Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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Computations: Subtracting 20 from each observation and discarding the zeroes. We
obtain the signs
\u2212 + + \u2212 + + \u2212 + + +
for which n = 10 and x = 7. Therefore, the P -value is
P = P (X \u2265 7 | p = 1/2) =
10\u2211
x=7
b(x; 10, 1/2)
= 1\u2212
6\u2211
x=0
b(x; 10, 1/2) = 1\u2212 0.8281 = 0.1719 > 0.05.
Decision: Do not reject H0.
16.2 The hypotheses
H0 : µ\u2dc = 12
H1 : µ\u2dc 6= 12.
\u3b1 = 0.02.
Test statistic: binomial variable X with p = 1/2.
Computations: Replacing each value above and below 12 by the symbol \u201c+\u201d and \u201c\u2212\u201d,
respectively, and discarding the two values which equal to 12. We obtain the sequence
\u2212 + \u2212 \u2212 + + + + \u2212 + + \u2212 + + \u2212 +
257
258 Chapter 16 Nonparametric Statistics
for which n = 16, x = 10 and n/2 = 8. Therefore, the P -value is
P = 2P (X \u2265 10 | p = 1/2) = 2
16\u2211
x=10
b(x; 16, 1/2)
= 2(1\u2212
9\u2211
x=0
b(x; 16, 1/2)) = 2(1\u2212 0.7728) = 0.4544 > 0.02.
Decision: Do not reject H0.
16.3 The hypotheses
H0 : µ\u2dc = 2.5
H1 : µ\u2dc 6= 2.5.
\u3b1 = 0.05.
Test statistic: binomial variable X with p = 1/2.
Computations: Replacing each value above and below 2.5 by the symbol \u201c+\u201d and \u201c\u2212\u201d,
respectively. We obtain the sequence
\u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 + + \u2212 + \u2212 \u2212 \u2212 \u2212 \u2212
for which n = 16, x = 3. Therefore, µ = np = (16)(0.5) = 8 and \u3c3 =
\u221a
(16)(0.5)(0.5) =
2. Hence z = (3.5\u2212 8)/2 = \u22122.25, and then
P = 2P (X \u2264 3 | p = 1/2) \u2248 2P (Z < \u22122.25) = (2)(0.0122) = 0.0244 < 0.05.
Decision: Reject H0.
16.4 The hypotheses
H0 : µ\u2dc1 = µ\u2dc2
H1 : µ\u2dc1 < µ\u2dc2.
\u3b1 = 0.05.
Test statistic: binomial variable X with p = 1/2.
Computations: After replacing each positive difference by a \u201c+\u201d symbol and negative
difference by a \u201c\u2212\u201d symbol, respectively, and discarding the two zero differences, we
have n = 10 and x = 2. Therefore, the P -value is
P = P (X \u2264 2 | p = 1/2) =
2\u2211
x=0
b(x; 10, 1/2) = 0.0547 > 0.05.
Decision: Do not reject H0.
Solutions for Exercises in Chapter 16 259
16.5 The hypotheses
H0 : µ\u2dc1 \u2212 µ\u2dc2 = 4.5
H1 : µ\u2dc1 \u2212 µ\u2dc2 < 4.5.
\u3b1 = 0.05.
Test statistic: binomial variable X with p = 1/2.
Computations: We have n = 10 and x = 4 plus signs. Therefore, the P -value is
P = P (X \u2264 4 | p = 1/2) =
4\u2211
x=0
b(x; 10, 1/2) = 0.3770 > 0.05.
Decision: Do not reject H0.
16.6 The hypotheses
H0 : µ\u2dcA = µ\u2dcB
H1 : µ\u2dcA 6= µ\u2dcB.
\u3b1 = 0.05.
Test statistic: binomial variable X with p = 1/2.
Computations: We have n = 14 and x = 12. Therefore, µ = np = (14)(1/2) = 7 and
\u3c3 =
\u221a
(14)(1/2)(1/2) = 1.8708. Hence, z = (11.5\u2212 7)/1.8708 = 2.41, and then
P = 2P (X \u2265 12 | p = 1/2) = 2P (Z > 2.41) = (2)(0.0080) = 0.0160 < 0.05.
Decision: Reject H0.
16.7 The hypotheses
H0 : µ\u2dc2 \u2212 µ\u2dc1 = 8
H1 : µ\u2dc2 \u2212 µ\u2dc1 < 8.
\u3b1 = 0.05.
Test statistic: binomial variable X with p = 1/2.
Computations: We have n = 13 and x = 4. Therefore, µ = np = (13)(1/2) = 6.5 and
\u3c3 =
\u221a
(13)(1/2)(1/2) = 1.803. Hence, z = (4.5\u2212 6.5)/1.803 = \u22121.11, and then
P = P (X \u2265 4 | p = 1/2) = P (Z < \u22121.11) = 0.1335 > 0.05.
