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# Solucionario Walpole 8 ED

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```AF , and BD, that have largest SS values and pool the other 2-way interactions to the
error term. An ANOVA can be obtained.
250 Chapter 15 2k Factorial Experiments and Fractions
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
A
B
C
D
E
F
AC
AF
BD
Error
1
1
1
1
1
1
1
1
1
6
81.54
166.54
5.64
4.41
40.20
1678.54
978.75
625.00
429.53
219.18
0.37
0.76
0.03
0.02
0.18
7.66
4.47
2.85
1.96
0.5643
0.4169
0.8778
0.8918
0.6834
0.0325
0.0790
0.1423
0.2111
Total 15
Main effect F , the location of detection, appears to be the only significant effect. The
AC interaction, which is aliased with BE, has a P -value closed to 0.05.
15.31 To get all main effects and two-way interactions in the model, this is a saturated design,
with no degrees of freedom left for error. Hence, we first get all SS of these effects and
pick the 2-way interactions with large SS values, which are AD, AE, BD and BE.
An ANOVA table is obtained.
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
A
B
C
D
E
AE
BD
BE
Error
1
1
1
1
1
1
1
1
1
6
388, 129.00
277, 202.25
4, 692.25
9, 702.25
1, 806.25
1, 406.25
462.25
1, 156.25
961.00
108.25
3, 585.49
2, 560.76
43.35
89.63
16.69
12.99
4.27
10.68
8.88
< 0.0001
< 0.0001
0.0006
< 0.0001
0.0065
0.0113
0.0843
0.0171
0.0247
Total 15
All main effects, plus AD, BD and BE two-way interactions, are significant at 0.05
level.
15.32 Consider a 24 design with letters A, B, C, and D, with design points
{(1), a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd, abcd}
. Using E = ABCD, we have the following design:
{e, a, b, c, d, abe, ace, ade, bce, bde, cde, abc, abd, acd, bcd, abcde}.
Solutions for Exercises in Chapter 15 251
15.33 Begin with a 23 with design points
{(1), a, b, c, ab, ac, bc, abc}.
Now, use the generator D = AB, E = AC, and F = BC. We have the following
result:
{def, af, be, cd, abd, ace, bcf, abcdef}.
15.34 We can use the D = AB, E = \u2212AC and F = BC as generators and obtain the result:
{df, aef, b, cde, abde, ac, bcef, abcdf}.
15.35 Here are all the aliases
B\u2261AD\u2261CF\u2261CDE\u2261AEF \u2261 \u2261ABCE \u2261BDEF \u2261ABCDF ;
D\u2261AB\u2261EF\u2261BCE\u2261ACF \u2261 \u2261BCDF \u2261ACDE \u2261ABDEF ;
E\u2261AC\u2261DF\u2261ABF \u2261BCD\u2261 \u2261ABDE \u2261BCEF \u2261ACDEF ;
F\u2261BC\u2261DE\u2261ACD\u2261ABE\u2261 \u2261ACEF \u2261ABDF \u2261BCDEF.
15.36 (a) The defining relation is ABC = \u2212I.
(b) A = \u2212BC, B = \u2212AC, and C = \u2212AB.
(c) The mean squares for A, B, and C are 1.50, 0.34, and 5.07, respectively. So,
factor C, the amount of grain refiner, appears to be most important.
(d) Low level of C.
(e) All at the \u201clow\u201d level.
(f) A hazard here is that the interactions may play significant roles. The following
are two interaction plots.
1
1
0.
5
1.
0
1.
5
2.
0
2.
5
3.
0
3.
5
4.
0
AB Interaction
A
y
2
2
\u22121 1
B
1
2
\u22121
1
1
1
0.
5
1.
0
1.
5
2.
0
2.
5
3.
0
3.
5
4.
0
BC Interaction
B
y
2
2
\u22121 1
C
1
2
\u22121
1
252 Chapter 15 2k Factorial Experiments and Fractions
15.37 When the variables are centered and scaled, the fitted model is
y\u2c6 = 12.7519 + 4.7194x1 + 0.8656x2 \u2212 1.4156x3.
The lack-of-fit test results in an f -value of 81.58 with P -value < 0.0001. Hence, higher-
order terms are needed in the model.
15.38 The ANOVA table for the regression model looks like the following.
Coefficients Degrees of Freedom
Intercept
\u3b21
\u3b22
\u3b23
\u3b24
\u3b25
Two-factor interactions
Lack of fit
Pure error
1
1
1
1
1
1
10
16
32
Total 63
15.39 The defining contrasts are
AFG, CEFG, ACDF, BEG, BDFG, CDG, BCDE, ABCDEFG, DEF, ADEG.
15.40 Begin with the basic line for N = 24; permute as described in Section 15.12 until 18
columns are formed.
15.41 The fitted model is
y\u2c6 =190, 056.67 + 181, 343.33x1 + 40, 395.00x2 + 16, 133.67x3 + 45, 593.67x4
\u2212 29, 412.33x5 + 8, 405.00x6.
