STAT 503, Fall 2005
Homework Solution 8
Due: Nov. 15, 2005
7-10 Consider the fill height deviation experiment in Problem 6-18. Suppose that each replicate was run
on a separate day. Analyze the data assuming that the days are blocks.
Minitab output is on page 2 and 3.
Design Expert Output
Response:
Fill Deviation
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Block
1.00
1
1.00
Model
70.75
4
17.69
A
36.00
1
36.00
B
20.25
1
20.25
C
12.25
1
12.25
AB
2.25
1
2.25
Residual
6.25
10
0.62
Cor Total
78.00
15
F
Value
Prob > F
28.30
57.60
32.40
19.60
3.60
< 0.0001
< 0.0001
0.0002
0.0013
0.0870
significant
The Model F-value of 28.30 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Values of "Prob > F" less than 0.0500 indicate model terms are significant.
In this case A, B, C are significant model terms.
The analysis is very similar to the original analysis in chapter 6. The same effects are significant.
7-11 Consider the fill height deviation experiment in Problem 6-18. Suppose that only four runs could be
made on each shift. Set up a design with ABC confounded in replicate 1 and AC confounded in replicate 2.
Analyze the data and commend on your findings.
Minitab output is on page 4 and 5.
Design Expert Output
Response:
Fill Deviation
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Block
1.50
3
0.50
Model
70.75
4
17.69
A
36.00
1
36.00
B
20.25
1
20.25
C
12.25
1
12.25
AB
2.25
1
2.25
Residual
5.75
8
0.72
Cor Total
78.00
15
F
Value
Prob > F
24.61
50.09
28.17
17.04
3.13
0.0001
0.0001
0.0007
0.0033
0.1148
significant
The Model F-value of 24.61 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Values of "Prob > F" less than 0.0500 indicate model terms are significant.
In this case A, B, C are significant model terms.
The analysis is very similar to the original analysis of Problem 6-18 and that of problem 7-10. The AB
interaction is less significant in this scenario.
8-1
STAT 503, Fall 2005
7.10
Homework Solution 8
Due: Nov. 15, 2005
Full model fitting
Factorial Fit: Fill Height versus Block, Carbonation, Pressure, Speed
Estimated Effects and Coefficients for Fill Height Deviation (coded units)
Term
Constant
Block
Carbonation
Pressure
Speed
Carbonation*Pressure
Carbonation*Speed
Pressure*Speed
Carbonation*Pressure*Speed
S = 0.755929
Effect
3.0000
2.2500
1.7500
0.7500
0.2500
0.5000
0.5000
R-Sq = 94.87%
Coef
1.0000
-0.2500
1.5000
1.1250
0.8750
0.3750
0.1250
0.2500
0.2500
SE Coef
0.1890
0.1890
0.1890
0.1890
0.1890
0.1890
0.1890
0.1890
0.1890
T
5.29
-1.32
7.94
5.95
4.63
1.98
0.66
1.32
1.32
P
0.001
0.227
0.000
0.001
0.002
0.088
0.529
0.227
0.227
R-Sq(adj) = 89.01%
Analysis of Variance for Fill Height Deviation (coded units)
Source
Blocks
Main Effects
2-Way Interactions
3-Way Interactions
Residual Error
Total
DF
1
3
3
1
7
15
Seq SS
1.000
68.500
3.500
1.000
4.000
78.000
Adj SS
1.000
68.500
3.500
1.000
4.000
Adj MS
1.0000
22.8333
1.1667
1.0000
0.5714
F
1.75
39.96
2.04
1.75
P
0.227
0.000
0.197
0.227
Estimated Coefficients for Fill Height Deviation using data in uncoded units
Term
Constant
Block
Carbonation
Pressure
Speed
Carbonation*Pressure
Carbonation*Speed
Pressure*Speed
