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4 EXAMPLE: AVERAGE DAILY GAINS AND BIRTH WEIGHT Û COMMON SLOPE

4 EXAMPLE: AVERAGE DAILY GAINS AND BIRTH WEIGHT Û COMMON SLOPE

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C0317ch05 frame Page 131 Monday, June 25, 2001 10:08 PM



Two-Way Treatment Structure and Analysis of Covariance



131



TABLE 5.1

Average Daily Gains (ADG lb/day) and Birth Weights (lb) of Hereford

Calves for Example 5.4

Control (0 mg)

Drug

Female



Male



Birth_wt

69

70

74

66

74

78

78

64

76

69

70

72

60

80

71

77



adg

2.58

2.60

2.13

2.00

1.98

2.31

2.30

2.19

2.43

2.36

2.93

2.58

2.27

3.11

2.42

2.66



Low (2.5 mg)

Birth_wt

78

73

83

79

66

63

82

75

71

81

66

69

70

76

79

78



adg

2.92

2.51

2.78

2.91

2.18

2.25

2.91

2.42

2.52

2.54

2.95

2.46

3.13

2.72

3.41

3.43



Med (5.0 mg)

Birth_wt

77

85

77

61

76

74

75

65

69

69

71

68

63

61

64

73



adg

3.01

3.13

2.75

2.36

2.71

2.79

2.84

2.59

3.14

2.91

2.53

3.09

2.86

2.88

2.75

2.91



High (7.5 mg)

Birth_wt

67

84

83

73

66

63

66

83

77

79

76

72

66

84

79

83



adg

2.94

2.80

2.85

2.44

2.28

2.70

2.70

2.85

2.73

3.17

2.92

2.85

2.47

3.28

3.13

2.73



TABLE 5.2

Analysis of Variance Table with Effects for Both Intercepts and Slopes

proc glm data=common; class Drug Sex;

model ADG=drug sex drug*sex Birth_wt Birth_wt*drug Birth_wt*sex;

Source

Model

Error

Corrected Total



df

15

48

63



SS

4.1363

2.7993

6.9357



MS

0.2758

0.0583



FValue

4.73



ProbF

0.0000



Source

drug

Sex

drug*Sex

Birth_wt

Birth_wt*drug

Birth_wt*Sex

Birth_wt*drug*Sex



df

3

1

3

1

3

1

3



SS (Type III)

0.0527

0.0037

0.3044

0.7576

0.0546

0.0001

0.2966



MS

0.0176

0.0037

0.1015

0.7576

0.0182

0.0001

0.0989



FValue

0.30

0.06

1.74

12.99

0.31

0.00

1.70



ProbF

0.8245

0.8012

0.1714

0.0007

0.8163

0.9690

0.1805



Birth_wt. The estimated slope is 0.0217 indicating for each 1 lb increase in Birth_wt,

there is an estimated 0.0217 lb/day increase in ADG. The significance levels corresponding to the intercept effects are 0.0001, 0.0001, and 0.6584 respectively for

Sex, Drug, and Sex*Drug. Thus there are important Sex and Drug effects, but there

© 2002 by CRC Press LLC



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Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 5.3

Analysis of Variance Table with Effects for the Intercepts

and a Common Slope Model

proc glm data=common; class Drug Sex;

model ADG=sex drug drug*sex Birth_wt/solution;

Source

Model

Error

Corrected total



df

8

55

63



SS

3.7583

3.1774

6.9357



MS

0.4698

0.0578



FValue

8.13



ProbF

0.0000



Source

Sex

drug

drug*Sex

Birth_wt



df

1

3

3

1



SS (Type III)

1.0366

1.5517

0.0932

1.0964



MS

1.0366

0.5172

0.0311

1.0964



FValue

17.94

8.95

0.54

18.98



ProbF

0.0001

0.0001

0.6584

0.0001



Estimate

0.0217



StdErr

0.0050



tValue

4.36



Probt

0.0001



Parameter

Birth_wt



is not a significant interaction between the levels of Drug and levels of SEX.

Therefore, the main effects of the treatments need to be compared. The PROC GLM

code for fitting a means model to the intercepts and a common slope is in Table 5.4.

