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Ch06
Design Of Experiments (Texas Tech University)
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Design Of Experiments (Texas Tech University)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-1 Chapter 6 The 2
k
Factorial Design Solutions 6.1.
In a 2
4
factorial design, the number of degrees of freedom for the model, assuming the complete factorial model, is (a)
7 (b)
5 (c)
6 (d)
11 (e)
12 (f)
None of the above 6.2.
A 2
3
factorial is replicated twice. The number of pure error or residual degrees of freedom are (a)
4 (b)
12 (c)
15 (d)
2 (e)
8 (f)
None of the above 6.3.
A 2
3
factorial is replicated twice. The ANOVA indicates that all main effects are significant but the interactions are not significant. The interaction terms are dropped from the model. The number of residual degrees of freedom for the reduced model are (a)
12
(b)
8 (c)
16 (d)
14 (e)
10
(f)
None of the above 6.4.
A 2
3
factorial is replicated three times. The ANOVA indicates that all main effects are significant, but two of the interactions are not significant. The interaction terms are dropped from the model. The number of residual degrees of freedom for the reduced model are (a)
12 (b)
14 (c)
6 (d)
10 (e)
8
(f)
None of the above
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-2 6.5.
An engineer is interested in the effects of cutting speed (
A
), tool geometry (
B
), and cutting angle on the life (in hours) of a machine tool. Two levels of each factor are chosen, and three replicates of a 2
3
factorial design are run. The results are as follows: Treatment
Replicate
A B C Combination
I
II
III
- - - (1)
22
31
25
+ - - a 32
43
29
- + - b 35
34
50
+ + - ab 55
47
46
- - + c 44
45
38
+ - + ac 40
37
36
- + + bc 60
50
54
+ + + abc 39
41
47
(a) Estimate the factor effects. Which effects appear to be large? From the normal probability plot of effects below, factors B
, C
, and the AC
interaction appear to be significant. (b) Use the analysis of variance to confirm your conclusions for part (a). The analysis of variance confirms the significance of factors B
, C
, and the AC
interaction. Design Expert Output Response:
Life
in hours
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F Model 1612.67 7 230.38 7.64 0.0004 significant A 0.67 1 0.67 0.022 0.8837 B 770.67 1 770.67 25.55 0.0001 C 280.17 1 280.17 9.29 0.0077 AB 16.67 1 16.67 0.55 0.4681 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-3 AC 468.17 1 468.17 15.52 0.0012 BC 48.17 1 48.17 1.60 0.2245 ABC 28.17 1 28.17 0.93 0.3483 Pure Error 482.67 16 30.17 Cor Total 2095.33 23 The Model F-value of 7.64 implies the model is significant. There is only a 0.04% chance that a "Model F-Value" this large could occur due to noise. The reduced model ANOVA is shown below. Factor A
was included to maintain hierarchy. Design Expert Output Response:
Life
in hours
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 1519.67 4 379.92 12.54 < 0.0001 significant A
0.67
1
0.67
0.022
0.8836
B
770.67
1
770.67
25.44
< 0.0001
C
280.17
1
280.17
9.25
0.0067
AC
468.17
1
468.17
15.45
0.0009
Residual 575.67 19 30.30 Lack of Fit
93.00
3
31.00
1.03
0.4067
not significant
Pure Error
482.67
16
30.17
Cor Total 2095.33 23 The Model F-value of 12.54 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Effects B, C
and AC
are significant at 1%. (c) Write down a regression model for predicting tool life (in hours) based on the results of this experiment. 40.8333
0.1667
5.6667
3.4167
4.4167
ijk
A
B
C
A
C
y
x
x
x
x x
=
+
+
+
+
Design Expert Output Coefficient
Standard
95% CI
95% CI
Factor
Estimate
DF
Error
Low
High
VIF
Intercept 40.83 1 1.12 38.48 43.19 A-Cutting Speed 0.17 1 1.12 -2.19 2.52 1.00 B-Tool Geometry 5.67 1 1.12 3.31 8.02 1.00 C-Cutting Angle 3.42 1 1.12 1.06 5.77 1.00 AC -4.42 1 1.12 -6.77 -2.06 1.00 Final Equation in Terms of Coded Factors:
Life = +40.83 +0.17 * A +5.67 * B +3.42 * C -4.42 * A * C Final Equation in Terms of Actual Factors:
Life = +40.83333 +0.16667 * Cutting Speed +5.66667 * Tool Geometry +3.41667 * Cutting Angle -4.41667 * Cutting Speed * Cutting Angle Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-4 The equation in part (c) and in the given in the computer output form a “hierarchical” model, that is, if an interaction is included in the model, then all of the main effects referenced in the interaction are also included in the model. (d) Analyze the residuals. Are there any obvious problems? There is nothing unusual about the residual plots. (e) Based on the analysis of main effects and interaction plots, what levels of A
, B
, and C
would you recommend using? Since B
has a positive effect, set B
at the high level to increase life. The AC
interaction plot reveals that life would be maximized with C
at the high level and A
at the low level. 6.6.
Reconsider part (c) of Problem 6.5. Use the regression model to generate response surface and contour plots of the tool life response. Interpret these plots. Do they provide insight regarding the desirable operating conditions for this process? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-5 The response surface plot and the contour plot in terms of factors A
and C
with B
at the high level are shown below. They show the curvature due to the AC
interaction. These plots make it easy to see the region of greatest tool life. 6.7.
Find the standard error of the factor effects and approximate 95 percent confidence limits for the factor effects in Problem 6.5. Do the results of this analysis agree with the conclusions from the analysis of variance? ( )
2
(
)
2
3 2
1
1
30.17
2.24
2
3 2
effect
k
SE
S
n
-
-
=
=
=
Variable
Effect
A
0.333
B
11.333
*
AB
-1.667
C
6.833
*
AC
-8.833
*
BC
-2.833
ABC
-2.167
The 95% confidence intervals for factors B, C
and AC
do not contain zero. This agrees with the analysis of variance approach. 6.8.
Plot the factor effects from Problem 6.5 on a graph relative to an appropriately scaled t
distribution. Does this graphical display adequately identify the important factors? Compare the conclusions from this plot with the results from the analysis of variance. 30.17
3.17
3
E
MS
S
n
=
=
=
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-6 This method identifies the same factors as the analysis of variance. 6.9.
A router is used to cut locating notches on a printed circuit board. The vibration level at the surface of the board as it is cut is considered to be a major source of dimensional variation in the notches. Two factors are thought to influence vibration: bit size (
A
) and cutting speed (
B
). Two bit sizes (1/16 and 1/8 inch) and two speeds (40 and 90 rpm) are selected, and four boards are cut at each set of conditions shown below. The response variable is vibration measured as a resultant vector of three accelerometers (
x
, y
, and z
) on each test circuit board. Treatment
Replicate
A B Combination
I
II
III
IV
- - (1)
18.2
18.9
12.9
14.4 + - a 27.2
24.0
22.4
22.5 - + b 15.9
14.5
15.1
14.2 + + ab 41.0
43.9
36.3
39.9 (a) Analyze the data from this experiment. Design Expert Output Response:
Vibration
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 1638.11 3 546.04 91.36 < 0.0001 significant A
1107.23
1
1107.23
185.25
< 0.0001
B
227.26
1
227.26
38.02
< 0.0001
AB
303.63
1
303.63
50.80
< 0.0001
Residual 71.72 12 5.98 Lack of Fit
0.000
0
Pure Error
71.72
12
5.98
Cor Total 1709.83 15 The Model F-value of 91.36 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. (b) Construct a normal probability plot of the residuals, and plot the residuals versus the predicted vibration level. Interpret these plots. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-7 There is nothing unusual about the residual plots. (c) Draw the AB
interaction plot. Interpret this plot. What levels of bit size and speed would you recommend for routine operation? To reduce the vibration, use the smaller bit. Once the small bit is specified, either speed will work equally well, because the slope of the curve relating vibration to speed for the small tip is approximately zero. The process is robust to speed changes if the small bit is used. 6.10.
Reconsider the experiment described in Problem 6.5. Suppose that the experimenter only performed the eight trials from replicate I. In addition, he ran four center points and obtained the following response values: 36, 40, 43, 45. (a) Estimate the factor effects. Which effects are large? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-8 Effects B, C,
and
AC appear to be large. (b) Perform an analysis of variance, including a check for pure quadratic curvature. What are your conclusions? (
)
( )(
)(
)
2
2
8
4
40.875
41.000
0.0417
8
4
F
C
F
C
PureQuadratic
F
C
n n
y
y
SS
n
n
-
-
=
=
=
+
+
Design Expert Output Response:
Life
in hours
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 1048.88 7 149.84 9.77 0.0439 significant A 3.13 1 3.13 0.20 0.6823 B 325.13 1 325.13 21.20 0.0193 C 190.12 1 190.12 12.40 0.0389 AB 6.13 1 6.13 0.40 0.5722 AC 378.12 1 378.12 24.66 0.0157 BC 55.12 1 55.12 3.60 0.1542 ABC 91.12 1 91.12 5.94 0.0927 Curvature 0.042 1 0.042 2.717E-003 0.9617 not significant Pure Error 46.00 3 15.33 Cor Total 1094.92 11 The Model F-value of 9.77 implies the model is significant. There is only a 4.39% chance that a "Model F-Value" this large could occur due to noise. The "Curvature F-value" of 0.00 implies the curvature (as measured by difference between the average of the center points and the average of the factorial points) in the design space is not significant relative to the noise. There is a 96.17% chance that a "Curvature F-value" this large could occur due to noise. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-9 Design Expert Output Response:
Life
in hours
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 896.50 4 224.13 7.91 0.0098 significant A 3.13 1 3.13 0.11 0.7496 B 325.12 1 325.12 11.47 0.0117 C 190.12 1 190.12 6.71 0.0360 AC 378.12 1 378.12 13.34 0.0082 Residual 198.42 7 28.35 Lack of Fit 152.42 4 38.10 2.49 0.2402 not significant Pure Error 46.00 3 15.33 Cor Total 1094.92 11 The Model F-value of 7.91 implies the model is significant. There is only a 0.98% chance that a "Model F-Value" this large could occur due to noise. Effects B, C
and AC
are significant at 5%. There is no effect of curvature. (c) Write down an appropriate model for predicting tool life, based on the results of this experiment. Does this model differ in any substantial way from the model in Problem 6.1, part (c)? The model shown in the Design Expert
output below does not differ substantially from the model in Problem 6.5, part (c). Design Expert Output Final Equation in Terms of Coded Factors:
Life = +40.88 +0.62 * A +6.37 * B +4.87 * C -6.88 * A * C (d) Analyze the residuals. (e) What conclusions would you draw about the appropriate operating conditions for this process? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-10 To maximize life run with B
at the high level, A
at the low level and C
at the high level 6.11.
An experiment was performed to improve the yield of a chemical process. Four factors were selected, and two replicates of a completely randomized experiment were run. The results are shown in the following table: Treatment Replicate Replicate Treatment Replicate Replicate Combination I II Combination I II (1) 90 93 d 98 95 a 74 78 ad 72 76 b 81 85 bd 87 83 ab 83 80 abd 85 86 c 77 78 cd 99 90 ac 81 80 acd 79 75 bc 88 82 bcd 87 84 abc 73 70 abcd 80 80 (a) Estimate the factor effects. Design Expert Output
Term Effect SumSqr % Contribtn Model Intercept Error A -9.0625 657.031 40.3714 Error B -1.3125 13.7812 0.84679 Error C -2.6875 57.7813 3.55038 Error D 3.9375 124.031 7.62111 Error AB 4.0625 132.031 8.11267 Error AC 0.6875 3.78125 0.232339 Error AD -2.1875 38.2813 2.3522 Error BC -0.5625 2.53125 0.155533 Error BD -0.1875 0.28125 0.0172814 Error CD 1.6875 22.7812 1.3998 Error ABC -5.1875 215.281 13.228 Error ABD 4.6875 175.781 10.8009 Error ACD -0.9375 7.03125 0.432036 Error BCD -0.9375 7.03125 0.432036 Error ABCD 2.4375 47.5313 2.92056 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-11 (b) Prepare an analysis of variance table, and determine which factors are important in explaining yield. Design Expert Output
Response:
yield
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 1504.97 15 100.33 13.10 < 0.0001 significant A
657.03
1
657.03
85.82
< 0.0001
B
13.78
1
13.78
1.80
0.1984
C
57.78
1
57.78
7.55
0.0143
D
124.03
1
124.03
16.20
0.0010
AB
132.03
1
132.03
17.24
0.0007
AC
3.78
1
3.78
0.49
0.4923
AD
38.28
1
38.28
5.00
0.0399
BC
2.53
1
2.53
0.33
0.5733
BD
0.28
1
0.28
0.037
0.8504
CD
22.78
1
22.78
2.98
0.1038
ABC
215.28
1
215.28
28.12
< 0.0001
ABD
175.78
1
175.78
22.96
0.0002
ACD
7.03
1
7.03
0.92
0.3522
BCD
7.03
1
7.03
0.92
0.3522
ABCD
47.53
1
47.53
6.21
0.0241
Residual 122.50 16 7.66 Lack of Fit
0.000
0
Pure Error
122.50
16
7.66
Cor Total 1627.47 31 The Model F-value of 13.10 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, C, D, AB, AD, ABC, ABD, ABCD are significant model terms. =
, and =
therefore, factors A
and D
and interactions AB, ABC
, and ABD
are significant at 1%. Factor C
and interactions AD
and ABCD are significant at 5%. (c) Write down a regression model for predicting yield, assuming that all four factors were varied over the range from –1 to +1 (in coded units). Model with hierarchy maintained: Design Expert Output
Final Equation in Terms of Coded Factors:
yield = +82.78 -4.53 * A -0.66 * B -1.34 * C +1.97 * D +2.03 * A * B +0.34 * A * C -1.09 * A * D -0.28 * B * C -0.094 * B * D +0.84 * C * D -2.59 * A * B * C +2.34 * A * B * D -0.47 * A * C * D -0.47 * B * C * D +1.22 * A * B * C * D Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-12 Model without hierarchy terms: Design Expert Output
Final Equation in Terms of Coded Factors:
yield = +82.78 -4.53 * A -1.34 * C +1.97 * D +2.03 * A * B -1.09 * A * D -2.59 * A * B * C +2.34 * A * B * D +1.22 * A * B * C * D Confirmation runs might be run to see if the simpler model without hierarchy is satisfactory. (d) Does the residual analysis appear satisfactory? There appears to be one large residual both in the normal probability plot and in the plot of residuals versus predicted. (e) Two three-factor interactions, ABC
and ABD
, apparently have large effects. Draw a cube plot in the factors A
, B
, and C
with the average yields shown at each corner. Repeat using the factors A
, B
, and D
. Do these two plots aid in data interpretation? Where would you recommend that the process be run with respect to the four variables? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-13 Run the process at A
low B
low, C
low and D
high. 6.12.
A bacteriologist is interested in the effects of two different culture media and two different times on the growth of a particular virus. She performs six replicates of a 2
2
design, making the runs in random order. Analyze the bacterial growth data that follow and draw appropriate conclusions. Analyze the residuals and comment on the model’s adequacy. Culture Medium
Time 1
2
21
22
25
26
12 hr 23
28
24
25
20
26
29
27
37
39
31
34
18 hr 38
38
29
33
35
36
30
35
Design Expert Output
Response:
Virus growth
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 691.46 3 230.49 45.12 < 0.0001 significant A
9.38
1
9.38
1.84
0.1906
B
590.04
1
590.04
115.51
< 0.0001
AB
92.04
1
92.04
18.02
0.0004
Residual 102.17 20 5.11 Lack of Fit
0.000
0
Pure Error
102.17
20
5.11
Cor Total 793.63 23 The Model F-value of 45.12 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 B, AB are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-14 Growth rate is affected by factor B
(Time) and the AB
interaction (Culture medium and Time). There is some very slight indication of inequality of variance shown by the small decreasing funnel shape in the plot of residuals versus predicted. 6.13.
