Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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(r \u2212 1)\u3c32 + r
\u2211
k
\u3c4 2k .
13.39 The hypotheses are
H0 : \u3c41 = \u3c42 = \u3c43 = \u3c44 = 0, professor effects are zero
H1 : At least one of the \u3c4i\u2019s is not equal to zero.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Time Periods
Courses
Professors
Error
474.50
252.50
723.50
287.50
3
3
3
6
158.17
84.17
241.17
47.92
5.03
Total 1738.00 15
P -value= 0.0446. Decision: Reject H0; grades are affected by different professors.
200 Chapter 13 One-Factor Experiments: General
13.40 The hypotheses are
H0 : \u3c4A = \u3c4B = \u3c4C = \u3c4D = \u3c4E = 0, color additive effects are zero
H1 : At least one of the \u3c4i\u2019s is not equal to zero.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Workers
Days
Additives
Error
12.4344
14.7944
3.9864
9.2712
4
4
4
12
3.1086
3.6986
0.9966
0.7726
1.29
Total 40.4864 24
P -value= 0.3280. Decision: Do not reject H0; color additives could not be shown to
have an effect on setting time.
13.41 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0, dye effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Amounts
Plants
Error
1238.8825
53.7004
101.2433
2
1
20
619.4413
53.7004
5.0622
122.37
Total 1393.8263 23
P -value< 0.0001. Decision: Reject H0; color densities of fabric differ significantly for
three levels of dyes.
13.42 (a) After a transformation g(y) =
\u221a
y, we come up with an ANOVA table as follows.
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Materials
Error
7.5123
6.2616
2
27
3.7561
0.2319
16.20
Total 13.7739 29
Solutions for Exercises in Chapter 13 201
(b) The P -value< 0.0001. Hence, there is significant difference in flaws among three
materials.
(c) A residual plot is given below and it does show some heterogeneity of the variance
among three treatment levels.
1.0 1.5 2.0 2.5 3.0
\u2212
0.
5
0.
0
0.
5
1.
0
material
re
si
du
al
(d) The purpose of the transformation is to stabilize the variances.
(e) One could be the distribution assumption itself. Once the data is transformed, it
is not necessary that the data would follow a normal distribution.
(f) Here the normal probability plot on residuals is shown.
\u22122 \u22121 0 1 2
\u2212
0.
5
0.
0
0.
5
1.
0
Theoretical Quantiles
Sa
m
pl
e 
Qu
an
tile
s
It appears to be close to a straight
line. So, it is likely that the transformed data are normally distributed.
13.43 (a) The hypotheses are
H0 : \u3c3
2
\u3b1 = 0,
H1 : \u3c3
2
\u3b1 6= 0
\u3b1 = 0.05.
Computation:
202 Chapter 13 One-Factor Experiments: General
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Operators
Error
371.8719
99.7925
3
12
123.9573
8.3160
14.91
Total 471.6644 15
P -value= 0.0002. Decision: Reject H0; operators are different.
(b) \u3c3\u2c62 = 8.316 and \u3c3\u2c62\u3b1 =
123.9573\u22128.3160
4
= 28.910.
13.44 The model is yij = µ+ Ai +Bj + \u1ebij . Hence
y¯.j = µ+ A¯. +Bj + \u1eb¯.j , and y¯.. = µ+ A¯. + B¯. + \u1eb¯...
Therefore,
SSB = k
b\u2211
j=1
(y¯.j \u2212 y¯..)2 = k
b\u2211
j=1
[(Bj \u2212 B¯.) + (\u1eb¯.j \u2212 \u1eb¯..)]2,
and
E(SSB) = k
b\u2211
j=1
E(B2j )\u2212 kbE(B¯2. ) + k
b\u2211
j=1
E(\u1eb¯2.j)\u2212 kbE(\u1eb¯2..)
= kb\u3c32\u3b2 \u2212 k\u3c32\u3b2 + b\u3c32 \u2212 \u3c32 = (b\u2212 1)\u3c32 + k(b\u2212 1)\u3c32\u3b2.
13.45 (a) The hypotheses are
H0 : \u3c3
2
\u3b1 = 0,
H1 : \u3c3
2
\u3b1 6= 0
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatments
Blocks
Error
23.238
45.283
27.937
3
4
12
7.746
11.321
2.328
3.33
Total 96.458 19
P -value= 0.0565. Decision: Not able to show a significant difference in the ran-
dom treatments at 0.05 level, although the P -value shows marginal significance.
(b) \u3c32\u3b1 =
7.746\u22122.328
5
= 1.084, and \u3c32\u3b2 =
11.321\u22122.328
4
= 2.248.
13.46 From the model
yijk = µ+ Ai +Bj + Tk + \u1ebij ,
Solutions for Exercises in Chapter 13 203
we have
y¯..k = µ+ A¯. + B¯. + Tk + \u1eb¯..k, and y¯... = µ+ A¯. + B¯. + T¯. + \u1eb¯....
Hence,
SSTr = r
\u2211
k
(y¯..k \u2212 y¯...)2 = r
\u2211
k
[(Tj \u2212 T¯.) + (\u1eb¯..k \u2212 \u1eb¯...)]2,
and
E(SSTr) = r
\u2211
k
E(T 2k )\u2212 r2E(T¯ 2. ) + r
\u2211
k
E(\u1eb¯2..k)\u2212 r2E(\u1eb¯2...)
= r2\u3c32\u3c4 \u2212 r\u3c32\u3c4 + r\u3c32 \u2212 \u3c32 = (r \u2212 1)(\u3c32 + r\u3c32\u3b2).
13.47 (a) The matrix is
A =
\uf8ee\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8f0
bk b b · · · b k k · · · k
b b 0 · · · 0 1 1 · · · 1
b 0 b · · · 0 1 1 · · · 1
...