Carbonation*Pressure*Speed
Coef
-225.500
-0.250000
21.0000
7.80000
1.08000
-0.750000
-0.105000
-0.0400000
0.00400000
Alias Structure
I
Blocks =
Carbonation
Pressure
Speed
Carbonation*Pressure
Carbonation*Speed
Pressure*Speed
Carbonation*Pressure*Speed
8-2
STAT 503, Fall 2005
7.10
Homework Solution 8
Due: Nov. 15, 2005
Final model fitting
Factorial Fit: Fill Height versus Block, Carbonation, Pressure, Speed
Estimated Effects and Coefficients for Fill Height Deviation (coded units)
Term
Constant
Block
Carbonation
Pressure
Speed
Carbonation*Pressure
S = 0.790569
Effect
3.0000
2.2500
1.7500
0.7500
R-Sq = 91.99%
Coef
1.0000
-0.2500
1.5000
1.1250
0.8750
0.3750
SE Coef
0.1976
0.1976
0.1976
0.1976
0.1976
0.1976
T
5.06
-1.26
7.59
5.69
4.43
1.90
P
0.000
0.235
0.000
0.000
0.001
0.087
R-Sq(adj) = 87.98%
Analysis of Variance for Fill Height Deviation (coded units)
Source
Blocks
Main Effects
2-Way Interactions
Residual Error
Total
DF
1
3
1
10
15
Seq SS
1.000
68.500
2.250
6.250
78.000
Adj SS
1.000
68.500
2.250
6.250
Adj MS
1.0000
22.8333
2.2500
0.6250
F
1.60
36.53
3.60
P
0.235
0.000
0.087
Unusual Observations for Fill Height Deviation
Obs
15
StdOrder
15
Fill
Height
Deviation
6.00000
Fit
4.62500
SE Fit
0.48412
Residual
1.37500
St Resid
2.20R
R denotes an observation with a large standardized residual.
Estimated Coefficients for Fill Height Deviation using data in uncoded units
Term
Constant
Block
Carbonation
Pressure
Speed
Carbonation*Pressure
Coef
9.6250
-0.250000
-2.62500
-1.20000
0.0350000
0.150000
Alias Structure
I
Blocks =
Carbonation
Pressure
Speed
Carbonation*Pressure
8-3
STAT 503, Fall 2005
7.11
Homework Solution 8
Due: Nov. 15, 2005
Full model fitting
* NOTE * This design is not orthogonal.
Factorial Fit: Fill Height versus Block, Carbonation, Pressure, Speed
Estimated Effects and Coefficients for Fill Height Deviation (coded units)
Term
Constant
Block 1
Block 2
Block 3
Carbonation
Pressure
Speed
Carbonation*Pressure
Carbonation*Speed
Pressure*Speed
Carbonation*Pressure*Speed
S = 0.866025
R-Sq = 95.19%
Effect
3.0000
2.2500
1.7500
0.7500
0.5000
0.5000
0.5000
Coef
1.0000
-0.2500
-0.2500
0.5000
1.5000
1.1250
0.8750
0.3750
0.2500
0.2500
0.2500
SE Coef
0.2165
0.4841
0.4841
0.4841
0.2165
0.2165
0.2165
0.2165
0.3062
0.2165
0.3062
T
4.62
-0.52
-0.52
1.03
6.93
5.20
4.04
1.73
0.82
1.15
0.82
P
0.006
0.628
0.628
0.349
0.001
0.003
0.010
0.144
0.451
0.300
0.451
R-Sq(adj) = 85.58%
Analysis of Variance for Fill Height Deviation (coded units)
Source
Blocks
Main Effects
2-Way Interactions
3-Way Interactions
Residual Error
Total
DF
3
3
3
1
5
15
Seq SS
1.5000
68.5000
3.7500
0.5000
3.7500
78.0000
Adj SS
1.2500
68.5000
3.7500
0.5000
3.7500
Adj MS
0.4167
22.8333
1.2500
0.5000
0.7500
F
0.56
30.44
1.67
0.67
P
0.667
0.001
0.288
0.451
Estimated Coefficients for Fill Height Deviation using data in uncoded units
Term
Constant
Block 1
Block 2
Block 3
Carbonation
Pressure
Speed
Carbonation*Pressure
Carbonation*Speed
Pressure*Speed
Carbonation*Pressure*Speed
Coef
-213.125
-0.250000
-0.250000
0.500000
19.8750
7.8000
1.02500
-0.75000
-0.100000
-0.0400000
0.00400000
* NOTE * There is partial confounding, no alias table was printed.
