Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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an ANOVA analysis, we obtain the P -value as 0.0121. Hence, the loom
variance component is significantly different from 0 at level 0.05.
(c) The suspicion is supported by the data.
13.54 The hypotheses are
H0 : µ1 = µ2 = µ3 = µ4,
H1 : At least two of the µi\u2019s are not equal.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Garlon levels
Error
3.7289
9.5213
3
12
1.2430
0.7934
1.57
Total 13.2502 15
P -value= 0.2487. Decision: Do not reject H0; there is insufficient evidence to claim
that the concentration levels of Garlon would impact the heights of shoots.
206 Chapter 13 One-Factor Experiments: General
13.55 Bartlett\u2019s statistic is b = 0.8254. Conclusion: do not reject homogeneous variance
assumption.
13.56 The hypotheses are
H0 : \u3c4A = \u3c4B = \u3c4C = \u3c4D = \u3c4E = 0, ration effects are zero
H1 : At least one of the \u3c4i\u2019s is not zero.
\u3b1 = 0.01.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Lactation Periods
Cows
Rations
Error
245.8224
353.1224
89.2624
21.3392
4
4
4
12
61.4556
88.2806
22.3156
1.7783
12.55
Total 709.5464 24
P -value= 0.0003. Decision: Reject H0; different rations have an effect on the daily
milk production.
13.57 It can be shown that y¯C = 76.16, y¯1 = 81.20, y¯2 = 81.54 and y¯3 = 80.98. Since this is
a one-sided test, we find d0.01(3, 16) = 3.05 and\u221a
2s2
n
=
\u221a
(2)(3.52575)
5
= 1.18756.
Hence
d1 =
81.20\u2212 76.16
1.18756
= 4.244, d2 =
81.54\u2212 76.16
1.18756
= 4.532, d3 =
80.98\u2212 76.16
1.18756
= 4.059,
which are all larger than the critical value. Hence, significantly higher yields are
obtained with the catalysts than with no catalyst.
13.58 (a) The hypotheses for the Bartlett\u2019s test are
H0 : \u3c3
2
A = \u3c3
2
B = \u3c3
2
C = \u3c3
2
D,
H1 : The variances are not all equal.
\u3b1 = 0.05.
Critical region: We have n1 = n2 = n3 = n4 = 5, N = 20, and k = 4. Therefore,
we reject H0 when b < b4(0.05, 5) = 0.5850.
Computation: sA = 1.40819, sB = 2.16056, sC = 1.16276, sD = 0.76942 and
hence sp = 1.46586. From these, we can obtain that b = 0.7678.
Decision: Do not reject H0; there is no sufficient evidence to conclude that the
variances are not equal.
Solutions for Exercises in Chapter 13 207
(b) The hypotheses are
H0 : µA = µB = µC = µD,
H1 : At least two of the µi\u2019s are not equal.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Laboratories
Error
85.9255
34.3800
3
16
28.6418
2.1488
13.33
Total 120.3055 19
P -value= 0.0001. Decision: Reject H0; the laboratory means are significantly
different.
(c) The normal probability plot is given as follows:
\u22122 \u22121 0 1 2
\u2212
2
\u2212
1
0
1
2
Theoretical Quantiles
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13.59 The hypotheses for the Bartlett\u2019s test are
H0 : \u3c3
2
1 = \u3c3
2
2 = \u3c3
2
3 = \u3c3
2
4 ,
H1 : The variances are not all equal.
\u3b1 = 0.01.
Critical region: We have n1 = n2 = n3 = 4, n4 = 9 N = 21, and k = 4. Therefore, we
reject H0 when
b < b4(0.01, 4, 4, 4, 9)
=
(4)(0.3475) + (4)(0.3475) + (4)(0.3475) + (9)(0.6892)
21
= 0.4939.
Computation: s21 = 0.41709, s
2
2 = 0.93857, s
2
3 = 0.25673, s
2
4 = 1.72451 and hence
s2p = 1.0962. Therefore,
b =
[(0.41709)3(0.93857)3(0.25763)3(1.72451)8]1/17
1.0962
= 0.79.
208 Chapter 13 One-Factor Experiments: General
Decision: Do not reject H0; the variances are not significantly different.
13.60 The hypotheses for the Cochran\u2019s test are
H0 : \u3c3
2
A = \u3c3
2
B = \u3c3
2
C ,
H1 : The variances are not all equal.
