–1 d=1’ intercept 1 d 1 a*b –1 b*c 1 a*b*c 1

x 21.4 z 1.0875;

estimate ‘a=1,b=–1,c=1 d=1’ intercept 1 d 1 a*b –1 b*c –1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=1,b=1,c=–1 d=1’ intercept 1 d 1 a*b 1 b*c –1 a*b*c

–1 x 21.4 z 1.0875;

estimate ‘a=1,b=1,c= 1d=1’ intercept 1 d 1 a*b 1 b*c 1 a*b*c

1 x 21.4 z 1.0875;

estimate ‘a=1 minus a=–1,@b=–1,c=–1’ a*b –2 a*b*c 2;

estimate ‘a=1 minus a=–1,@b=–1,c= 1’ a*b –2 a*b*c –2;

estimate ‘a=1 minus a=–1,@b= 1,c=–1’ a*b 2 a*b*c –2;

estimate ‘a=1 minus a=–1,@b= 1,c= 1’ a*b 2 a*b*c 2;

estimate ‘b=1 minus b=–1,@a=–1,c=–1’ a*b –2 b*c –2 a*b*c 2;

estimate ‘b=1 minus b=–1,@a=–1,c= 1’ a*b –2 b*c 2 a*b*c –2;

estimate ‘b=1 minus b=–1,@a= 1,c=–1’ a*b 2 b*c –2 a*b*c –2;

estimate ‘b=1 minus b=–1,@a= 1,c= 1’ a*b 2 b*c 2 a*b*c 2;

estimate ‘c=1 minus c=–1,@a=–1,b=–1’ b*c –2 a*b*c 2;

estimate ‘c=1 minus c=–1,@a=–1,b= 1’ b*c 2 a*b*c –2;

estimate ‘c=1 minus c=–1,@a= 1,b=–1’ b*c –2 a*b*c –2;

estimate ‘c=1 minus c=–1,@a= 1,b= 1’ b*c 2 a*b*c 2;

Note: Results are in Table 17.15.