8-4
STAT 503, Fall 2005
7.11
Homework Solution 8
Due: Nov. 15, 2005
Final model fitting
Factorial Fit: Fill Height versus Block, Carbonation, Pressure, Speed
Estimated Effects and Coefficients for Fill Height Deviation (coded units)
Term
Constant
Block 1
Block 2
Block 3
Carbonation
Pressure
Speed
Carbonation*Pressure
S = 0.847791
Effect
3.0000
2.2500
1.7500
0.7500
Coef
1.0000
-0.5000
-0.0000
0.2500
1.5000
1.1250
0.8750
0.3750
R-Sq = 92.63%
SE Coef
0.2119
0.3671
0.3671
0.3671
0.2119
0.2119
0.2119
0.2119
T
4.72
-1.36
-0.00
0.68
7.08
5.31
4.13
1.77
P
0.002
0.210
1.000
0.515
0.000
0.001
0.003
0.115
R-Sq(adj) = 86.18%
Analysis of Variance for Fill Height Deviation (coded units)
Source
Blocks
Main Effects
2-Way Interactions
Residual Error
Total
DF
3
3
1
8
15
Seq SS
1.500
68.500
2.250
5.750
78.000
Adj SS
1.500
68.500
2.250
5.750
Adj MS
0.5000
22.8333
2.2500
0.7187
F
0.70
31.77
3.13
P
0.580
0.000
0.115
Estimated Coefficients for Fill Height Deviation using data in uncoded units
Term
Constant
Block 1
Block 2
Block 3
Carbonation
Pressure
Speed
Carbonation*Pressure
Coef
9.6250
-0.500000
0.000000
0.250000
-2.62500
-1.20000
0.0350000
0.150000
Alias Structure
I
Blocks = Carbonation*Speed + Carbonation*Pressure*Speed
Carbonation
Pressure
Speed
Carbonation*Pressure
8-5
STAT 503, Fall 2005
Homework Solution 8
Due: Nov. 15, 2005
7-16 Consider the 26 design in eight blocks of eight runs each with ABCD, ACE, and ABEF as the
independent effects chosen to be confounded with blocks. Generate the design. Find the other effects
confound with blocks.
Block 1
Block 2
Block 3
Block 4
Block 5
Block 6
Block 7
Block 8
b
acd
ce
abde
abcf
df
aef
bcdef
abc
d
ae
bcde
bf
acdf
cef
abdef
a
bcd
abce
de
cf
abdf
bef
acdef
c
abd
be
acde
af
bcdf
abcef
def
ac
bd
abe
cde
f
abcdf
bcef
adef
(1)
abcd
bce
ade
acf
bdf
abef
cdef
bc
ad
e
abcde
abf
cdf
acef
bdef
ab
cd
ace
bde
bcf
adf
ef
abcdef
The factors that are confounded with blocks are ABCD, ABEF, ACE, BDE, CDEF, BCF, and ADF.
8-6 R.D. Snee (“Experimenting with a Large Number of Variables,” in Experiments in Industry: Design,
Analysis and Interpretation of Results, by R.D. Snee, L.B. Hare, and J.B. Trout, Editors, ASQC, 1985)
describes an experiment in which a 25-1 design with I=ABCDE was used to investigate the effects of five
factors on the color of a chemical product. The factors are A = solvent/reactant, B = catalyst/reactant, C =
temperature, D = reactant purity, and E = reactant pH. The results obtained were as follows:
e=
a=
b=
abe =
c=
ace =
bce =
abc =
-0.63
2.51
-2.68
1.66
2.06
1.22
-2.09
1.93
d=
ade =
bde =
abd =
cde =
acd =
bcd =
abcde =
6.79
5.47
3.45
5.68
5.22
4.38
4.30
4.05
(a) Prepare a normal probability plot of the effects. Which effects seem active?
Factors A, B, D, and the AB, AD interactions appear to be active.
