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11 EXAMPLE: EFFECT OF DRUGS ON HEART RATE

11 EXAMPLE: EFFECT OF DRUGS ON HEART RATE

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20



Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 9.7

PROC MIXED Code to Use REML Estimates of the Variance

Components to Provide Combined Within and Between Block

Estimates of the Models Parameters with Results

proc mixed data=long method=reml;

class herb block;

model y=herb om*herb/solution noint;

contrast ‘b1=b2’ om*herb 1 -1;

random block;

lsmeans herb/diff at om=0.9 cl; lsmeans herb/diff at om=1.0 cl;

lsmeans herb/diff at om=1.1 cl; lsmeans herb/diff at om=1.2 cl;

lsmeans herb/diff at om=1.3 cl; lsmeans herb/diff at om=1.4 cl;

lsmeans herb/diff at om=1.5 cl;

CovParm

block

Residual



Estimate

0.1065

0.2128



Effect

herb

om*herb



NumDF

2

2



DenDF

3

3



FValue

186.47

86.47



ProbF

0.0007

0.0022



Effect

herb

herb

om*herb

om*herb



herb

1

2

1

2



Estimate

12.0000

25.4433

15.5107

4.4609



StdErr

1.4829

1.3379

1.1813

1.1044



df

3

3

3

3



NumDF

1



DenDF

3



FValue

62.41



ProbF

0.0042



Label

b1=b2



tValue

8.09

19.02

13.13

4.04



Probt

0.0039

0.0003

0.0010

0.0273



assigned drug 1 and HR2 and IHR2 are the heart rates and initial heart rates for

those assigned to drug 2.

The variable dif_HR is the difference of the heart rates for drug 1 and drug 2

and the variable dif_IHR is the same difference for the initial heart rates. The

variables sum_HR and sum_IHR are the sums of the respective values within each

block and negIHR2 is the negative of IHR2. The main conclusion from the analysis

of this data set will be that there is not sufficient evidence to conclude the slopes

are unequal. This part of the analysis is accomplished by using PROC MIXED.

Table 9.9 contains the PROC MIXED code to fit the unequal slopes models to the

data set. The contrast statement is used to extract the statistic to test the equal slopes

hypothesis. The last part of Table 9.9 contains the equal slopes test statistic, which

has a significance level of 0.8643, indicating there is not sufficient evidence to

conclude the slopes are unequal.

With that step out of the way, the next step is to fit a common slopes model.

The PROC GLM code in Table 9.10 fits the within block common slope model of

Section 9.7. The estimate of the within block variance component is σˆ ε2 = 12.5364/2 =

6.2682. The second part of Table 9.10 is the inverse of Z0′ Z0, which is used to

© 2002 by CRC Press LLC



Two Treatments in a Randomized Complete Block Design Structure



21



Two Herbicides in RCB



Soy Beans, Bu/acre



40



35

30

25

20



0.90



1.10



1.30



1.50



Organic Matter

Herb 1 Model

Herb 2 95% low



Herb 2 Model

Herb 1 95% Up



Herb 1 95% low

Herb 2 95% Up



FIGURE 9.1 Plot of estimated herbicide models with nonsimultaneous 95% confidence

bands.



Difference of Herbicide Models

6



Herb 1 - Herb 2



4

2

0

-2

-4

-6



0.90



1.10



1.30



1.50



Organic Matter

Difference



95% Lower



95% Upper



FIGURE 9.2 Graph of the difference of the two herbicide models with nonsimultaneous 95%

confidence band.

© 2002 by CRC Press LLC



22



Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 9.8

Data for Effect of Drugs of Heart Rate

data heart; input pair HR1 IHR1 HR2 IHR2;

dif_HR=HR1-HR2; sum_HR=HR1+HR2;dif_IHR=IHR1-IHR2; Sum_IHR=IHR1+IHR2;

negIHR2=-IHR2;