An industrial engineer employed by a beverage bottler is interested in the effects of two different types of 32-ounce bottles on the time to deliver 12-bottle cases of the product. The two bottle types are glass and plastic. Two workers are used to perform a task consisting of moving 40 cases of the product 50 feet on a standard type of hand truck and stacking the cases in a display. Four replicates of a 2
2
factorial design are performed, and the times observed are listed in the following table. Analyze the data and draw the appropriate conclusions. Analyze the residuals and comment on the model’s adequacy. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-15 Worker
Bottle Type
1
1
2
2
Glass 5.12
4.89
6.65
6.24
4.98
5.00
5.49
5.55
Plastic 4.95
4.43
5.28
4.91
4.27
4.25
4.75
4.71
Design Expert Output
Response:
Times
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 4.86 3 1.62 13.04 0.0004 significant A
2.02
1
2.02
16.28
0.0017
B
2.54
1
2.54
20.41
0.0007
AB
0.30
1
0.30
2.41
0.1463
Residual 1.49 12 0.12 Lack of Fit
0.000
0
Pure Error
1.49
12
0.12
Cor Total 6.35 15 The Model F-value of 13.04 implies the model is significant. There is only a 0.04% 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 are significant model terms. There is some indication of non-constant variance in this experiment. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-16 6.14.
In problem 6.13, the engineer was also interested in potential fatigue differences resulting from the two types of bottles. As a measure of the amount of effort required, he measured the elevation of heart rate (pulse) induced by the task. The results follow. Analyze the data and draw conclusions. Analyze the residuals and comment on the model’s adequacy. Worker
Bottle Type
1
1
2
2
Glass 39
45
20
13
58
35
16
11
Plastic 44
35
13
10
42
21
16
15
Design Expert Output
Response:
Pulse
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 2784.19 3 928.06 16.03 0.0002 significant A
2626.56
1
2626.56
45.37
< 0.0001
B
105.06
1
105.06
1.81
0.2028
AB
52.56
1
52.56
0.91
0.3595
Residual 694.75 12 57.90 Lack of Fit
0.000
0
Pure Error
694.75
12
57.90
Cor Total 3478.94 15 The Model F-value of 16.03 implies the model is significant. There is only a 0.02% 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 are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-17 There is an indication that one worker exhibits greater variability than the other. 6.15.
Calculate approximate 95 percent confidence limits for the factor effects in Problem 6.14. Do the results of this analysis agree with the analysis of variance performed in Problem 6.14? (
)
2
(
)
2
2
2
1
1
57.90
3.80
2
4 2
effect
k
SE
S
n
-
-
=
=
=
Variable
Effect
C.I.
A –25.625 ±
3.80(1.96) = ±
7.448 B –5.125 ±
3.80(1.96) = ±
7.448 AB 3.625 ±
3.80(1.96) = ±
7.448 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-18 The 95% confidence interval for factor A does not contain zero. This agrees with the analysis of variance approach. 6.16.
An article in the AT&T Technical Journal
(March/April 1986, Vol. 65, pp. 39-50) describes the application of two-level factorial designs to integrated circuit manufacturing. A basic processing step is to grow an epitaxial layer on polished silicon wafers. The wafers mounted on a susceptor are positioned inside a bell jar, and chemical vapors are introduced. The susceptor is rotated and heat is applied until the epitaxial layer is thick enough. An experiment was run using two factors: arsenic flow rate (
A
) and deposition time (
B
). Four replicates were run, and the epitaxial layer thickness was measured (in mm). The data are shown below: Replicate
Factor
Levels A B I II
III
IV
Low (–)
High (+) – – 14.037 16.165
13.972
13.907
A 55%
59% + – 13.880 13.860
14.032
13.914
– + 14.821 14.757
14.843
14.878
B Short
Long + + 14.888 14.921
14.415
14.932
(10 min)
(15 min) (a) Estimate the factor effects. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -0.31725 0.40259 6.79865 Error B 0.586 1.37358 23.1961 Error AB 0.2815 0.316969 5.35274 Error Lack Of Fit 0 0 Error Pure Error 3.82848 64.6525 (b) Conduct an analysis of variance. Which factors are important? From the analysis of variance shown below, no factors appear to be important. Factor B
is only marginally interesting with an F
-value of 4.31. Design Expert Output
Response:
Thickness
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 2.09 3 0.70 2.19 0.1425 not significant A
0.40
1
0.40
1.26
0.2833
B
1.37
1
1.37
4.31
0.0602
AB
0.32
1
0.32
0.99
0.3386
Residual 3.83 12 0.32 Lack of Fit
0.000
0
Pure Error
3.83
12
0.32
Cor Total 5.92 15 The "Model F-value" of 2.19 implies the model is not significant relative to the noise. There is a 14.25 % 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 there are no significant model terms. (c) Write down a regression equation that could be used to predict epitaxial layer thickness over the region of arsenic flow rate and deposition time used in this experiment. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-19 Design Expert Output
Final Equation in Terms of Coded Factors:
Thickness = +14.51 -0.16 * A +0.29 * B +0.14 * A * B Final Equation in Terms of Actual Factors:
Thickness = +37.62656 -0.43119 * Flow Rate -1.48735 * Dep Time +0.028150 * Flow Rate * Dep Time (d) Analyze the residuals. Are there any residuals that should cause concern? Observation #2 falls outside the groupings in the normal probability plot and the plot of residual versus predicted. (e) Discuss how you might deal with the potential outlier found in part (d). One approach would be to replace the observation with the average of the observations from that experimental cell. Another approach would be to identify if there was a recording issue in the original data. The first analysis below replaces the data point with the average of the other three. The second analysis assumes that the reading was incorrectly recorded and should have been 14.165. The analysis with the run associated with standard order 2 replaced with the average of the remaining three runs in the cell, 13.972, is shown below. Design Expert Output
Response:
Thickness
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 2.97 3 0.99 53.57 < 0.0001 significant A 7.439E-003 1 7.439E-003 0.40 0.5375 B 2.96 1 2.96 160.29 < 0.0001 AB 2.176E-004 1 2.176E-004 0.012 0.9153 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-20 Pure Error 0.22 12 0.018 Cor Total 3.19 15 The Model F-value of 53.57 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 B are significant model terms. Final Equation in Terms of Coded Factors:
Thickness = +14.38 -0.022 * A +0.43 * B +3.688E-003 * A * B Final Equation in Terms of Actual Factors:
Thickness = +13.36650 -0.020000 * Flow Rate +0.12999 * Dep Time +7.37500E-004 * Flow Rate * Dep Time A new outlier is present and should be investigated. Analysis with the run associated with standard order 2 replaced with the value 14.165: Design Expert Output
Response:
Thickness
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 2.82 3 0.94 45.18 < 0.0001 significant A 0.018 1 0.018 0.87 0.3693 B 2.80 1 2.80 134.47 < 0.0001 AB 3.969E-003 1 3.969E-003 0.19 0.6699 Pure Error 0.25 12 0.021 Cor Total 3.07 15 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-21 The Model F-value of 45.18 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 B are significant model terms. Final Equation in Terms of Coded Factors:
Thickness = +14.39 -0.034 * A +0.42 * B +0.016 * A * B Final Equation in Terms of Actual Factors:
Thickness = +15.50156 -0.056188 * Flow Rate -0.012350 * Dep Time +3.15000E-003 * Flow Rate * Dep Time Another outlier is present and should be investigated. 6.17. Continuation of Problem 6.16.
Use the regression model in part (c) of Problem 6.16 to generate a response surface contour plot for epitaxial layer thickness. Suppose it is critically important to obtain layer thickness of 14.5 mm. What settings of arsenic flow rate and deposition time would you recommend? Arsenic flow rate may be set at any of the experimental levels, while the deposition time should be set at 12.4 minutes. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-22 6.18. Continuation of Problem 6.17.
How would your answer to Problem 6.17 change if arsenic flow rate was more difficult to control in the process than the deposition time? Running the process at a high level of Deposition Time there is no change in thickness as flow rate changes. 6.19.
A nickel-titanium alloy is used to make components for jet turbine aircraft engines. Cracking is a potentially serious problem in the final part, as it can lead to non-recoverable failure. A test is run at the parts producer to determine the effects of four factors on cracks. The four factors are pouring temperature (
A
), titanium content (
B
), heat treatment method (
C
), and the amount of grain refiner used (
D
). Two replicated of a 2
4
design are run, and the length of crack (in μ
m) induced in a sample coupon subjected to a standard test is measured. The data are shown below: Treatment
Replicate
Replicate A B C D Combination
I
II
–
–
–
–
(1)
7.037
6.376
+ –
–
–
a 14.707
15.219
– + –
–
b 11.635
12.089
+ + –
–
ab 17.273
17.815
–
–
+ –
c 10.403
10.151
+ – + –
ac 4.368
4.098
– + + –
bc 9.360
9.253
+ + + –
abc 13.440
12.923
–
–
–
+
d 8.561
8.951
+ –
–
+
ad 16.867
17.052
– + –
+
bd 13.876
13.658
+ + –
+
abd 19.824
19.639
–
–
+ +
cd 11.846
12.337
+ – + +
acd 6.125
5.904
– + + +
bcd 11.190
10.935
+ + + +
abcd 15.653
15.053
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-23 (a) Estimate the factor effects. Which factors appear to be large? From the half normal plot of effects shown below, factors A
, B
, C
, D
, AB
, AC
, and ABC
appear to be large. Design Expert Output
Term Effect SumSqr % Contribtn Model Intercept Model A 3.01888 72.9089 12.7408 Model B 3.97588 126.461 22.099 Model C -3.59625 103.464 18.0804 Model D 1.95775 30.6623 5.35823 Model AB 1.93412 29.9267 5.22969 Model AC -4.00775 128.496 22.4548 Error AD 0.0765 0.046818 0.00818145 Error BC 0.096 0.073728 0.012884 Error BD 0.04725 0.0178605 0.00312112 Error CD -0.076875 0.0472781 0.00826185 Model ABC 3.1375 78.7512 13.7618 Error ABD 0.098 0.076832 0.0134264 Error ACD 0.019125 0.00292613 0.00051134 Error BCD 0.035625 0.0101531 0.00177426 Error ABCD 0.014125 0.00159613 0.000278923
(b) Do any of the factors affect cracking? Use α
=0.05. The Design Expert output below identifies factors A
, B
, C
, D
, AB
, AC
, and ABC
as significant. Design Expert Output
Response:
Crack Lengthin mm x 10^-2
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 570.95 15 38.06 468.99 < 0.0001 significant A
72.91
1
72.91
898.34
< 0.0001
B
126.46
1
126.46
1558.17
< 0.0001
C
103.46
1
103.46
1274.82
< 0.0001
D
30.66
1
30.66
377.80
< 0.0001
AB
29.93
1
29.93
368.74
< 0.0001
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-24 AC
128.50
1
128.50
1583.26
< 0.0001
AD
0.047
1
0.047
0.58
0.4586
BC
0.074
1
0.074
0.91
0.3547
BD
0.018
1
0.018
0.22
0.6453
CD
0.047
1
0.047
0.58
0.4564
ABC
78.75
1
78.75
970.33
< 0.0001
ABD
0.077
1
0.077
0.95
0.3450
ACD
2.926E-003
1
2.926E-003
0.036
0.8518
BCD
0.010
1
0.010
0.13
0.7282
ABCD
1.596E-003
1
1.596E-003
0.020
0.8902
Residual 1.30 16 0.081 Lack of Fit
0.000
0
Pure Error
1.30
16
0.081
Cor Total 572.25 31 The Model F-value of 468.99 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, D, AB, AC, ABC are significant model terms. (c) Write down a regression model that can be used to predict crack length as a function of the significant main effects and interactions you have identified in part (b). Design Expert Output Final Equation in Terms of Coded Factors:
Crack Length= +11.99 +1.51 *A +1.99 *B -1.80 *C +0.98 *D +0.97 *A*B -2.00 *A*C +1.57 * A * B * C
(d) Analyze the residuals from this experiment. There is nothing unusual about the residuals. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-25 (e) Is there an indication that any of the factors affect the variability in cracking? By calculating the range of the two readings in each cell, we can also evaluate the effects of the factors on variation. The following is the normal probability plot of effects: It appears that the AB
and CD
interactions could be significant. The following is the ANOVA for the range data: Design Expert Output
Response:
Range
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 0.29 2 0.14 11.46 0.0014 significant AB 0.13 1 0.13 9.98 0.0075 CD 0.16 1 0.16 12.94 0.0032 Residual 0.16 13 0.013 Cor Total 0.45 15 The Model F-value of 11.46 implies the model is significant. There is only a 0.14% 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 AB, CD are significant model terms. Final Equation in Terms of Coded Factors:
Range = +0.37 +0.089 * A * B +0.10 * C * D (f) What recommendations would you make regarding process operations? Use interaction and/or main effect plots to assist in drawing conclusions. From the interaction plots, choose A
at the high level and B
at the low level. In each of these plots, D
can be at either level. From the main effects plot of C
, choose C
at the high level. Based on the range analysis, with C
at the high level, D
should be set at the low level. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-26 From the analysis of the crack length data: Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-27 From the analysis of the ranges: 6.20. Continuation of Problem 6.19.
One of the variables in the experiment described in Problem 6.19, heat treatment method (
C
), is a categorical variable. Assume that the remaining factors are continuous. (a) Write two regression models for predicting crack length, one for each level of the heat treatment method variable. What differences, if any, do you notice in these two equations? Design Expert Output
Final Equation in Terms of Actual Factors Heat Treat Method -1 Crack Length = +13.78619 +3.51331 * Pour Temp +1.93994 * Titanium Content +0.97888 * Grain Refiner -0.60169 * Pour Temp * Titanium Content Heat Treat Method 1 Crack Length = +10.18994 -0.49444 * Pour Temp +2.03594 * Titanium Content +0.97888 * Grain Refiner +2.53581 * Pour Temp * Titanium Content Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-28 (b) Generate appropriate response surface contour plots for the two regression models in part (a). (c) What set of conditions would you recommend for the factors A
, B
and D
if you use heat treatment method C = +? High level of A
, low level of B,
and low level of D
. (d) Repeat part (c) assuming that you wish to use heat treatment method C = –. Low level of A
, low level of B,
and low level of D
. 6.21.
An experimenter has run a single replicate of a 2
4
design. The following effect estimates have been calculated: A
= 76.95
AB
= -51.32
ABC
= -2.82
B
= -67.52
AC
= 11.69
ABD
= -6.50
C
= -7.84
AD
= 9.78
ACD
= 10.20
D
= -18.73
BC
= 20.78
BCD
= -7.98
BD
= 14.74
ABCD
= -6.25
CD
= 1.27
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-29 (a) Construct a normal probability plot of these effects. The plot from Minitab follows. (b) Identify a tentative model, based on the plot of the effects in part (a). ˆ
38.475
33.76
25.66
A
B
A
B
y
Intercept
x
x
x x
=
+
-
-
6.22.
The effect estimates from a 2
4
factorial design are as follows: ABCD
= -1.5138, ABC
= -
1.2661, ABD
= -0.9852, ACD
= -0.7566, BCD
= -0.4842, CD
= -0.0795, BD
= -0.0793, AD
= 0.5988, BC
= 0.9216, AC
= 1.1616, AB
= 1.3266, D
= 4.6744, C
= 5.1458, B
= 8.2469, and A
= 12.7151. Are you comfortable with the conclusions that all main effects are active? The upper right 4 dots are the four main effects. Since they do not follow the rest of the data, the normal probability plot shows that they are active. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-30 6.23.
The effect estimates from a 2
4
factorial experiment are listed here. Are any of the effects significant? ABCD
= -2.5251
AD
= -1.6564
BCD
= 4.4054
AC
= 1.1109
ACD
= -0.4932
AB
= -10.5229
ABD
= -5.0842
D
= -6.0275
ABC
= -5.7696
C
= -8.2045
CD
= 4.6707
B
= -6.5304
BD
= -4.6620
A
= -0.7914
BC
= -0.7982
No effects appear to be significant. 6.24.
