STAT 503, Fall 2005
Homework Solution 7
Due: Nov. 8, 2005
6-18 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.
Run
1
2
3
4
5
6
7
8
A
+
+
+
+
Coded Factors
B
C
+
+
+
+
+
+
+
+
Fill Height Deviation
Replicate 1
Replicate 2
-3
0
-1
2
-1
2
1
6
-1
1
0
3
0
1
1
5
Factor Levels
Low (-1)
High (+1)
10
12
25
30
200
250
A (%)
B (psi)
C (b/m)
(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 Plot
Fill Deviation
A: Carbonation
B: Pressure
C: Speed
Half Normal plot
99
97
A
Half Norm al % probability
95
90
B
85
C
80
AB
70
60
40
20
0
0.00
0.75
1.50
2.25
3.00
|Effect|
Design Expert Output
Response:
Fill Deviation
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
70.75
4
17.69
F
Value
26.84
7-1
Prob > F
< 0.0001
significant
STAT 503, Fall 2005
A
B
C
AB
Residual
Lack of Fit
Pure Error
Cor Total
36.00
20.25
12.25
2.25
7.25
2.25
5.00
78.00
Homework Solution 7
1
1
1
1
11
3
8
15
36.00
20.25
12.25
2.25
0.66
0.75
0.63
Due: Nov. 8, 2005
54.62
30.72
18.59
3.41
< 0.0001
0.0002
0.0012
0.0917
1.20
0.3700
not significant
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.
Mean
C.V.
PRESS
0.81
1.00
81.18
15.34
R-Squared
Adj R-Squared
Pred R-Squared
Adeq Precision
0.9071
0.8733
0.8033
15.424
(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.
Normal plot of residuals
Residuals vs. Predicted
1.125
99
0.5625
90
80
Res iduals
Norm al % probability
95
70
50
30
0
2
20
10
-0.5625
5
1
-1.125
-1.125
-0.5625
0
0.5625
1.125
-2.13
Res idual
-0.38
1.38
Predicted
7-2
3.13
4.88
STAT 503, Fall 2005
Homework Solution 7
Due: Nov. 8, 2005
Residuals vs. Carbonation
1.125
0.5625
0.5625
Res iduals
Res iduals
Residuals vs. Run
1.125
0
2
0
2
-0.5625
-0.5625
-1.125
-1.125
1
4
7
10
13
16
2
10
11
Run Num ber
Carbonation
Residuals vs. Pressure
Residuals vs. Speed
1.125
1.125
0.5625
0.5625
Res iduals
Res iduals
12
0
3
-0.5625
2
0
2
2
-0.5625
2
-1.125
-1.125
25
26
27
28
29
30
200
Pres s ure
208
217
225
233
242
250
Speed
(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
+1.13
+0.88
+0.38
=
*A
*B
*C
*A*B
Final Equation in Terms of Actual Factors
Fill Deviation =
+9.62500
-2.62500 * Carbonation
-1.20000 * Pressure
7-3
STAT 503, Fall 2005
Homework Solution 7
Due: Nov. 8, 2005
+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.
Fill Deviation
2
30.00
2
Fill Deviation
2
30.00
3
2
4.5
2.5
4
2
3.5
28.75
28.75
1.5
3
B: Pres s ure
B: Pres s ure
1
0.5
27.50
-0.5
0
2.5
27.50
2
1
1.5
-1
0.5
26.25
26.25
-1.5
0
2
-2
2
2
25.00
2
25.00
10.00
10.50
11.00
11.50
12.00
10.00
10.50
A: Carbonation
11.00
11.50
12.00
A: Carbonation
Speed set at 200 b/mSpeed 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
Source
Squares
DF
Square
Model
68.50
3
22.83
A
36.00
1
36.00
B
20.25
1
20.25
C
12.25
1
12.25
Residual
9.50
12
0.79
Lack of Fit
4.50
4
1.13
Pure Error
5.00
8
0.63
Cor Total
78.00
15
F
Value
28.84
45.47
25.58
15.47
Prob > F
< 0.0001
< 0.0001
0.0003
0.0020
1.80
0.2221
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
7-4
significant
not significant
STAT 503, Fall 2005
Mean
C.V.