© 2002 by CRC Press LLC



29



30



Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 17.15

Results from the Estimate Statements in Table 17.14 with

Adjusted Means and Comparisons of Adjusted Means

for the Levels of a, b, c, and d

Parameter

d=1

d=–1

d=1 minus d=–1

a=–1,b=–1,c=–1 d=–1

a=–1,b=–1,c=1 d=–1

a=–1,b=1,c=–1 d=–1

a=–1,b=1,c= 1 d=–1

a=–1,b=–1,c=–1 d=1

a=–1,b=–1,c=1 d=1

a=–1,b=1,c=–1 d=1

a=–1,b=1,c= 1 d=1

a=1,b=–1,c=–1 d=–1

a=1,b=–1,c=1 d=–1

a=1,b=1,c=–1 d=–1

a=1,b=1,c= 1 d=–1

a=1,b=–1,c=–1 d=1

a=1,b=–1,c=1 d=1

a=1,b=1,c=–1 d=1

a=1,b=1,c= 1d=1

a=1 minus a=–1,@b=–1,c=–1

a=1 minus a=–1,@b=–1,c= 1

a=1 minus a=–1,@b= 1,c=–1

a=1 minus a=–1,@b= 1,c= 1

b=1 minus b=–1,@a=–1,c=–1

b=1 minus b=–1,@a=–1,c= 1

b=1 minus b=–1,@a= 1,c=–1

b=1 minus b=–1,@a= 1,c= 1

c=1 minus c=–1,@a=–1,b=–1

c=1 minus c=–1,@a=–1,b= 1

c=1 minus c=–1,@a= 1,b=–1

c=1 minus c=–1,@a= 1,b= 1



Estimate

57.1953

68.9547

–11.7593

74.5079

77.7039

63.4014

60.2054

62.7486

65.9446

51.6421

48.4461

55.9668

67.6400

81.9425

70.2693

44.2075

55.8807

70.1832

58.5100

–18.5411

–10.0639

18.5411

10.0639

–11.1065

–17.4984

25.9756

2.6293

3.1960

–3.1960

11.6732

–11.6732



StdErr

1.8783

1.8783

2.8999

2.7640

3.0334

3.0538

2.9549

3.0538

2.9549

2.7640

3.0334

3.2021

2.7637

2.6827

3.0788

2.6827

3.0788

3.2021

2.7637

3.7769

3.8599

3.7769

3.8599

4.4519

4.6642

4.5595

4.4858

3.8430

3.8430

3.7997

3.7997



tValue

30.45

36.71

–4.06

26.96

25.62

20.76

20.37

20.55

22.32

18.68

15.97

17.48

24.47

30.54

22.82

16.48

18.15

21.92

21.17

–4.91

–2.61

4.91

2.61

–2.49

–3.75

5.70

0.59

0.83

–0.83

3.07

–3.07



Probt

0.0000

0.0000

0.0029

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0008

0.0284

0.0008

0.0284

0.0342

0.0045

0.0003

0.5722

0.4271

0.4271

0.0133

0.0133



time and porosity data for each of the treatment combinations is in Figure 17.19.

This plot does not provide any indication that there is a relationship between the

abrasion time and porosity, but that relationship is confounded with the factorial

treatment effects. From prior experience, the relationship between resistance to

abrasion and porosity can depend on the method of application, i.e., there may be

different slopes for spray and brush. The possible model that uses a linear relationship

between the mean of the abrasion time and porosity is

y ijkm = µ ijkm + βi x ijkm + ε ijkm for i = −1, 1, j = −1, 1, k = −1, 1, m = −1, 1,

© 2002 by CRC Press LLC



Analysis of Covariance for Nonreplicated Experiments



31



TABLE 17.16

Times to Wear through the Paint Layer on Bricks from

a Four-Way Factorial Treatment Structure with Porsity

of the Brick as a Possible Covariate

a

–1

–1

–1

–1

–1

–1

–1

–1

1

1

1

1

1

1

1

1



b

–1

–1

–1

–1

1

1

1

1

–1

–1

–1

–1

1

1

1

1



c

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1



d

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1



Time

13.11

11.13

8.32

5.40

16.42

13.17

10.71

10.64

7.41

3.70

12.21

8.34

9.76

8.63

14.01

13.88



Porosity

0.10

0.27

0.10

0.10

0.01

0.15

0.10

0.16

0.14

0.02

0.29

0.21

0.18

0.20

0.09

0.19



Porosity a = –1 (xx)

0.10

0.27

0.10

0.10

0.01

0.15

0.10

0.16

0

0

0

0

0

0

0

0



Porosity a = 1 (zz)

0

0

0

0

0

0

0

0

0.14

0.02

0.29

0.21

0.18

0.20

0.09

0.19



Grinding Time of Brick's Surface

2.5



*



Probit



2.0

1.5

1.0

0.5

0.0



* *

*

*

***

**

**

0.0



0.5



*



A



*



1.0



*



D



1.5



AC



B



2.0



2.5



3.0



ABS(Time)



FIGURE 17.18 Half-normal probability plot of the grinding time data.



where i denotes the level of A; j, the level of B; k, the level of C; m, the level of

D; and βi is the slope for the ith level of A. The last two columns of Table 17.16 are

the values of the covariate (porosity) separated into two columns so that the model

© 2002 by CRC Press LLC



32



Analysis of Messy Data, Volume III: Analysis of Covariance



Grinding Time of Brick's Surface

20



Time



16



*b



abc



12



bc



8

4



abcd

(1) bd



*ad



**

*

c*

*cd



*



*



*a



*d *ac



bcd



* ab*



abd

* * acd



0

0.00



0.05



0.10



0.15



0.20



0.25



0.30



Porosity



FIGURE 17.19 Plot of the grinding time and porosity data factorial effects.



can be expressed as with two covariates (as in Section 17.9) so there are different

slopes for spray and brush. The possible model is

y ijkm = µ ijkm + β −1x −1 jkm + β1x1 jkm + ε ijkm for i = −1, 1, j = −1, 1, k = −1, 1, m = −1, 1

where x–1jkm and x1jkm are the porosity values for a = –1 and a = 1, respectively as

displayed in Table 17.16.