Consider a variation of the bottle filling experiment from Example 5.3. Suppose that only two levels of carbonation are used so that the experiment is a 2
3
factorial design with two replicates. The data are shown below. Coded Factors Fill Height Deviation Run A B C Replicate 1 Replicate 2 1 – – – -3 -1 2 + – – 0 1 3 – + – -1 0 4 + + – 2 3 5 – – + -1 0 6 + – + 2 1 7 – + + 1 1 8 + + + 6 5 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-31 Factor Levels Low (–1) High (+1) A
(%) 10 12 B
(psi) 25 30 C
(b/m) 200 250 (a) Analyze the data from this experiment. Which factors significantly affect fill height deviation? The half normal probability plot of effects shown below identifies the factors A
, B
, and C
as being significant and the AB
interaction as being marginally significant. The analysis of variance in the Design Expert output below confirms that factors A
, B
, and C
are significant and the AB
interaction is marginally significant. Design Expert Output
Response:
Fill Deviation
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 70.75 4 17.69 26.84 < 0.0001 significant A 36.00 1 36.00 54.62 < 0.0001 B 20.25 1 20.25 30.72 0.0002 C 12.25 1 12.25 18.59 0.0012 AB 2.25 1 2.25 3.41 0.0917 Residual 7.25 11 0.66 Lack of Fit 2.25 3 0.75 1.20 0.3700 not significant Pure Error 5.00 8 0.63 Cor Total 78.00 15 The Model F-value of 26.84 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. Std. Dev. 0.81 R-Squared 0.9071 Mean 1.00 Adj R-Squared 0.8733 C.V. 81.18 Pred R-Squared 0.8033 PRESS 15.34 Adeq Precision 15.424 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-32 (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy? The residual plots below do not identify any violations to the assumptions. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-33 (c) Obtain a model for predicting fill height deviation in terms of the important process variables. Use this model to construct contour plots to assist in interpreting the results of the experiment. The model in both coded and actual factors are shown below. Design Expert Output
Final Equation in Terms of Coded Factors Fill Deviation = +1.00 +1.50 * A +1.13 * B +0.88 * C +0.38 * A * B Final Equation in Terms of Actual Factors Fill Deviation = +9.62500 -2.62500 * Carbonation -1.20000 * Pressure +0.035000 * Speed +0.15000 * Carbonation * Pressure The following contour plots identify the fill deviation with respect to carbonation and pressure. The plot on the left sets the speed at 200 b/m while the plot on the right sets the speed at 250 b/m. Assuming a faster bottle speed is better, settings in pressure and carbonation that produce a fill deviation near zero can be found in the lower left hand corner of the contour plot on the right. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-34 Speed set at 200 b/m Speed set at 250 b/m (d) In part (a), you probably noticed that there was an interaction term that was borderline significant. If you did not include the interaction term in your model, include it now and repeat the analysis. What difference did this make? If you elected to include the interaction term in part (a), remove it and repeat the analysis. What difference does this make? The following analysis of variance, residual plots, and contour plots represent the model without the interaction. As in the original analysis, the residual plots do not identify any concerns with the assumptions. The contour plots did not change significantly either. The interaction effect is small relative to the main effects. Design Expert Output
Response:
Fill Deviation
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 68.50 3 22.83 28.84 < 0.0001 significant A 36.00 1 36.00 45.47 < 0.0001 B 20.25 1 20.25 25.58 0.0003 C 12.25 1 12.25 15.47 0.0020 Residual 9.50 12 0.79 Lack of Fit 4.50 4 1.13 1.80 0.2221 not significant Pure Error 5.00 8 0.63 Cor Total 78.00 15 The Model F-value of 28.84 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. Std. Dev. 0.89 R-Squared 0.8782 Mean 1.00 Adj R-Squared 0.8478 C.V. 88.98 Pred R-Squared 0.7835 PRESS 16.89 Adeq Precision 15.735 Final Equation in Terms of Coded Factors Fill Deviation = +1.00 +1.50 * A Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-35 +1.13 * B +0.88 * C Final Equation in Terms of Actual Factors Fill Deviation = -35.75000 +1.50000 * Carbonation +0.45000 * Pressure +0.035000 * Speed Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-36 Speed set at 200 b/m Speed set at 250 b/m 6.25.
I am always interested in improving my golf scores. Since a typical golfer uses the putter for about 35-45% of his or her strokes, it seems logical that in improving one’s putting score is a logical and perhaps simple way to improve a golf score (“The man who can putt is a match for any man.” – Willie Parks, 1864-1925, two-time winner of the British Open). An experiment was conducted to study the effects of four factors on putting accuracy. The design factors are length of putt, type of putter, breaking putt vs. straight putt, and level versus downhill putt. The response variable is distance from the ball to the center of the cup after the ball comes to rest. One golfer performs the experiment, a 2
4
factorial design with seven replicates was used, and all putts were made in random order. The results are as follows. Design Factors Distance from cup (replicates) Length of putt (ft) Type of putter Break of putt Slope of putt 1 2 3 4 5 6 7 10 Mallet Straight Level 10.0 18.0 14.0 12.5 19.0 16.0 18.5 30 Mallet Straight Level 0.0 16.5 4.5 17.5 20.5 17.5 33.0 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-37 (a) Analyze the data from this experiment. Which factors significantly affect putting performance? The half normal probability plot of effects identifies only factors A
and B
, length of putt and type of putter, as having a potentially significant effect on putting performance. The analysis of variance with only these significant factors is presented as well and confirms significance. Design Expert Output with Only Factors
A
and B
Response:
Distance from cup
ANOVA for Selected Factorial Model
Analysis of variance table [Terms added sequentially (first to last)]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 1305.29 2 652.65 7.69 0.0008 significant A 917.15 1 917.15 10.81 0.0014 B 388.15 1 388.15 4.57 0.0347 Residual 9248.94 109 84.85 Lack of Fit 933.15 13 71.78 0.83 0.6290 not significant Pure Error 8315.79 96 86.62 Cor Total 10554.23 111 The Model F-value of 7.69 implies the model is significant. There is only a 0.08% chance that a "Model F-Value" this large could occur due to noise. 10 Cavity-back Straight Level 4.0 6.0 1.0 14.5 12.0 14.0 5.0 30 Cavity-back Straight Level 0.0 10.0 34.0 11.0 25.5 21.5 0.0 10 Mallet Breaking Level 0.0 0.0 18.5 19.5 16.0 15.0 11.0 30 Mallet Breaking Level 5.0 20.5 18.0 20.0 29.5 19.0 10.0 10 Cavity-back Breaking Level 6.5 18.5 7.5 6.0 0.0 10.0 0.0 30 Cavity-back Breaking Level 16.5 4.5 0.0 23.5 8.0 8.0 8.0 10 Mallet Straight Downhill 4.5 18.0 14.5 10.0 0.0 17.5 6.0 30 Mallet Straight Downhill 19.5 18.0 16.0 5.5 10.0 7.0 36.0 10 Cavity-back Straight Downhill 15.0 16.0 8.5 0.0 0.5 9.0 3.0 30 Cavity-back Straight Downhill 41.5 39.0 6.5 3.5 7.0 8.5 36.0 10 Mallet Breaking Downhill 8.0 4.5 6.5 10.0 13.0 41.0 14.0 30 Mallet Breaking Downhill 21.5 10.5 6.5 0.0 15.5 24.0 16.0 10 Cavity-back Breaking Downhill 0.0 0.0 0.0 4.5 1.0 4.0 6.5 30 Cavity-back Breaking Downhill 18.0 5.0 7.0 10.0 32.5 18.5 8.0 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-38 Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B, are significant model terms. Std. Dev. 9.21 R-Squared 0.1237 Mean 12.30 Adj R-Squared 0.1076 C.V. 74.90 Pred R-Squared 0.0748 PRESS 9765.06 Adeq Precision 6.266 (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy? The residual plots for the model containing only the significant factors A
and B
are shown below. The normality assumption appears to be violated. Also, as a golfer might expect, there is a slight inequality of variance with regards to the length of putt. A square root transformation is applied which corrects the violations. The analysis of variance and corrected residual plots are also presented. Finally, an effects plot identifies a 10 foot putt and the cavity-back putter reduce the mean distance from the cup. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-39 Design Expert Output with Only Factors
A
and B and a Square Rot Transformation
Response: Distance from cupTransform:Square root Constant: 0 ANOVA for Selected Factorial Model
Analysis of variance table [Terms added sequentially (first to last)]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 37.26 2 18.63 7.85 0.0007 significant A 21.61 1 21.61 9.11 0.0032 B 15.64 1 15.64 6.59 0.0116 Residual 258.63 109 2.37 Lack of Fit 30.19 13 2.32 0.98 0.4807 not significant Pure Error 228.45 96 2.38 Cor Total 295.89 111 The Model F-value of 7.85 implies the model is significant. There is only a 0.07% 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, are significant model terms. Std. Dev. 1.54 R-Squared 0.1259 Mean 3.11 Adj R-Squared 0.1099 C.V. 49.57 Pred R-Squared 0.0771 PRESS 273.06 Adeq Precision 6.450 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-40 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-41 6.26.
Semiconductor manufacturing processes have long and complex assembly flows, so matrix marks and automated 2d-matrix readers are used at several process steps throughout factories. Unreadable matrix marks negatively affect factory run rates, because manual entry of part data is required before manufacturing can resume. A 2
4
factorial experiment was conducted to develop a 2d-
matrix laser mark on a metal cover that protects a substrate mounted die. The design factors are A
= laser power (9W, 13W), B
= laser pulse frequency (4000 Hz, 12000 Hz), C
= matrix cell size (0.07 in, 0.12 in), and D
= writing speed (10 in/sec, 20 in/sec), and the response variable is the unused error correction (UEC). This is a measure of the unused portion of the redundant information embedded in the 2d matrix. A UEC of 0 represents the lowest reading that still results in a decodable matrix while a value of 1 is the highest reading. A DMX Verifier was used to measure UEC. The data from this experiment are shown below. Standard Order Run Order Laser Power Pulse Frequency Cell Size Writing Speed UEC 8 1 1 1 1 -1 0.80 10 2 1 -1 -1 1 0.81 12 3 1 1 -1 1 0.79 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-42 9 4 -1 -1 -1 1 0.60 7 5 -1 1 1 -1 0.65 15 6 -1 1 1 1 0.55 2 7 1 -1 -1 -1 0.98 6 8 1 -1 1 -1 0.67 16 9 1 1 1 1 0.69 13 10 -1 -1 1 1 0.56 5 11 -1 -1 1 -1 0.63 14 12 1 -1 1 1 0.65 1 13 -1 -1 -1 -1 0.75 3 14 -1 1 -1 -1 0.72 4 15 1 1 -1 -1 0.98 11 16 -1 1 -1 1 0.63 (a) Analyze the data from this experiment. Which factors significantly affect UEC? The normal probability plot of effects identifies A
, C
, D
, and the AC
interaction as significant. The Design Expert output including the analysis of variance confirms the significance and identifies the corresponding model. Contour plots identify factors A
and C
with B
held constant at zero and D
toggled from -1 to +1. Design Expert Output
Response:
UEC
ANOVA for Selected Factorial Model
Analysis of variance table [Terms added sequentially (first to last)]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 0.24 4 0.059 35.51 < 0.0001 significant A 0.10 1 0.10 61.81 < 0.0001 C 0.070 1 0.070 42.39 < 0.0001 D 0.051 1 0.051 30.56 0.0002 AC 0.012 1 0.012 7.30 0.0206 Residual 0.018 11 1.657E-003 Cor Total 0.25 15 The Model F-value of 35.51 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. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-43 In this case A, C, D, AC are significant model terms. Std. Dev. 0.041 R-Squared 0.9281 Mean 0.72 Adj R-Squared 0.9020 C.V. 5.68 Pred R-Squared 0.8479 PRESS 0.039 Adeq Precision 17.799 Final Equation in Terms of Coded Factors UEC = +0.72 +0.080 * A -0.066 * C -0.056 * D -0.027 * A * C Final Equation in Terms of Actual Factors UEC = +0.71625 +0.080000 * Laser Power -0.066250 * Cell Size -0.056250 * Writing Speed -0.027500 * Laser Power * Cell Size (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy? The residual plots appear acceptable with the exception of run 8, standard order 6. This value should be verified by the engineer. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-44 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-45 6.27. Reconsider the experiment in Problem 6.26. Suppose that four center points are available, and the UEC response at these four runs is 0.98, 0.95, 0.93 and 0.96, respectively. Reanalyze the experiment incorporating a test for curvature into the analysis. What conclusions can you draw? What recommendations would you make to the experimenters? As with the results of problem 6.26, factors A
, C
, D
, and the AC
interaction remain significant. However, the CD
interaction and curvature are significant as well. The curvature is the strongest effect; unfortunately, we are not able to determine which factor(s) have a quadratic term. We recommend that the engineer augment the experiment with additional experimental runs, such as axial points and a couple of extra center points for blocking purposes. These extra runs will determine the pure quadratic effects and allow us to fit a second order model. Design Expert Output
Response:
UEC
ANOVA for Selected Factorial Model
Analysis of variance table [Terms added sequentially (first to last)]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 0.24 5 0.048 45.07 < 0.0001 significant A 0.10 1 0.10 95.77 < 0.0001 C 0.070 1 0.070 65.68 < 0.0001 D 0.051 1 0.051 47.35 < 0.0001 AC 0.012 1 0.012 11.32 0.0051 CD 5.625E-003 1 5.625E-003 5.26 0.0391 Curvature 0.18 1 0.18 170.59 < 0.0001 significant Residual 0.014 13 1.069E-003 Lack of Fit 0.013 10 1.260E-003 2.91 0.2057 not significant Pure Error 1.300E-003 3 4.333E-004 Cor Total 0.44 19 The Model F-value of 45.07 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, C, D, AC, CD are significant model terms. Std. Dev. 0.033 R-Squared 0.9455 Mean 0.76 Adj R-Squared 0.9245 C.V. 4.28 Pred R-Squared 0.9209 PRESS 0.035 Adeq Precision 20.936 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-46 Intercept 0.72 1 8.175E-003 0.70 0.73 A-Laser Power 0.080 1 8.175E-003 0.062 0.098 1.00 C-Cell Size -0.066 1 8.175E-003 -0.084 -0.049 1.00 D-Writing Speed -0.056 1 8.175E-003 -0.074 -0.039 1.00 AC -0.028 1 8.175E-003 -0.045 -9.839E-003 1.00 CD 0.019 1 8.175E-003 1.089E-003 0.036 1.00 Center Point 0.24 1 0.018 0.20 0.28 1.00 6.28. A company markets its products by direct mail. An experiment was conducted to study the effects of three factors on the customer response rate for a particular product. The three factors are A
= type of mail used (3
rd
class, 1
st
class), B
= type of descriptive brochure (color, black-and-white), and C
= offer price ($19.95, $24.95). The mailings are made to two groups of 8,000 randomly selected customers, with 1,000 customers in each group receiving each treatment combination. Each group of customers is considered as a replicate. The response variable is the number of orders placed. The experimental data is shown below. Coded Factors Number of Orders Run A B C Replicate 1 Replicate 2 1 – – – 50 54 2 + – – 44 42 3 – + – 46 48 4 + + – 42 43 5 – – + 49 46 6 + – + 48 45 7 – + + 47 48 8 + + + 56 54 Factor Levels Low (-1) High (+1) A
(class) 3rd 1st B
(type) BW Color C
($) $19.95 $24.95 (a) Analyze the data from this experiment. Which factors significantly affect the customer response rate? The half normal probability plot of effects identifies the two factor interactions, AB
, AC
, BC
, and factors A
and C
as significant. Factor B
is not significant; however, remains in the model to satisfy the hierarchal principle. The analysis of variance confirms the significance of two factor interactions and factor C
. Factor A
is marginally significant. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-47 Design Expert Output
Response:
Number of orders
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 241.75 6 40.29 12.95 0.0006 significant A 12.25 1 12.25 3.94 0.0785 B 2.25 1 2.25 0.72 0.4171 C 36.00 1 36.00 11.57 0.0079 AB 42.25 1 42.25 13.58 0.0050 AC 100.00 1 100.00 32.14 0.0003 BC 49.00 1 49.00 15.75 0.0033 Residual 28.00 9 3.11 Lack of Fit 4.00 1 4.00 1.33 0.2815 not significant Pure Error 24.00 8 3.00 Cor Total 269.75 15 The Model F-value of 12.95 implies the model is significant. There is only a 0.06% 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 C, AB, AC, BC are significant model terms. Std. Dev. 1.76 R-Squared 0.8962 Mean 47.63 Adj R-Squared 0.8270 C.V. 3.70 Pred R-Squared 0.6719 PRESS 88.49 Adeq Precision 10.286 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 47.63 1 0.44 46.63 48.62 A-Class -0.88 1 0.44 -1.87 0.12 1.00 B-Type 0.37 1 0.44 -0.62 1.37 1.00 C-Price 1.50 1 0.44 0.50 2.50 1.00 AB 1.63 1 0.44 0.63 2.62 1.00 AC 2.50 1 0.44 1.50 3.50 1.00 BC 1.75 1 0.44 0.75 2.75 1.00 (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy? The residual plots below do not identify model inadequacy. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-48 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-49 (c) What would you recommend to the company? Based on the interaction plots below, we recommend 3
rd
class mail, black-and-white brochures, and an offered price of $19.95 would achieve the greatest number of orders. If the offered price must be $24.95, then the 1
st
class mail with color brochures is recommended. 6.29.