pair

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20



HR1

60

45

73

63

66

51

77

80

69

53

58

55

52

86

60

59

53

74

64

66



IHR1

85

67

87

83

87

67

99

105

87

67

68

68

77

104

77

68

75

97

83

92



HR2

56

69

60

60

76

69

64

57

86

71

62

79

69

78

65

94

63

62

76

82



IHR2

78

94

68

76

89

83

80

67

104

81

72

90

87

90

84

105

74

85

97

103



dif_HR

4

–24

13

3

–10

–18

13

23

–17

–18

–4

–24

–17

8

–5

–35

–10

12

–12

–16



sum_HR

116

114

133

123

142

120

141

137

155

124

120

134

121

164

125

153

116

136

140

148



dif_IHR

7

–27

19

7

–2

–16

19

38

–17

–14

–4

–22

–10

14

–7

–37

1

12

–14

–11



Sum_IHR

163

161

155

159

176

150

179

172

191

148

140

158

164

194

161

173

149

182

180

195



negIHR2

–78

–94

–68

–76

–89

–83

–80

–67

–104

–81

–72

–90

–87

–90

–84

–105

–74

–85

–97

–103



compute Σˆ w . The within block estimate of the common slope is 0.8245 and the

estimate of the difference of the intercepts is –4.0617. Table 9.11 contains the PROC

GLM code to fit the between block common slope model to the block totals. The

estimate of the block variance component is σˆ 2b = (69.5030/2 – σˆ ε2 )/2 = 14.24166.

The second part of the table contains the inverse of M1′ M1, which can be used to

compute Σˆ b . The between block estimate of the slope is 0.7465 and the estimate of

the sum of the intercepts is 8.0531. The PROC IML code in Tables 9.16 can be used

to compute the combined within and between block parameter estimates. Table 9.12

contains the PROC MIXED code to carry out the combined analysis of the data set.

The combined estimate of the slope is 0.8148. Part five of Table 9.12 contains the

estimated adjusted means, which have estimated standard error of 1.0149. The last

part of Table 9.12 contains the estimated difference of the two regression models

(they are parallel) with estimate difference of –4.0925 and estimated standard error

of 0.8031. The test for comparing the two adjusted means has a significance level

of 0.0001, indicating the two drugs are different.

Finally, Table 9.13 contains the PROC MIXED code to use REML to estimate

the variance components. The estimates are σˆ ε2 = 6.2414 and σˆ b2 = 13.7478. The

© 2002 by CRC Press LLC



Two Treatments in a Randomized Complete Block Design Structure



TABLE 9.9

PROC MIXED Code to Fit the Unequal Slopes Model

to the Drug Data Set with Results

proc mixed data=long method=reml;

class drug pair;

model hr=drug ihr*drug/solution noint;

contrast ‘b1=b2’ ihr*drug 1 –1;

random pair;

CovParm

pair

Residual



Estimate

13.8550

6.4962



Effect

drug

ihr*drug



NumDF

2

2



DenDF

17

17



FValue

0.38

182.02



ProbF

0.6904

0.0000



Effect

drug

drug

ihr*drug

ihr*drug



drug

1

2

1

2



Estimate

–4.2987

1.1552

0.8217

0.8054



StdErr

5.0154

5.7988

0.0598

0.0669



df

17

17

17

17



NumDF

1



DenDF

17



FValue

0.03



ProbF

0.8643



Label

b1=b2



tValue

–0.86

0.20

13.74

12.04



Probt

0.4033

0.8445

0.0000

0.0000



TABLE 9.10

PROC GLM Code to Fit the Within Block Common Slopes

Model to the Heart Rate Data with Results

PROC GLM data=heart;

MODEL DIF_hr=dif_IHR/SOLUTION INVERSE;*--model

for differences;

Parameter

Intercept

dif_IHR



Intercept

0.0516

0.0005



dif_IHR

0.0005

0.0002



Source

Model

Error

Corr Total



df

1

18

19



SS

4220.5451

225.6549

4446.2000



MS

4220.5451

12.5364



FValue

336.66



Source

dif_IHR



df

1



SS (III)

4220.5451



MS

4220.5451



FValue

336.66



Estimate

–4.0617

0.8245



StdErr

0.8047

0.0449



tValue

–5.05

18.35



Probt

0.0001

0.0000



Parameter

Intercept

dif_IHR



© 2002 by CRC Press LLC



ProbF

0.0000



23



24



Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 9.11

PROC GLM Code to Fit the Between Common Slopes Model

to the Heart Rate Data with Results

PROC GLM data=heart;

MODEL SUM_hr=sum_ihr/SOLUTION INVERSE;*--model for pair sums;

Parameter

Intercept

Sum_IHR



Intercept

5.8312

–0.0345



Sum_IHR

–0.0345

0.0002



Source

Model

Error

Corr Total



df

1

18

19



SS

2704.7453

1251.0547

3955.8000



MS

2704.7453

69.5030



FValue

38.92



ProbF

0.0000



Source

Sum_IHR



df

1



SS (III)

