Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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\u3b2 = P (6 \u2264 X \u2264 12 | p = 0.5) = 0.9963\u2212 0.1509 = 0.8454.
\u3b2 = P (6 \u2264 X \u2264 12 | p = 0.7) = 0.8732\u2212 0.0037 = 0.8695.
(c) This test procedure is not good for detecting differences of 0.1 in p.
10.5 (a) \u3b1 = P (X < 110 | p = 0.6) + P (X > 130 | p = 0.6) = P (Z < \u22121.52) + P (Z >
1.52) = 2(0.0643) = 0.1286.
(b) \u3b2 = P (110 < X < 130 | p = 0.5) = P (1.34 < Z < 4.31) = 0.0901.
\u3b2 = P (110 < X < 130 | p = 0.7) = P (\u22124.71 < Z < \u22121.47) = 0.0708.
(c) The probability of a Type I error is somewhat high for this procedure, although
Type II errors are reduced dramatically.
10.6 (a) \u3b1 = P (X \u2264 3 | p = 0.6) = 0.0548.
(b) \u3b2 = P (X > 3 | p = 0.3) = 1\u2212 0.6496 = 0.3504.
\u3b2 = P (X > 3 | p = 0.4) = 1\u2212 0.3823 = 0.6177.
\u3b2 = P (X > 3 | p = 0.5) = 1\u2212 0.1719 = 0.8281.
121
122 Chapter 10 One- and Two-Sample Tests of Hypotheses
10.7 (a) \u3b1 = P (X \u2264 24 | p = 0.6) = P (Z < \u22121.59) = 0.0559.
(b) \u3b2 = P (X > 24 | p = 0.3) = P (Z > 2.93) = 1\u2212 0.9983 = 0.0017.
\u3b2 = P (X > 24 | p = 0.4) = P (Z > 1.30) = 1\u2212 0.9032 = 0.0968.
\u3b2 = P (X > 24 | p = 0.5) = P (Z > \u22120.14) = 1\u2212 0.4443 = 0.5557.
10.8 (a) n = 12, p = 0.7, and \u3b1 = P (X > 11) = 0.0712 + 0.0138 = 0.0850.
(b) n = 12, p = 0.9, and \u3b2 = P (X \u2264 10) = 0.3410.
10.9 (a) n = 100, p = 0.7, µ = np = 70, and \u3c3 =
\u221a
npq =
\u221a
(100)(0.7)(0.3) = 4.583.
Hence z = 82.5\u221270
4.583
= 0.3410. Therefore,
\u3b1 = P (X > 82) = P (Z > 2.73) = 1\u2212 0.9968 = 0.0032.
(b) n = 100, p = 0.9, µ = np = 90, and \u3c3 =
\u221a
npq =
\u221a
(100)(0.9)(0.1) = 3. Hence
z = 82.5\u221290
3
= \u22122.5. So,
\u3b2 = P (X \u2264 82) = P (X < \u22122.5) = 0.0062.
10.10 (a) n = 7, p = 0.4, \u3b1 = P (X \u2264 2) = 0.4199.
(b) n = 7, p = 0.3, \u3b2 = P (X \u2265 3) = 1\u2212 P (X \u2264 2) = 1\u2212 0.6471 = 0.3529.
10.11 (a) n = 70, p = 0.4, µ = np = 28, and \u3c3 =
\u221a
npq = 4.099, with z = 23.5\u221228
4.099
= \u22121.10.
Then \u3b1 = P (X < 24) = P (Z < \u22121.10) = 0.1357.
(b) n = 70, p = 0.3, µ = np = 21, and \u3c3 =
\u221a
npq = 3.834, with z = 23.5\u221221
3.834
= 0.65
Then \u3b2 = P (X \u2265 24) = P (Z > 0.65) = 0.2578.
10.12 (a) n = 400, p = 0.6, µ = np = 240, and \u3c3 =
\u221a
npq = 9.798, with
z1 =
259.5\u2212 240
9.978
= 1.990, and z2 =
220.5\u2212 240
9.978
= \u22121.990.
Hence,
\u3b1 = 2P (Z < \u22121.990) = (2)(0.0233) = 0.0466.
(b) When p = 0.48, then µ = 192 and \u3c3 = 9.992, with
z1 =
220.5\u2212 192
9.992
= 2.852, and z2 =
259.5\u2212 192
9.992
= 6.755.
Therefore,
\u3b2 = P (2.852 < Z < 6.755) = 1\u2212 0.9978 = 0.0022.
10.13 From Exercise 10.12(a) we have µ = 240 and \u3c3 = 9.798. We then obtain
z1 =
214.5\u2212 240
9.978
= \u22122.60, and z2 = 265.5\u2212 240
9.978
= 2.60.
Solutions for Exercises in Chapter 10 123
So
\u3b1 = 2P (Z < \u22122.60) = (2)(0.0047) = 0.0094.
Also, from Exercise 10.12(b) we have µ = 192 and \u3c3 = 9.992, with
z1 =
214.5\u2212 192
9.992
= 2.25, and z2 =
265.5\u2212 192
9.992
= 7.36.
Therefore,
\u3b2 = P (2.25 < Z < 7.36) = 1\u2212 0.9878 = 0.0122.