Consider the single replicate of the 2
4
design in Example 6.2. Suppose we had arbitrarily decided to analyze the data assuming that all three- and four-factor interactions were negligible. Conduct this analysis and compare your results with those obtained in the example. Do you think that it is a good idea to arbitrarily assume interactions to be negligible even if they are relatively high-
order ones? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-50 Design Expert Output
Response:
Filtration Rate
in A/min
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 5603.13 10 560.31 21.92 0.0016 significant A 1870.56 1 1870.56 73.18 0.0004 B 39.06 1 39.06 1.53 0.2713 C 390.06 1 390.06 15.26 0.0113 D 855.56 1 855.56 33.47 0.0022 AB 0.063 1 0.063 2.445E-003 0.9625 AC 1314.06 1 1314.06 51.41 0.0008 AD 1105.56 1 1105.56 43.25 0.0012 BC 22.56 1 22.56 0.88 0.3906 BD 0.56 1 0.56 0.022 0.8879 CD 5.06 1 5.06 0.20 0.6749 Residual 127.81 5 25.56 Cor Total 5730.94 15 The Model F-value of 21.92 implies the model is significant. There is only a 0.16% 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, C, D, AC, AD are significant model terms. This analysis of variance identifies the same effects as the normal probability plot of effects approach used in Example 6.2. In general, it is not a good idea to arbitrarily pool interactions. Use the normal probability plot of effect estimates as a guide in the choice of which effects to tentatively include in the model. 6.30.
An experiment was run in a semiconductor fabrication plant in an effort to increase yield. Five factors, each at two levels, were studied. The factors (and levels) were A
= aperture setting (small, large), B
= exposure time (20% below nominal, 20% above nominal), C
= development time (30 s, 45 s), D
= mask dimension (small, large), and E
= etch time (14.5 min, 15.5 min). The unreplicated 2
5
design shown below was run. (1) = 7 d = 8
e = 8
de =
6
a = 9 ad = 10
ae =
12
ade =
10 b = 34 bd =
32
be =
35
bde =
30 ab = 55 abd =
50
abe =
52
abde =
53 c = 16 cd =
18
ce =
15
cde =
15 ac = 20 acd =
21
ace =
22
acde =
20 bc = 40 bcd =
44
bce =
45
bcde =
41 abc = 60 abcd =
61
abce =
65
abcde =
63 (a) Construct a normal probability plot of the effect estimates. Which effects appear to be large? From the normal probability plot of effects shown below, effects A
, B
, C
, and the AB
interaction appear to be large. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-51 (b) Conduct an analysis of variance to confirm your findings for part (a). Design Expert Output
Response:
Yield
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 11585.13 4 2896.28 991.83 < 0.0001 significant A
1116.28
1
1116.28
382.27
< 0.0001
B
9214.03
1
9214.03
3155.34
< 0.0001
C
750.78
1
750.78
257.10
< 0.0001
AB
504.03
1
504.03
172.61
< 0.0001
Residual 78.84 27 2.92 Cor Total 11663.97 31 The Model F-value of 991.83 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, AB are significant model terms. (c) Write down the regression model relating yield to the significant process variables. Design Expert Output
Final Equation in Terms of Actual Factors:
Aperture small Yield = +0.40625 +0.65000 * Exposure Time +0.64583 * Develop Time Aperture large Yield = +12.21875 +1.04688 * Exposure Time +0.64583 * Develop Time (d) Plot the residuals on normal probability paper. Is the plot satisfactory? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-52 There is nothing unusual about this plot. (e) Plot the residuals versus the predicted yields and versus each of the five factors. Comment on the plots. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-53 The plot of residual versus exposure time shows some very slight inequality of variance. There is no strong evidence of a potential problem. (f) Interpret any significant interactions. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-54 Factor A
does not have as large an effect when B
is at its low level as it does when B
is at its high level. (g) What are your recommendations regarding process operating conditions? To achieve the highest yield, run B
at the high level, A
at the high level, and C
at the high level. (h) Project the 2
5
design in this problem into a 2
k
design in the important factors. Sketch the design and show the average and range of yields at each run. Does this sketch aid in interpreting the results of this experiment? This cube plot aids in interpretation. The strong AB
interaction and the large positive effect of C
are clearly evident. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-55 6.31. Continuation of Problem 6.30. Suppose that the experimenter had run four runs at the center points in addition to the 32 trials in the original experiment. The yields obtained at the center point runs were 68, 74, 76, and 70. (a) Reanalyze the experiment, including a test for pure quadratic curvature. Because aperture and mask dimension are not continuous variables, the four center points were split amongst these two factors as follows. The sum of squares for the curvature can be estimated with the following equation and is confirmed with the analysis of variance shown in the Design Expert output. (
)
(
)(
)(
)
2
2
32
4
30.53125
72
6114.337
32
4
F
C
F
C
PureQuadratic
F
C
n n
y
y
SS
n
n
-
-
=
=
=
+
+
Design Expert Output
Response:
Yield
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 11461.09 4 2865.27 353.92 < 0.0001 significant A
992.25
1
992.25
122.56
< 0.0001
B
9214.03
1
9214.03
1138.12
< 0.0001
C
750.78
1
750.78
92.74
< 0.0001
AB
504.03
1
504.03
62.26
< 0.0001
Curvature 6114.34 1 6114.34 755.24 < 0.0001 significant Residual 242.88 30 8.10 Cor Total 17818.31 35 The Model F-value of 353.92 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, AB are significant model terms. (b) Discuss what your next step would be. Add axial points for factors B
and C
along with four more center points to fit a second-order model and satisfy blocking concerns. 6.32.
In a process development study on yield, four factors were studied, each at two levels: time (
A
), concentration (
B
), pressure (
C
), and temperature (
D
). A single replicate of a 2
4
design was run, and the resulting data are shown in the following table: Aperture Mask
Dimension Yield Small Small 68 Large Small 74 Small Large 76 Large Large 70 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-56 Actual Run Run Yield
Factor Levels Number Order A B C D (lbs)
Low (-) High (+) 1 5 – – –
–
12
A
(h)
2.5
3.0 2 9 + – –
–
18
B
(%)
14
18 3 8 – + –
–
13
C
(psi)
60
80 4 13 + + –
–
16
D
(
°
C) 225
250 5 3 – – +
–
17
6 7 + – +
–
15
7 14 – + +
–
20
8 1 + + +
–
15
9 6 – – –
+
10
10 11 + – –
+
25
11 2 – + –
+
13
12 15 + + –
+
24
13 4 – – +
+
19
14 16 + – +
+
21
15 10 – + +
+
17
16 12 + + +
+
23
(a) Construct a normal probability plot of the effect estimates. Which factors appear to have large effects? A, C, D
and the AC
and AD
interactions appear to have large effects. (b) Conduct an analysis of variance using the normal probability plot in part (a) for guidance in forming an error term. What are your conclusions? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-57 Design Expert Output
Response:
Yield
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 275.50 5 55.10 33.91 < 0.0001 significant A
81.00
1
81.00
49.85
< 0.0001
C
16.00
1
16.00
9.85
0.0105
D
42.25
1
42.25
26.00
0.0005
AC
72.25
1
72.25
44.46
< 0.0001
AD
64.00
1
64.00
39.38
< 0.0001
Residual 16.25 10 1.62 Cor Total 291.75 15 The Model F-value of 33.91 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, C, D, AC, AD are significant model terms. (c) Write down a regression model relating yield to the important process variables.
Design Expert Output
Final Equation in Terms of Coded Factors:
Yield = +17.38 +2.25 *A +1.00 *C +1.63 *D -2.13 *A*C +2.00 *A*D Final Equation in Terms of Actual Factors:
Yield = +209.12500 -83.50000 * Time +2.43750 * Pressure -1.63000 * Temperature -0.85000 * Time * Pressure +0.64000 * Time * Temperature
(d) Analyze the residuals from this experiment. Does your analysis indicate any potential problems? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-58 There is nothing unusual about the residual plots. (e) Can this design be collapsed into a 2
3
design with two replicates? If so, sketch the design with the average and range of yield shown at each point in the cube. Interpret the results. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-59 6.33. Continuation of Problem 6.32.
Use the regression model in part (c) of Problem 6.32 to generate a response surface contour plot of yield. Discuss the practical purpose of this response surface plot. The response surface contour plot shows the adjustments in the process variables that lead to an increasing or decreasing response. It also displays the curvature of the response in the design region, possibly indicating where robust operating conditions can be found. Two response surface contour plots for this process are shown below. These were formed from the model written in terms of the original design variables. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-60 6.34.
The scrumptious brownie experiment. The author is an engineer by training and a firm believer in learning by doing. I have taught experimental design for many years to a wide variety of audiences and have always assigned the planning, conduct, and analysis of an actual experiment to the class participants. The participants seem to enjoy this practical experience and always learn a great deal from it. This problem uses the results of an experiment performed by Gretchen Krueger at Arizona State University. There are many different ways to bake brownies. The purpose of this experiment was to determine how the pan material, the brand of brownie mix, and the stirring method affect the scrumptiousness of brownies. The factor levels were Factor Low (–)
High (+)
A
= pan material
Glass
Aluminum
B
= stirring method
Spoon
Mixer
C
= brand of mix
Expensive Cheap
The response variable was scrumptiousness, a subjective measure derived from a questionnaire given to the subjects who sampled each batch of brownies. (The questionnaire dealt with such issues as taste, appearance, consistency, aroma, and so forth.) An eight-person test panel sampled each batch and filled out the questionnaire. The design matrix and the response data are shown below: Brownie Test
Panel
Results
Batch A B C 1
2
3
4
5
6
7 8 1 – – – 11
9
10
10
11
10
8 9 2 + – – 15
10
16
14
12
9
6 15 3 – + – 9
12
11
11
11
11
11 12 4 + + – 16
17
15
12
13
13
11 11 5 – – + 10
11
15
8
6
8
9 14 6 + – + 12
13
14
13
9
13
14 9 7 – + + 10
12
13
10
7
7
17 13 8 + + + 15
12
15
13
12
12
9 14 (a) Analyze the data from this experiment as if there were eight replicates of a 2
3
design. Comment on the results.
Design Expert Output
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 93.25 7 13.32 2.20 0.0475 significant A
72.25
1
72.25
11.95
0.0010
B
18.06
1
18.06
2.99
0.0894
C
0.063
1
0.063
0.010
0.9194
AB
0.062
1
0.062
0.010
0.9194
AC
1.56
1
1.56
0.26
0.6132
BC
1.00
1
1.00
0.17
0.6858
ABC
0.25
1
0.25
0.041
0.8396
Residual 338.50 56 6.04 Lack of Fit
0.000
0
Pure Error
338.50
56
6.04
Cor Total 431.75 63 The Model F-value of 2.20 implies the model is significant. There is only a 4.75% 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 are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-61 In this analysis, A
, the pan material and B
, the stirring method, appear to be significant. There are 56 degrees of freedom for the error, yet only eight batches of brownies were cooked, one for each recipe. (b) Is the analysis in part (a) the correct approach? There are only eight batches; do we really have eight replicates of a 2
3
factorial design? The different rankings by the taste-test panel are not replicates, but repeat observations by different testers on the same batch of brownies. It is not a good idea to use the analysis in part (a) because the estimate of error may not reflect the batch-to-batch variation. (c) Analyze the average and standard deviation of the scrumptiousness ratings. Comment on the results. Is this analysis more appropriate than the one in part (a)? Why or why not? Design Expert Output
Response:
Average
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 11.28 2 5.64 76.13 0.0002 significant A
9.03
1
9.03
121.93
0.0001
B
2.25
1
2.25
30.34
0.0027
Residual 0.37 5 0.074 Cor Total 11.65 7 The Model F-value of 76.13 implies the model is significant. There is only a 0.02% 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 are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-62 Design Expert Output
Response:
Stdev
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 6.05 3 2.02 9.77 0.0259 significant A
0.24
1
0.24
1.15
0.3432
C
0.91
1
0.91
4.42
0.1034
AC
4.90
1
4.90
23.75
0.0082
Residual 0.82 4 0.21 Cor Total 6.87 7 The Model F-value of 9.77 implies the model is significant. There is only a 2.59% 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 AC are significant model terms. Variables A
and B
affect the mean rank of the brownies. Note that the AC
interaction affects the standard deviation of the ranks. This is an indication that both factors A
and C
have some effect on the variability in the ranks. It may also indicate that there is some inconsistency in the taste test panel members. For the analysis of both the average of the ranks and the standard deviation of the ranks, the mean square error is now determined by pooling apparently unimportant effects. This is a more accurate estimate of error than obtained assuming that all observations were replicates. 6.35.
An experiment was conducted on a chemical process that produces a polymer. The four factors studied were temperature (
A
), catalyst concentration (
B
), time (
C
), and pressure (
D
). Two responses, molecular weight and viscosity, were observed. The design matrix and response data are shown below: Actual Run Run Molecular
Factor Levels
Number Order A B C D Weight
Viscosity
Low (
–
) High (+)
1 18 – – – –
2400
1400
A
(
°
C) 100 120 2 9 + – – –
2410
1500
B
(%)
4 8 3 13 – + – –
2315
1520
C
(min)
20 30 4 8 + + – –
2510
1630
D
(psi)
60 75 5 3 – – + –
2615
1380
6 11 + – + –
2625
1525
7 14 – + + –
2400
1500
8 17 + + + –
2750
1620
9 6 – – – +
2400
1400
10 7 + – – +
2390
1525
11 2 – + – +
2300
1500
12 10 + + – +
2520
1500
13 4 – – + +
2625
1420
14 19 + – + +
2630
1490
15 15 – + + +
2500
1500
16 20 + + + +
2710
1600
17 1 0 0 0 0
2515
1500
18 5 0 0 0 0
2500
1460
19 16 0 0 0 0
2400
1525
20 12 0 0 0 0
2475
1500
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-63 (a) Consider only the molecular weight response. Plot the effect estimates on a normal probability scale. What effects appear important? A, C
and the AB
interaction appear to be important. (b) Use an analysis of variance to confirm the results from part (a). Is there an indication of curvature? A,C
and the AB
interaction are significant. While the main effect of B
is not significant, it could be included to preserve hierarchy in the model. There is no indication of quadratic curvature. Design Expert Output
Response:
Molecular Wt
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 2.809E+005 3 93620.83 73.00 < 0.0001 significant A
61256.25
1
61256.25
47.76
< 0.0001
C
1.620E+005
1
1.620E+005
126.32
< 0.0001
AB
57600.00
1
57600.00
44.91
< 0.0001
Curvature 3645.00 1 3645.00 2.84 0.1125 not significant Residual 19237.50 15 1282.50 Lack of Fit
11412.50
12
951.04
0.36
0.9106
not significant
Pure Error
7825.00
3
2608.33
Cor Total 3.037E+00519 The Model F-value of 73.00 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, C, AB are significant model terms. (c) Write down a regression model to predict molecular weight as a function of the important variables. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-64 Design Expert Output
Final Equation in Terms of Coded Factors:
Molecular Wt = +2506.25 +61.87 * A +100.63 * C +60.00 * A * B (d) Analyze the residuals and comment on model adequacy. There are two residuals that appear to be large and should be investigated.