The first step in the analysis is to compute the estimates of the factorial effects

for three variables: time(y), porosity for a = –1(xx), and porosity for a = 1(zz).

Table 17.17 contains the PROC GLM code to fit the full four-way factorial treatment

structure model to each of the three variables and the respective estimates of the

factorial effects are displayed in the lower portion of the table. Since the two

covariates are constructed from a single covariate, two-dimensional scatter plots

were used to look for the possible non-null effects. Figure 17.20 contains the scatter

plot for a = –1 and Figure 17.21 contains the scatter plot for a = 1. The visual

indication from each of the scatter plots is that B, D, and AC are non-null effects

with possibly A also being considered as non-null. Using the possible non-null

effects and the two covariates, the PROC GLM code in Table 17.18 fits the model

y ijkm = α + A i + B j + D m + ACik + β −1xx ijkm + β1 zz ijkm + ε ijkm

to the data set. The estimate of the variance is 0.7984. The significance levels

corresponding to testing the hypothesis that the factorial effects are zero, i.e., A = 0,

B = 0, D = 0, and AC = 0, are 0.0893, 0.0000, 0.0020, and 0.0000, respectively. The

significance levels corresponding testing the slopes are zero, i.e., β–1 = 0 and β1 =

0, are 0.5984 and 0.1783, indicating the covariates are possibly not needed in the

© 2002 by CRC Press LLC



Analysis of Covariance for Nonreplicated Experiments



33



TABLE 17.17

PROC GLM Code to Fit Complete Four-Way Factorial Effects Model

and Estimates for Time (y), the Porosity for a = –1 (xx) and Porosity

for a = 1 (zz)

proc glm data=values;

model y xx zz=a b c a*b a*c b*c a*b*c d a*d b*d a*b*d c*d

a*c*d b*c*d a*b*c*d/solution;

Parameter

Intercept

a

b

c

a*b

a*c

b*c

a*b*c

d

a*d

b*d

a*b*d

c*d

a*c*d

b*c*d

a*b*c*d



Estimate (Time)

10.4275

–0.6850

1.7250

0.0113

0.1025

2.3563

0.1463

–0.1388

–1.0663

–0.0388

0.4938

0.2963

0.1925

–0.0875

0.3300

–0.1850



Estimate (Porosity a = –1)

0.0613

–0.0613

–0.0093

–0.0039

0.0093

0.0039

0.0164

–0.0164

0.0241

–0.0241

0.0017

–0.0017

–0.0152

0.0152

0.0054

–0.0054



Estimate (Porosity a = 1)

0.0819

0.0819

–0.0002

0.0147

–0.0002

0.0147

–0.0282

–0.0282

–0.0046

–0.0046

0.0206

0.0206

0.0077

0.0077

0.0016

0.0016



Grinding Time of Brick's Surface

2.4



*



Treat Effects For Time



1.8



*



AC



B



1.2

0.6



** * * *

*

* * **

*



0.0

-0.6



*



A



*



-1. 2

-0.07



-0.03



0.01



D

0.05



Treat Effects For Porosity a= -1



FIGURE 17.20 Plot of the factorial effects of grinding time by factorial effects of porosity

at level a = –1.