2704.7453



MS

2704.7453



FValue

38.92



ProbF

0.0000



Parameter

Intercept

Sum_IHR



Estimate

8.0531

0.7465



StdErr

20.1317

0.1197



tValue

0.40

6.24



Probt

0.6938

0.0000



combined estimate of the slope is 0.8146. The fifth part of Table 9.13 contains the

estimated adjusted means, which are very similar to those in Table 9.12, but the

estimates of the standard deviations are a little smaller when computed using the

REML estimates of the variance components. Again the combined estimates of the

model’s parameters depend on the values of the variance components, which depend

on the method of variance component estimation.



9.12 SUMMARY

This chapter presents a very interesting problem that occurs when blocking or some

other random effects are used in the model where there are covariates to be considered. Information about the slopes of the model can come from within the intersection of all of the random effects as well as from between the levels of each of the

random effects. This is an important problem for split-plot and repeated measures

types of designs. These topics are to be discussed in the following chapters. Finally,

the combining of within and between block estimates occurs when the mixed models

analysis is used.



© 2002 by CRC Press LLC



Two Treatments in a Randomized Complete Block Design Structure



25



TABLE 9.12

PROC MIXED Code to Carry Out the Combined Within and Between

Block Analysis for the Common Slopes Model

proc mixed data=long method=reml;

class drug pair;

model hr=drug ihr/solution noint;

random pair;

lsmeans drug/diff;

parms (14.24166) (6.26819)/hold=1,2;

CovParm

pair

Residual



Estimate

014.2417

006.2682



Effect

drug

ihr



NumDF

2

1



DenDF

18

18



FValue

13.84

375.21



ProbF

0.0002

0.0000



Effect

drug

drug

ihr



drug

1

2



Estimate

–3.7381

0.3544

0.8148



StdErr

3.6010

3.7304

0.0421



df

18

18

18



tValue

–1.04

0.10

19.37



Probt

0.3130

0.9254

0.0000



Effect

drug

drug



drug

1

2



Estimate

64.5037

68.5963



StdErr

1.0149

1.0149



df

18

18



tValue

63.56

67.59



Probt

0.0000

0.0000



Effect

drug



drug

1



_drug

2



Estimate

–4.0925



StdErr

0.8031



df

18



tValue

–5.10



© 2002 by CRC Press LLC



Probt

0.0001



26



Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 9.13

PROC MIXED Code to Use REML As a Method to Estimate

the Variance Components for Use in the Combined Within

and Between Block Analysis with Results

proc mixed data=long method=reml;

class drug pair;

model hr=drug ihr/solution noint;

random pair;

lsmeans drug/diff;

CovPar m

pair

Residu al



Estimate

13.7478

6.2414



Effect

drug

ihr



NumDF

2

1



DenDF

18

18



FValue

13.90

377.81



ProbF

0.0002

0.0000



Effect

drug

drug

ihr



drug

1

2



Estimate

–3.7204

0.3728

0.8146



StdErr

3.5851

3.7140

0.0419



df

18

18

18



tValue

–1.04

0.10

19.44



Probt

0.3131

0.9212

0.0000



Effect

drug

drug



drug

1

2



Estimate

64.5034

68.5966



StdErr

1.0020

1.0020



df

18

18



tValue

64.38

68.46



Probt

0.0000

0.0000



Effect

drug



drug

1



_drug

2



Estimate

–4.0932



StdErr

0.8013



df

18



tValue

–5.11



© 2002 by CRC Press LLC



Probt

0.0001



Two Treatments in a Randomized Complete Block Design Structure



TABLE 9.14

PROC IML Code to Combine Within and Between Block Information

and Finish the Computations for the Herbicide Example

proc iml;

bw={–13.82302281, 16.95662402, 5.645132210};

sbw=.479144725*{7.5830055428 –4.265247955 1.7827637526,

–4.265247955

10.78674209 7.63433986,1.7827637526 7.63433986 9.42682557};

bb={38.48854914, 13.83912245, 5.32201593};

sbb=1.08378847*{7.5830055428 –4.265247955 –1.7827637526, –4.265247955

10.78674209 –7.63433986,–1.7827637526 –7.63433986 9.42682557};

print bw sbw bb sbb;

hw={1 –1 0 0 , 0 0 1 0 , 0 0 0 1};

hb={1 1 0 0 , 0 0 1 0 , 0 0 0 1};

isbw=inv(sbw); isbb=inv(sbb);

bc=inv(hw`*isbw*hw+hb`*isbb*hb)*(hw`*isbw*bw+hb`*isbb*bb);

sbc=inv(hw`*isbw*hw+hb`*isbb*hb);

dia=diag(sbc); dia=dia*{1 1 1 1}`;

dia=sqrt(dia); si=inv(sbc);

print bc dia; print bc sbc si;