(e) Repeat parts (a) – (d) using the viscosity response. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-65 Design Expert Output
Response:
Viscosity
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 70362.50 2 35181.25 35.97 < 0.0001 significant A
37056.25
1
37056.25
37.88
< 0.0001
B
33306.25
1
33306.25
34.05
< 0.0001
Curvature 61.25 1 61.25 0.063 0.8056 not significant Residual 15650.00 16 978.13 Lack of Fit
13481.25
13
1037.02
1.43
0.4298
not significant
Pure Error
2168.75
3
722.92
Cor Total 86073.75 19 The Model F-value of 35.97 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 are significant model terms. Final Equation in Terms of Coded Factors:
Viscosity = +1500.62 +48.13 * A +45.63 * B There is one large residual that should be investigated. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-66 6.36. Continuation of Problem 6.35.
Use the regression models for molecular weight and viscosity to answer the following questions. (a) Construct a response surface contour plot for molecular weight. In what direction would you adjust the process variables to increase molecular weight? Increase temperature, catalyst and time. (b) Construct a response surface contour plot for viscosity. In what direction would you adjust the process variables to decrease viscosity? Decrease temperature and catalyst. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-67 (c) What operating conditions would you recommend if it was necessary to produce a product with a molecular weight between 2400 and 2500, and the lowest possible viscosity? Set the temperature between 100 and 105, the catalyst between 4 and 5%, and the time at 24.5 minutes. The pressure was not significant and can be set at conditions that may improve other results of the process such as cost. 6.37.
Consider the single replicate of the 2
4
design in Example 6.2. Suppose that we ran five points at the center (0, 0, 0, 0) and observed the following responses: 93, 95, 91, 89, and 96. Test for curvature in this experiment. Interpret the results. Design Expert Output
Response:
Filtration Rate
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares – Type III]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 5535.81 5 1107.16 68.01 < 0.0001 significant A
1870.56
1
1870.56
114.90
< 0.0001
C
390.06
1
390.06
23.96
0.0002
D
855.56
1
855.56
52.55
< 0.0001
AC
1314.06
1
1314.06
80.71
< 0.0001
AD
1105.56
1
1105.56
67.91
< 0.0001
Curvature 1969.50 1 1969.50 120.97 < 0.0001 significant Residual 227.92 14 16.28 Lack of Fit
195.12
10
19.51
2.38
0.2093
not significant
Pure Error
32.80
4
8.20
Cor Total 7733.24 20 The Model F-value of 68.01 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, C, D, AC, AD are significant model terms. With an F
-value of 120.97 and a corresponding P
-value of < 0.0001, curvature is significant. 6.38.
A missing value in a 2
k
factorial.
It is not unusual to find that one of the observations in a 2
k
design is missing due to faulty measuring equipment, a spoiled test, or some other reason. If the Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-68 design is replicated n
times (
n > 1) some of the techniques discussed in Chapter 5 can be employed. However, for an unreplicated factorial (
n – 1) some other method must be used. One logical approach is to estimate the missing value with a number that makes the highest-order interaction contrast zero. Apply this technique to the experiment in Example 6.2 assuming that run ab
is missing. Compare the results with the results of Example 6.2. Treatment Combination Response Response * ABCD ABCD A B C D (1) 45 45
1
-1
-1
-1 -1 a 71 -71
-1
1
-1
-1 -1 b 48 -48
-1
-1
1
-1 -1 ab missing missing * 1
1
1
1
-1 -1 c 68 -68
-1
-1
-1
1 -1 ac 60 60
1
1
-1
1 -1 bc 80 80
1
-1
1
1 -1 abc 65 -65
-1
1
1
1 -1 d 43 -43
-1
-1
-1
-1 1 ad 100 100
1
1
-1
-1 1 bd 45 45
1
-1
1
-1 1 abd 104 -104
-1
1
1
-1 1 cd 75 75
1
-1
-1
1 1 acd 86 -86
-1
1
-1
1 1 bcd 70 -70
-1
-1
1
1 1 abcd 96 96
1
1
1
1 1 Contrast (ABCD) = Missing – 54 = 0 Missing = 54 Substitute the value 54 for the missing run at ab
. From the effects list and half normal plot shown below, factors A
, C
, D
, AC
, and AD
appear to be large; the same result as found in Example 6.2. Design Expert Output
Term Effect SumSqr % Contribtn Model Intercept Model A 20.25 1640.25 27.5406 Model B 1.75 12.25 0.205684 Model C 11.25 506.25 8.50019 Model D 16 1024 17.1935 Model AB -1.25 6.25 0.104941 Model AC -16.75 1122.25 18.8431 Model AD 18 1296 21.7605 Model BC 3.75 56.25 0.944465 Model BD 1 4 0.067162 Model CD -2.5 25 0.419762 Model ABC 3.25 42.25 0.709398 Model ABD 5.5 121 2.03165 Model ACD -3 36 0.604458 Model BCD -4 64 1.07459 Model ABCD 0 0 0 Lenth's ME 11.5676 Lenth's SME 23.4839 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-69 6.39.
An engineer has performed an experiment to study the effect of four factors on the surface roughness of a machined part. The factors (and their levels) are A
= tool angle (12 degrees, 15 degrees), B
= cutting fluid viscosity (300, 400), C
= feed rate (10 in/min, 15 in/min), and D
= cutting fluid cooler used (no, yes). The data from this experiment (with the factors coded to the usual -1, +1 levels) are shown below. Run A
B
C
D
Surface Roughness
1 –
–
–
–
0.00340 2 + –
–
–
0.00362 3 – + –
–
0.00301 4 + + –
–
0.00182 5 –
–
+
–
0.00280 6 + – +
–
0.00290 7 – + +
–
0.00252 8 + + +
–
0.00160 9 –
–
–
+ 0.00336 10 + –
–
+ 0.00344 11 – + –
+ 0.00308 12 + + –
+ 0.00184 13 –
–
+
+ 0.00269 14 + – +
+ 0.00284 15 – + +
+ 0.00253 16 + + +
+ 0.00163 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-70 (a) Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model. (b) Fit the model identified in part (a) and analyze the residuals. Is there any indication of model inadequacy? Design Expert Output
Response:
Surface Roughness
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 6.406E-006 4 1.601E-006 114.97 < 0.0001 significant A
8.556E-007
1
8.556E-007
61.43
< 0.0001
B
3.080E-006
1
3.080E-006
221.11
< 0.0001
C
1.030E-006
1
1.030E-006
73.96
< 0.0001
AB
1.440E-006
1
1.440E-006
103.38
< 0.0001
Residual 1.532E-007 11 1.393E-008 Cor Total 6.559E-006 15 The Model F-value of 114.97 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, AB are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-71 The plot of residuals versus predicted shows a slight “u-shaped” appearance in the residuals, and the plot of residuals versus tool angle shows an outward-opening funnel. (c) Repeat the analysis from parts (a) and (b) using 1/
y
as the response variable. Is there and indication that the transformation has been useful? The plots of the residuals are more representative of a model that does not violate the constant variance assumption. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-72 Design Expert Output
Response:
Surface RoughnessTransform:
Inverse
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 2.059E+005 4 51472.28 1455.72 < 0.0001 significant A
42610.92
1
42610.92
1205.11
< 0.0001
B
89386.27
1
89386.27
2527.99
< 0.0001
C
18762.29
1
18762.29
530.63
< 0.0001
AB
55129.62
1
55129.62
1559.16
< 0.0001
Residual 388.94 11 35.36 Cor Total 2.063E+00515 The Model F-value of 1455.72 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, AB are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-73 (d) Fit a model in terms of the coded variables that can be used to predict the surface roughness. Convert this prediction equation into a model in the natural variables. Design Expert Output
Final Equation in Terms of Coded Factors:
1.0/(Surface Roughness) = +397.81 +51.61 * A +74.74 * B +34.24 * C +58.70 * A * B 6.40.
Resistivity on a silicon wafer is influenced by several factors. The results of a 2
4
factorial experiment performed during a critical process step is shown below. Run A B C D Resistivity 1 – – – – 1.92 2 + – – – 11.28 3 – + – – 1.09 4 + + – – 5.75 5 – – + – 2.13 6 + – + – 9.53 7 – + + – 1.03 8 + + + – 5.35 9 – – – + 1.60 10 + – – + 11.73 11 – + – + 1.16 12 + + – + 4.68 13 – – + + 2.16 14 + – + + 9.11 15 – + + + 1.07 16 + + + + 5.30 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-74 (a) Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model. (b) Fit the model identified in part (a) and analyze the residuals. Is there any indication of model inadequacy? The normal probability plot of residuals is not satisfactory. The plots of residual versus predicted, residual versus factor A
, and the residual versus factor B
are funnel shaped indicating non-constant variance. Design Expert Output
Response:
Resistivity
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 214.22 3 71.41 148.81 < 0.0001 significant A
159.83
1
159.83
333.09
< 0.0001
B
36.09
1
36.09
75.21
< 0.0001
AB
18.30
1
18.30
38.13
< 0.0001
Residual 5.76 12 0.48 Cor Total 219.98 15 The Model F-value of 148.81 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, AB are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-75 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-76 (c) Repeat the analysis from parts (a) and (b) using ln(
y
) as the response variable. Is there any indication that the transformation has been useful? Design Expert Output
Response:
Resistivity
Transform:
Natural log
Constant:
0.000
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 12.15 2 6.08 553.44 < 0.0001 significant A
10.57
1
10.57
962.95
< 0.0001
B
1.58
1
1.58
143.94
< 0.0001
Residual 0.14 13 0.011 Cor Total 12.30 15 The Model F-value of 553.44 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 are significant model terms. The transformed data no longer indicates that the AB
interaction is significant. A simpler model has resulted from the log transformation. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-77 The residual plots are much improved. (d) Fit a model in terms of the coded variables that can be used to predict the resistivity. Design Expert Output
Final Equation in Terms of Coded Factors:
Ln(Resistivity) = +1.19 +0.81 * A -0.31 * B Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-78 6.41. Continuation of Problem 6.40.
Suppose that the experiment had also run four center points along with the 16 runs in Problem 6.40. The resistivity measurements at the center points are: 8.15, 7.63, 8.95, 6.48. Analyze the experiment again incorporating the center points. What conclusions can you draw now? Design Expert Output
Response:
Resistivity
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 214.22 3 71.41 119.35 < 0.0001 significant A
159.83
1
159.83
267.14
< 0.0001
B
36.09
1
36.09
60.32
< 0.0001
AB
18.30
1
18.30
30.58
< 0.0001
Curvature 31.19 1 31.19 52.13 < 0.0001 significant Residual 8.97 15 0.60 Lack of Fit
5.76
12
0.48
0.45
0.8632
not significant
Pure Error
3.22
3
1.07
Cor Total 254.38 19 The Model F-value of 119.35 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, AB are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-79 Because of the funnel shaped residual versus predicted plot, the analysis was repeated with the natural log transformation. Design Expert Output
Response:
Resistivity
Transform:
Natural log
Constant:
0.000
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 12.15 2 6.08 490.37 < 0.0001 significant A
10.57
1
10.57
853.20
< 0.0001
B
1.58
1
1.58
127.54
< 0.0001
Curvature 2.38 1 2.38 191.98 < 0.0001 significant Residual 0.20 16 0.012 Lack of Fit
0.14
13
0.011
0.59
0.7811
not significant
Pure Error
0.056
3
0.019
Cor Total 14.73 19 The Model F-value of 490.37 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-80 Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms. The "Curvature F-value" of 191.98 implies there is significant curvature (as measured by difference between the average of the center points and the average of the factorial points) in the design space. There is only a 0.01% chance that a "Curvature F-value" this large could occur due to noise.