The results in Table 5.4 provide the estimates of the parameters for the regression

model. Each of the Sex*Drug combinations has an intercept and there is a common

slope for the covariate, Birth_wt.

Since there is no interaction between the levels of sex and the levels of drug for

the intercepts, a regression model can be constructed for each sex and for each level

of drug. Table 5.5 contains the estimate statements used to compute the estimates

of the intercepts of the regression models for each sex by averaging over the levels

of drug and the estimates of the intercepts for each level of drug by averaging over

the levels of sex. For example, the regression model to describe the mean of the

ADG values as a function of birth weight for females and for males is ADGfemale =

0.997 + 0.0217 Birth_wt and ADGmale = 1.252 + 0.0217 Birth_wt. Figure 5.3 contains

a graph of the estimated regression lines for each sex. The models for each level of

drug are ADGControl = 0.875 + 0.0217 Birth_wt, ADGLow = 1.144 + 0.0217 Birth_wt,

ADGMed = 1.302 + 0.0217 Birth_wt, and ADGHigh = 1.177 + 0.0217 Birth_wt.

Figure 5.4 contains a graph of the estimated regression lines for each level of drug.

Tables 5.6 and 5.7 contain the adjusted means (from main effects regression lines

evaluated at the average birth weight of the experiment, 73 lb) for the main effects

of Drug and main effects of SEX, respectively. The vertical line at 73 in Figure 5.4

indicates the locations of the adjusted means or LSMEANS. The mean ADG for

males calves is significantly larger than the mean ADG for female calves. The mean

ADG for the calves on the Control is significantly less than the means of the three

non-zero levels of drug.



© 2002 by CRC Press LLC



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Two-Way Treatment Structure and Analysis of Covariance



133



TABLE 5.4

Analysis of Variance Table with Intercepts Expressed as Means

proc glm data=common; class Drug Sex;

model ADG=drug*sex Birth_wt/solution noint;

Source

Model

Error

Uncorrected Total



df

9

55

64



SS

471.2908

3.1774

474.4682



MS

52.3656

0.0578



FValue

906.44



ProbF

0.0000



Source

drug*Sex

Birth_wt



df

8

1



SS

3.5166

1.0964



MS

0.4396

1.0964



FValue

7.61

18.98



ProbF

0.0000

0.0001



Estimate

0.7105

1.0388

1.1118

1.2429

0.9889

1.2983

1.1758

1.4277

0.0217



StdErr

0.3660

0.3672

0.3732

0.3920

0.3817

0.3762

0.3762

0.3449

0.0050



tValue

1.94

2.83

2.98

3.17

2.59

3.45

3.12

4.14

4.36



Probt

0.0573

0.0065

0.0043

0.0025

0.0122

0.0011

0.0028

0.0001

0.0001



Parameter

Control (0 mg) Female

Control (0 mg) Male

High (7.5 mg) Female

High (7.5 mg) Male

Low (2.5 mg) Female

Low (2.5 mg) Male

Med (5.0 mg) Female

Med (5.0 mg) Male

Birth_wt



TABLE 5.5

Estimates of the Intercepts for the Sex Models

and for the Drug Models

estimate

estimate

estimate

estimate

estimate

estimate

Parameter

int Female

int Male

int 0 mg

int 2.5 mg

int 5.0 mg

int 7.5 mg



‘int

‘int

‘int

‘int

‘int

‘int



Female’ drug*sex 1 0 1 0 1 0 1 0 /divisor=4;

Male’ drug*sex 0 1 0 1 0 1 0 1 /divisor=4;

0 mg’ drug*sex 1 1 0 0 0 0 0 0 /divisor=2;

2.5 mg’ drug*sex 0 0 0 0 1 1 0 0 /divisor=2;

5.0 mg’ drug*sex 0 0 0 0 0 0 1 1/divisor=2;

7.5 mg’ drug*sex 0 0 1 1 0 0 0 0 /divisor=2;

