Homework #3 Solution Exercise 3.1
tc A B AB SumY (1) -1 -1 +1 64.4 a +1 -1 -1 96.1 b -1 +1 -1 59.7
ab +1 +1 +1 Sums of squares for effects SS =
161.1
(contrast)2 4n
, n = 4
∑ 2
∑2(y) 2
SST= y − =10797- 4n
ANOVA table
AB Error Total
(381.3) 16
=1710
Source A
df 1
304 1 72 12 1710 15
MS F
F0.05, 1, 12 = 4.75 all effects are significant Interaction plot
50
40
Y 30
20
10
0
-1 Factor A 1
B(+1) B(-1)
SS 1107
1107
B 227 1 227 37.8
Contrasts:
A = 133.1
B = 60.3 AB = 69.7
304 50.7 6
184.5
Response is vibration smaller values are preferred Best conditions:
A(-1) bit size = 1/16 inch B(+1) speed = 90 rpm
Exercise 3.1 (continued)
contrast
Effect estimates: 2n , A = 16.64, B = 7.54, AB=8.71
Regression model:
yˆ = 381.3+16.64x +7.54x +8.71x x 1212
16 2 2 2 where x1 = A−(3/32) x2 = B−65
1/32 25 For A=1/8, x1 =+1
25 ≥ 32.15 + 8.125×2 −0.88 ≥ x2
−0.88≥ B−65 25
43 ≥ B
So,
yˆ = 381.3+16.64(+1)+7.54x +8.71(+1)x 22
162 22 yˆ = 3 2 . 1 5 + 8 . 1 2 5 x 2
Exercise 3.9
MTB > WOPEN “E:\Kurt\Documents\ise525\Blackboard\mm3- 9_2018.DAT”;
SUBC> FTYPE;
SUBC> TEXT;
SUBC> FIELD;
SUBC> COMMA;
SUBC> TDELIMITER;
SUBC> DOUBLEQUOTE;
SUBC> DECSEP;
SUBC> PERIOD;
SUBC> DATA;
SUBC> IGNOREBLANKROWS;
SUBC> EQUALCOLUMNS;
ANOVA: Y versus A, B, C, D Factor Information
Factor Type Levels
A Fixed 2
B Fixed 2
C Fixed 2
D Fixed 2
Values -1, 1 -1, 1 -1, 1 -1, 1
Analysis of Variance for Y
SHEET 1;
VNAMES -1;
FIRST 1;
NROWS 48.
Source DF
A 1
B 1
C 1
D 1
A*B 1
A*C 1
A*D 1
B*C 1 B*D 1 C*D 1 A*B*C 1 A*B*D 1 A*C*D 1 B*C*D 1 A*B*C*D 1
SS MS F P 16.0210 16.0210 1481.52 0.000 2.8193 2.8193 260.71 0.000 14.2365 14.2365 1316.50 0.000 0.0343 0.0343 3.17 0.084 0.1502 0.1502 13.89 0.001 1.4073 1.4073 130.14 0.000 0.5132 0.5132 47.45 0.000 0.1967 0.1967 18.19 0.000 0.4862 0.4862 44.96 0.000 0.0775 0.0775 7.17 0.012 0.2232 0.2232 20.64 0.000 0.0023 0.0023 0.22 0.645 0.0012 0.0012 0.11 0.738 0.2536 0.2536 23.45 0.000 0.0090 0.0090 0.84 0.367
Comment 1: 2(a)
The highlighted effects are significant. To understand impact on response, focus interpretation on high-order effects ABC and BCD.
Worksheet was saved on Mon Jan 29 2018
SUBC>
SUBC>
SUBC>
SUBC>
Retrieving worksheet from file: ‘E:\Kurt\Documents\ise525\Blackboard\mm3- 9_2018.DAT’
Results for: mm3-9_2018.DAT
MTB > ANOVA ‘Y’ = A | B | ‘C’ | D; SUBC> MeansA|B|’C’|D; SUBC> GNormalplot;
SUBC> NoDGraphs;
SUBC> MTB >
GVars ‘A’-‘D’.