The curvature test indicates that the model has significant pure quadratic curvature. 6.42. The book by Davies (
Design and Analysis of Industrial Experiments
) describes an experiment to study the yield of isatin. The factors studies and their levels are as follows: Factor Low (
–
)
High (+)
A: Acid strength (%)
87
93
B: Reaction time (min)
15
30
C: Amount of acid (ml)
35
45
D: Reaction temperature (°C)
60
70
The data from the 2
4
factorial is shown in Table P6.11. Table P6.11 A B C D Yield -1 -1 -1 -1 6.08 1 -1 -1 -1 6.04 -1 1 -1 -1 6.53 1 1 -1 -1 6.43 -1 -1 1 -1 6.31 1 -1 1 -1 6.09 -1 1 1 -1 6.12 1 1 1 -1 6.36 -1 -1 -1 1 6.79 1 -1 -1 1 6.68 -1 1 -1 1 6.73 1 1 -1 1 6.08 -1 -1 1 1 6.77 1 -1 1 1 6.38 -1 1 1 1 6.49 1 1 1 1 6.23 (a) Fit a main-effects-only model to the data from this experiment. Are any of the main effects significant? Temperature appears to be significant. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-81 Design Expert Output
Response 1 Yield ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 0.47 4 0.12 2.26 0.1287 not significant A-Acid strength
0.15
1
0.15
2.81
0.1221
B-Reaction time
1.806E-003
1
1.806E-003
0.035
0.8558
C-Amount of acid
0.023
1
0.023
0.45
0.5181
D-Temperature
0.30
1
0.30
5.75
0.0354
Residual 0.57 11 0.052 Cor Total 1.04 15 The "Model F-value" of 2.26 implies the model is not significant relative to the noise. There is a 12.87 % 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 D are significant model terms. (b) Analyze the residuals. Are there any indications of model inadequacy or violation of the assumptions? The residual plots appear to be acceptable. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-82 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-83 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-84 (c) Find an equation for predicting the yield of isatin over the design space. Express the equation in both coded and engineering units. Design Expert Output
Final Equation in Terms of Coded Factors:
Yield = +6.38 -0.096 * A -0.011 * B -0.038 * C +0.14 * D Design Expert Output
Final Equation in Terms of Coded Factors:
Yield = +7.80813 -0.031875 * Acid strength -1.41667E-003 * Reaction time -7.62500E-003 * Amount of acid +0.027375 * Reaction temperature Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-85 (d) Is there any indication that adding interactions to the model would improve the results that you have obtained? The normal probability of effect shown below suggests that the BD
and other interactions may improve the results. The BD
interaction plot shows why the interaction is strong, yet the main effect B
is not significant. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-86 6.43. An article in Quality and Reliability Engineering International
(2010, Vol. 26, pp. 223-233) presents a 2
5
factorial design. The experiment is shown in the following table: A B C D E y -1 -1 -1 -1 -1 8.11 1 -1 -1 -1 -1 5.56 -1 1 -1 -1 -1 5.77 1 1 -1 -1 -1 5.82 -1 -1 1 -1 -1 9.17 1 -1 1 -1 -1 7.8 -1 1 1 -1 -1 3.23 1 1 1 -1 -1 5.69 -1 -1 -1 1 -1 8.82 1 -1 -1 1 -1 14.23 -1 1 -1 1 -1 9.2 1 1 -1 1 -1 8.94 -1 -1 1 1 -1 8.68 1 -1 1 1 -1 11.49 -1 1 1 1 -1 6.25 1 1 1 1 -1 9.12 -1 -1 -1 -1 1 7.93 1 -1 -1 -1 1 5 -1 1 -1 -1 1 7.47 1 1 -1 -1 1 12 -1 -1 1 -1 1 9.86 1 -1 1 -1 1 3.65 -1 1 1 -1 1 6.4 1 1 1 -1 1 11.61 -1 -1 -1 1 1 12.43 1 -1 -1 1 1 17.55 -1 1 -1 1 1 8.87 1 1 -1 1 1 25.38 -1 -1 1 1 1 13.06 1 -1 1 1 1 18.85 -1 1 1 1 1 11.78 1 1 1 1 1 26.05 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-87 (a) Analyze the data from this experiment. Identify the significant factors and interactions. The half normal plot of effects below identifies the significant factors and interactions. Design Expert Output
Response 1 y ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 879.62 11 79.97 49.15 < 0.0001 significant A-A
83.56
1
83.56
51.36
< 0.0001
B-B
0.060
1
0.060
0.037
0.8492
D-D
285.78
1
285.78
175.66
< 0.0001
E-E
153.17
1
153.17
94.15
< 0.0001
AB
48.93
1
48.93
30.08
< 0.0001
AD
88.88
1
88.88
54.63
< 0.0001
AE
33.76
1
33.76
20.75
0.0002
BE
52.71
1
52.71
32.40
< 0.0001
DE
61.80
1
61.80
37.99
< 0.0001
ABE
44.96
1
44.96
27.64
< 0.0001
ADE
26.01
1
26.01
15.99
0.0007
Residual 32.54 20 1.63 Cor Total 912.16 31 The Model F-value of 49.15 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, D, E, AB, AD, AE, BE, DE, ABE, ADE are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-88 (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy or violations of the assumptions? The residual plots below do not identify any concerns with model adequacy or the violations of the assumptions. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-89 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-90 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-91 (c) One of the factors from this experiment does not seem to be important. If you drop this factor, what type of design remains? Analyze the data using the full factorial model for only the four active factors. Compare your results with those obtained in part (a). The resulting experimental design is a replicated 2
4
full factorial design. The ANOVA is shown below. The factor names in the output below were modified to match the factor names in the original problem. The same factors are significant below as were significant in the original analysis. Design Expert Output
Response 1 y ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 888.80 15 59.25 40.58 < 0.0001 significant A-A
83.56
1
83.56
57.23
< 0.0001
B-B
0.060
1
0.060
0.041
0.8414
D-D
285.78
1
285.78
195.74
< 0.0001
E-E
153.17
1
153.17
104.91
< 0.0001
AB
48.93
1
48.93
33.51
< 0.0001
AD
88.88
1
88.88
60.88
< 0.0001
AE
33.76
1
33.76
23.13
0.0002
BD
5.778E-003
1
5.778E-003
3.958E-003
0.9506
BE
52.71
1
52.71
36.10
< 0.0001
DE
61.80
1
61.80
42.33
< 0.0001
ABD
3.82
1
3.82
2.61
0.1255
ABE
44.96
1
44.96
30.79
< 0.0001
ADE
26.01
1
26.01
17.82
0.0006
BDE
0.050
1
0.050
0.035
0.8549
ABDE
5.31
1
5.31
3.63
0.0747
Pure Error 23.36 16 1.46 Cor Total 912.16 31 The Model F-value of 40.58 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, C, D, AB, AC, AD, BD, CD, ABD, ACD are significant model terms. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-92 (d) Find the settings of the active factors that maximize the predicted response. The cube plot below, with factors E
set at +1 and C
set at 0, identifies the maximum predicted response with the remaining factors, A
, B
, and D
all set at +1. 6.44. A paper in the Journal of Chemical Technology and Biotechnology
(“Response Surface Optimization of the Critical Media Components for the Production of Surfactin,” 1997, Vol. 68, pp. 263-270) describes the use of a designed experiment to maximize surfactin production. A portion of the data from this experiment is shown in the table below. Surfactin was assayed by an indirect method, which involves measurement of surface tension of the diluted broth samples. Relative surfactin concentrations were determined by serially diluting the broth until the critical micelle concentration (CMC) was reached. The dilution at which the surface tension starts rising abruptly was denoted by CMC
-1
and was considered proportional to the amount of surfactant present in the original sample. Run Glucose (g dm
-3
) NH
4
NO
3
(g dm
-3
) FeSO
4
(g dm
-3
x 10
-4
) MnSO
4
(g dm
-3
x 10
-2
) y (CMC)
-1
1 20.00 2.00 6.00 4.00 23 2 60.00 2.00 6.00 4.00 15 3 20.00 6.00 6.00 4.00 16 4 60.00 6.00 6.00 4.00 18 5 20.00 2.00 30.00 4.00 25 6 60.00 2.00 30.00 4.00 16 7 20.00 6.00 30.00 4.00 17 8 60.00 6.00 30.00 4.00 26 9 20.00 2.00 6.00 20.00 28 10 60.00 2.00 6.00 20.00 16 11 20.00 6.00 6.00 20.00 18 12 60.00 6.00 6.00 20.00 21 13 20.00 2.00 30.00 20.00 36 14 60.00 2.00 30.00 20.00 24 15 20.00 6.00 30.00 20.00 33 16 60.00 6.00 30.00 20.00 34 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-93 (a) Analyze the data from this experiment. Identify the significant factors and interactions. The half normal probability plot of effects, followed by the ANOVA, identify the significant factors and interactions. Although factor B
is not significant, the AB
interaction is. Design Expert Output
Response 1 y ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 680.50 6 113.42 20.73 < 0.0001 significant A-Glucose
42.25
1
42.25
7.72
0.0214
B-NH4NO3
0.000
1
0.000
0.000
1.0000
C-FeSO4
196.00
1
196.00
35.82
0.0002
D-MnSO4
182.25
1
182.25
33.30
0.0003
AB
196.00
1
196.00
35.82
0.0002
CD
64.00
1
64.00
11.70
0.0076
Residual 49.25 9 5.47 Cor Total 729.75 15 The Model F-value of 20.73 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, C, D, AB, CD are significant model terms. (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy or violations of the assumptions? The residual plots below do not identify any concerns with model adequacy or the violations of the assumptions. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-94 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-95 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-96 (c) What conditions would optimize the surfactin production? The response, y, is maximized when factor A
is 20, B
is 2, C
is 30, and D
is 20. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-97 6.45. Continuation of Problem 6.44.
The experiment in Problem 6.44 actually included six center points. The responses at these conditions were 35, 35, 35, 36, 36, and 34. Is there any indication of curvature in the response function? Are additional experiments necessary? What would you recommend doing now? Curvature appears to be very significant with a p
value less than 0.0001. Axial runs, along with additional center point runs to identify blocking effects, should be run. Design Expert Output
Response 1 y ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 713.00 8 89.12 54.61 < 0.0001 significant A-Glucose
42.25
1
42.25
25.89
0.0003
B-NH4NO3
0.000
1
0.000
0.000
1.0000
C-FeSO4
196.00
1
196.00
120.10
< 0.0001
D-MnSO4
182.25
1
182.25
111.68
< 0.0001
AB
196.00
1
196.00
120.10
< 0.0001
AD
12.25
1
12.25
7.51
0.0179
BC
20.25
1
20.25
12.41
0.0042
CD
64.00
1
64.00
39.22
< 0.0001
Curvature 659.28 1 659.28 403.98 < 0.0001 significant Residual 19.58 12 1.63 Lack of Fit
16.75
7
2.39
4.22
0.0660
not significant
Pure Error
2.83
5
0.57
Cor Total 1391.86 21 6.46. An article in the Journal of Hazardous Materials
(“Feasibility of Using Natural Fishbone Apatite as a Substitute for Hydroxyapatite in Remediating Aqueous Heavy Metals,” Vol. 69, Issue 2, 1999, pp. 187-197) describes an experiment to study the suitability of fishbone, a natural, apatite rich substance, as a substitute for hydroxyapatite in the sequestering of aqueous divalent heavy metal ions. Direct comparison of hydroxyapatite and fishbone apatite was performed using a three-factor two-level full factorial design. Apatite (30 or 60 mg) was added to 100mL deionized water and gently Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-98 agitated overnight in a shaker. The pH was then adjusted to 5 or 7 using nitric acid. Sufficient concentration of lead nitrate solution was added to each flask to result in a final volume of 200 mL and a lead concentration of 0.483 or 2.41 mM, respectively. The experiment was a 2
3
replicated twice and it was performed for both fishbone and synthetic apatite. Results are shown below: Fishbone Hydroxyapatite Apatite pH Pb Pb,mM pH Pb,mM pH + + + 1.82 5.22 0.11 3.49 + + + 1.81 5.12 0.12 3.46 + + – 0.01 6.84 0.00 5.84 + + – 0.00 6.61 0.00 5.90 + – + 1.11 3.35 0.80 2.70 + – + 1.04 3.34 0.76 2.74 + – – 0.00 5.77 0.03 3.36 + – – 0.01 6.25 0.05 3.24 – + + 2.11 5.29 1.03 3.22 – + + 2.18 5.06 1.05 3.22 – + – 0.03 5.93 0.00 5.53 – + – 0.05 6.02 0.00 5.43 – – + 1.70 3.39 1.34 2.82 – – + 1.69 3.34 1.26 2.79 – – – 0.05 4.50 0.06 3.28 – – – 0.05 4.74 0.07 3.28 (a) Analyze the lead response for fishbone apatite. What factors are important? As shown below, all main effects and interactions are significant. Design Expert Output
Response 1 Fishbone Pb ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 12.19 7 1.74 2629.41 < 0.0001 significant A-Apatite
0.27
1
0.27
400.34
< 0.0001
B-pH
0.35
1
0.35
525.43
< 0.0001
C-Pb
10.99
1
10.99
16587.51
< 0.0001
AB
0.023
1
0.023
33.96
0.0004
AC
0.19
1
0.19
285.62
< 0.0001
BC
0.36
1
0.36
543.40
< 0.0001
ABC
0.020
1
0.020
29.58
0.0006
Pure Error 5.300E-003 8 6.625E-004 Cor Total 12.20 15 Coefficient
Standard
95% CI
95% CI
Factor
Estimate
df
Error
Low
High
VIF
Intercept 0.85 1 6.435E-003 0.84 0.87 A-Apatite -0.13 1 6.435E-003 -0.14 -0.11 1.00 B-pH 0.15 1 6.435E-003 0.13 0.16 1.00 C-Pb 0.83 1 6.435E-003 0.81 0.84 1.00 AB 0.038 1 6.435E-003 0.023 0.052 1.00 AC -0.11 1 6.435E-003 -0.12 -0.094 1.00 BC 0.15 1 6.435E-003 0.14 0.16 1.00 ABC 0.035 1 6.435E-003 0.020 0.050 1.00 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-99 (b) Analyze the residuals from this response and comment on model adequacy. The normal plot identifies slightly thicker tails in the distribution of the residuals. The plots of residuals vs. predicted and residuals vs. the Pb effect identify nonconstant variance. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-100 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-101 (c) Analyze the pH response for fishbone apatite. What factors are important? The AB
and ABC
interactions are only moderately significant; all other main effects and interactions are significant. Design Expert Output
Response 1 Fishbone pH ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 21.09 7 3.01 116.29 < 0.0001 significant A-Apatite
1.12
1
1.12
43.17
0.0002
B-pH
8.14
1
8.14
314.08
< 0.0001
C-Pb
9.84
1
9.84
379.98
< 0.0001
AB
0.098
1
0.098
3.77
0.0881
AC
1.17
1
1.17
45.23
0.0001
BC
0.61
1
0.61
23.64
0.0013
ABC
0.11
1
0.11
4.14
0.0763
Pure Error 0.21 8 0.026 Cor Total 21.30 15 Coefficient
Standard
95% CI
95% CI
Factor
Estimate
df
Error
Low
High
VIF
Intercept 5.05 1 0.040 4.96 5.14 A-Apatite 0.26 1 0.040 0.17 0.36 1.00 B-pH 0.71 1 0.040 0.62 0.81 1.00 C-Pb -0.78 1 0.040 -0.88 -0.69 1.00 AB -0.078 1 0.040 -0.17 0.015 1.00 AC -0.27 1 0.040 -0.36 -0.18 1.00 BC 0.20 1 0.040 0.10 0.29 1.00 ABC 0.082 1 0.040 -0.011 0.17 1.00 (d) Analyze the residuals from this response and comment on model adequacy. Although the normal probability plot is acceptable, the plots of residuals vs. predicted identifies non-
constant variance. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-102 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-103 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-104 (e) Analyze the lead response for hydroxyapatite apatite. What factors are important? As shown below, all main effects and interactions are significant. Design Expert Output
Response 1 Hydroxyapatite Pb ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 4.01 7 0.57 1018.21 < 0.0001 significant A-Apatite
0.54
1
0.54
960.40
< 0.0001
B-pH
0.27
1
0.27
471.51
< 0.0001
C-Pb
2.45
1
2.45
4354.18
< 0.0001
AB
0.036
1
0.036
64.18
< 0.0001
AC
0.50
1
0.50
896.18
< 0.0001
BC
0.17
1
0.17
298.84
< 0.0001
ABC
0.046
1
0.046
82.18
< 0.0001
Pure Error 4.500E-003 8 5.625E-004 Cor Total 4.01 15 Coefficient
Standard
95% CI
95% CI
Factor
Estimate
df
Error
Low
High
VIF