Estimate

0.997

1.252

0.875

1.144

1.302

1.177



StdErr

0.367

0.363

0.362

0.374

0.355

0.378



tValue

2.72

3.45

2.42

3.06

3.66

3.12



Probt

0.0088

0.0011

0.0189

0.0035

0.0006

0.0029



Thus far, the animal scientist has determined a common slope model using birth

weight as a covariate is appropriate, that there is an effect of the drug (Table 5.7

shows the mean ADG of all non-zero levels of the Drug are significantly larger than



© 2002 by CRC Press LLC



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Analysis of Messy Data, Volume III: Analysis of Covariance



Models for Males and Females



ADG (Ibs)



3.5500

2.9500



*

*



2.3500

1.7500

58.00



68.00



78.00



88.00



Birth Weight (Ibs)

Females



Males



LS MEAN



FIGURE 5.3 Plot of estimated regression models for each level of sex with least squares means.



Models for Each Drug



ADG (Ibs)



3.5500

2.9500



**

*



2.3500

1.7500

58.00



68.00



78.00



88.00



Birth Weight (Ibs)



Control (0 mg)

High (7.5 mg)



Low (2.5 mg)



Med (5.0 mg)



LS MEAN



FIGURE 5.4 Plot of estimated regression models for each level of drug with least squares

means.



the mean of the control. The mean of MED dose is larger than the means of all the

other doses, but not significantly), that males gain significantly faster than females

(see Table 5.6), and that there is no interaction between the levels of SEX and Drug.

Finally, the animal scientist wants to estimate the optimal dose, i.e., estimate

that dose of the compound which should produce the maximum ADG. To accomplish

this objective, the adjusted means and the corresponding standard errors were

© 2002 by CRC Press LLC



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Two-Way Treatment Structure and Analysis of Covariance



135



TABLE 5.6

Least Squares Means and Comparisons

for the Levels of Sex

Sex

Female

Male



LSMean

2.5752

2.8304



StdErr

0.0425

0.0425



ProbtDiff

0.0001



TABLE 5.7

Least Squares Means and Comparisons for the Levels of Drug

Drug

Control (0 mg)

High (7.5 mg)

Low (2.5 mg)

Med (5.0 mg)



LSMean

2.453

2.756

2.722

2.880



StdErr

0.060

0.061

0.060

0.061



LSMean #

1

2

3

4



_1



_2

0.0009



0.0009

0.0028

0.0000



0.6930

0.1629



_3

0.0028

0.6930



_4

0.0000

0.1629

0.0747



0.0747



TABLE 5.8

Results of Fitting Quadratic Regression Model

to Drug Least Squares Means

data lsmeans; input xlsmean drug stderr @@;

wt=1/(stderr**2); drug2=drug**2;

datalines;

2.4536 0 .06037 2.7225 2.5 .06048

2.8806 5.0 .06129 2.7562 7.5 .06102

proc reg data=lsmeans; weight wt; model

xlsmean=drug drug2;

Source

Model

Error

Corr Total



df

2

1

3



SS

25.9274

0.3979

26.3253



MS

12.9637

0.3979



FValue

32.58



ProbF

0.1229



Variable

Intercept

drug

drug2



df

1

1

1



Estimate

2.4452

0.1604

–0.0157



StdErr

0.0371

0.0240

0.0031



tValue

65.85

6.69

–5.12



Probt

0.0097

0.0945

0.1229



obtained from Table 5.7 and used to construct a data set to which a quadratic

regression model was fit using the reciprocal of the square of the standard errors as

weights. The code and data are in Table 5.8. The results of fitting the quadratic

regression model are in Table 5.8 and the estimated regression model is

ADG = 2.4452 + 0.1604 DRUG –0.0157 DRUG2

© 2002 by CRC Press LLC



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Analysis of Messy Data, Volume III: Analysis of Covariance



Models for LSMEANS vs Level of Drug



ADG (Ibs)



3.55

2.95

2.35



*



*



*



*



1.75

0.0



2.5



5.0



7.5



Level of Drug mg

LSMEAN



Model



xxx



Optimum Dose



FIGURE 5.5 Plot of quadratic regression model for drug least squares means with predicted

maximum.