PRESS
Homework Solution 7
1.00
88.98
16.89
Adj R-Squared
Pred R-Squared
Adeq Precision
Due: Nov. 8, 2005
0.8478
0.7835
15.735
Final Equation in Terms of Coded Factors
Fill Deviation
+1.00
+1.50
+1.13
+0.88
=
*A
*B
*C
Final Equation in Terms of Actual Factors
Fill Deviation
-35.75000
+1.50000
+0.45000
+0.035000
=
* Carbonation
* Pressure
* Speed
Normal plot of residuals
Residuals vs. Predicted
1.5
99
0.8125
90
80
Res iduals
Norm al % probability
95
70
50
30
0.125
20
10
2
-0.5625
5
1
-1.25
-1.25
-0.5625
0.125
0.8125
1.5
-2.50
Res idual
-0.75
1.00
2.75
4.50
Predicted
Residuals vs. Run
Residuals vs. Carbonation
1.5
1.5
0.8125
0.8125
Res iduals
Res iduals
2
0.125
0.125
-0.5625
-0.5625
-1.25
-1.25
1
4
7
10
13
16
3
10
Run Num ber
11
Carbonation
7-5
12
STAT 503, Fall 2005
Homework Solution 7
Due: Nov. 8, 2005
Residuals vs. Speed
1.5
0.8125
0.8125
Res iduals
Res iduals
Residuals vs. Pressure
1.5
2
0.125
2
0.125
2
-0.5625
2
2
2
-0.5625
2
2
2
2
-1.25
-1.25
25
26
27
28
29
30
200
208
217
Pres s ure
233
242
250
Speed
Fill Deviation
30.00 2
225
2
2.5
Fill Deviation
30.00 2
2
4
2
3.5
28.75
28.75
1.5
3
0
27.50
B: Pressure
B: Pressure
1
0.5
-0.5
2.5
1.5
27.50
2
1
-1
0.5
-1.5
26.25
26.25
0
-2
-0.5
25.00 2
10.00
2
10.50
11.00
11.50
25.00 2
12.00
10.00
2
10.50
11.00
11.50
A: Carbonation
A: Carbonation
Speed set at 200 b/m
Speed set at 250 b/m
7-6
12.00
STAT 503, Fall 2005
Homework Solution 7
Due: Nov. 8, 2005
6-21 Reconsider the experiment in Problem 6-20. 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 6-20, 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
Model
0.24
5
0.048
45.07
A
0.10
1
0.10
95.77
C
0.070
1
0.070
65.68
D
0.051
1
0.051
47.35
AC
0.012
1
0.012
11.32
CD
5.625E-003 1
5.625E-003
5.26
Curvature
0.18
1
0.18
170.59
Residual
0.014
13
1.069E-003
Lack of Fit
0.013
10
1.260E-003
2.91
Pure Error
1.300E-003 3
4.333E-004
Cor Total
0.44
19
Prob > F
< 0.0001
< 0.0001
< 0.0001
< 0.0001
0.0051
0.0391
< 0.0001
0.2057
significant
significant
not significant
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.
Mean
C.V.
PRESS
Coefficient
Factor
Intercept
A-Laser Power
C-Cell Size
D-Writing Speed
AC
CD
Center Point
0.033
0.76
4.28
0.035
R-Squared
Adj R-Squared
Pred R-Squared
Adeq Precision
Standard
Estimate
0.72
0.080
-0.066
-0.056
-0.028
0.019
0.24
95% CI
DF
1
1
1
1
1
1
1
0.9455
0.9245
0.9209
20.936
95% CI
Error
8.175E-003
8.175E-003
8.175E-003
8.175E-003
8.175E-003
8.175E-003
0.018
Low
0.70
0.062
-0.084
-0.074
-0.045
1.089E-003
0.20
7-7
High
0.73
0.098
-0.049
-0.039
-9.839E-003
0.036
0.28
VIF
1.00
1.00
1.00
1.00
1.00
1.00
STAT 503, Fall 2005
Homework Solution 7
Due: Nov. 8, 2005
6-22 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 (3rd class, 1st 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.