* compute test of b1=b2;

j={1, 1};

h={0 0 1 0, 0 0 0 1};

betas=h*bc; vbetas=h*sbc*h`;

iv=inv(vbetas); uc=betas`*(iv–iv*j*inv(j`*iv*j)*j`*iv)*betas;

print uc;



TABLE 9.15

PROC IML Code to Construct Confidence Band about Difference

of the Herbicide Models

x={1 –1 .9 –.9, 1 –1 1 –1, 1 –1 1.1 –1.1, 1 –1 1.2 –1.2,

1 –1 1.3 –1.3, 1 –1 1.4 –1.4, 1 –1 1.5 –1.5, 1 –1 1.6 –1.6};

dif=x*bc; vardif=x*sbc*x`;

dia=diag(vardif);

dia=dia*{1 1 1 1 1 1 1 1}`;

dia=sqrt(dia);

*dif is difference and dia is standard error;

print dif dia x;

*compute adjusted means at x=0.9 and x=1.6;

x={1 0 .9 0, 0 1 0 .9, 1 0 1.6 0 , 0 1 0 1.6};

dif=x*bc; vardif=x*sbc*x`;

dia=diag(vardif); dia=dia*{1 1 1 1 }`;

dia=sqrt(dia);

*dif is adjusted mean and dia is standard error;

print dif dia x;



© 2002 by CRC Press LLC



27



28



Analysis of Messy Data, Volume III: Analysis of Covariance



TABLE 9.16

PROC IML Code to Combine Within and Between Pair Estimates

of the Slopes and Test the Equality of Slopes Using the Combined

Estimates for the Drug Data Set

Proc IML;

bw={–7.336075 , .842023 , .8030000};

sbw=13.1665*{5.9293706 –.031022 0.0390255 , –.031022 .0003302

–.000045 , 0.0390255 –.000045 .0004133 };

bb={7.136951325, .713955852, .788653490};

sbb=73.08841*{5.9293706 –.031022 –0.0390255 , –.031022 .0003302

.0000455 , –0.0390255 .0000455 .0004133665};

print bw sbw bb sbb;

hw={1 –1 0 0 , 0 0 1 0 , 0 0 0 1}; hb={1 1 0 0 , 0 0 1 0 , 0 0 0 1};

isbw=inv(sbw); isbb=inv(sbb);

bc=inv(hw`*isbw*hw+hb`*isbb*hb)*(hw`*isbw*bw+hb`*isbb*bb);

sbc=inv(hw`*isbw*hw+hb`*isbb*hb);

dia=diag(sbc); dia=dia*{1 1 1 1}`; dia=sqrt(dia);

si=inv(sbc); print bc dia;

*test b1=b2;

j={1, 1}; h={0 0 1 0, 0 0 0 1};

betas=h*bc; vbetas=h*sbc*h`;

iv=inv(vbetas); uc=betas`*(iv–iv*j*inv(j`*iv*j)*j`*iv)*betas;

print uc;



TABLE 9.17

PROC IML Code to Combine Between and Within Block Information

for Common Slope Model for the Drug Data Set

Proc IML; bw={–4.061747085, .824454036};

sbw=12.53638*{0.0516491658 0.0005153 , 0.0005153 0.0001610513};

bb={8.053121780, 0.746548527};

sbb=69.50304*{5.8312178034 –.034514733, –.034514733 .0002060581};

print bw sbw bb sbb;

hw={1 –1 0 , 0 0 1}; hb={1 1 0 , 0 0 1};

isbw=inv(sbw); isbb=inv(sbb);

bc=inv(hw`*isbw*hw+hb`*isbb*hb)*(hw`*isbw*bw+hb`*isbb*bb);

sbc=inv(hw`*isbw*hw+hb`*isbb*hb);

dia=diag(sbc); dia=dia*{1 1 1}`;

dia=sqrt(dia); si=inv(sbc);

print bc dia; print bc sbc si;

