STAT 503, Fall 2005
Homework Solution 5
Due: Oct. 18, 2005
5-1 The yield of a chemical process is being studied. The two most important variables are thought to be
the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with
two replicates is performed. The yield data follow:
Temperature
150
160
170
200
90.4
90.2
90.1
90.3
90.5
90.7
Pressure
215
90.7
90.6
90.5
90.6
90.8
90.9
230
90.2
90.4
89.9
90.1
90.4
90.1
(a) Analyze the data and draw conclusions. Use  = 0.05.
Both pressure (A) and temperature (B) are significant, the interaction is not.
Design Expert Output
Response:Surface Finish
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Source
Squares
DF
Model
1.14
8
A
0.77
2
B
0.30
2
AB
0.069
4
Residual
0.16
9
Lack of Fit
0.000
0
Pure Error
0.16
9
Cor Total
1.30
17
Mean
Square
0.14
0.38
0.15
0.017
0.018
F
Value
8.00
21.59
8.47
0.97
0.018
The Model F-value of 8.00 implies the model is significant. There is only a 0.26% 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.
Values greater than 0.1000 indicate the model terms are not significant.
If there are many insignificant model terms (not counting those required to support hierarchy),
model reduction may improve your model.
(b) Prepare appropriate residual plots and comment on the model’s adequacy.
The residual plots show no serious deviations from the assumptions.
5-1
Prob > F
0.0026
0.0004
0.0085
0.4700
significant
STAT 503, Fall 2005
Homework Solution 5
Due: Oct. 18, 2005
Residuals vs. Predicted
Normal plot of residuals
0.15
99
95
90
Res iduals
Norm al % probability
0.075
-4.26326E-014
-0.075
80
70
50
30
20
10
5
1
-0.15
90.00
90.21
90.43
90.64
90.85
-0.15
Predicted
-0.075
-4.26326E-014
0.075
0.15
Res idual
Residuals vs. Temperature
Residuals vs. Pressure
0.15
0.15
2
2
0.075
0.075
Res iduals
Res iduals
3
-4.26326E-014
-4.26326E-014
-0.075
-0.075
2
2
-0.15
2
-0.15
1
2
3
1
Tem perature
2
Pres sure
(c) Under what conditions would you operate this process?
5-2
3
STAT 503, Fall 2005
Homework Solution 5
DESIGN-EXPERT Plot
Yield
Due: Oct. 18, 2005
Interaction Graph
Tem perature
91.0008
X = A: Pressure
Y = B: Temperature
Design Points
90.7129
Yield
2
B1 150
B2 160
B3 170
90.425
2
90.1371
2
89.8492
200
215
230
Pres s ure
Set pressure at 215 and Temperature at the high level, 170 degrees C, as this gives the highest yield.
The standard analysis of variance treats all design factors as if they were qualitative. In this case, both
factors are quantitative, so some further analysis can be performed. In Section 5-5, we show how response
curves and surfaces can be fit to the data from a factorial experiment with at least one quantative factor.
Since both factors in this problem are quantitative and have three levels, we can fit linear and quadratic
effects of both temperature and pressure, exactly as in Example 5-5 in the text. The Design-Expert output,
including the response surface plots, now follows.
Design Expert Output
Response:Surface Finish
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Source
Squares
DF
Model
1.13
5
A
0.10
1
B
0.067
1
A2
0.67
1
B2
0.23
1
AB
0.061
1
Residual
0.17
12
Mean
Square
0.23
0.10
0.067
0.67
0.23
0.061
0.014
F
Value
16.18
7.22
4.83
47.74
16.72
4.38
Lack of Fit 7.639E-003
not significant
Pure Error
Cor Total1.30
0.16
17
9
3
Prob > F
< 0.0001
0.0198
0.0483
< 0.0001
0.0015
0.0582
2.546E-003
0.018
The Model F-value of 16.18 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 A, B, A2, B2 are significant model terms.
Values greater than 0.1000 indicate the model terms are not significant.
If there are many insignificant model terms (not counting those required to support hierarchy),
model reduction may improve your model.
Std. Dev. 0.12R-Squared0.8708
Mean 90.41Adj R-Squared0.8170
5-3
significant
0.14
0.9314
STAT 503, Fall 2005
C.V.