Intercept 0.42 1 5.929E-003 0.40 0.43 A-Apatite -0.18 1 5.929E-003 -0.20 -0.17 1.00 B-pH -0.13 1 5.929E-003 -0.14 -0.12 1.00 C-Pb 0.39 1 5.929E-003 0.38 0.40 1.00 AB -0.048 1 5.929E-003 -0.061 -0.034 1.00 AC -0.18 1 5.929E-003 -0.19 -0.16 1.00 BC -0.10 1 5.929E-003 -0.12 -0.089 1.00 ABC -0.054 1 5.929E-003 -0.067 -0.040 1.00 (f) Analyze the residuals from this response and comment on model adequacy. The normal plot identifies slightly thicker tails in the distribution of the residuals. The plots of residuals vs. predicted and residuals vs. the effects identifies non-constant variance. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-105 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-106 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-107 (g) Analyze the pH response for hydroxyapatite apatite. What factors are important? The ABC
interaction is not significant; all of the main effects and two factor interactions are significant. Design Expert Output
Response 1 Hydroxyapatite pH ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 20.44 7 2.92 1487.66 < 0.0001 significant A-Apatite
0.084
1
0.084
42.85
0.0002
B-pH
8.82
1
8.82
4494.73
< 0.0001
C-Pb
8.15
1
8.15
4153.39
< 0.0001
AB
0.13
1
0.13
64.22
< 0.0001
AC
0.014
1
0.014
7.34
0.0267
BC
3.24
1
3.24
1650.96
< 0.0001
ABC
2.250E-004
1
2.250E-004
0.11
0.7436
Pure Error 0.016 8 1.963E-003 Cor Total 20.45 15 Coefficient
Standard
95% CI
95% CI
Factor
Estimate
df
Error
Low
High
VIF
Intercept 3.77 1 0.011 3.74 3.79 A-Apatite 0.072 1 0.011 0.047 0.098 1.00 B-pH 0.74 1 0.011 0.72 0.77 1.00 C-Pb -0.71 1 0.011 -0.74 -0.69 1.00 AB 0.089 1 0.011 0.063 0.11 1.00 AC -0.030 1 0.011 -0.056 -4.461E-003 1.00 BC -0.45 1 0.011 -0.48 -0.42 1.00 ABC -3.750E-003 1 0.011 -0.029 0.022 1.00 (h) Analyze the residuals from this response and comment on model adequacy. The only potential concern with the residual plots is the non-constant variance shown in the plot of residuals vs. pH. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-108 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-109 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-110 (i) What differences do you see between fishbone and hydroxyapatite apatite? The authors of this paper concluded that fishbone apatite was comparable to hydroxyapatite apatite. Because the fishbone apatite is cheaper, it was recommended for adoption. Do you agree with these conclusions? The authors of the journal article did not show their analysis for this experiment. When comparing the Fishbone and Hydroxyapatite models main effects and interactions for the Pb and pH responses, we might disagree with the authors. A more effective approach to understand the differences between Fishbone and Hydroxyapatite would be to include this as a factor in the experimental design. The modified table is shown below followed by the analysis. Apatite pH Pb Type Pb,mM pH + + + Fishbone 1.82 5.22 + + + Fishbone 1.81 5.12 + + – Fishbone 0.01 6.84 + + – Fishbone 0.00 6.61 + – + Fishbone 1.11 3.35 + – + Fishbone 1.04 3.34 + – – Fishbone 0.00 5.77 + – – Fishbone 0.01 6.25 – + + Fishbone 2.11 5.29 – + + Fishbone 2.18 5.06 – + – Fishbone 0.03 5.93 – + – Fishbone 0.05 6.02 – – + Fishbone 1.70 3.39 – – + Fishbone 1.69 3.34 – – – Fishbone 0.05 4.50 – – – Fishbone 0.05 4.74 + + + Hydroxyapatite 0.11 3.49 + + + Hydroxyapatite 0.12 3.46 + + – Hydroxyapatite 0.00 5.84 + + – Hydroxyapatite 0.00 5.90 + – + Hydroxyapatite 0.80 2.70 + – + Hydroxyapatite 0.76 2.74 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-111 + – – Hydroxyapatite 0.03 3.36 + – – Hydroxyapatite 0.05 3.24 – + + Hydroxyapatite 1.03 3.22 – + + Hydroxyapatite 1.05 3.22 – + – Hydroxyapatite 0.00 5.53 – + – Hydroxyapatite 0.00 5.43 – – + Hydroxyapatite 1.34 2.82 – – + Hydroxyapatite 1.26 2.79 – – – Hydroxyapatite 0.06 3.28 – – – Hydroxyapatite 0.07 3.28 The ANOVA below identifies factor D
, the type of apatite, as being very significant. Design Expert Output
Response 1 Pb Response ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 17.73 15 1.18 1929.32 < 0.0001 significant A-Apatite
0.78
1
0.78
1275.51
< 0.0001
B-pH
2.813E-003
1
2.813E-003
4.59
0.0478
C-Pb
11.91
1
11.91
19440.33
< 0.0001
D-Type
1.52
1
1.52
2485.73
< 0.0001
AB
8.000E-004
1
8.000E-004
1.31
0.2699
AC
0.66
1
0.66
1070.22
< 0.0001
AD
0.024
1
0.024
39.51
< 0.0001
BC
0.018
1
0.018
29.47
< 0.0001
BD
0.61
1
0.61
996.76
< 0.0001
CD
1.53
1
1.53
2500.00
< 0.0001
ABC
2.813E-003
1
2.813E-003
4.59
0.0478
ABD
0.058
1
0.058
94.37
< 0.0001
ACD
0.038
1
0.038
61.73
< 0.0001
BCD
0.51
1
0.51
832.73
< 0.0001
ABCD
0.063
1
0.063
102.88
< 0.0001
Pure Error 9.800E-003 16 6.125E-004 Cor Total 17.74 31 The residual plots below identify concerns, so a power transformation with lambda of 0.7 was applied to the Pb response. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-112 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-113 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-114 Design Expert Output
Response 1 Pb Response Transform: Power Lambda: 0.7 Constant: 0 ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 12.10 15 0.81 1844.15 < 0.0001 significant A-Apatite
0.49
1
0.49
1123.40
< 0.0001
B-pH
0.014
1
0.014
32.86
< 0.0001
C-Pb
9.02
1
9.02
20612.51
< 0.0001
D-Type
0.75
1
0.75
1712.43
< 0.0001
AB
3.973E-003
1
3.973E-003
9.08
0.0082
AC
0.29
1
0.29
661.73
< 0.0001
AD
0.024
1
0.024
55.70
< 0.0001
BC
5.077E-003
1
5.077E-003
11.61
0.0036
BD
0.38
1
0.38
877.65
< 0.0001
CD
0.73
1
0.73
1670.33
< 0.0001
ABC
0.011
1
0.011
25.97
0.0001
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-115 ABD
0.049
1
0.049
112.97
< 0.0001
ACD
0.067
1
0.067
152.27
< 0.0001
BCD
0.21
1
0.21
472.40
< 0.0001
ABCD
0.057
1
0.057
131.40
< 0.0001
Pure Error 6.999E-003 16 4.374E-004 Cor Total 12.11 31 There are no concerns with the residual plots shown below. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-116 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-117 The analysis for the pH response is shown below. Design Expert Output
Response 2 pH Response ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 54.62 15 3.64 261.32 < 0.0001 significant A-Apatite
0.91
1
0.91
65.15
< 0.0001
B-pH
16.95
1
16.95
1216.47
< 0.0001
C-Pb
17.96
1
17.96
1288.54
< 0.0001
D-Type
13.09
1
13.09
939.72
< 0.0001
AB
9.031E-004
1
9.031E-004
0.065
0.8023
AC
0.72
1
0.72
51.89
< 0.0001
AD
0.29
1
0.29
21.14
0.0003
BC
0.52
1
0.52
37.15
< 0.0001
BD
6.903E-003
1
6.903E-003
0.50
0.4916
CD
0.040
1
0.040
2.86
0.1100
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-118 ABC
0.049
1
0.049
3.50
0.0796
ABD
0.22
1
0.22
15.99
0.0010
ACD
0.46
1
0.46
33.24
< 0.0001
BCD
3.33
1
3.33
239.31
< 0.0001
ABCD
0.059
1
0.059
4.21
0.0569
Pure Error 0.22 16 0.014 Cor Total 54.84 31 Non-constant variance is identified in the residual plots below, so an inverse square root transformation was applied. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-119 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-120 Design Expert Output
Response 2 pH Response Transform: Inverse Sqrt Constant: 0 ANOVA for selected factorial model Analysis of variance table [Partial sum of squares - Type III] Sum of
Mean
F
p-value
Source
Squares
df
Square
Value
Prob > F
Model 0.17 15 0.011 483.43 < 0.0001 significant A-Apatite
1.418E-003
1
1.418E-003
60.44
< 0.0001
B-pH
0.055
1
0.055
2346.06
< 0.0001
C-Pb
0.054
1
0.054
2303.65
< 0.0001
D-Type
0.045
1
0.045
1918.39
< 0.0001
AB
1.417E-005
1
1.417E-005
0.60
0.4484
AC
9.441E-004
1
9.441E-004
40.23
< 0.0001
AD
3.287E-004
1
3.287E-004
14.00
0.0018
BC
4.687E-005
1
4.687E-005
2.00
0.1768
BD
8.263E-004
1
8.263E-004
35.21
< 0.0001
CD
3.302E-004
1
3.302E-004
14.07
0.0017
ABC
3.443E-004
1
3.443E-004
14.67
0.0015
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-121 ABD
7.072E-004
1
7.072E-004
30.14
< 0.0001
ACD
7.593E-004
1
7.593E-004
32.36
< 0.0001
BCD
0.010
1
0.010
437.95
< 0.0001
ABCD
4.034E-005
1
4.034E-005
1.72
0.2083
Pure Error 3.755E-004 16 2.347E-005 Cor Total 0.17 31 There are no concerns with the residual plots shown below. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-122 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-123 In summary, there is a difference between the Fishbone Apatite and the synthetic Hydroxyapatite. 6.47. Often the fitted regression model from a 2
k
factorial design is used to make predictions at points of interest in the design space. (a) Find the variance of the predicted response ˆ
y
at the point x
1
, x
2
, . . . , x
k
in the design space. Hint: Remember that the x
’s are coded variables, and assume a 2
k
design with an equal number of replicates n
at each design point so that the variance of a regression coefficient β
is 2
2
k
n
σ
and that the covariance between any pair of regression coefficients is zero. Let’s assume that the model can be written as follows: 0
1
1
2
2
ˆ
ˆ
ˆ
ˆ
ˆ( )=
...
p
p
y
x
x
x
β
β
β
β
+
+
+
+
x
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-124 where x
’ = [ x
1
, x
2
, . . . , x
k
] are the values of the original variables in the design at the point of interest where a prediction is required, and the variables in the model x
1
, x
2
, . . . , x
p
potentially include interaction terms among the original k
variables. Now the variance of the predicted response is β
β
β
β
β
β
β
β
σ
=
=
+
+
+
+
=
+
+
+
+
⎛
⎞
=
+
⎜
⎟
⎜
⎟
⎝
⎠
∑
0
1
1
2
2
0
1
1
2
2
2
2
1
ˆ
ˆ
ˆ
ˆ
ˆ
[ (
)]
(
...
)
ˆ
ˆ
ˆ
ˆ
(
)
(
)
(
)
...
(
)
1
2
p
p
p
p
p
i
k
i
V y
V
x
x
x
V
V
x
V
x
V
x
x
n
x
This result follows because the design is orthogonal and all model parameter estimates have the same variance. Remember that some of the x
’s involved in this equation are potentially interaction terms. (b) Use the result of part (a) to find an equation for a 100(1–
α
)% confidence interval on the true mean response at the point 1
x
,
2
x
,… , k
x
in the design space. The confidence interval is α
α
-
≤
≤
+
/2,
/2,
ˆ
ˆ
ˆ
ˆ
(
)
[ (
)]
(
)
(
)
[ (
)]
E
E
df
df
y
t
V y
y
y
t
V y
x
x
x
x
x
where df
E
is the number of degrees of freedom used to estimate σ
2
and the estimate of σ
2
has been used in computing the variance of the predicted value of the response at the point of interest. 6.48. Hierarchical Models.
Several times we have utilized the hierarchy principle in selecting a model; that is, we have included non-significant terms in a model because they were factors involved in significant higher-order terms. Hierarchy is certainly not an absolute principle that must be followed in all cases. To illustrate, consider the model resulting from Problem 6.5, which required that a non-significant main effect be included to achieve hierarchy. Using the data from Problem 6.5: (a) Fit both the hierarchical model and the non-hierarchical model. Design Expert Output for Hierarchial Model Response:
Life
in hours
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 1519.67 4 379.92 12.54 < 0.0001 significant A
0.67
1
0.67
0.022
0.8836
B
770.67
1
770.67
25.44
< 0.0001
C
280.17
1
280.17
9.25
0.0067
AC
468.17
1
468.17
15.45
0.0009
Residual 575.67 19 30.30 Lack of Fit
93.00
3
31.00
1.03
0.4067
not significant
Pure Error
482.67
16
30.17
Cor Total 2095.33 23 The Model F-value of 12.54 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 B, C, AC are significant model terms. Std. Dev. 5.50 R-Squared 0.7253 Mean 40.83 Adj R-Squared 0.6674 C.V. 13.48 Pred R-Squared 0.5616 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-125 PRESS 918.52 Adeq Precision 10.747 The "Pred R-Squared" of 0.5616 is in reasonable agreement with the "Adj R-Squared" of 0.6674. A difference greater than 0.20 between the "Pred R-Squared" and the "Adj R-Squared" indicates a possible problem with your model and/or data. Design Expert Output for Non-Hierarchical Model Response:
Life
in hours
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Prob > F
Model 1519.00 3 506.33 17.57 < 0.0001 significant B
770.67
1
770.67
26.74
< 0.0001
C
280.17
1
280.17
9.72
0.0054
AC
468.17
1
468.17
16.25
0.0007
Residual 576.33 20 28.82 Lack of Fit
93.67
4
23.42
0.78
0.5566
not significant
Pure Error
482.67
16
30.17
Cor Total 2095.33 23 The Model F-value of 17.57 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 B, C, AC are significant model terms. The "Lack of Fit F-value" of 0.78 implies the Lack of Fit is not significant relative to the pure error. There is a 55.66% chance that a "Lack of Fit F-value" this large could occur due to noise. Non-significant lack of fit is good -- we want the model to fit. Std. Dev. 5.37 R-Squared 0.7249 Mean 40.83 Adj R-Squared 0.6837 C.V. 13.15 Pred R-Squared 0.6039 PRESS 829.92 Adeq Precision 12.320 The "Pred R-Squared" of 0.6039 is in reasonable agreement with the "Adj R-Squared" of 0.6837. A difference greater than 0.20 between the "Pred R-Squared" and the "Adj R-Squared" indicates a possible problem with your model and/or data. (b) Calculate the PRESS statistic, the adjusted R
2
and the mean square error for both models. The PRESS and R
2
are in the Design Expert Output above. The PRESS is smaller for the non-
hierarchical model than the hierarchical model suggesting that the non-hierarchical model is a better predictor. (c) Find a 95 percent confidence interval on the estimate of the mean response at a cube corner (
x
1
= x
2
= x
3
= ±
1. Hint: Use the result of Problem 6.40. Design Expert Output Prediction
SE Mean
95% CI low
95% CI high
SE Pred
95% PI low
95% PI high
Life
27.45 2.18 22.91 31.99 5.79 15.37 39.54 Life
36.17 2.19 31.60 40.74 5.80 24.07 48.26 Life
38.67 2.19 34.10 43.24 5.80 26.57 50.76 Life
47.50 2.19 42.93 52.07 5.80 35.41 59.59 Life
43.00 2.19 38.43 47.57 5.80 30.91 55.09 Life
34.17 2.19 29.60 38.74 5.80 22.07 46.26 Life
54.33 2.19 49.76 58.90 5.80 42.24 66.43 Life
45.50 2.19 40.93 50.07 5.80 33.41 57.59 (d) Based on the analyses you have conducted, which model would you prefer? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-126 Notice that PRESS is smaller and the adjusted R
2
is larger for the non-hierarchical model. This is an indication that strict adherence to the hierarchy principle isn’t always necessary. Note also that the confidence interval is shorter for the non-hierarchical model. 6.49. Suppose that you want to run a 2
3
factorial design. The variance of an individual observation is expected to be about 4. Suppose that you want the length of a 95% confidence interval on any effect to be less than or equal to 1.5. How many replicates of the design do you need to run? With the equations for the se
(Effect) and 100(1–
α
)% confidence interval on the effects shown below, we can iteratively estimate the number of replicates. From the table of iterations, 14 replicates are required. (
)
Effect
2
2
k
se
S
n
=
(
)
(
)
α
-
±
n se(Effect) t
(0.025,
N
–
p
) 95% CI Length 12 0.408 1.987 1.623 13 0.392 1.985 1.557 14 0.378 1.983 1.499 15 0.365 1.981 1.447 6.50. Suppose that a full 2
4
factorial uses the following levels: Factor Low (–) High(+) A: Acid strength (%) 85 95 B: Reaction time (min) 15 35 C: Amount of acid (ml) 35 45 D: Reaction temperature (
C) 60 80 The fitted model from this experiment is 1
2
3
4
1
2
1
3
ˆ
24
16
34
12
6
10
16
y
x
x
x
x
x x
x x
=
+
-
+
+
-
+
Predict the response at the following points: (a) A
= 89, B
= 20, C
= 38, D
= 66 The coded values are: x
1
= –0.2, x
2
= –0.5, x
3
= –0.4, x
4
= –0.4 (
)
(
)
(
)
(
)
(
)(
)
(
)(
)
ˆ
24
16
0.2
34
0.5
12
0.4
6
0.4
10
0.2
0.5
16
0.2
0.4
30.88
y
=
+
-
-
-
+
-
+
-
-
-
-
+
-
-
=
(b) A = 90, B = 16, C = 40, D = 70 The coded values are: x
1
= 0, x
2
= –0.9, x
3
= 0, x
4
= 0 ( )
(
)
( )
( )
( )(
)
( )( )
=
+
-
-
+
+
-
-
+
=
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-127 (c) A = 87, B = 28, C = 42, D = 61 The coded values are: x
1
= –0.6, x
2
= 0.3, x
3
= 0.4, x
4
= –0.9 (
)
(
)
(
)
(
)
(
)(
)
(
)(
)
ˆ
24
16
0.6
34 0.3
12 0.4
6
0.9
10
0.6
0.3
16
0.6
0.4
1.56
y
=
+
-
-
+
+
-
-
-
+
-
=
(d) A = 90, B = 27, C = 37, D = 69 The coded values are: x
1
= 0, x
2
= 0.2, x
3
= –0.6, x
4
= –0.1 ( )
(
)
(
)
(
)
( )(
)
( )(
)
=
+
-
+
-
+
-
-
+
-
=
6.51.