© 2002 by CRC Press LLC



34



Analysis of Messy Data, Volume III: Analysis of Covariance



Grinding Time of Brick's Surface

2.4



*



Treat Effects For Time



1.8



AC



B



*



1.2

0.6

0.0



*

*



-0.6



* * **

* ** * *

*



-1. 2

-0.03



A



D



0.00



0.03



0.06



*

0.09



Treat Effects For Porosity a= 1



FIGURE 17.21 Plot of the factorial effects of grinding time by factorial effects of porosity

at level a = 1.



analysis. Using the stepwise regression approach, no additional effects were identified as non-null. The model was not simplified for the remainder of the analysis in

this section, but the covariate part of the model should be simplified and then evaluate

the possibility of declaring A as a null effect should be evaluated. To continue with

the analysis, estimate statements in Table 17.19 were used to provide adjusted means

for the factorial effects remaining in the model. The first line provides a test of the

equal slopes hypothesis for the levels of A. Since the covariate has been split into

two parts, porosity at a = –1 or xx is used when evaluating means that involve a = –1

and porosity at a = 1 or zz is used when evaluating the means that involve a = 1.

The next two lines provide estimates of the effect of A at b = 1, c = –1, and d = 1

at a porosity of 0.1. The first statement includes “zz 0.1” for level a = 1 and the

second statement includes “xx 0.1” for level a = –1. The third estimate statement is

a comparison of the above two means or a comparison of a = 1 and a = –1 at b = 1,

c = –1, and d = 1 which includes “zz .1 and xx – .1” as this estimate statement is

constructed by subtracting the first two estimate statements. Since the slopes are

different for the levels of A, the means for a = 1 and a = –1 need to be evaluated at

three or more values of the covariate. In this case, the means were evaluated at

porosity values of 0.1, 0.2, and 0.3. Also, since AC is included in the model, the

means for the levels of A were evaluated and compared at each level of C for the

three values of porosity where b = 1 and d = 1. The levels C were evaluated and

compared at each level of A for a porosity value of 0.2 and b = 1 and d = 1. The

levels of B were evaluated compared at a = 1, c = 1, d = 1, and porosity = 0.2 and

the levels of D were evaluated and compared at a = 1, b = 1, c = 1, and porosity of

0.2. The results of the estimate statements are in Table 17.20 where the first partition



© 2002 by CRC Press LLC



Analysis of Covariance for Nonreplicated Experiments



35



TABLE 17.18

PROC GLM Code to Fit the Final Model to the Time Data

with Different Slopes for Each Level of a

proc glm data=values; model y=a b d a*c xx

zz/solution;

Source

Model

Error

Corrected Total



df

6

9



SS

164.1078

7.1853

15



MS

27.3513

0.7984

171.2931



FValue

34.26



ProbF

0.0000



Source

a

b

d

a*c

xx

zz



df

1

1

1

1

1

1



SS(III)

2.8941

46.8610

14.6537

74.9340

0.2379

1.7015



MS

2.8941

46.8610

14.6537

74.9340

0.2379

1.7015



FValue

3.62

58.70

18.35

93.86

0.30

2.13



ProbF

0.0893

0.0000

0.0020

0.0000

0.5984

0.1783



Estimate

9.7430

–1.0117

1.7533

–1.1083

2.2541

2.9172

6.1707



StdErr

0.5216

0.5314

0.2289

0.2587

0.2327

5.3446

4.2268



tValue

18.68

–1.90

7.66

–4.28

9.69

0.55

1.46



Probt

0.0000

0.0893

0.0000

0.0020

0.0000

0.5984

0.1783



Parameter

Intercept

a

b

d

a*c

xx

zz



contains a test of the equality of the slopes, the second partition contains the results

for evaluating and comparing the levels of A, the third partition contains the results

for evaluating and comparing the levels of C, and the last two partitions contains

the results for evaluating and comparing the levels of B and D, respectively. All of

the comparisons of the pairs of means are significantly different. The estimates of

the parameters of the model in Table 17.18 indicate that the maximum time is

predicted to occur at a = –1, b = 1, c = –1, and d = –1 since the estimates of the

slopes for a and d are negative and for b and ac are positive.



