Problema 8.6: RD Snee (“Experimenting with a large number

Problema 8.6:
R.D. Snee (“Experimenting with a large number of variables,” in experiments in Industry:
Design, Analysis and Interpretation of Results, by R. D. Snee, L.B. Hare, and J. B. Trout,
Editors, ASQC, 1985) describes an experiment in which a 25-1 design with I = ABCDE was used
to investigate the effects of five factors on the color of a chemical product. The factors are A=
solvent/reactant, B = catalyst/reactant, C = temperature, D = reactant purity, and E = reactant pH.
The results obtained were as follows:
Tratamientos
Resultados
e
a
b
abe
c
ace
bce
abc
d
ade
bde
abd
cde
acd
bcd
abcde
-0.63
2.51
-2.68
1.66
2.06
1.22
-2.09
1.93
6.79
5.47
3.45
5.68
5.22
4.38
4.30
4.05
a) Prepare a normal probability plot of the effects. Which effects seem active?
Normal Probability Plot of the Effects
(response is Results, Alpha = .05)
99
Effect Type
Not Significant
Significant
D
95
90
F actor
A
B
C
D
E
Percent
80
70
60
50
40
30
N ame
A
B
C
D
E
20
10
5
1
-2
-1
0
1
2
3
4
5
Effect
Lenth's PSE = 0.76125
El efecto significativo es la D = pureza del reactante.
b) Calculate the residuals. Construct a normal probability plot of the residuals and plot the
residuals versus the fitted values. Comment on the plots.
R e s idua l P lots for R e s ults
No rm al P ro b ab ilit y P lo t o f t h e R esid u als
R esid u als V ersu s t h e Fit t ed V alu es
99
2
Residual
Percent
90
50
10
1
-4
-2
0
Re sidual
2
0
-2
4
0.0
H ist o g ram o f t h e R esid u als
3.6
4.8
2
2
Residual
Frequency
2.4
F itte d V alue
R esid u als V ersu s t h e O rd er o f t h e D at a
3
1
0
1.2
0
-2
-3
-2
-1
0
Re sidual
1
2
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16
O bse r v a tion O r de r
Los residuales muestran un comportamiento aproximadamente normal y al compararlos con los
fitted values no se ve un cambio en la varianza poco significativo.
c) If any factors are negligible, collapse the 25-1 design into a full factorial in the active
factors. Comment on the resulting design, and interpret the results.
Factorial Fit: Respuesta versus D
Estimated Effects and Coefficients for Respuesta (coded units)
Term
Effect
Constant
D
4.420
S = 1.61556
Coef
SE Coef
T
P
2.708
0.4039
6.70
0.000
2.210
0.4039
5.47
0.000
R-Sq = 68.14%
R-Sq(adj) = 65.86%
Analysis of Variance for Respuesta (coded units)
Source
DF
Seq SS
Adj SS
Adj MS
F
P
1
78.146
78.1456
78.146
29.94
0.000
Residual Error
14
36.541
36.5407
2.610
Pure Error
14
36.541
36.5407
2.610
15
114.686
Main Effects
Total
Unusual Observations for Respuesta
Obs
StdOrder
Respuesta
Fit
SE Fit
Residual
3
3
-2.68000
0.49750
0.57119
-3.17750
St Resid
-2.10R
R denotes an observation with a large standardized residual.
Alias Structure
I
D
Residual Plots for Respuesta
Normal Probability Plot of the Residuals
Residuals Versus the Fitted Values
99
2
Residual
Percent
90
50
10
1
-4
-2
0
Residual
2
0
-2
4
0.0
Histogram of the Residuals
3.6
4.8
2
2
Residual
Frequency
2.4
Fitted Value
Residuals Versus the Order of the Data
3
1
0
1.2
-3
-2
-1
0
Residual
1
2
0
-2
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16
Observation Order
Problema 8.8:
An article in Industrial and Engineering Chemistry uses a 25-2 design to investigate the effect of
A = condensation temperature, B = amount of material 1, C = solvent volume, D = condensation
time, and E = amount of material 2 on yield. The results obtained are as follows:
Tratamientos
Resultados
e
ab
ad
bc
cd
ace
bde
abcde
23.2
15.5
16.9
16.2
23.8
23.4
16.8
18.1
a) Verify that the design generators used were I = ACE and I = BDE.
Factors:
Runs:
Blocks:
5
8
1
Base Design:
Replicates:
Center pts (total):
3, 8
1
0
Resolution:
Fraction:
III
1/4
* NOTE * Some main effects are confounded with two-way interactions.
Design Generators: D = ABC, E = AC
Alias Structure (up to order 3)
I + ACE + BDE
A + CE + BCD + ABDE
B + DE + ACD + ABCE
C + AE + ABD + BCDE
D + BE + ABC + ACDE
E + AC + BD + ABCDE
AB + CD + ADE + BCE
AD + BC + ABE + CDE
ABCD
Los generadores si son ACE y BDE.
b) Write down the complete defining relation and the aliases for this design.
A = CE = BCD = ABDE
B = DE = ACD = ABCE
C = AE = ABD = BCDE
D = BE = ABC = ACDE
E = AC = BD = ABCDE
AB = CD = ADE = BCE
AD = BC = ABE = CDE
ABCD
c) Estimate the main effects.