8-6
STAT 503, Fall 2005
Homework Solution 8
DESIGN-EXPERT Plot
Color
Normal plot
Solvent/Reactant
Catalyst/Reactant
Temperature
Reactant Purity
Reactant pH
99
D
95
90
Norm al % probability
A:
B:
C:
D:
E:
Due: Nov. 15, 2005
A
AB
80
70
50
30
20
10
B
5
AD
1
-1.36
0.09
1.53
2.98
4.42
Effect
Design Expert Output
Term
Effect
Model
Intercept
Model
A
1.31
Model
B
-1.34
Error
C
-0.1475
Model
D
4.42
Error
E
-0.8275
Model
AB
1.275
Error
AC
-0.7875
Model
AD
-1.355
Error
AE
0.3025
Error
BC
0.1675
Error
BD
0.245
Error
BE
0.2875
Error
CD
-0.7125
Error
CE
-0.24
Error
DE
0.0875
Lenth's ME 1.95686
Lenth's SME 3.9727
SumSqr
% Contribtn
6.8644
7.1824
0.087025
78.1456
2.73902
6.5025
2.48062
7.3441
0.366025
0.112225
0.2401
0.330625
2.03063
0.2304
0.030625
Design Expert Output
Response:
Color
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
106.04
5
21.21
A
6.86
1
6.86
B
7.18
1
7.18
D
78.15
1
78.15
AB
6.50
1
6.50
AD
7.34
1
7.34
Residual
8.65
10
0.86
Cor Total
114.69
15
5.98537
6.26265
0.0758809
68.1386
2.38828
5.66981
2.16297
6.40364
0.319153
0.0978539
0.209354
0.288286
1.77059
0.200896
0.0267033
F
Value
24.53
7.94
8.31
90.37
7.52
8.49
Prob > F
< 0.0001
0.0182
0.0163
< 0.0001
0.0208
0.0155
The Model F-value of 24.53 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Std. Dev.
Mean
C.V.
PRESS
0.93
2.71
34.35
22.14
Coefficient
R-Squared
Adj R-Squared
Pred R-Squared
Adeq Precision
Standard
0.9246
0.8869
0.8070
14.734
95% CI
8-7
95% CI
significant
STAT 503, Fall 2005
Homework Solution 8
Factor
Estimate
Intercept
2.71
A-Solvent/Reactant 0.66
B-Catalyst/Reactant-0.67
D-Reactant Purity 2.21
AB
0.64
AD
-0.68
DF
1
1
1
1
1
1
Error
0.23
0.23
0.23
0.23
0.23
0.23
Low
2.19
0.14
-1.19
1.69
0.12
-1.20
Due: Nov. 15, 2005
High
3.23
1.17
-0.15
2.73
1.16
-0.16
VIF
1.00
1.00
1.00
1.00
1.00
Final Equation in Terms of Coded Factors:
Color
+2.71
+0.66
-0.67
+2.21
+0.64
-0.68
=
*A
*B
*D
*A*B
*A*D
Final Equation in Terms of Actual Factors:
Color
+2.70750
+0.65500
-0.67000
+2.21000
+0.63750
-0.67750
=
* Solvent/Reactant
* Catalyst/Reactant
* Reactant Purity
* Solvent/Reactant * Catalyst/Reactant
* Solvent/Reactant * Reactant Purity
(b) Calculate the residuals. Construct a normal probability plot of the residuals and plot the residuals
versus the fitted values. Comment on the plots.
Design Expert Output
Diagnostics Case Statistics
Standard Actual
Predicted
Order
Value
Value
1
-0.63
0.47
2
2.51
1.86
3
-2.68
-2.14
4
1.66
1.80
5
2.06
0.47
6
1.22
1.86
7
-2.09
-2.14
8
1.93
1.80
9
6.79
6.25
10
5.47
4.93
11
3.45
3.63
12
5.68
4.86
13
5.22
6.25
14
4.38
4.93
15
4.30
3.63
16
4.05
4.86
Residual
-1.10
0.65
-0.54
-0.14
1.59
-0.64
0.053
0.13
0.54
0.54
-0.18
0.82
-1.03
-0.55
0.67
-0.81
Leverage
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
0.375
8-8
Student
Residual
-1.500
0.881
-0.731
-0.187
2.159
-0.874
0.071
0.180
0.738
0.738
-0.248
1.112
-1.398
-0.745
0.908
-1.105
Cook's
Distance
0.225
0.078
0.053
0.003
0.466
0.076
0.001
0.003
0.054
0.054
0.006
0.124
0.195
0.055
0.082
0.122
Outlier
t
-1.616
0.870
-0.713
-0.178
2.804
-0.863
0.068
0.171
0.720
0.720
-0.236
1.127
-1.478
-0.727
0.899
-1.119
Run
Order
2
6
14
11
8
15
10
3
4
5
16
12
9
1
13
7
STAT 503, Fall 2005
Homework Solution 8
Due: Nov. 15, 2005
Normal plot of residuals
Residuals vs. Predicted
1.5875
95
90
0.915
80
70
Res iduals
Norm al % probability
99
50
30
20
0.2425
10
5
-0.43
1
-1.1025
-1.1025
-0.43
0.2425
0.915
1.5875
-2.14
Res idual
-0.05
2.05
4.15
6.25
Predicted
The residual plots are satisfactory.
