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# Solucionario Walpole 8 ED

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```ad
bd
d
ac
bc
abd
a
b
cd
abcd
L1 = 0 L1 = 1 L1 = 0 L1 = 1
L2 = 0 L2 = 0 L2 = 1 L2 = 1
Since (ABC)(ABD) = A2B2CD = CD (mod 2), then CD is the other effect con-
founded.
15.16 (a) L1 = \u3b31+ \u3b32+ \u3b34+ \u3b35, L2 = \u3b31+ \u3b35. We find that the following treatment combi-
nations are in the principal block (L1 = 0, L2 = 0): (1), c, ae, bd, ace, abde, abcde.
The other blocks are constructed by multiplying the treatment combinations in
the principal block modulo 2 by a, b, and ab, respectively, to give the following
blocking arrangement:
Block 1 Block 2 Block 3 Block 4
(1)
c
ae
bd
ace
bcd
abde
abcde
a
ac
e
abd
ce
abcd
bde
bcde
b
bc
abe
d
abce
cd
acde
ab
abc
bce
bce
acd
de
cde
(b) (ABDE)(AE) = BD (mod 2). Therefore BD is also confounded with days.
(c) Yates\u2019 technique gives the following sums of squares for the main effects:
SSA = 21.9453, SSB = 40.2753, SSC = 2.4753,
SSD = 7.7028, SSE = 1.0878.
15.17 L1 = \u3b31 + \u3b32 + \u3b33, L2 = \u3b31 + \u3b32.
Block Block Block
1 2 1 2 1 2
abc
a
b
c
ab
ac
bc
(1)
abc
a
b
c
ab
ac
bc
(1)
(1)
c
ab
abc
a
b
ac
bc
Rep 1 Rep 2 Rep 3
ABC Confounded ABC Confounded AB Confounded
Solutions for Exercises in Chapter 15 245
For treatment combination (1) we find L1 (mod 2) = 0 and L2 (mod 2) = 0. For
treatment combination a we find L1 (mod 2) = 1 and L2 (mod 2) = 1. Replicate 1 and
Replicate 2 have L1 = 0 in one block and L1 = 1 in the other. Replicate 3 has L2 = 0
in one block and L2 = 1 in the other.
Analysis of Variance
Source of Variation Degrees of Freedom
Blocks
A
B
C
AB
AC
BC
ABC
Error
5
1
1
1
1
\u2032
1
1
1
\u2032
11
Total 23
Relative information on ABC = 1
3
and relative information on AB = 2
3
.
15.18 (a) The ANOVA table is shown here.
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
Operators
A
B
C
D
Error
1
1
1
1
1
10
0.1225
4.4100
3.6100
9.9225
2.2500
2.8423
0.04
1.55
1.27
3.49
0.79
0.2413
0.2861
0.0912
0.3945
Total 15
None of the main effects is significant at 0.05 level.
(b) ABC is confounded with operators since all treatments with positive signs in the
ABC contrast are in one block and those with negative signs are in the other
block.
15.19 (a) One possible design would be:
Machine
1
2
3
4
(1)
a
c
d
ab
b
abc
abd
ce
ace
e
cde
abce
bce
abe
abcde
acd
cd
ac
bde
abde
bcde
be
de
acde
ae
bcd
abcd
bd
bc
(b) ABD, CDE, and ABCE.
246 Chapter 15 2k Factorial Experiments and Fractions
15.20 (a) y\u2c6 = 43.9 + 1.625x1 \u2212 8.625x2 + 0.375x3 + 9.125x1x2 + 0.625x1x3 + 0.875x2x3.
(b) The Lack-of-fit test results in a P -value of 0.0493. There are possible quadratic
terms missing in the model.
15.21 (a) The P -values of the regression coefficients are:
Parameter Intercept x1 x2 x3 x1x2 x1x3 x2x3 x1x2x3
P -value < 0.0001 0.5054 0.0772 0.0570 0.0125 0.0205 0.7984 0.6161
and s2 = 0.57487 with 4 degrees of freedom. So x2, x3, x1x2 and x1x3 are impor-
tant in the model.
(b) t =
y¯f\u2212y¯C\u221a
s2(1/nf+1/nC)
= 52.075\u221249.275\u221a
(0.57487)(1/8+1/4)
= 6.0306. Hence the P -value = 0.0038 for
testing quadratic curvature. It is significant.
(c) Need one additional design point different from the original ones.
15.22 (a) No.
(b) It could be as follows.
Machine
1
2
3
4
(1)
a
b
d
e
abe
bc
abc
c
bcd
abce
bce
ace
abcde
acd
cd
abcd
ac
abd
bd
ab
cde
acde
bcde
ce
bde
abde
de
be
ADE, BCD and ABCE are confounded with blocks.
(c) Partial confounding.
15.23 To estimate the quadratic terms, it might be good to add points in the middle of the
edges. Hence (\u22121, 0), (0,\u22121), (1, 0), and (0, 1) might be added.
15.24 The alias for each effect is obtained by multiplying each effect by the defining contrast
and reducing the exponents modulo 2.