To determine the level of drug that is estimated to yield the maximum response,

differentiate the model with respect to DRUG, set the derivative equal to zero, and

solve for the resulting equation for the level of drug estimated to produce the

maximum response, providing

DRUGMax =

–(0.1604)/(2*(-0.0157)) = 5.11 mg

This stationary point is a maximum since the coefficient of DRUG2 is negative.

Figure 5.5 is a graph of the least squares means, the estimated quadratic regression

model, and the estimated level of Drug producing the maximum response from the

quadratic model.



5.5 EXAMPLE: ENERGY FROM WOOD OF DIFFERENT

TYPES OF TREES — SOME UNEQUAL SLOPES

A forester and a chemical engineer studied the amount of energy produced by

burning wood blocks from different species of trees (four levels) in different types

of stoves (three levels). It is suspected that the moisture content (%) of a block of

wood affects the amount of energy (kilogram/calorie denoted by kg/cal) produced

by burning a block of wood. Table 5.9 contains the data where ten blocks (uniform

dimensions) of each wood type were burned in each stove type. For several runs

complications occurred and the data were not usable; thus the unequal sample sizes

occurred. The wood from osage orange and red oak are considered to be hard woods

and the wood from white pine and black walnut are are considered to be soft woods.



© 2002 by CRC Press LLC



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Two-Way Treatment Structure and Analysis of Covariance



137



TABLE 5.9

Energy Produced from Blocks of Wood with Different Moisture

Contents Burned in Three Different Types of Stoves

Osage Orange



Red Oak



Stove

Type A



Moist

8.70

10.20

15.60

16.20

17.10

10.50

19.00

13.00

7.20



Energy

7.33

7.58

6.91

7.08

6.82

7.63

6.68

7.64

7.52



Moist

16.50

16.80

7.20

11.40

9.90



Energy

5.72

6.54

6.09

6.87

6.53



Type B



13.10

9.40

11.10

17.60

14.70

18.40



7.41

6.66

7.01

5.69

6.65

6.24



11.30

12.60

13.90

12.40

20.40

12.10



5.62

5.37

5.05

5.25

5.77

5.43



Type C



14.20

10.80

10.40

15.80

10.10

15.90

14.30

18.90

20.90



6.33

6.01

6.62

5.97

5.86

6.19

5.30

6.38

5.32



12.50

17.90

11.20

13.20

16.90

12.40

8.70

10.90

13.90



4.81

4.48

5.92

4.60

4.63

4.82

5.50

5.27

4.12



White Pine

Moist

8.60

7.60

7.40

17.20

9.20

13.30

21.00

17.50

13.30

7.00

15.40

16.60

16.20

8.70

19.50



Energy

2.71

2.99

3.29

1.58

2.36

2.47

1.12

1.35

1.88

2.51

2.64

2.68

3.76

4.50

1.66



15.80

12.90

14.80

8.20

16.20

11.60

18.30



1.59

2.06

1.66

2.71

1.28

2.61

0.34



Black Walnut

Moist

15.50

13.70

18.70

11.30

7.30

15.90



Energy

1.87

2.39

1.14

2.42

3.23

1.81



10.20

12.80

9.20

13.60

16.50

11.80

18.60

14.20

8.30

18.20

12.60

13.10

9.40

8.80

18.00

20.90



3.43

2.79

3.92

2.67

2.38

2.49

1.92

2.77

4.65

3.51

5.13

4.46

4.86

5.49

3.25

2.85



Table 5.10 contains the PROC MIXED code to fit a model with a means model

for the intercepts and an effects model for the slopes. The significance level for the

moist*wood*stove term is 0.0842. For this problem, the conclusion is that there is

not an important three-way interaction. The next step is to remove moist*wood*stove

from the model and fit a model with just the two-way interaction terms, moist*wood

and moist*stove. For this model, the significance level corresponding to the

moist*stove term is .3267, indicating it is not an important term (results not shown).

Table 5.11 contains the results of fitting a model with unequal slopes for each level

of wood and the estimates of those slopes. The slopes for the two hard woods seem

to be smaller (closer to zero) than the slopes for the two soft woods. In an attempt

to simplify the model, pairwise comparisons of the slopes were accomplished by

using a set of estimate statements.



© 2002 by CRC Press LLC



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