Decision: Do not reject H0.
16.8 The hypotheses
H0 : µ\u2dc = 20
H1 : µ\u2dc > 20.
\u3b1 = 0.05.
Critical region: w\u226411 for n = 10.
Computations:
260 Chapter 16 Nonparametric Statistics
di \u22123 12 5 \u22125 8 5 \u22128 15 6 4
Rank 1 9 4 4 7.5 4 7.5 10 6 2
Therefore, w=12.5.
Decision: Do not reject H0.
16.9 The hypotheses
H0 : µ\u2dc = 12
H1 : µ\u2dc 6= 12.
\u3b1 = 0.02.
Critical region: w\u226420 for n = 15.
Computations:
di \u22123 1 \u22122 \u22121 6 4 1 2 \u22121 3 \u22123 1 2 \u22121 2
Rank 12 3.5 8.5 3.5 15 14 3.5 8.5 3.5 12 12 3.5 8.5 3.5 8.5
Now, w=43 and w+ = 77, so that w = 43.
Decision: Do not reject H0.
16.10 The hypotheses
H0 : µ\u2dc1 \u2212 µ\u2dc2 = 0
H1 : µ\u2dc1 \u2212 µ\u2dc2 < 0.
\u3b1 = 0.02.
Critiral region: w+ \u2264 1 for n = 5.
Computations:
Pair 1 2 3 4 5
di \u22125 \u22122 1 \u22124 2
Rank 5 2.5 1 4 2.5
Therefore, w+ = 3.5.
Decision: Do not reject H0.
16.11 The hypotheses
H0 : µ\u2dc1 \u2212 µ\u2dc2 = 4.5
H1 : µ\u2dc1 \u2212 µ\u2dc2 < 4.5.
\u3b1 = 0.05.
Critiral region: w+ \u2264 11.
Computations:
Solutions for Exercises in Chapter 16 261
Woman 1 2 3 4 5 6 7 8 9 10
di \u22121.5 5.4 3.6 6.9 5.5 2.7 2.3 3.4 5.9 0.7
di \u2212 d0 \u22126.0 0.9 \u22120.9 2.4 1.0 \u22121.8 \u22122.2 \u22121.1 1.4 \u22123.8
Rank 10 1.5 1.5 8 3 6 7 4 5 9
Therefore, w+ = 17.5.
Decision: Do not reject H0.
16.12 The hypotheses
H0 : µ\u2dcA \u2212 µ\u2dcB = 0
H1 : µ\u2dcA \u2212 µ\u2dcB > 0.
\u3b1 = 0.01.
Critiral region: z > 2.575.
Computations:
Day 1 2 3 4 5 6 7 8 9 10
di 2 6 3 5 8 \u22123 8 1 6 \u22123
Rank 4 15.5 7.5 13 19.5 7.5 19.5 1.5 15.5 7.5
Day 11 12 13 14 15 16 17 18 19 20
di 4 6 6 2 \u22124 3 7 1 \u22122 4
Rank 11 15.5 15.5 4 11 7.5 18 1.5 4 11
Now w = 180, n = 20, µW+ = (20)(21)/4 = 105, and \u3c3W+ =
\u221a
(20)(21)(41)/24 =
26.786. Therefore, z = (180\u2212 105)/26.786 = 2.80
Decision: Reject H0; on average, Pharmacy A fills more prescriptions than Pharmacy
B.
16.13 The hypotheses
H0 : µ\u2dc1 \u2212 µ\u2dc2 = 8
H1 : µ\u2dc1 \u2212 µ\u2dc2 < 8.
\u3b1 = 0.05.
Critiral region: z < \u22121.645.
Computations:
di 6 9 3 5 8 9 4 10
di \u2212 d0 \u22122 1 \u22125 \u22123 0 1 \u22124 2
Rank 4.5 1.5 10.5 7.5 \u2212 1.5 9 4.5
di 8 2 6 3 1 6 8 11
di \u2212 d0 0 \u22126 \u22122 \u22125 \u22127 \u22122 0 3
Rank \u2212 12 4.5 10.5 13 4.5 \u2212 7.5
262 Chapter 16 Nonparametric Statistics
Discarding zero differences, we have w+ = 15, n = 13, µW+ = (13)(14)/4 = 45, 5, and
\u3c3W+ =
\u221a
(13)(14)(27)/24 = 15.309. Therefore, z = (15\u2212 45.5)/14.309 = \u22122.13
Decision: Reject H0; the average increase is less than 8 points.
16.14 The hypotheses
H0 : µ\u2dcA \u2212 µ\u2dcB = 0
H1 : µ\u2dcA \u2212 µ\u2dcB 6= 0.
\u3b1 = 0.05.