Estimated Effects and Coefficients for Respuesta (coded units)
Term
Constant
A
B
C
D
E
Effect
-1.525
-5.175
2.275
-0.675
2.275
S = 2.22626
Coef
19.238
-0.763
-2.587
1.138
-0.337
1.138
SE Coef
0.7871
0.7871
0.7871
0.7871
0.7871
0.7871
R-Sq = 88.95%
T
24.44
-0.97
-3.29
1.45
-0.43
1.45
P
0.002
0.435
0.081
0.285
0.710
0.285
R-Sq(adj) = 61.34%
Normal Probability Plot of the Effects
(response is Respuesta, Alpha = .05)
99
Effect Ty pe
Not Significant
Significant
95
90
F actor
A
B
C
D
E
Percent
80
70
60
50
40
30
N ame
A
B
C
D
E
20
10
5
1
-7.5
-5.0
-2.5
0.0
Effect
2.5
5.0
Lenth's PSE = 2.7375
No se ve ningun factor significativo sin embargo se verifico que factor tiene un efecto mayor
sobre la respuesta con el siguiente grafico:
Main Effects Plot (data means) for Respuesta
A
22
B
C
Mean of Respuesta
20
18
16
-1
1
-1
D
22
1
-1
1
E
20
18
16
-1
1
-1
1
Se observa que B es el que mas impacta la respuesta.
d) Prepare an analysis of variance table. Verify that the AB and AD interactions are
available to use as error.
Analysis of Variance for Respuesta (coded units)
Source
Main Effects
2-Way Interactions
Residual Error
Total
DF
5
2
0
7
Seq SS
79.826
9.913
*
89.739
Adj SS
79.826
9.913
*
Adj MS
15.965
4.956
*
F
*
*
P
*
*
Effects Plot for Respuesta
Alias Structure
I + A*C*E + B*D*E + A*B*C*D
A + C*E + B*C*D + A*B*D*E
B + D*E + A*C*D + A*B*C*E
C + A*E + A*B*D + B*C*D*E
D + B*E + A*B*C + A*C*D*E
E + A*C + B*D + A*B*C*D*E
A*D + B*C + A*B*E + C*D*E
A*B + C*D + A*D*E + B*C*E
Dejando las interacciones mencionadas por fuera no se obtiene aun grados de libertad para el
error. Sabiendo que B es el que mas impacta, se procedió a realizar una anova incluyendo solo el
factor B:
Factorial Fit: Respuesta versus B
Estimated Effects and Coefficients for Respuesta (coded units)
Term
Constant
B
Effect
Coef
19.238
-2.588
-5.175
S = 2.45552
SE Coef
0.8682
0.8682
R-Sq = 59.69%
T
22.16
-2.98
P
0.000
0.025
R-Sq(adj) = 52.97%
Analysis of Variance for Respuesta (coded units)
Source
Main Effects
Residual Error
Pure Error
Total
DF
1
6
6
7
Seq SS
53.5612
36.1775
36.1775
89.7388
Adj SS
53.5612
36.1775
36.1775
Adj MS
53.561
6.030
6.030
F
8.88
P
0.025
Unusual Observations for Respuesta
Obs
7
StdOrder
2
Respuesta
16.9000
Fit
21.8250
SE Fit
1.2278
Residual
-4.9250
St Resid
-2.32R
R denotes an observation with a large standardized residual.
* NOTE * Normal and Pareto effects plots require at least 3 terms.
Alias Structure
I
B
e) Plot the residuals versus the fitted values. Also construct a normal probability plot of the
residuals.
Residual Plots for Respuesta
Normal Probability Plot of the Residuals
Residuals Versus the Fitted Values
99
Residual
Percent
90
50
0.0
-2.5
10
1
-5.0
-5.0
-2.5
0.0
Residual
2.5
5.0
16
18
20
22
Fitted Value
Histogram of the Residuals
Residuals Versus the Order of the Data
1.5
Residual
Frequency
2.0
1.0
0.0
-2.5
0.5
0.0
-5.0
-5
-4
-3
-2
-1
Residual
0
1
2
1
2
3
4
5
6
Observation Order
7
8
Se observa un comportamiento aproximadamente normal, pero parece que a mayor valor en la
respuesta mayor varianza en los residuales.
Problema 13.9:
An experiment was performed to investigate the capability of a measurement system. Ten parts
were randomly selected, and two randomly selected operators measured each part three times.
The tests were made in random order, and the data below resulted.
Part number
Operator 1
Operator 2
1
2
3
1
2
3
1
50
49
50
50
48
51
2
52
52
51
51
51
51
3
53
50
50
54
52
51
4
49
51
50
48
50
51
5
48
49
48
48
49
48
6
52
50
50
52
50
50
7
51
51
51
51
50
50
8
52
50
49
53
48
50
9
50
51
50
51
48
49
10
47
46
49
46
47
48
General Linear Model: results versus operador, Part number
Factor
operador
Part number
Type
random
random
Levels
2
10
Values
1, 2
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Analysis of Variance for results, using Adjusted SS for Tests
Source
operador
Part number
operador*Part number
Error
Total
DF
1
9
9
40
59
Seq SS
0.417
99.017
5.417
60.000
164.850
Adj SS
0.417
99.017
5.417
60.000
Adj MS
0.417
11.002
0.602
1.500
F
0.69
18.28
0.40
P
0.427
0.000
0.927
Normal Probability Plot of the Residuals
(response is results)
99.9
99
Percent
95
90
80
70
60
50
40
30
20
10
5
1
0.1
-3
-2
-1
0
Residual
1
2
3
Según lo anterior la interacción de las partes con el operador no es significativa.
b) Find point estimates of the variance components using the analysis of variance method.
σ2 = MSE = 1.5
σ2τβ = (MSAB – MSE) / n = (0.6018 - 1.5) / 3 < 0, asumiendo σ2τβ = 0
σ2τ = (MSB – MSAB) / an = (11.0018 - 0.6018) / 2*3 = 1.733
σ2β = (MSA – MSAB) / bn < 0, asumiendo σ2τ = 0