Homework Solution 5
Due: Oct. 18, 2005
0.13Pred R-Squared0.6794
PRESS
0.42
Factor
Intercept
A-Pressure
B-Temperature
A2
B2
AB
Adeq Precision
Coefficient
Estimate
90.52
-0.092
0.075
-0.41
0.24
-0.087
11.968
Standard
Error
0.062
0.034
0.034
0.059
0.059
0.042
DF
1
1
1
1
1
1
95% CI
Low
90.39
-0.17
6.594E-004
-0.54
0.11
-0.18
95% CI
High
90.66
-0.017
0.15
-0.28
0.37
3.548E-003
VIF
1.00
1.00
1.00
1.00
1.00
Final Equation in Terms of Coded Factors:
Yield
+90.52
-0.092
+0.075
-0.41
+0.24
-0.087
=
*A
*B
* A2
* B2
*A*B
Final Equation in Terms of Actual Factors:
Yield
+48.54630
+0.86759
-0.64042
-1.81481E-003
+2.41667E-003
-5.83333E-004
=
* Pressure
* Temperature
* Pressure2
* Temperature2
* Pressure * Temperature
2
Yield
2
170.00
2
90.8
90.7
90.6
91
90.8
90.2
90.5
2
160.00
2
90.5
90.4 90.3
90.6
90.12
90.4
90.4
Yield
B: Tem perature
165.00
90.3
90.2
90
155.00
170.00
90.6
2
2
200.00
2
207.50
150.00
200.00
207.50
215.00
222.50
165.00
230.00
160.00
215.00
155.00B: Temperature
222.50
A: Pressure
A: Pres sure
230.00
150.00
5-5 Johnson and Leone (Statistics and Experimental Design in Engineering and the Physical Sciences,
Wiley 1977) describe an experiment to investigate the warping of copper plates. The two factors studied
were the temperature and the copper content of the plates. The response variable was a measure of the
amount of warping. The data were as follows:
Temperature (°C)
50
75
100
40
17,20
12,9
16,12
Copper
60
16,21
18,13
18,21
5-4
Content (%)
80
24,22
17,12
25,23
100
28,27
27,31
30,23
STAT 503, Fall 2005
Homework Solution 5
125
21,17
23,21
Due: Oct. 18, 2005
23,22
29,31
(a) Is there any indication that either factor affects the amount of warping? Is there any interaction
between the factors? Use  = 0.05.
Both factors, copper content (A) and temperature (B) affect warping, the interaction does not.
Design Expert Output
Response: Warping
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Source
Squares
DF
Model
968.22
15
A
698.34
3
B
156.09
3
AB
113.78
9
Residual
108.50
16
Lack of Fit
0.000
0
Pure Error
108.50
16
Cor Total
1076.72
31
Mean
Square
64.55
232.78
52.03
12.64
6.78
F
Value
9.52
34.33
7.67
1.86
Prob > F
< 0.0001
< 0.0001
0.0021
0.1327
significant
6.78
The Model F-value of 9.52 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.
(b) Analyze the residuals from this experiment.
There is nothing unusual about the residual plots.
Residuals vs. Predicted
Normal plot of residuals
3.5
95
90
1.75
80
70
Res iduals
Norm al % probability
99
-1.06581E-014
50
30
20
10
5
-1.75
1
-3.5
-3.5
-1.75
-1.06581E-014
1.75
10.50
3.5
15.38
20.25
Predicted
Res idual
5-5
25.13
30.00
STAT 503, Fall 2005
Homework Solution 5
Due: Oct. 18, 2005
Residuals vs. Copper Content
Residuals vs. Temperature
3.5
1.75
3.5
1.75
2
Res iduals
Res iduals
2
-1.06581E-014
-1.06581E-014
2
-1.75
2
-1.75
-3.5
-3.5
1
2
3
4
1
Copper Content
2
3
4
Tem perature
(c) Plot the average warping at each level of copper content and compare them to an appropriately scaled t
distribution. Describe the differences in the effects of the different levels of copper content on
warping. If low warping is desirable, what level of copper content would you specify?
Design Expert Output
Factor
Name
A
Copper Content
B
Temperature
Level
40
Average
Low Level
40
50
High Level
100
125
Warping
Prediction
15.5
SE Mean
0.92
95% CI low
13.55
95% CI high
17.45
Factor
A
B
Name
Copper Content
Temperature
Level
60
Average
Low Level
40
50
High Level
100
125
Warping
Prediction
18.875
SE Mean
0.92
95% CI low
16.92
95% CI high
20.83
Factor
A
B
Name
Copper Content
Temperature
Level
80
Average
Low Level
40
50
High Level
100
125
Warping
Prediction
21
SE Mean
0.92
95% CI low
19.05
95% CI high
22.95
Factor
A
B
Name
Copper Content
Temperature
Level
100
Average
Low Level
40
50
High Level
100
125
Warping
Prediction
28.25
SE Mean
0.92
95% CI low
26.30
95% CI high
30.20
Use a copper content of 40 for the lowest warping.
S
MSE
6.78125

 0.92
b
8
5-6
SE Pred
2.76
95% PI low
9.64
95% PI high
21.36
SE Pred
2.76
95% PI low
13.02
95% PI high
24.73
SE Pred
2.76
95% PI low
15.14
95% PI high
26.86
SE Pred
2.76
95% PI low
22.39
95% PI high
34.11
STAT 503, Fall 2005
Homework Solution 5
Due: Oct. 18, 2005
Scaled t Distribution
Cu=40
Cu=60
15.0
Cu=80
18.0
21.0
Cu=100
24.0
27.0
Warping
(d) Suppose that temperature cannot be easily controlled in the environment in which the copper plates are
to be used. Does this change your answer for part (c)?