*compare the intercepts–like comparing LSMEANS for parallel lines model;

x={1 –1 0}; dif=x*bc; vardif=x*sbc*x`;

dia=diag(vardif); dia=dia*{1 }`; dia=sqrt(dia);

print dif dia x;

Compute adjusted means for x=70, 100 and 83.75;

x={1 0 70, 0 1 70, 1 0 100 , 0 1 100, 1 0 83.75, 0 1 83.75};

dif=x*bc; vardif=x*sbc*x`; dia=diag(vardif); dia=dia*{1 1 1 1 1 1}`;

dia=sqrt(dia); print dif dia x; run;



© 2002 by CRC Press LLC



Two Treatments in a Randomized Complete Block Design Structure



29



REFERENCES

Ash, Katherine (1982). Use of Inter-Block Information in Analysis of Covariance and Some

Repeated Measures Covariance Models, Ph.D. dissertation, Department of Statistics,

Kansas State University, Manhattan.

Fergen, Brian (1997). On the Mixed-Models Estimator, Ph.D. dissertation, Department of

Statistics, Kansas State University, Manhattan.

John, P. W. M. (1971). Statistical Design and Analysis of Experiments, New York: MacMillan.

Littell, R., Milliken, G. A., Stroup, W., Wolfinger, R. (1996). SAS System for Mixed Models,

Cary NC: SAS Institute Inc.

Milliken, G. A. and Johnson, D. E. (1992). Analysis of Messy Data, Volume I: Design

Experiments, London: Chapman & Hall.



EXERCISES

EXERCISE 9.1: A boy and a girl were selected from each of 16 sixth grade classes

to determine if there were gender differences on the ability to perform on an

engineering and mathematics aptitude test. Each student’s IQ was obtained as a

possible covariate. Provide the estimates of the between block model parameters

and the within block model parameters. Determine if the slopes are similar or

different. Provide a combined within and between block analysis and make the

necessary comparisons between the regression line for the boys and the regression

line for the girls. School is the blocking factor.



Data for Exercise 9.1

School

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16



© 2002 by CRC Press LLC



Male_score

83

76

81

85

70

65

65

82

79

81

67

72

83

79

63

84



Male_IQ

120

115

126

123

95

93

94

124

119

116

89

102

125

111

88

127



Female_score

75

55

62

59

68

78

74

76

88

59

86

81

63

86

79

71



Female_IQ

113

96

102

96

105

117

115

115

127

95

125

120

103

124

119

109



30



Analysis of Messy Data, Volume III: Analysis of Covariance



EXERCISE 9.2: A chemical engineer designed a study to evaluate the ability of

a wafer cleaning machine to remove particles from the surface of a silicon wafer

used in the manufacture of semi-conductors. The specific study was to see if the

machine cleaned two types of wafers differently after each had been exposed to a

processing step that leaves particles on the surface. The particles must be removed

before continuing with the development. The cleaning machine had two slots, so

two wafers could be cleaned at the same time or during one run. The process was

to randomly select a wafer of each type from a lot of wafers and then randomly

assign the two wafers to the two wafer slots in the machine. The number of particles

on the surface of each wafer was determined before the cleaning process started and

was to be used as a possible covariate. After the cleaning process, the number of

particles remaining on the surface was determined. Use the following data to obtain

the within and between block or run estimates of the parameters of the models and

then obtain the combined estimates. Determine if there is evidence that the two

slopes are unequal. Complete the analysis by comparing the resulting regression

models for the two wafer types.



Data set for Exercise 9.2

Run

1

2

3

4

5

6

7

8

9

10

11

12



A_pre_part

214

159

179

160

188

225

218

221

259

155

252

235



A_rem_part

56

36

43

44

51

56

52

56

63

45

62

58



B_pre_part

236

219

255

245

154

148

153

248

234

222

137

179



B_rem_part

77

63

73

80

53

39

42

75

68

73

48

58



EXERCISE 9.3: Use change from baseline (pre_part – rem_part) in Exercise 9.2

and use the randomized complete block analysis of variance to compare the wafer

types. Were the conclusions similar?

EXERCISE 9.4: Use the change from baseline (IHR – HR) in Section 9.11 and

use the randomized complete block analysis of variance to compare the two drugs.

Are the conclusions similar?



© 2002 by CRC Press LLC



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