17.11 SUMMARY

A methodology is described that incorporates covariates into the analysis of nonreplicated factorial or fractional factorial treatment structures. The process depends

on being able to separate the non-null factorial effects from the null effects. The

slopes corresponding to the covariates are estimated from the set of factorial effects

declared to be null effects, computed for the dependent variables as well as for each

covariate. A scatter plot of the estimated factorial effects from the response variable



© 2002 by CRC Press LLC



36



Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 17.19

Estimate Statements to Provide Adjusted Means and Comparisons of Adjusted

Means for Model with Unequal Slopes for the Covariate for Each Level of a

estimate ‘compare slopes’ xx 1 zz –1;

estimate ‘a= 1 b=1 c=–1 d=1 x=.1’ intercept 1 a 1 b 1 d 1 a*c –1 zz .1;

estimate ‘a=–1 b=1 c=–1 d=1 x=.1’ intercept 1 a –1 b 1 d 1 a*c 1 xx .1;

estimate ‘a=1 minus a=–1 at c=–1 x=.1’ a 2 a*c –2 zz .1 xx –.1;

estimate ‘a= 1 b=1 c=–1 d=1 x=.2’ intercept 1 a 1 b 1 d 1 a*c –1 zz .2;

estimate ‘a=–1 b=1 c=–1 d=1 x=.2’ intercept 1 a –1 b 1 d 1 a*c 1 xx .2;

estimate ‘a=1 minus a=–1 at c=–1 x=.2’ a 2 a*c –2 zz .2 xx –.2;

estimate ‘a= 1 b=1 c=–1 d=1 x=.3’ intercept 1 a 1 b 1 d 1 a*c –1 zz .3;

estimate ‘a=–1 b=1 c=–1 d=1 x=.3’ intercept 1 a –1 b 1 d 1 a*c 1 xx .3;

estimate ‘a=1 minus a=–1 at c=–1 x=.3’ a 2 a*c –2 zz .3 xx –.3;

estimate ‘a= 1 b=1 c=1 d=1 x=.1’ intercept 1 a 1 b 1 d 1 a*c 1 zz .1;

estimate ‘a=–1 b=1 c=1 d=1 x=.1’ intercept 1 a –1 b 1 d 1 a*c –1 xx .1;

estimate ‘a=1 minus a=–1 at c=1 x=.1’ a 2 a*c 2 zz .1 xx –.1;

estimate ‘a= 1 b=1 c=1 d=1 x=.2’ intercept 1 a 1 b 1 d 1 a*c 1 zz .2;

estimate ‘a=–1 b=1 c=1 d=1 x=.2’ intercept 1 a –1 b 1 d 1 a*c –1 xx .2;

estimate ‘a=1 minus a=–1 at c=1 x=.2’ a 2 a*c 2 zz .2 xx –.2;

estimate ‘a= 1 b=1 c=1 d=1 x=.3’ intercept 1 a 1 b 1 d 1 a*c 1 zz .3;

estimate ‘a=–1 b=1 c=1 d=1 x=.3’ intercept 1 a –1 b 1 d 1 a*c –1 xx .3;

estimate ‘a=1 minus a=–1 at c=1 x=.3’ a 2 a*c 2 zz .3 xx –.3;

estimate ‘c=1 at a=1 b=1 d=1 x=.2’ intercept 1 a 1 b 1 d 1 a*c 1 xx .2;

estimate ‘c=–1 at a=1 b=1 d=1 x=.2’ intercept 1 a 1 b 1 d 1 a*c –1 xx .2;

estimate ‘c=1 minus c=–1@a=1’ a*c 2;

estimate ‘c=1 at a=–1 b=1 d=1 x=.2’ intercept 1 a –1 b 1 d 1 a*c –1 zz .2;

estimate ‘c=–1 at a=–1 b=1 d=1 x=.2’ intercept 1 a –1 b 1 d 1 a*c 1 zz .2;

estimate ‘c=1 minus c=–1@a=–1’ a*c –2;