A \u2261CDE, AB \u2261BCDE, BD\u2261ABCE, B \u2261ABCDE, AC \u2261DE,
AE\u2261CD, ABD\u2261BCE, E \u2261ACD, BC \u2261ABDE, ABE\u2261BCD,
15.25 (a) With BCD as the defining contrast, we have L = \u3b32+\u3b33+\u3b34. The
1
2
fraction cor-
responding to L = 0 (mod 2 is the principal block: {(1), a, bc, abc, bd, abd, cd, acd}.
(b) To obtain 2 blocks for the 1
2
fraction the interaction ABC is confounded using
L = \u3b31 + \u3b32 + \u3b33:
Solutions for Exercises in Chapter 15 247
Block 1 Block 2
(1)
bc
abd
acd
a
abc
bd
cd
(c) Using BCD as the defining contrast we have the following aliases:
A\u2261ABCD, AB\u2261ACD, B\u2261CD, AC\u2261ABD,
Since AD andABC are confounded with blocks there are only 2 degrees of freedom
for error from the unconfounded interactions.
Analysis of Variance
Source of Variation Degrees of Freedom
Blocks
A
B
C
D
Error
1
1
1
1
1
2
Total 7
15.26 With ABCD and BDEF as defining contrasts, we have
L1 = \u3b31 + \u3b32 + \u3b33 + \u3b34, L2 = \u3b32 + \u3b34 + \u3b35 + \u3b36.
The following treatment combinations give L1 = 0, L2 = 0 (mod 2) and thereby suffice
as the 1
4
fraction:
{(1), ac, bd, abcd, abe, bce, ade, abf, bcf, adf, cdf, ef, acef, bdef, abcdef}.
The third defining contrast is given by
(ABCD)(BDEF ) = AB2CD2EF = ACEF (mod 2).
The effects that are aliased with the six main effects are:
A\u2261BCD \u2261ABDEF\u2261CEF, B\u2261ACD \u2261DEF\u2261ABCEF,
C\u2261ABD \u2261BCDEF\u2261AEF, D\u2261ABC \u2261BEF\u2261ACDEF,
E\u2261ABCDE\u2261BDF \u2261ACF, F\u2261ABCDF\u2261BDE\u2261ACE.
248 Chapter 15 2k Factorial Experiments and Fractions
15.27 (a) With ABCE and ABDF , and hence (ABCE)(ABDF ) = CDEF as the defining
contrasts, we have
L1 = \u3b31 + \u3b32 + \u3b33 + \u3b35, L2 = \u3b31 + \u3b32 + \u3b34 + \u3b36.
The principal block, for which L1 = 0, and L2 = 0, is as follows:
{(1), ab, acd, bcd, ce, abce, ade, bde, acf, bcf, df, abdf, aef, bef, cdef, abcdef}.
(b) The aliases for each effect are obtained by multiplying each effect by the three
defining contrasts and reducing the exponents modulo 2.
A \u2261BCE \u2261BDF \u2261ACDEF, B \u2261ACE \u2261ADF \u2261BCDEF,
C \u2261ABE \u2261ABCDF\u2261DEF, D \u2261ABCDE\u2261ABF \u2261CEF,
E \u2261ABC \u2261ABDEF\u2261CDF, F \u2261ABCEF\u2261ABD \u2261CDE,
AB \u2261CE \u2261DF \u2261ABCDEF, AC \u2261BE \u2261BCDF\u2261ADEF,
AD \u2261BCDE\u2261BF \u2261ACEF, AE \u2261BC \u2261BDEF\u2261ACDF,
AF \u2261BCEF\u2261BD \u2261ACDE, CD \u2261ABDE \u2261ABCF \u2261EF,
DE \u2261ABCD\u2261ABEF \u2261CF, BCD\u2261ADE \u2261ACF \u2261BEF,
DCF\u2261AEF \u2261ACD \u2261BDE, .
Since E and F do not interact and all three-factor and higher interactions are
negligible, we obtain the following ANOVA table:
Source of Variation Degrees of Freedom
A
B
C
D
E
F
AB
AC
BC
BD
CD
Error
1
1
1
1
1
1
1
1
1
1
1
1
3
Total 15
15.28 The ANOVA table is shown here and the error term is computed by pooling all the
interaction effects. Factor E is the only significant effect, at level 0.05, although the
decision on factor G is marginal.
Solutions for Exercises in Chapter 15 249
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
A
B
C
D
E
F
G
Error
1
1
1
1
1
1
1
8
1.44
4.00
9.00
5.76
16.00
3.24
12.96
2.97
0.48
1.35
3.03
1.94
5.39
1.09
4.36
0.5060
0.2793
0.1199
0.2012
0.0488
0.3268
0.0701
Total 15
15.29 All two-factor interactions are aliased with each other. So, assuming that two-factor
as well as higher order interactions are negligible, a test on the main effects is given in
the ANOVA table.
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
A
B
C
D
Error
1
1
1
1
3
6.125
0.605
4.805
0.245
1.053
5.81
0.57
4.56
0.23
0.0949
0.5036
0.1223
0.6626
Total 7
Apparently no main effects is significant at level 0.05. Comparatively factors A and C
are more significant than the other two. Note that the degrees of freedom on the error
term is only 3, the test is not very powerful.
15.30 Two-factor interactions are aliased with each other. There are total 7 two-factor in-
teractions that can be estimated. Among those 7, we picked the three, which are AC,```