Error 32 0.3460 0.0108 Total 47 36.7777
Model Summary
S R-sq 0.103990 99.06%
R-sq(adj) 98.62%
Means
ANY -1 24 0.55112 1 24 1.70658
BNY -1 24 0.88650 1 24 1.37121
CNY -1 24 0.58425 1 24 1.67346
DNY -1 24 1.10213 1 24 1.15558
A*BN Y
-1 -1 12 -1 1 12 1-1 12 1 1 12
0.25283 0.84942 1.52017 1.89300
A*CN Y
-1 -1 12 -1 1 12 1-1 12 1 1 12
0.17775 0.92450 0.99075 2.42242
A*DN Y
-1 -1 12 -1 1 12 1-1 12 1 1 12
0.42100 0.68125 1.78325 1.62992
B*CN Y
-1 -1 12 -1 1 12 1-1 12 1 1 12
0.40592 1.36708 0.76258 1.97983
B*DN Y
-1 -1 12 -1 1 12 1-1 12 1 1 12
0.96042 0.81258 1.24383 1.49858
C*DN Y
-1 -1 12 -1 1 12 1-1 12 1 1 12
0.51733 0.65117 1.68692 1.66000
A*B*C N Y -1-1-1 6 0.01167 -1-11 6 0.49400 -11-1 6 0.34383 -1 1 1 6 1.35500 1-1-1 6 0.80017 1-11 6 2.24017 11-1 6 1.18133 1 1 1 6 2.60467
Comment 2: 2(b)
A*B*D N Y -1-1-1 6 0.23033 -1-11 6 0.27533 -11-1 6 0.61167 -1 1 1 6 1.08717 1-1-1 6 1.69050 1-11 6 1.34983 11-1 6 1.87600 1 1 1 6 1.91000
This treatment combination produces minimum crack length due to ABC interaction.
A*C*D N Y -1-1-1 6 0.01250 -1-11 6 0.34300 -11-1 6 0.82950 -1 1 1 6 1.01950 1-1-1 6 1.02217 1-11 6 0.95933 11-1 6 2.54433 1 1 1 6 2.30050
B*C*D N Y -1-1-1 6 0.51233 -1-11 6 0.29950 -11-1 6 1.40850 -1 1 1 6 1.32567 1-1-1 6 0.52233 1-11 6 1.00283 11-1 6 1.96533 1 1 1 6 1.99433
Comment 3: 2(b)
This treatment combination produces minimum crack length due to BCD.
Overall, recommend A(-1) B(-1) C(-1) D(+1)
A*B*C*D N
Y 0.01300 0.01033 0.44767 0.54033 0.01200 0.67567 1.21133 1.49867 1.01167 0.58867 2.36933 2.11100 1.03267 1.33000 2.71933 2.49000
-1-1-1-1 -1-1-1 1 -1-1 1-1 -1-111 -1 1-1-1 -11-11 -111-1
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
-1111 1-1-1-1 1-1-11 1-11-1 1-1 1 1 11-1-1 1 1-1 1 1 1 1-1 1111
Percent
Normplot of Residuals for Y
99
95 90
80 70
60 50 40 30
20
10 5
1
-0.3
-0.2 -0.1 0.0 0.1 Residual
0.2 0.3
Comment 4: 2(c)
Normal Probability Plot (response is Y)
The graph above shows that the residuals are not normally distributed. The plotted points appear as an S-curve, rather than a straight line. This indicates that the p-values in the ANOVA are suspect. Even so, the most significant effects (p-value = 0.000) are still easily identified.
The graphs below show that the variance of the residuals is approximately equal across all factor level groupings. The vertical spread of the points is similar in all cases. This indicates that the equal variance ANOVA assumption is valid.
Residual Residual
Residuals from Y vs A
0.3 0.2 0.1 0.0
-0.1 -0.2 -0.3
Residuals from Y vs B
0.3 0.2 0.1 0.0
-0.1 -0.2 -0.3
Residuals Versus A (response is Y)
-1.0 -0.5 0.0 0.5 1.0 A
Residuals Versus B (response is Y)
-1.0 -0.5 0.0 0.5 1.0 B
Residual Residual
Residuals from Y vs C
0.3 0.2 0.1 0.0
-0.1 -0.2 -0.3
Residuals from Y vs D
0.3 0.2 0.1 0.0
-0.1 -0.2 -0.3
Residuals Versus C (response is Y)
-1.0 -0.5 0.0 0.5 1.0 C
Residuals Versus D (response is Y)
-1.0 -0.5 0.0 0.5 1.0 D