\u3b1 = 0.01.
Critical region: g > 0.6912.
Computation: s2A = 29.5667, s
2
B = 10.8889, s
2
C = 3.2111, and hence
\u2211
s2i = 43.6667.
Now, g = 29.5667
43.6667
= 0.6771.
Decision: Do not reject H0; the variances are not significantly different.
13.61 The hypotheses for the Bartlett\u2019s test are
H0 : \u3c3
2
1 = \u3c3
2
2 = \u3c3
2
3,
H1 : The variances are not all equal.
\u3b1 = 0.05.
Critical region: reject H0 when
b < b4(0.05, 9, 8, 15) =
(9)(0.7686) + (8)(0.7387) + (15)(0.8632)
32
= 0.8055.
Computation: b = [(0.02832)
8(0.16077)7(0.04310)14 ]1/29
0.067426
= 0.7822.
Decision: Reject H0; the variances are significantly different.
13.62 (a) The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = 0,
H1 : At least one of the \u3b1i\u2019s is not zero.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Diets
Blocks
Error
822.1360
17.1038
106.3597
3
5
15
274.0453
3.4208
7.0906
38.65
Total 945.5995 23
with P -value< 0.0001.
Decision: Reject H0; diets do have a significant effect on mean percent dry matter.
Solutions for Exercises in Chapter 13 209
(b) We know that y¯C = 35.8483, y¯F = 36.4217, y¯T = 45.1833, y¯A = 49.6250, and\u221a
2s2
n
=
\u221a
(2)(7.0906)
6
= 1.5374.
Hence,
dAmmonia =
49.6250\u2212 35.8483
1.5374
= 8.961,
dUrea Feeding =
36.4217\u2212 35.8483
1.5374
= 0.3730,
dUrea Treated =
45.1833\u2212 35.8483
1.5374
= 6.0719.
Using the critical value d0.05(3, 15) = 2.61, we obtain that only \u201cUrea Feeding\u201d is
not significantly different from the control, at level 0.05.
(c) The normal probability plot for the residuals are given below.
\u22122 \u22121 0 1 2
\u2212
4
\u2212
2
0
2
4
Theoretical Quantiles
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13.63 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0,
H1 : At least one of the \u3b1i\u2019s is not zero.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Diet
Error
0.32356
0.20808
2
12
0.16178
0.01734
9.33
Total 0.53164 14
210 Chapter 13 One-Factor Experiments: General
with P -value= 0.0036.
Decision: Reject H0; zinc is significantly different among the diets.
13.64 (a) The gasoline manufacturers would want to apply their results to more than one
model of car.
(b) Yes, there is a significant difference in the miles per gallon for the three brands
of gasoline.
(c) I would choose brand C for the best miles per gallon.
13.65 (a) The process would include more than one stamping machine and the results might
differ with different machines.
(b) The mean plot is shown below.
1
1
1
3.
2
3.
4
3.
6
3.
8
4.
0
4.
2
4.
4
Material
N
um
be
r o
f G
as
ke
ts
2
2
2
cork plastic rubber
(c) Material 1 appears to be the best.
(d) Yes, there is interaction. Materials 1 and 3 have better results with machine 1
but material 2 has better results with machine 2.
13.66 (a) The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0,
H1 : At least one of the \u3b1i\u2019s is not zero.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Paint Types
Error
227875.11
336361.83
2
15
113937.57
22424.12
5.08
Total 564236.94 17
with P -value= 0.0207.
Decision: Reject H0 at level 0.05; the average wearing quality differs significantly
for three paints.
(b) Using Tukey\u2019s test, it turns out the following.
Solutions for Exercises in Chapter 13 211
y¯1. y¯3. y¯2.
197.83 419.50 450.50
Types 2 and 3 are not significantly different, while Type 1 is significantly different
from Type 2.
(c) We plot the residual plot and the normal probability plot for the residuals as
follows.
1.0 1.5 2.0 2.5 3.0
\u2212
20
0
\u2212
10
0
0
10
0
20
0
Type
R
es
id
ua
l
\u22122 \u22121 0 1 2
\u2212
20
0
\u2212
10
0
0
10
0
20
0
Theoretical Quantiles
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It appears that the heterogeneity in variances may be violated, as is the normality
assumption.
(d) We do a log transformation of the data, i.e., y
\u2032
= log(y). The ANOVA result