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# Solucionario Walpole 8 ED

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```15.6 15.9 15.9 16.3 16.3 16.3 16.3 16.5 16.8 17.2
Rank 11.5 13\u2217 14.5\u2217 14.5 17.5\u2217 17.5\u2217 17.5 17.5 20 21\u2217 22
Observation 17.4 17.7 17.7 18.1 18.1 18.3 18.6 18.6 18.6 19.1 20.0
Rank 23 24.5 24.5\u2217 26.5\u2217 26.5 28 30 30\u2217 30 32 33
Now w1 = 181.5 and u1 = 181.5\u2212 (12)(13)/2 = 103.5. Then with µU1 = (21)(12)/2 =
126 and \u3c3U1 =
\u221a
(21)(12)(34)/12 = 26.721, we find z = (103.5\u2212 126)/26.721 = \u22120.84.
Decision: Do not reject H0.
16.21 The hypotheses
H0 : Operating times for all three calculators are equal.
H1 : Operating times are not all equal.
\u3b1 = 0.01.
Critiral region: h > \u3c720.01 = 9.210 with v = 2 degrees of freedom.
Computations:
Solutions for Exercises in Chapter 16 265
Ranks for Calculators
A B C
4 8.5 15
12 7 18
1 13 10
2 11 16
6 8.5 14
r1 = 25 5 17
3 r3 = 90
r2 = 56
Now h = 12
(18)(19)
[
252
5
+ 56
2
7
+ 90
2
6
]
\u2212 (3)(19) = 10.47.
Decision: Reject H0; the operating times for all three calculators are not equal.
16.22 Kruskal-Wallis test (Chi-square approximation)
h =
12
(32)(33)
[
210.52
9
+
1892
8
+
128.52
15
]
\u2212 (3)(33) = 20.21.
\u3c720.05 = 5.991 with 2 degrees of freedom. So, we reject H0 and claim that the mean
sorptions are not the same for all three solvents.
16.23 The hypotheses
H0 : Sample is random.
H1 : Sample is not random.
\u3b1 = 0.1.
Test statistics: V , the total number of runs.
Computations: for the given sequence we obtain n1 = 5, n2 = 10, and v = 7. Therefore,
from Table A.18, the P -value is
P = 2P (V \u2264 7 when H0 is true) = (2)(0.455) = 0.910 > 0.1
Decision: Do not reject H0; the sample is random.
16.24 The hypotheses
H0 : Fluctuations are random.
H1 : Fluctuations are not random.
\u3b1 = 0.05.
Test statistics: V , the total number of runs.
Computations: for the given sequence we find x\u2dc = 0.021. Replacing each measurement
266 Chapter 16 Nonparametric Statistics
by the symbol \u201c+\u201d if it falls above 0.021 and by the symbol \u201c\u2212\u201d if it falls below 0.021
and omitting the two measurements that equal 0.021, we obtain the sequence
\u2212 \u2212 \u2212 \u2212 \u2212 + + + + +
for which n1 = 5, n2 = 5, and v = 2. Therefore, the P -value is
P = 2P (V \u2264 2 when H0 is true) = (2)(0.008) = 0.016 < 0.05
Decision: Reject H0; the fluctuations are not random.
16.25 The hypotheses
H0 : µA = µB
H1 : µA > µB.
\u3b1 = 0.01.
Test statistics: V , the total number of runs.
Computations: from Exercise 16.17 we can write the sequence
B B B B B B A B B A A B A A A A A A
for which n1 = 9, n2 = 9, and v = 6. Therefore, the P -value is
P = P (V \u2264 6 when H0 is true) = 0.044 > 0.01
Decision: Do not reject H0.
16.26 The hypotheses
H0 : Defectives occur at random.
H1 : Defectives do not occur at random.
\u3b1 = 0.05.
Critical region: z < \u22121.96 or z > 1.96.