...
...
. . .
...
...
...
. . .
...
b 0 0 · · · b 1 1 · · · 1
k 1 1 · · · k 0 0 · · · 0
k 1 1 · · · 0 k 0 · · · 0
...
...
...
. . .
...
...
...
. . .
...
k 1 1 · · · 0 0 0 · · · k
\uf8f9\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fb
,
where b = number of blocks and k = number of treatments. The vectors are
b
\u2032
= (µ, \u3b11, \u3b12, · · · , \u3b1k, \u3b21, \u3b22, · · · , \u3b2b)\u2032, and
g
\u2032
= (T.., T1., T2., · · · , Tk., T.1, T.2, · · · , T.b)\u2032 .
(b) Solving the system Ab = g with the constraints
k\u2211
i=1
\u3b1i = 0 and
b\u2211
j=1
\u3b2j = 0, we
have
µ\u2c6 = y¯..,
\u3b1\u2c6i = y¯i. \u2212 y¯.., for i = 1, 2, . . . , k,
\u3b2\u2c6j = y¯.j \u2212 y¯.., for j = 1, 2, . . . , b.
Therefore,
R(\u3b11, \u3b12, . . . , \u3b1k, \u3b21, \u3b22, . . . , \u3b2b) = b
\u2032
g \u2212 T
2
..
bk
=
k\u2211
i=1
T 2i.
b
+
b\u2211
j=1
T 2.j
k
\u2212 2T
2
..
bk
.
204 Chapter 13 One-Factor Experiments: General
To find R(\u3b21, \u3b22, . . . , \u3b2b | \u3b11, \u3b12, . . . , \u3b1k) we first find R(\u3b11, \u3b12, . . . , \u3b1k). Setting
\u3b2j = 0 in the model, we obtain the estimates (after applying the constraint
k\u2211
i=1
\u3b1i = 0)
µ\u2c6 = y¯.., and \u3b1\u2c6i = y¯i. \u2212 y¯.., for i = 1, 2, . . . , k.
The g vector is the same as in part (a) with the exception that T.1, T.2, . . . , T.b do
not appear. Thus one obtains
R(\u3b11, \u3b12, . . . , \u3b1k) =
k\u2211
i=1
T 2i.
b
\u2212 T
2
..
bk
and thus
R(\u3b21, \u3b22, . . . , \u3b2b | \u3b11, \u3b12, . . . , \u3b1k) = R(\u3b11, \u3b12, . . . , \u3b1k, \u3b21, \u3b22, . . . , \u3b2b)
\u2212 R(\u3b11, \u3b12, . . . , \u3b1k) =
b\u2211
j=1
T 2.j
k
\u2212 T
2
..
bk
= SSB.
13.48 Since
1\u2212 \u3b2 = P
[
F (3, 12) >
3.49
1 + (4)(1.5)
]
= P [F (12, 3) < 2.006] < 0.95.
Hence we do not have large enough samples. We then find, by trial and error, that
n = 16 is sufficient since
1\u2212 \u3b2 = P
[
F (3, 60) >
2.76
1 + (16)(1.5)
]
= P [F (60, 3) < 9.07] > 0.95.
13.49 We know \u3c62 = b
4\u2211
i=1
\u3b12i
4\u3c32
= b
2
, when
4\u2211
i=1
\u3b12i
\u3c32
= 2.0.
If b = 10, \u3c6 = 2.24; v1 = 3 and v2 = 27 degrees of freedom.
If b = 9, \u3c6 = 2.12; v1 = 3 and v2 = 24 degrees of freedom.
If b = 8, \u3c6 = 2.00; v1 = 3 and v2 = 21 degrees of freedom.
From Table A.16 we see that b = 9 gives the desired result.
13.50 For the randomized complete block design we have
E(S21) = E
(
SSA
k \u2212 1
)
= \u3c32 + b
k\u2211
i=1
\u3b12i
k \u2212 1 .
Therefore,
\u3bb =
v1[E(S
2
1)]
2\u3c32
\u2212 v1
2
=
(k \u2212 1)
[
\u3c32 + b
k\u2211
i=1
\u3b12i /(k \u2212 1)
]
2\u3c32
\u2212 k \u2212 1
2
= b
k\u2211
i=1
\u3b12i
2\u3c32
,
Solutions for Exercises in Chapter 13 205
and then
\u3c62 =
E(S21)\u2212 \u3c32
\u3c32
· v1
v1 + 1
=
[\u3c32 + b
k\u2211
i=1
\u3b12i /(k \u2212 1)]\u2212 \u3c32
\u3c32
· k \u2212 1
k
= b
k\u2211
i=1
\u3b12i
k\u3c32
.
13.51 (a) The model is yij = µ+ \u3b1i + \u1ebij , where \u3b1i \u223c n(0, \u3c32\u3b1).
(b) Since s2 = 0.02056 and s21 = 0.01791, we have \u3c3\u2c6
2 = 0.02056 and
s21\u2212s2
10
=
0.01791\u22120.02056
10
= \u22120.00027, which implies \u3c3\u2c62\u3b1 = 0.
13.52 (a) The P -value of the test result is 0.7830. Hence, the variance component of pour
is not significantly different from 0.
(b) We have the ANOVA table as follows:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Pours
Error
0.08906
1.02788
4
20
0.02227
0.05139
0.43
Total 1.11694 24
Since
s21\u2212s2
5
= 0.02227\u22120.05139
5
< 0, we have \u3c3\u2c62\u3b1 = 0.
13.53 (a) yij = µ+ \u3b1i + \u1ebij , where \u3b1i \u223c n(x; 0, \u3c32\u3b1).
(b) Running