(c) If any factors are negligible, collapse the 2 5-1 design into a full factorial in the active factors. Comment
on the resulting design, and interpret the results.
The design becomes two replicates of a 23 in the factors A, B and D. When re-analyzing the data in three
factors, D becomes labeled as C.
Design Expert Output
Response:
Color
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
106.51
7
15.22
A
6.86
1
6.86
B
7.18
1
7.18
C
78.15
1
78.15
AB
6.50
1
6.50
AC
7.34
1
7.34
BC
0.24
1
0.24
ABC
0.23
1
0.23
Residual
8.18
8
1.02
Lack of Fit
0.000
0
Pure Error
8.18
8
1.02
Cor Total
114.69
15
F
Value
14.89
6.72
7.03
76.46
6.36
7.19
0.23
0.23
Prob > F
0.0005
0.0320
0.0292
< 0.0001
0.0357
0.0279
0.6409
0.6476
significant
The Model F-value of 14.89 implies the model is significant. There is only
a 0.05% chance that a "Model F-Value" this large could occur due to noise.
Std. Dev.
Mean
C.V.
PRESS
1.01
2.71
37.34
32.71
Coefficient
Factor
Estimate
Intercept
2.71
A-Solvent/Reactant 0.66
B-Catalyst/Reactant-0.67
C-Reactant Purity 2.21
AB
0.64
AC
-0.68
R-Squared
Adj R-Squared
Pred R-Squared
Adeq Precision
DF
1
1
1
1
1
1
Standard
Error
0.25
0.25
0.25
0.25
0.25
0.25
0.9287
0.8663
0.7148
11.736
95% CI
Low
2.12
0.072
-1.25
1.63
0.055
-1.26
8-9
95% CI
High
3.29
1.24
-0.087
2.79
1.22
-0.095
VIF
1.00
1.00
1.00
1.00
1.00
STAT 503, Fall 2005
BC
ABC
Homework Solution 8
0.12
-0.12
1
1
0.25
0.25
-0.46
-0.70
Due: Nov. 15, 2005
0.71
0.46
1.00
1.00
Final Equation in Terms of Coded Factors:
Color
+2.71
+0.66
-0.67
+2.21
+0.64
-0.68
+0.12
-0.12
=
*A
*B
*C
*A*B
*A*C
*B*C
*A*B*C
Final Equation in Terms of Actual Factors:
Color
+2.70750
+0.65500
-0.67000
+2.21000
+0.63750
-0.67750
+0.12250
-0.12000
=
* Solvent/Reactant
* Catalyst/Reactant
* Reactant Purity
* Solvent/Reactant * Catalyst/Reactant
* Solvent/Reactant * Reactant Purity
* Catalyst/Reactant * Reactant Purity
* Solvent/Reactant * Catalyst/Reactant * Reactant Purity
8-8 An article in Industrial and Engineering Chemistry (“More on Planning Experiments to Increase
Research Efficiency,” 1970, pp. 60-65) uses a 25-2 design to investigate the effect of A = condensation, B =
amount of material 1, C = solvent volume, D = condensation time, and E = amount of material 2 on yield.
The results obtained are as follows:
e=
ab =
23.2
15.5
ad =
bc =
16.9
16.2
cd =
ace =
23.8
23.4
bde =
abcde =
16.8
18.1
(a) Verify that the design generators used were I = ACE and I = BDE.
A
+
+
+
+
B
+
+
+
+
C
+
+
+
+
D=BE
+
+
+
+
E=AC
+
+
+
+
e
ad
bde
ab
cd
ace
bc
abcde
(b) Write down the complete defining relation and the aliases for this design.
I=BDE=ACE=ABCD.
A
B
C
D
E
AB
AD
(BDE)
(BDE)
(BDE)
(BDE)
(BDE)
(BDE)
(BDE)
=ABDE
=DE
=BCDE
=BE
=BD
=ADE
=ABE
A
B
C
D
E
AB
AD
(ACE)
(ACE)
(ACE)
(ACE)
(ACE)
(ACE)
(ACE)
=CE
=ABCE
=AE
=ACDE
=AC
=BCE
=CDE
(c) Estimate the main effects.