An article in Quality and Reliability Engineering International
(2010, Vol. 26, pp. 223-233) presents a factorial design. The experiment is shown in Table P6.15. A B C D E y -1 -1 -1 -1 -1 8.11 1 -1 -1 -1 -1 5.56 -1 1 -1 -1 -1 5.77 1 1 -1 -1 -1 5.82 -1 -1 1 -1 -1 9.17 1 -1 1 -1 -1 7.8 -1 1 1 -1 -1 3.23 1 1 1 -1 -1 5.69 -1 -1 -1 1 -1 8.82 1 -1 -1 1 -1 14.23 -1 1 -1 1 -1 9.2 1 1 -1 1 -1 8.94 -1 -1 1 1 -1 8.68 1 -1 1 1 -1 11.49 -1 1 1 1 -1 6.25 1 1 1 1 -1 9.12 -1 -1 -1 -1 1 7.93 1 -1 -1 -1 1 5 -1 1 -1 -1 1 7.47 1 1 -1 -1 1 12 -1 -1 1 -1 1 9.86 1 -1 1 -1 1 3.65 -1 1 1 -1 1 6.4 1 1 1 -1 1 11.61 -1 -1 -1 1 1 12.43 1 -1 -1 1 1 17.55 -1 1 -1 1 1 8.87 1 1 -1 1 1 25.38 -1 -1 1 1 1 13.06 1 -1 1 1 1 18.85 -1 1 1 1 1 11.78 1 1 1 1 1 26.05 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-128 (a) Analyze the data from this experiment. Identify the significant factors and interactions. Minitab Output The half-normal plot shows that no 4-factor or 5-factor interactions are significant, so they can be dropped from the set of terms and be used to estimate error. Factors A, D, E
, and the AD, DE, BE, AB, AE, ABE
, ADE
interactions appear to be important. The model can now be fit. But notice that there are several of the 2 and 3-factor interactions that are not significant. They can be removed for a better model fit; see part (c). Minitab Output Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Model 25 900.849 36.034 19.11 0.001 Linear 5 522.578 104.516 55.43 0.000 A 1 83.560 83.560 44.31 0.001 B 1 0.060 0.060 0.03 0.864 C 1 0.005 0.005 0.00 0.962 D 1 285.784 285.784 151.56 0.000 E 1 153.169 153.169 81.23 0.000 2-Way Interactions 10 290.219 29.022 15.39 0.002 A*B 1 48.931 48.931 25.95 0.002 A*C 1 0.000 0.000 0.00 0.995 A*D 1 88.878 88.878 47.13 0.000 A*E 1 33.764 33.764 17.91 0.005 B*C 1 1.221 1.221 0.65 0.452 B*D 1 0.006 0.006 0.00 0.958 B*E 1 52.711 52.711 27.95 0.002 C*D 1 0.000 0.000 0.00 0.989 C*E 1 2.910 2.910 1.54 0.260 D*E 1 61.799 61.799 32.77 0.001 3-Way Interactions 10 88.051 8.805 4.67 0.036 A*B*C 1 2.005 2.005 1.06 0.342 A*B*D 1 3.816 3.816 2.02 0.205 A*B*E 1 44.959 44.959 23.84 0.003 A*C*D 1 0.129 0.129 0.07 0.803 A*C*E 1 2.148 2.148 1.14 0.327 A*D*E 1 26.010 26.010 13.79 0.010 B*C*D 1 2.983 2.983 1.58 0.255 Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-129 B*C*E 1 0.935 0.935 0.50 0.508 B*D*E 1 0.050 0.050 0.03 0.875 C*D*E 1 5.017 5.017 2.66 0.154 Error 6 11.314 1.886 Total 31 912.162 Model Summary S R-sq R-sq(adj) R-sq(pred) 1.37319 98.76% 93.59% 64.72% (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy or violations of the assumptions? The residual plots look good; no indication of model inadequacy. Minitab Output (c) One of the factors from this experiment does not seem to be important. If you drop this factor, what type of design remains? Analyze the data using the full factorial model for only the four active factors. Compare your results with those obtained in part (a). Factor C
and all of its interactions are not important. The design becomes a replicated 2
4
. The degrees of freedom increase to 17 when Factor C
has been removed and R
2
adjusted and R
2
predicted have increased. There are a couple of 2 and 3-factor interaction that can be removed at this point; BD, ABD
and BDE
. B
is not significant by as a main effect, but is part of several interactions, so leave it in the model. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-130 Minitab Output Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Model 11 879.625 79.966 49.15 0.000 Linear 4 522.574 130.643 80.30 0.000 A 1 83.560 83.560 51.36 0.000 B 1 0.060 0.060 0.04 0.849 D 1 285.784 285.784 175.66 0.000 E 1 153.169 153.169 94.15 0.000 2-Way Interactions 5 286.082 57.216 35.17 0.000 A*B 1 48.931 48.931 30.08 0.000 A*D 1 88.878 88.878 54.63 0.000 A*E 1 33.764 33.764 20.75 0.000 B*E 1 52.711 52.711 32.40 0.000 D*E 1 61.799 61.799 37.99 0.000 3-Way Interactions 2 70.969 35.484 21.81 0.000 A*B*E 1 44.959 44.959 27.64 0.000 A*D*E 1 26.010 26.010 15.99 0.001 Error 20 32.538 1.627 Total 31 912.162 Model Summary S R-sq R-sq(adj) R-sq(pred) 1.27549 96.43% 94.47% 90.87% (d) Find settings of the active factors that maximize the predicted response. The optimum is found at Factors A
, B
, D
and E
all set to the high level. The predicted response at these settings is 25.7063 Minitab Output Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-131 6.52.
Consider the 2
3
experiment shown in the figure below: Process Variables Coded Variables Run Temp (
C) Pressure (psing) Conc (g/l) x
1
x
2
x
3
Yield,
y
1 120 40 15 -1 -1 -1 32 2 160 40 15 1 -1 -1 46 3 120 80 15 -1 1 -1 57 4 160 80 15 1 1 -1 65 5 120 40 30 -1 -1 1 36 6 160 40 30 1 -1 1 48 7 120 80 30 -1 1 1 57 8 160 80 30 1 1 1 68 9 140 60 22.5 0 0 0 50 10 140 60 22.5 0 0 0 44 11 140 60 22.5 0 0 0 53 12 140 60 22.5 0 0 0 56 When running a designed experiment, it is sometimes difficult to reach and hold the precise factors required by the design. Small discrepancies are not important, but large ones are potentially of more concern. To illustrate, the experiment presented in Table P6.16 below shows a variation of the 2
3
design above, where many of the test combinations are not exactly the ones specified in the design. Most of the difficulty seems to have occurred with the temperature variable. Table P6.16 Process Variables Coded Variables Run Temp (
C) Pressure (psing) Conc (g/l) x
1
x
2
x
3
Yield,
y
1 120 41 14 -0.75 -0.95 -1.133 32 2 158 40 15 0.9 -1 -1 46 3 121 82 15 -0.95 1.1 -1 57 4 160 80 15 1 1 -1 65 5 118 39 33 -1.1 -1.05 1.14 36 6 163 40 30 1.15 -1 1 48 7 122 80 30 -0.9 1 1 57 8 165 83 30 1.25 1.15 1 68 9 140 60 22.5 0 0 0 50 10 140 60 22.5 0 0 0 44 11 140 60 22.5 0 0 0 53 12 140 60 22.5 0 0 0 56 Fit a first-order model to both the original data and the data in Table P6.16. Compare the inference from the two models. What conclusions can you draw from this simple example? Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-132 Minitab Output Analysis of Variance Source DF Seq SS Contribution Adj SS Adj MS F-Value P-Value Regression 3 1166.38 92.86% 1166.38 388.792 34.70 0.000 X1 1 253.13 20.15% 253.13 253.125 22.59 0.001 X2 1 903.12 71.90% 903.12 903.125 80.61 0.000 X3 1 10.13 0.81% 10.13 10.125 0.90 0.370 Error 8 89.63 7.14% 89.63 11.203 Lack-of-Fit 5 10.88 0.87% 10.88 2.175 0.08 0.990 Pure Error 3 78.75 6.27% 78.75 26.250 Total 11 1256.00 100.00% Model Summary S R-sq R-sq(adj) PRESS R-sq(pred) 3.34711 92.86% 90.19% 130.230 89.63% Coefficients Term Coef SE Coef 95% CI T-Value P-Value VIF Constant 51.000 0.966 (48.772, 53.228) 52.78 0.000 X1 5.63 1.18 ( 2.90, 8.35) 4.75 0.001 1.00 X2 10.62 1.18 ( 7.90, 13.35) 8.98 0.000 1.00 X3 1.13 1.18 ( -1.60, 3.85) 0.95 0.370 1.00 Regression Equation yield = 51.000 + 5.63 X1 + 10.62 X2 + 1.13 X3 Minitab Output for Table XXX (2
nd
set of data) Analysis of Variance Source DF Seq SS Contribution Adj SS Adj MS F-Value P-Value Regression 3 1153.38 91.83% 1153.38 384.460 29.97 0.000 X1 1 270.47 21.53% 234.78 234.780 18.30 0.003 X2 1 872.70 69.48% 874.31 874.315 68.16 0.000 X3 1 10.21 0.81% 10.21 10.210 0.80 0.398 Error 8 102.62 8.17% 102.62 12.828 Lack-of-Fit 5 23.87 1.90% 23.87 4.774 0.18 0.951 Pure Error 3 78.75 6.27% 78.75 26.250 Total 11 1256.00 100.00% Model Summary S R-sq R-sq(adj) PRESS R-sq(pred) 3.58156 91.83% 88.77% 170.624 86.42% Coefficients Term Coef SE Coef 95% CI T-Value P-Value VIF Constant 50.52 1.04 (48.13, 52.91) 48.76 0.000 X1 5.37 1.26 ( 2.48, 8.27) 4.28 0.003 1.00 X2 10.13 1.23 ( 7.30, 12.96) 8.26 0.000 1.00 X3 1.09 1.22 (-1.73, 3.91) 0.89 0.398 1.00 Regression Equation yield = 50.52 + 5.37 X1 + 10.13 X2 + 1.09 X3 The Minitab regression output for both experiments will model the first order model. There is not much difference between the regression equations. Note that x
2
is the most important variable and there is very little deviation from the original design in x
2
. Because of the deviations in x
1
, there are now 9 distinct levels of x
1
and not enough degrees of freedom to model the interactions. Downloaded by Omar Hassan (ayssar.omar.borham@hotmail.com)
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-133 6.53.
In two-level designs, the expected value of a non-significant factor effect is zero. True
False 6.54.
A half-normal plot of factor effects plots the expected normal percentile versus the effect estimate. True False
6.55.
In an unreplicated design, the degrees of freedom associated with the ‘pure error’ component of error are zero. True False 6.56.
In a replicated design (16 runs) the estimated of the model intercept is equal to one-half of the total of all 16 runs. True False
6.57.
Adding center runs to a design affects the estimate of the intercept term but not the estimates of any other factor effects. True
False 6.58.
The mean square for pure error in a replicated factorial design can get smaller if non-significant terms are added to a model. True False
6.59.
A factorial design is a D-optimal design for fitting a first-order model. True
False 6.60.
If a D-optimal design algorithm is used to create a 12-run design for fitting a first-order model in three variables with all three two-factor interactions, the algorithm will construct a factorial with 4 center runs. True False
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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments
, Wiley, NY 6-134 6.61.
Suppose that you want to replicate 2 of the 8 runs in a 2
3
factorial design. How many ways are there to choose the 2 runs to replicate? 8 choose 2 = 8!/(8-2)!2!=28. However, the better way to choose is to pick a fraction. Suppose that you decide to replicate the run with all three factors at the high level and the run with all three factors at the low level. (a) Is the resulting design orthogonal? No. In order to be orthogonal, all of the column of +1 and –1 sum to zero. The column of the AB
, AC
and BC
interaction no longer sum to zero. (b) What are the relative variance of the model coefficients if the main effects plus two-factor interaction model is to fit the data from this design? Per the table below from Design Expert, the variance inflation of the main effects is now 1.07, versus 1.00 for the 8 run, orthogonal design, and 1.05 versus 1.00 for the two-factor interactions. Design Expert Output
Term StdErr
1
VIF
Ri-Squared
2 Std. Dev.
A 0.33
1.07
0.0667
54.6 %
B 0.33
1.07
0.0667
54.6 %
C 0.33
1.07
0.0667
54.6 %
AB 0.33
1.05
0.0476
53.9 %
AC 0.33
1.05
0.0476
53.9 %
BC 0.33
1.05
0.0476
53.9 %
(c) What is the power for detecting effects of two standard deviations in magnitude? From the Design Expert output, the power is 54.6% for the main effects and 53.9 for the two-factor interaction. 6.62.
The display below summarizes the results of analyzing a factorial design. Term Effect Estimate Sum of Squares % Contribution A ?
6.25
3.25945
B 5.25
110.25
57.4967
C 3.5
49
25.5541
D 0.75
?
1.1734
AB 0.75
2.25
1.1734
AC -0.5
1
0.521512
AD 0.75
2.25
1.1734
BC 1.5
9
?
BD 0.25
0.25
0.130378
CD 0.5
1
0.521512
ABC -1
4
2.08605
ABD ?
2.25
1.1734
ACD -0.5
?
0.521512
BCD 0
0
0
ABCD -0.5
1
?
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(a) (b) Solutions from
Fill in the mi
Construct a n
and C
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Which factors s
Experiments
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Wiley, NY ive? Factors B
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Calculating molarity using solute moles
A chemist prepares a solution of magnesium fluoride (MgF,) by measuring out 0.00744 µmol of magnesium fluoride into a 500. mL volumetric flask and filling
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O STOICHIOMETRY
Finding mole ratios from chemical formulae
This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar):
CH,CO,H
An analytical chemist has determined by measurements that there are 42. moles of carbon in a sample of acetic acid. How many moles of hydrogen are in the
sample?
Round your answer to 2 significant digits.
mol
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O STOICHIOMETRY
Finding mole ratios from chemical formulae
This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar):
CH,CO,H
An analytical chemist has determined by measurements that there are 42. moles of carbon in a sample of acetic acid. How many moles of hydrogen are in the
sample?
Round your answer to 2 significant digits.
mol
Check
Explanation
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Chem 1120: Intro to Gen Chem for the Allied Health Sciences.
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time
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0
A chemistry graduate student is studying the rate of this reaction:
2SO3 (g) →2SO₂ (g) + O₂(g)
He fills a reaction vessel with SO3 and measures its concentration as the reaction proceeds:
10.
I Don't Know
20.
30.
40.
[SO3]
College information
0.0600M
0.0351M
0.0206M
0.0121 M
b Home | bartleby
0.00706 M
Use this data to answer the following questions.
Write the rate law for this reaction.
Calculate the value of the rate constant k.
Round your answer to 2 significant digits. Also be
sure your answer has the correct unit symbol.
Submit
x +
rate = k
0
Question 13
0
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O THERMOCHEMISTRY
Understanding the definitions of heat and work
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The
cylinder has a piston that moves in or out, as necessary, to keep a constant pressure on the mixture of 1 atm. The
cylinder is also submerged in a large insulated water bath. (See sketch at right.)
From previous experiments, this chemical reaction is known to release 141. kJ of energy.
The position of the piston is monitored, and it is determined from this data that the piston does 358. kJ of work on the
system during the reaction.
O exothermic
Is the reaction exothermic or endothermic?
O…
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O CHEMICAL REACTIONS
Writing a chemical equation from a description of the reaction
Solid calcium carbonate (CACO,) decomposes into carbon dioxide gas and solid calcium oxide .
Write a balanced chemical equation for this reaction.
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O ACIDS AND BASES
Predicting acid or base strength from the conjugate
News
species
H₂O
Explanation
Bro
Order these chemical species by increasing pH of an 0.1 M aqueous solution of each. That is, imagine making an 0.1 M solution of each species. Select 1 next
to the species that makes the solution with the lowest pH. Select 2 next to the species that makes the solution with the next higher pH, and so on.
H₂0*
HF
Notice that some of the rankings have been filled in for you already. Also notice that water is on the list. For that particular case, just compare the pH of pure
water to the pH of the other solutions.
Note for advanced students: for all charged species, you may assume the necessary counterions act as neither acids nor bases.
CH₂CH₂COOH
b Answered: O ACIDS AND
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Using specific heat capacity to find temperature change
Ava v
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.70 kg of water at 26.0 °C. During the reaction 98.8 kJ of
heat flows out of the bath and into the flask.
-1
-1
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18 J•g ''K '. Round your
answer to 3 significant digits.
alo
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Explanation
Check
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Building Vocabulary
From the list below, choose the term that best completes each sentence.
matter
physical change
endothermic reaction
chemical change
chemistry
precipitate
exothermic reaction
7. Any change that alters a substance without changing it into another
substance is a(n)
8.
is anything that has mass and takes up space.
9. A reaction that releases energy in the form of heat is called a(n)
10. A(n)
absorbed.
11. A chemical reaction is also referred to as a(n)
is a reaction in which
energy
is
12. A(n)
a chemical reaction.
is a solid formed from a solution during
13.
is the study of the properties of matter and
how matter changes.
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