Use a copper of content of 40. This is the same as for part (c).
DESIGN-EXPERT Plot
Warping
Interaction Graph
Tem perature
32.7602
2
X = A: Copper Content
Y = B: Temperature
B1
B2
B3
B4
50
75
100
125
2
26.5051
Warping
Design Points
2
2
3
20.25
2
2
13.9949
2
7.73979
40
60
80
Copper Content
5-7
100
STAT 503, Fall 2005
Homework Solution 5
Due: Oct. 18, 2005
5-8 An experiment is conducted to study the influence of operating temperature and three types of faceplate glass in the light output of an oscilloscope tube. The following data are collected:
100
580
568
570
Temperature
125
1090
1087
1085
150
1392
1380
1386
2
550
530
579
1070
1035
1000
1328
1312
1299
3
546
575
599
1045
1053
1066
867
904
889
Glass Type
1
(a) Use  = 0.05 in the analysis. Is there a significant interaction effect? Does glass type or temperature
affect the response? What conclusions can you draw?
Design Expert Output
Response: Light Output
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Model
2.412E+006
8
3.015E+005 824.77
A
1.509E+005
2
75432.26
206.37
B
1.780E+006
1
1.780E+006 4869.13
B2
1.906E+005
1
1.906E+005 521.39
AB
2.262E+005
2
1.131E+005 309.39
AB2
64373.93
2
32186.96
88.06
Pure Error
6579.33
18
365.52
Cor Total
2.418E+006 26
Prob > F
< 0.0001
< 0.0001
< 0.0001
< 0.0001
< 0.0001
< 0.0001
significant
The Model F-value of 824.77 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Std. Dev. 19.12R-Squared0.9973
Mean 940.19Adj R-Squared 0.9961
C.V. 2.03Pred R-Squared0.9939
PRESS
14803.50
Adeq Precision
75.466
From the analysis of variance, both factors, Glass Type (A) and Temperature (B) are significant, as well as
the interaction (AB). B2 and AB2 interation terms are also significant. The interaction and pure quadratic
tems can be clearly seen in the plot shown below. For glass types 1 and 2 the temperature is fairly linear,
for glass type 3, there is a quadratic effect.
5-8
STAT 503, Fall 2005
Homework Solution 5
Interaction Graph
DESIGN-EXPERT Plot
Light Output
Due: Oct. 18, 2005
A: Glass T ype
1402.4
X = B: Temperature
Y = A: Glass Ty pe
1184.3
Light Output
A1 1
A2 2
A3 3
966.199
748.099
530
100.00
112.50
125.00
137.50
150.00
B: T emperature
(b) Fit an appropriate model relating light output to glass type and temperature.
The model, both coded and uncoded are shown in the Design Expert output below.
Design Expert Output
Final Equation in Terms of Coded Factors:
Light Output
+1059.00
+28.33
-24.00
+314.44
-178.22
+92.22
+65.56
+70.22
+76.22
=
* A[1]
* A[2]
*B
* B2
* A[1]B
* A[2]B
* A[1]B2
* A[2]B2
Final Equation in Terms of Actual Factors:
Glass Type
Light Output
-3646.00000
+59.46667
-0.17280
1
=
Glass Type
Light Output
-3415.00000
+56.00000
-0.16320
2
=
Glass Type
Light Output
-7845.33333
+136.13333
-0.51947
3
=
* Temperature
* Temperature2
* Temperature
* Temperature2
* Temperature
* Temperature2
(c) Analyze the residuals from this experiment. Comment on the adequacy of the models you have
considered.
5-9
STAT 503, Fall 2005
Homework Solution 5
Due: Oct. 18, 2005
The only concern from the residuals below is the inequality of variance observed in the residuals versus
glass type plot shown below.
Normal Plot of Residuals
Residuals vs. Predicted
35
99
17.5
90
80
70
Residuals
Normal % Probability
95
50
30
0
20
10
-17.5
5
1
-35
-35
-17.5
0
17.5
35
553.00
761.25
969.50
Residual
Residuals vs. Glass Type
Residuals vs. Temperature
35
17.5
17.5
Residuals
Residuals
1386.00
Predicted
35
0
0
-17.5
-17.5
-35
-35
1
1177.75
2
3
100
108
Glass T ype
117
125
133
142
150
T emperature
5-9 Consider the data in Problem 5-1. Fit an appropriate model to the response data. Use this model to
provide guidance concerning operating conditions for the process.
See the alternative analysis shown in Problem 5-1 part (c).
5-18 In Problem 5-1, suppose that we wish to reject the null hypothesis with a high probability if the
difference in the true mean yield at any two pressures is as great as 0.5. If a reasonable prior estimate of
the standard deviation of yield is 0.1, how many replicates should be run?
2 
n
2
2

25
5
naD 2
2b 2

n30.52
230.12
1  b  1
2
 12 .5n
2  abn  1
(3)(3)(1)

0.014
2 replications will be enough to detect the given difference.
5-10