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