Run
1
2
3
4
5
6
7
8
A
+
+
+
+
Coded Factors
B
+
+
+
+
Number of Orders
C
+
+
+
+
Replicate 1
Replicate 2
50
44
46
42
49
48
47
56
54
42
48
43
46
45
48
54
Factor Levels
Low (-1) High (+1)
3rd
1st
BW
Color
$19.95
$24.95
A (class)
B (type)
C ($)
(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.
Half Normal plot
DESIGN-EXPERT Plot
Number of orders
A: Class
B: Ty pe
C: Price
99
97
AC
Half Normal % probability
95
90
BC
85
AB
C
80
70
A
60
B
40
20
0
0.00
1.25
2.50
|Effect|
7-8
3.75
5.00
STAT 503, Fall 2005
Homework Solution 7
Design Expert Output
Response:
Number of orders
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
241.75
6
40.29
A
12.25
1
12.25
B
2.25
1
2.25
C
36.00
1
36.00
AB
42.25
1
42.25
AC
100.00
1
100.00
BC
49.00
1
49.00
Residual
28.00
9
3.11
Lack of Fit
4.00
1
4.00
Pure Error
24.00
8
3.00
Cor Total
269.75
15
Due: Nov. 8, 2005
F
Value
12.95
3.94
0.72
11.57
13.58
32.14
15.75
Prob > F
0.0006
0.0785
0.4171
0.0079
0.0050
0.0003
0.0033
1.33
0.2815
significant
not significant
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.
Mean
C.V.
PRESS
Coefficient
Factor
Intercept
A-Class
B-Type
C-Price
AB
AC
BC
1.76
47.63
3.70
88.49
R-Squared
Adj R-Squared
Pred R-Squared
Adeq Precision
Standard
Estimate
47.63
-0.88
0.37
1.50
1.63
2.50
1.75
95% CI
DF
1
1
1
1
1
1
1
0.8962
0.8270
0.6719
10.286
95% CI
Error
0.44
0.44
0.44
0.44
0.44
0.44
0.44
Low
46.63
-1.87
-0.62
0.50
0.63
1.50
0.75
High
48.62
0.12
1.37
2.50
2.62
3.50
2.75
VIF
1.00
1.00
1.00
1.00
1.00
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.
Normal Plot of Residuals
Residuals vs. Predicted
3
99
1.5
90
80
70
Residuals
Normal % Probability
95
50
30
0
20
10
-1.5
5
1
-3
-2.5
-1.375
-0.25
0.875
2
42.50
Residual
45.50
48.50
Predicted
7-9
51.50
54.50
STAT 503, Fall 2005
Homework Solution 7
Due: Nov. 8, 2005
Residuals vs. Class
3
1.5
1.5
Residuals
Residuals
Residuals vs. Run
3
0
2
2
0
2
2
-1.5
-1.5
-3
-3
1
4
7
10
13
16
1
2
Run Number
Class
Residuals vs. Type
Residuals vs. Price
3
2
2
1.5
Residuals
Residuals
1.5
3
2
0
2
3
0
2
2
2
-1.5
-1.5
-3
-3
1
2
19.95
T ype
21.20
22.45
23.70
24.95
Price
(c) What would you recommend to the company?
Based on the interaction plots below, we recommend 3rd 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 1st class mail with color brochures is recommended.
7-10
STAT 503, Fall 2005
Homework Solution 7
Interaction Graph
DESIGN-EXPERT Plot
Number of orders
Number of orders
X = A: Class
Y = C: Price
Design Points
Number of orders
52.5
C- 19.950
C+ 24.950
Actual Factor
B: Ty pe = BW
49
42
40.6338
1st
3rd
C: Price
56
X = B: Ty pe
Y = C: Price
Number of orders
52.5
49
3
45.5
42
BW
1st
A: Class
Interaction Graph
DESIGN-EXPERT Plot
Design Points
48.3169
44.4754
A: Class
Number of orders
52.1585
45.5
3rd
C- 19.950
C+ 24.950
Actual Factor
A: Class = 3rd
C: Price
56
Number of orders
X = A: Class
Y = B: Ty pe
B1 BW
B2 Color
Actual Factor
C: Price = 22.45
Interaction Graph
DESIGN-EXPERT Plot
B: T ype
56
Due: Nov. 8, 2005
Color
B: T ype
7-11