estimate ‘b=1 at a=1 c=1 d=1 x=.2’ intercept 1 a 1 b 1 d 1 a*c 1 xx .2;

estimate ‘b=–1 at a=1 c=1 d=1 x=.2’ intercept 1 a 1 b –1 d 1 a*c 1 xx .2;

estimate ‘b=1 minus b=–1’ b 2;

estimate ‘d=1 at a=1 b=1 c=1 x=.2’ intercept 1 a 1 b 1 d 1 a*c 1 xx .2;

estimate ‘d=–1 at a=1 b=1 c=1

x=.2’ intercept 1 a 1 b 1 d –1 a*c

1 xx .2;

estimate ‘d=1 minus d=–1’ d 2;



and the estimated factorial effects from each of the covariates is used to help identify

the non-null effects and the regression analysis is used to estimate the parameters

of the model corresponding to the non-null effects and the covariates. Stepwise

regression can be used to determine if additional factorial effects should be included

in the set of non-null effects. Adjusted means can be computed using the final model

to provide a method of comparing the means of the factors contained in the nonnull effects. Five examples were included to demonstrate possible cases for using

analysis of covariance to analyze nonreplicated treatment structures. These examples

show that the non-null effects can change from the unadjusted analysis to the

covariate adjusted analysis, and the use of covariates in the analyses can greatly

improve the estimate of the standard deviation.