Computations: n1 = 11, n2 = 17, and v = 13. Therefore,
µV =
(2)(11)(17)
28
+ 1 = 14.357,
\u3c32V =
(2)(11)(17)[(2)(11)(17)\u2212 11\u2212 17]
(282)(27)
= 6.113,
and hence \u3c3V = 2.472. Finally,
z = (13\u2212 14.357)/2.472 = \u22120.55.
Decision: Do not reject H0.
Solutions for Exercises in Chapter 16 267
16.27 The hypotheses
H0 : Sample is random.
H1 : Sample is not random.
\u3b1 = 0.05.
Critical region: z < \u22121.96 or z > 1.96.
Computations: we find x¯ = 2.15. Assigning \u201c+\u201d and \u201c\u2212\u201d signs for observations above
and below the median, respectively, we obtain n1 = 15, n2 = 15, and v = 19. Hence,
µV =
(2)(15)(15)
30
+ 1 = 16,
\u3c32V =
(2)(15)(15)[(2)(15)(15)\u2212 15\u2212 15]
(302)(29)
= 7.241,
which yields \u3c3V = 2.691. Therefore,
z = (19\u2212 16)/2.691 = 1.11.
Decision: Do not reject H0.
16.28 1\u2212 \u3b3 = 0.95, 1\u2212 \u3b1 = 0.85. From Table A.20, n = 30.
16.29 n = 24, 1\u2212 \u3b1 = 0.90. From Table A.20, 1\u2212 \u3b3 = 0.70.
16.30 1\u2212 \u3b3 = 0.99, 1\u2212 \u3b1 = 0.80. From Table A.21, n = 21.
16.31 n = 135, 1\u2212 \u3b1 = 0.95. From Table A.21, 1\u2212 \u3b3 = 0.995.
16.32 (a) Using the computations, we have
Student Test Exam di
L.S.A. 4 4 0
W.P.B. 10 2 8
R.W.K. 7 8 \u22121
J.R.L. 2 3 \u22121
J.K.L. 5 6.5 \u22121.5
D.L.P. 9 6.5 2.5
B.L.P. 3 10 \u22127
D.W.M. 1 1 0
M.N.M. 8 9 \u22121
R.H.S. 6 5 \u22121
rS = 1\u2212 (6)(125.5)
(10)(100\u2212 1) = 0.24.
268 Chapter 16 Nonparametric Statistics
(b) The hypotheses
H0 : \u3c1 = 0
H1 : \u3c1 > 0
\u3b1 = 0.025.
Critical region: rS > 0.648.
Decision: Do not reject H0.
16.33 (a) Using the following
Ranks Ranks
x y d x y d
1 6 \u22125 14 12 2
2 1 1 15 2 13
3 16 \u221213 16 6 10
4 9.5 \u22125.5 17 13.5 3.5
5 18.5 \u221213.5 18 13.5 4.5
6 23 \u221217 19 16 3
7 8 \u22121 20 23 \u22123
8 3 5 21 23 \u22122
9 9.5 \u22120.5 22 23 \u22121
10 16 \u22126 23 18.5 4.5
11 4 7 24 23 1
12 20 \u22128 25 6 19
13 11 2
we obtain rS = 1\u2212 (6)(1586.5)(25)(625\u22121) = 0.39.
(b) The hypotheses
H0 : \u3c1 = 0
H1 : \u3c1 6= 0
\u3b1 = 0.05.
Critical region: rS < \u22120.400 or rs > 0.400.
Decision: Do not reject H0.
16.34 The numbers come up as follows
Ranks Ranks Ranks
x y d x y d x y d
3 7 \u22124 4 6 \u22122 7 3 4
6 4.5 1.5 8 2 6 5 4.5 0.5
2 8 \u22126 1 9 \u22128 9 1 8
Solutions for Exercises in Chapter 16 269
\u2211
d2 = 238.5, rS = 1\u2212 (6)(238.5)
(9)(80)
= \u22120.99.
16.35 (a) We have the following table:
Weight Chest Size di Weight Chest Size di Weight Chest Size di
3 6 \u22123 1 1 0 8 8 0
9 9 0 4 2 2 7 3 4
2 4 \u22122 6 7 \u22121 5 5 0
rS = 1\u2212 (6)(34)
(9)(80)
= 0.72.