Design Expert Output
8-10
A
B
C
D
E
AB
AD
(ABCD)
(ABCD)
(ABCD)
(ABCD)
(ABCD)
(ABCD)
(ABCD)
=BCD
=ACD
=ABD
=ABC
=ABCDE
=CD
=BC
A=ABDE=CE=BCD
B=DE=ABCE=ACD
C=BCDE=AE=ABD
D=BE=ACDE=ABC
E=BD=AC=ABCDE
AB=ADE=BCE=CD
AD=ABE=CDE=BC
STAT 503, Fall 2005
Homework Solution 8
Term
Intercept
A
B
C
D
E
Model
Model
Model
Model
Model
Model
Effect
SumSqr
-1.525
-5.175
2.275
-0.675
2.275
Due: Nov. 15, 2005
% Contribtn
4.65125
53.5613
10.3512
0.91125
10.3513
5.1831
59.6858
11.5349
1.01545
11.5349
(d) Prepare an analysis of variance table. Verify that the AB and AD interactions are available to use as
error.
The analysis of variance table is shown below. Part (b) shows that AB and AD are aliased with other
factors. If all two-factor and three factor interactions are negligible, then AB and AD could be pooled as an
estimate of error.
Design Expert Output
Response:
Yield
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
79.83
5
15.97
A
4.65
1
4.65
B
53.56
1
53.56
C
10.35
1
10.35
D
0.91
1
0.91
E
10.35
1
10.35
Residual
9.91
2
4.96
Cor Total
89.74
7
F
Value
3.22
0.94
10.81
2.09
0.18
2.09
Prob > F
0.2537
0.4349
0.0814
0.2853
0.7098
0.2853
not significant
The "Model F-value" of 3.22 implies the model is not significant relative to the noise. There is a
25.37 % chance that a "Model F-value" this large could occur due to noise.
Std. Dev.
Mean
C.V.
PRESS
2.23
19.24
11.57
158.60
R-Squared
Adj R-Squared
Pred R-Squared
Adeq Precision
Coefficient
Factor
Estimate
Intercept
19.24
A-Condensation -0.76
B-Material 1
-2.59
C-Solvent
1.14
D-Time
-0.34
E-Material 2
1.14
DF
1
1
1
1
1
1
Standard
Error
0.79
0.79
0.79
0.79
0.79
0.79
0.8895
0.6134
-0.7674
5.044
95% CI
Low
15.85
-4.15
-5.97
-2.25
-3.72
-2.25
Final Equation in Terms of Coded Factors:
Yield
+19.24
-0.76
-2.59
+1.14
-0.34
+1.14
=
*A
*B
*C
*D
*E
Final Equation in Terms of Actual Factors:
Yield
+19.23750
-0.76250
-2.58750
+1.13750
-0.33750
+1.13750
=
* Condensation
* Material 1
* Solvent
* Time
* Material 2
8-11
95% CI
High
22.62
2.62
0.80
4.52
3.05
4.52
VIF
1.00
1.00
1.00
1.00
1.00
STAT 503, Fall 2005
Homework Solution 8
Due: Nov. 15, 2005
(e) Plot the residuals versus the fitted values. Also construct a normal probability plot of the residuals.
Comment on the results.
The residual plots are satisfactory.
Residuals vs. Predicted
Normal plot of residuals
1.55
99
Res iduals
Norm al % probability
0.775
-1.77636E-015
-0.775
95
90
80
70
50
30
20
10
5
1
-1.55
13.95
16.38
18.81
21.24
23.68
-1.55
-0.775
Predicted
-1.77636E-015
0.775
1.55
Res idual
8-10 Construct a 27-2 design by choosing two four-factor interactions as the independent generators.
Write down the complete alias structure for this design. Outline the analysis of variance table. What is the
resolution of this design?