© 2002 by CRC Press LLC



Analysis of Covariance for Nonreplicated Experiments



37



TABLE 17.20

Results from the Estimate Statements for Model

with the Covariate

Parameter

compare slopes



Estimate

–3.2534



StdErr

6.8889



tValue

–0.47



Probt

0.6480



a= 1 b=1 c=–1 d=1 x=.1

a=–1 b=1 c=–1 d=1 x=.1

a=1 minus a=–1 at c=–1 x=.1

a= 1 b=1 c=–1 d=1 x=.2

a=–1 b=1 c=–1 d=1 x=.2

a=1 minus a=–1 at c=–1 x=.2

a= 1 b=1 c=–1 d=1 x=.3

a=–1 b=1 c=–1 d=1 x=.3

a=1 minus a=–1 at c=–1 x=.3

a= 1 b=1 c=1 d=1 x=.1

a=–1 b=1 c=1 d=1 x=.1

a=1 minus a=–1 at c=1 x=.1

a= 1 b=1 c=1 d=1 x=.2

a=–1 b=1 c=1 d=1 x=.2

a=1 minus a=–1 at c=1 x=.2

a= 1 b=1 c=1 d=1 x=.3

a=–1 b=1 c=1 d=1 x=.3

a=1 minus a=–1 at c=1 x=.3



7.7392

13.9457

–6.2064

8.3563

14.2374

–5.8811

8.9734

14.5291

–5.5557

12.2475

9.4374

2.8101

12.8646

9.7291

3.1355

13.4816

10.0208

3.4608



0.5363

0.5473

0.6694

0.5555

0.5918

0.7864

0.8288

0.9860

1.3183

0.5961

0.5374

0.7498

0.5217

0.6176

0.7799

0.7392

1.0224

1.2662



14.43

25.48

–9.27

15.04

24.06

–7.48

10.83

14.74

–4.21

20.54

17.56

3.75

24.66

15.75

4.02

18.24

9.80

2.73



0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0000

0.0023

0.0000

0.0000

0.0046

0.0000

0.0000

0.0030

0.0000

0.0000

0.0231



c=1 at a=1 b=1 d=1 x=.2

c=–1 at a=1 b=1 d=1 x=.2

c=1 minus c=–1@a=1

c=1 at a=–1 b=1 d=1 x=.2

c=–1 at a=–1 b=1 d=1 x=.2

c=1 minus c=–1@a=–1



12.2139

7.7056

4.5083

10.3798

14.8881

–4.5083



1.3269

1.2933

0.4653

1.2839

1.2218

0.4653



9.20

5.96

9.69

8.08

12.19

–9.69



0.0000

0.0002

0.0000

0.0000

0.0000

0.0000



b=1 at a=1 c=1 d=1 x=.2

b=–1 at a=1 c=1 d=1 x=.2

b=1 minus b=–1



12.2139

8.7072

3.5066



1.3269

1.2557

0.4577



9.20

6.93

7.66



0.0000

0.0001

0.0000



d=1 at a=1 b=1 c=1 x=.2

d=–1 at a=1 b=1 c=1 x=.2

d=1 minus d=–1



12.2139

14.4305

–2.2166



1.3269

1.5473

0.5174



9.20

9.33

–4.28



0.0000

0.0000

0.0020



REFERENCES

Daniel, C. (1959). Use of Half-Normal Plots in Interpreting Factorial Two-Level Experiments.

Technometrics 1:311-341.

Milliken, G. A. and Johnson, D. E. (1989). Analysis of Messy Data, Volume II: Nonreplicated

Experiments. Chapman & Hall, New York.

SAS Institute Inc. (1989). SAS/STAT ® User’s Guide, Version 6, Fourth Edition, Volume 2,

Cary, NC.

© 2002 by CRC Press LLC



38



Analysis of Messy Data, Volume III: Analysis of Covariance



EXERCISES

EXERCISE 17.1: Use the data in the following table to carry out the analysis of

covariance on the single replication of the five-way treatment structure. Make any

necessary comparisons among the non-null effects.



Data for Exercise 17.1 (a Five-Way

Treatment Structure with y as the

Response Variable and x as the Covariate)

a

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1



© 2002 by CRC Press LLC



b

–1

–1

–1

–1

–1

–1

–1

–1

1

1

1

1

1

1

1

1

–1

–1

–1

–1

–1

–1

–1

–1

1

1

1

1

1

1

1

1



c

–1

–1

–1

–1

1

1

1

1

–1

–1

–1

–1

1

1

1

1

–1

–1

–1

–1

1

1

1

1

–1

–1

–1

–1

1

1

1

1



d

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1



e

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1



x

130

113

123

112

124

105

138

147

143

110

108

127

114

137

104

123

123

104

118

125

117

120

144

112

120

130

122

124

116

114

120

134



y

265

219

234

186

254

182

284

300

283

247

199

255

206

261

190

264

213

210

251

264

241

250

261

248

243

258

274

248

259

212

233

252



Analysis of Covariance for Nonreplicated Experiments



39



EXERCISE 17.2: Carry out a thorough analysis of covariance using the following

data set involving a four-way treatment structure. Make all necessary comparisons

among the selected non-null effects.



Data for Exercise 17.2 (Four-Way

Treatment Structure with Response

Variable y and Covariate x)

a

–1

–1

–1

–1

–1

–1

–1

–1

1

1

1

1

1

1

1

1



b

–1

–1

–1

–1

1

1

1

1

–1

–1

–1

–1

1

1

1

1



c

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1

–1

–1

1

1



d

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1

–1

1



x

1.18

2.57

4.51

3.85

1.38

4.53

1.08

1.37

1.54

2.88

2.45

4.16

1.33

2.99

2.53

2.73



y

24.66

28.65

35.56

37.16

27.34

37.73

24.46

23.85

27.98

32.27

33.88

39.25

28.86

34.99

30.62

30.45



EXERCISE 17.3: Compute adjusted means for the effects in the model of Section

17.7 and make necessary comparisons.

EXERCISE 17.4: Determine if there are other possible non-null effects for the

example of Section 17.8.

EXERCISE 17.5: Refine the model used in Section 17.9 and make necessary

comparisons among the adjusted means of the non-null effects remaining in the

model.

EXERCISE 17.6: Re-analyze the data in Section 17.10 and include only those

effects with coefficients that are significantly different from zero at p ≤ 0.05. Are

the conclusions from the reanalysis different than those reached using the analysis

in Section 17.10?



© 2002 by CRC Press LLC



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