(b) The hypotheses
H0 : \u3c1 = 0
H1 : \u3c1 > 0
\u3b1 = 0.025.
Critical region: rS > 0.683.
Decision: Reject H0 and claim \u3c1 > 0.
16.36 The hypotheses
H0 : \u3c1 = 0
H1 : \u3c1 6= 0
\u3b1 = 0.05.
Critical region: rS < \u22120.683 or rS > 0.683.
Computations:
Manufacture A B C D E F G H I
Panel rating 6 9 2 8 5 1 7 4 3
Price rank 5 1 9 8 6 7 2 4 3
di 1 8 \u22127 0 \u22121 \u22126 5 0 0
Therefore, rS = 1\u2212 (6)(176)(9)(80) = \u22120.47.
Decision: Do not reject H0.
16.37 (a)
\u2211
d2 = 24, rS = 1\u2212 (6)(24)(8)(63) = 0.71.
(b) The hypotheses
H0 : \u3c1 = 0
H1 : \u3c1 > 0
270 Chapter 16 Nonparametric Statistics
\u3b1 = 0.05.
Critical region: rS > 0.643.
Computations: rS = 0.71.
Decision: Reject H0, \u3c1 > 0.
16.38 (a)
\u2211
d2 = 1828, rS = 1\u2212 (6)(1828)(30)(899) = 0.59.
(b) The hypotheses
H0 : \u3c1 = 0
H1 : \u3c1 6= 0
\u3b1 = 0.05.
Critical region: rS < \u22120.364 or rS > 0.364.
Computations: rS = 0.59.
Decision: Reject H0, \u3c1 6= 0.
16.39 (a) The hypotheses
H0 : µA = µB
H1 : µA 6= µB
Test statistic: binomial variable X with p = 1/2.
Computations: n = 9, omitting the identical pair, so x = 3 and P -value is
P = P (X \u2264 3) = 0.2539.
Decision: Do not reject H0.
(b) w+ = 15.5, n = 9.
Decision: Do not reject H0.
16.40 The hypotheses:
H0 : µ1 = µ2 = µ3 = µ4.
H1 : At least two of the means are not equal.
\u3b1 = 0.05.
Critical region: h > \u3c720.05 = 7.815 with 3 degrees of freedom.
Computaions:
Ranks for the Laboratories
A B C D
7 18 2 12
15.5 20 3 10.5
13.5 19 4 13.5
8 9 1 15.5
6 10.5 5 17
r1 = 50 r2 = 76.5 r3 = 15 r4 = 68.5
Solutions for Exercises in Chapter 16 271
Now
h =
12
(20)(21)
[
502 + 76.52 + 152 + 68.52
5
]
\u2212 (3)(21) = 12.83.
Decision: Reject H0.
16.41 The hypotheses:
H0 : µ29 = µ54 = µ84.
H1 : At least two of the means are not equal.
Kruskal-Wallis test (Chi-squared approximation)
h =
12
(12)(13)
[
62
3
+
382
5
+
342
4
]
\u2212 (3)(13) = 6.37,
with 2 degrees of freedom. \u3c720.05 = 5.991.
Decision: rejectH0. Mean nitrogen loss is different for different levels of dietary protein.
Chapter 17
Statistical Quality Control
17.1 Let Y = X1 +X2 + · · ·+Xn. The moment generating function of a Poisson random
variable is given by MX(t) = e
µ(et\u22121). By Theorem 7.10,
MY (t) = e
µ1(et\u22121) · eµ2(et\u22121) · · · eµn(et\u22121) = e(µ1+µ2+···+µn)(et\u22121),
which we recognize as the moment generating function of a Poisson random variable
with mean and variance given by
n\u2211
i=1
µi.
17.2 The charts are shown as follows.
0 5 10 15 202.385
2.390
2.395
2.400
2.405
2.410
2.415
2.420
UCL
LCL
Sample
X\u2212
ba
r
0 5 10 15 200
0.003
0.006
0.009
0.012
0.015```