I=CDEF=ABCG=ABDEFG, Resolution IV
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
A
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
B
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
C
+
+
+
+
+
+
+
+
+
+
+
+
+
D
+
+
+
+
+
+
+
+
+
+
+
+
+
E
+
+
+
+
+
+
+
+
+
+
+
+
+
8-12
F=CDE
+
+
+
+
+
+
+
+
+
+
+
+
+
G=ABC
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
(1)
ag
bg
ab
cfg
acf
bcf
abcfg
df
adfg
bdfg
abdf
cdg
acd
bcd
abcdg
ef
aefg
befg
abef
ceg
ace
bce
abceg
de
adeg
bdeg
abde
cdefg
STAT 503, Fall 2005
Homework Solution 8
30
31
32
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Due: Nov. 15, 2005
+
acdef
bcdef
abcdefg
Alias Structure
A(CDEF)=
B(CDEF)=
C(CDEF)=
D(CDEF)=
E(CDEF)=
F(CDEF)=
G(CDEF)=
AB(CDEF)=
AC(CDEF)=
AD(CDEF)=
AE(CDEF)=
AF(CDEF)=
AG(CDEF)=
BD(CDEF)=
BE(CDEF)=
BF(CDEF)=
CD(CDEF)=
CE(CDEF)=
CF(CDEF)=
DG(CDEF)=
EG(CDEF)=
FG(CDEF)=
ACDEF
BCDEF
DEF
CEF
CDF
CDE
CDEFG
ABCDEF
ADEF
ACEF
ACDF
ACDE
ACDEFG
BCEF
BCDF
BCDE
EF
DF
DE
CEFG
CDFG
CDEG
A(ABCG)=
B(ABCG)=
C(ABCG)=
D(ABCG)=
E(ABCG)=
F(ABCG)=
G(ABCG)=
AB(ABCG)=
AC(ABCG)=
AD(ABCG)=
AE(ABCG)=
AF(ABCG)=
AG(ABCG)=
BD(ABCG)=
BE(ABCG)=
BF(ABCG)=
CD(ABCG)=
CE(ABCG)=
CF(ABCG)=
DG(ABCG)=
EG(ABCG)=
FG(ABCG)=
BCG
ACG
ABG
ABCDG
ABCEG
ABCFG
ABC
CG
BG
BCDG
BCEG
BCFG
BC
ACDG
ACEG
ACFG
ABDG
ABEG
ABFG
ABCD
ABCE
ABCF
A(ABDEFG)= BDEFG
B(ABDEFG)= ADEFG
C(ABDEFG)= ABCDEFG
D(ABDEFG)= ABEFG
E(ABDEFG)= ABDFG
F(ABDEFG)= ABDEG
G(ABDEFG)= ABDEF
AB(ABDEFG)= DEFG
AC(ABDEFG)= BCDEFG
AD(ABDEFG)= BEFG
AE(ABDEFG)= BDFG
AF(ABDEFG)= BDEG
AG(ABDEFG)= BDEF
BD(ABDEFG)= AEFG
BE(ABDEFG)= ADFG
BF(ABDEFG)= ADEG
CD(ABDEFG)= ABCEFG
CE(ABDEFG)= ABCDFG
CF(ABDEFG)= ABCDEG
DG(ABDEFG)= ABEF
EG(ABDEFG)= ABDF
FG(ABDEFG)= ABDE
Analysis of Variance Table
Source
A
B
C
D
E
F
G
AB=CG
AC=BG
AD
AE
AF
AG=BC
BD
BE
BF
CD=EF
CE=DF
CF=DE
DG
EG
FG
Error
Total
Degrees of Freedom
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
9
31
8-13
A=ACDEF=BCG=BDEFG
B=BCDEF=ACG=ADEFG
C=DEF=ABG=ABCDEFG
D=CEF=ABCDG=ABEFG
E=CDF=ABCEG=ABDFG
F=CDE=ABCFG=ABDEG
G=CDEFG=ABC=ABDEF
AB=ABCDEF=CG=DEFG
AC=ADEF=BG=BCDEFG
AD=ACEF=BCDG=BEFG
AE=ACDF=BCEG=BDFG
AF=ACDE=BCFG=BDEG
AG=ACDEFG=BC=BDEF
BD=BCEF=ACDG=AEFG
BE=BCDF=ACEG=ADFG
BF=BCDE=ACFG=ADEG
CD=EF=ABDG=ABCEFG
CE=DF=ABEG=ABCDFG
CF=DE=ABFG=ABCDEG
DG=CEFG=ABCD=ABEF
EG=CDFG=ABCE=ABDF
FG=CDEG=ABCF=ABDE