The t-tests are given as
Variable t P -value
Intercept
x1
x2
x3
x4
x5
x6
4.48
4.27
0.95
0.38
1.07
\u22120.69
0.20
0.0065
0.0079
0.3852
0.7196
0.3321
0.5194
0.8509
Only x1 and x2 are significant.
Solutions for Exercises in Chapter 15 253
15.42 An ANOVA table is obtained.
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
Polymer 1
Polymer 2
Polymer 1*Polymer 2
Error
1
1
1
4
172.98
180.50
1.62
0.17
1048.36
1093.94
9.82
< 0.0001
< 0.0001
0.0351
Total 7
All main effects and interactions are significant.
15.43 An ANOVA table is obtained.
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
Mode
Type
Mode*Type
Error
1
1
1
12
2, 054.36
4, 805.96
482.90
27.67
74.25
173.71
17.45
< 0.0001
< 0.0001
0.0013
Total 15
All main effects and interactions are significant.
15.44 Two factors at two levels each can be used with three replications of the experiment,
giving 12 observations. The requirement that there must be tests on main effects
and the interactions suggests that partial confounding be used The following design is
indicated:
Block Block Block
1 2 1 2 1 2
(1)
ab
a
b
a
ab
(1)
b
(1)
a
ab
a
Rep 1 Rep 2 Rep 3
15.45 Using the contrast method and compute sums of squares, we have
Source of Variation d.f. MS f
A
B
C
D
E
F
Error
1
1
1
1
1
1
8
0.0248
0.0322
0.0234
0.0676
0.0028
0.0006
0.0201
1.24
1.61
1.17
3.37
0.14
0.03
254 Chapter 15 2k Factorial Experiments and Fractions
15.46 With the defining contrasts ABCD, CDEFG, and BDF , we have
L1 = \u3b31 + \u3b32 + \u3b33 + \u3b34,
L2 = \u3b33 + \u3b34 + \u3b35 + \u3b36 + \u3b37.
L3 = \u3b32 + \u3b34 + \u3b36.
The principal block and the remaining 7 blocks are given by
Block 1 Block 2 Block 3 Block 4
(1), eg
abcd, bdg
cdef, abcdeg
bcefg, cdfg
acg, abef
ace, abfg
a, aeg
bcd, abdg
df, abcf
acdef, bcdeg
abde, defg
abcefg, acdfg
cg, bef
ce, bfg
b, beg
acd, dg
abdf, cf
bcdef, acdeg
de, abdefg
cefg, bcdfg
abcg, aef
abce, afg
c, ceg
abd, bcdg
acdf, bf
def, abdeg
bcde, acdefg
befg, dfg
ag, abcef
ae, abcfg
Block 5 Block 6 Block 7 Block 8
d, deg
abc, bg
af, bcdf
cef, abceg
be, aefg
bcdefg, cfg
acdg, abdef
acde, abdfg
e, g
abcde, bdeg
cdf, abcdg
bcfg, cdefg
aceg, abf
ac, abefg
f, efg
abcdf, bdfg
cde, abcdefg
bceg, cdg
acfg, abe
acef, abg
ab, abeg
bdf, acf
abcdef, cdeg
acefg, abcdfg
bcg, ef
bce, fg
The two-way interactions AB \u2261 CD, AC \u2261 BD, AD \u2261 BC, BD \u2261 F , BF \u2261 D and
DF \u2261 B.
15.47 A design (where L1 = L2 = L3 = L4 = 0 (mod 2) are used) is:
{(1), abcg, abdh, abef, acdf, aceh, adeg, afgh,
bcde, bcfh, bdfg, cdgh, cefg, defh, degh, abcdefgh}
15.48 In the four defining contrasts, BCDE, ACDF , ABCG, and ABDH , the length of
interactions are all 4. Hence, it must be a resolution IV design.
15.49 Assuming three factors the design is a 23 design with 4 center runs.
15.50 (a) Consider a 23\u22121III design with ABC \u2261 I as defining contrast. Then the design
points are
Solutions for Exercises in Chapter 15 255
x1 x2 x3
\u22121
1
\u22121
1
0
0
\u22121
\u22121
1
1
0
0
\u22121
1
1
\u22121
0
0
For the noncentral design points, x¯1 = x¯2 = x¯3 = 0 and x¯
2
1 = x¯
2
2 = x¯
2
3 = 1. Hence
E(y¯f\u2212 y¯0) = \u3b20+\u3b21x¯1+\u3b22x¯2+\u3b23x¯3+\u3b211x¯21+\u3b222x¯22+\u3b233x¯23\u2212\u3b20 = \u3b211+\u3b222+\u3b233.
(b) It is learned that the test for curvature that involves y¯f \u2212 y¯0 actually is testing
the hypothesis \u3b211 + \u3b222 + \u3b233 = 0.
Chapter 16
Nonparametric Statistics
16.1 The hypotheses
H0 : µ\u2dc = 20 minutes
H1 : µ\u2dc > 20 minutes.
\u3b1 = 0.05.
Test statistic: binomial variable X with p = 1/2.```