Critiral region: w \u2264 21 for n = 14.
Computations:
di 0.09 0.08 0.12 0.06 0.13 \u22120.06 0.12
Rank 7 5.5 10 2.5 12 2.5 10
di 0.11 0.12 \u22120.04 0.08 0.15 0.07 0.14
Rank 8 10 1 5.5 14 4 13
Hence, w+ = 101.5, w\u2212 = 3.5, so w = 3.5.
Decision: Reject H0; the different instruments lead to different results.
16.15 The hypotheses
H0 : µ\u2dcB = µ\u2dcA
H1 : µ\u2dcB < µ\u2dcA.
\u3b1 = 0.05.
Critiral region: n1 = 3, n2 = 6 so u1 \u2264 2.
Computations:
Original data 1 7 8 9 10 11 12 13 14
Rank 1 2\u2217 3\u2217 4 5\u2217 6 7 8 9
Now w1 = 10 and hence u1 = 10\u2212 (3)(4)/2 = 4
Decision: Do not reject H0; the claim that the tar content of brand B cigarettes is
lower than that of brand A is not statistically supported.
16.16 The hypotheses
H0 : µ\u2dc1 = µ\u2dc2
H1 : µ\u2dc1 < µ\u2dc2.
\u3b1 = 0.05.
Critiral region: u1 \u2264 2.
Computations:
Solutions for Exercises in Chapter 16 263
Original data 0.5 0.9 1.4 1.9 2.1 2.8 3.1 4.6 5.3
Rank 1\u2217 2 3 4\u2217 5 6\u2217 7\u2217 8 9
Now w1 = 18 and hence u1 = 18\u2212 (4)(5)/2 = 8
Decision: Do not reject H0.
16.17 The hypotheses
H0 : µ\u2dcA = µ\u2dcB
H1 : µ\u2dcA > µ\u2dcB.
\u3b1 = 0.01.
Critiral region: u2 \u2264 14.
Computations:
Original data 3.8 4.0 4.2 4.3 4.5 4.5 4.6 4.8 4.9
Rank 1\u2217 2\u2217 3\u2217 4\u2217 5.5\u2217 5.5\u2217 7 8\u2217 9\u2217
Original Data 5.0 5.1 5.2 5.3 5.5 5.6 5.8 6.2 6.3
Rank 10 11 12 13 14 15 16 17 18
Now w2 = 50 and hence u2 = 50\u2212 (9)(10)/2 = 5
Decision: Reject H0; calculator A operates longer.
16.18 The hypotheses
H0 : µ\u2dc1 = µ\u2dc2
H1 : µ\u2dc1 6= µ\u2dc2.
\u3b1 = 0.01.
Critiral region: u \u2264 27.
Computations:
Original data 8.7 9.3 9.5 9.6 9.8 9.8 9.8 9.9 9.9 10.0
Rank 1\u2217 2 3\u2217 4 6\u2217 6\u2217 6\u2217 8.5\u2217 8.5 10
Original Data 10.1 10.4 10.5 10.7 10.8 10.9 11.0 11.2 11.5 11.8
Rank 11\u2217 12 13\u2217 14 15\u2217 16 17\u2217 18\u2217 19 20
Here \u201c\u2217\u201d is for process 2. Now w1 = 111.5 for process 1 and w2 = 98.5 for process 2.
Therefore, u1 = 111.5\u2212 (10)(11)/2 = 56.5 and u2 = 98.5\u2212 (10)(11)/2 = 43.5, so that
u = 43.5.
Decision: Do not reject H0.
16.19 The hypotheses
H0 : µ\u2dc1 = µ\u2dc2
H1 : µ\u2dc1 6= µ\u2dc2.
264 Chapter 16 Nonparametric Statistics
\u3b1 = 0.05.
Critiral region: u \u2264 5.
Computations:
Original data 64 67 69 75 78 79 80 82 87 88 91 93
Rank 1 2 3\u2217 4 5\u2217 6 7\u2217 8 9\u2217 10 11\u2217 12
Now w1 = 35 and w2 = 43. Therefore, u1 = 35\u2212(5)(6)/2 = 20 and u2 = 43\u2212(7)(8)/2 =
15, so that u = 15.
Decision: Do not reject H0.
16.20 The hypotheses
H0 : µ\u2dc1 = µ\u2dc2
H1 : µ\u2dc1 6= µ\u2dc2.
\u3b1 = 0.05.
Critiral region: Z < \u22121.96 or z > 1.96.
Computations:
Observation 12.7 13.2 13.6 13.6 14.1 14.1 14.5 14.8 15.0 15.0 15.4
Rank 1\u2217 2 3.5\u2217 3.5 5.5\u2217 5.5 7 8 9.5\u2217 9.5 11.5\u2217
Observation 15.4