10.14 (a) n = 50, µ = 15, \u3c3 = 0.5, and \u3c3X¯ =
0.5\u221a
50
= 0.071, with z = 14.9\u221215
0.071
= \u22121.41.
Hence, \u3b1 = P (Z < \u22121.41) = 0.0793.
(b) If µ = 14.8, z = 14.9\u221214.8
0.071
= 1.41. So, \u3b2 = P (Z > 1.41) = 0.0793.
If µ = 14.9, then z = 0 and \u3b2 = P (Z > 0) = 0.5.
10.15 (a) µ = 200, n = 9, \u3c3 = 15 and \u3c3X¯ =
15
3
= 5. So,
z1 =
191\u2212 200
5
= \u22121.8, and z2 = 209\u2212 200
5
= 1.8,
with \u3b1 = 2P (Z < \u22121.8) = (2)(0.0359) = 0.0718.
(b) If µ = 215, then z \u2212 1 = 191\u2212215
5
= \u22124.8 and z2 = 209\u22122155 = \u22121.2, with
\u3b2 = P (\u22124.8 < Z < \u22121.2) = 0.1151\u2212 0 = 0.1151.
10.16 (a) When n = 15, then \u3c3X¯ =
15
5
= 3, with µ = 200 and n = 25. Hence
z1 =
191\u2212 200
3
= \u22123, and z2 = 209\u2212 200
3
= 3,
with \u3b1 = 2P (Z < \u22123) = (2)(0.0013) = 0.0026.
(b) When µ = 215, then z \u2212 1 = 191\u2212215
3
= \u22128 and z2 = 209\u22122153 = \u22122, with
\u3b2 = P (\u22128 < Z < \u22122) = 0.0228\u2212 0 = 0.0228.
10.17 (a) n = 50, µ = 5000, \u3c3 = 120, and \u3c3X¯ =
120\u221a
50
= 16.971, with z = 4970\u22125000
16.971
= \u22121.77
and \u3b1 = P (Z < \u22121.77) = 0.0384.
(b) If µ = 4970, then z = 0 and hence \u3b2 = P (Z > 0) = 0.5.
If µ = 4960, then z = 4970\u22124960
16.971
= 0.59 and \u3b2 = P (Z > 0.59) = 0.2776.
10.18 The OC curve is shown next.
124 Chapter 10 One- and Two-Sample Tests of Hypotheses
180 190 200 210 220
0.
0
0.
2
0.
4
0.
6
0.
8
OC curve
µ
Pr
ob
ab
ilit
y 
of
 a
cc
ep
tin
g 
th
e 
nu
ll h
yp
ot
he
sis
10.19 The hypotheses are
H0 : µ = 800,
H1 : µ 6= 800.
Now, z = 788\u2212800
40/
\u221a
30
= \u22121.64, and P -value= 2P (Z < \u22121.64) = (2)(0.0505) = 0.1010.
Hence, the mean is not significantly different from 800 for \u3b1 < 0.101.
10.20 The hypotheses are
H0 : µ = 5.5,
H1 : µ < 5.5.
Now, z = 5.23\u22125.5
0.24/
\u221a
64
= \u22129.0, and P -value= P (Z < \u22129.0) \u2248 0. The White Cheddar
Popcorn, on average, weighs less than 5.5oz.
10.21 The hypotheses are
H0 : µ = 40 months,
H1 : µ < 40 months.
Now, z = 38\u221240
5.8/
\u221a
64
= \u22122.76, and P -value= P (Z < \u22122.76) = 0.0029. Decision: reject
H0.
10.22 The hypotheses are
H0 : µ = 162.5 centimeters,
H1 : µ 6= 162.5 centimeters.
Now, z = 165.2\u2212162.5
6.9/
\u221a
50
= 2.77, and P -value= 2P (Z > 2.77) = (2)(0.0028) = 0.0056.
Decision: reject H0 and conclude that µ 6= 162.5.
Solutions for Exercises in Chapter 10 125
10.23 The hypotheses are
H0 : µ = 20, 000 kilometers,
H1 : µ > 20, 000 kilometers.
Now, z = 23,500\u221220,000
3900/
\u221a
100
= 8.97, and P -value= P (Z > 8.97) \u2248 0. Decision: reject H0 and
conclude that µ 6= 20, 000 kilometers.
10.24 The hypotheses are
H0 : µ = 8,
H1 : µ > 8.
Now, z = 8.5\u22128
2.25/
\u221a
225
= 3.33, and P -value= P (Z > 3.33) = 0.0004. Decision: Reject H0
and conclude that men who use TM, on average, mediate more than 8 hours per week.
10.25 The hypotheses are
H0 : µ = 10,
H1 : µ 6= 10.
\u3b1 = 0.01 and df = 9.
Critical region: t < \u22123.25 or t > 3.25.
Computation: t = 10.06\u221210
0.246/
\u221a
10
= 0.77.
Decision: Fail to reject H0.
10.26 The hypotheses are
H0 : µ = 220 milligrams,
H1 : µ > 220 milligrams.
\u3b1 = 0.01 and df = 9.
Critical region: t > 1.729.
Computation: t = 224\u2212220
24.5/
\u221a
20
= 4.38.
Decision: Reject H0 and claim µ > 220 milligrams.
10.27 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 > µ2.
Since sp =
\u221a
(29)(10.5)2+(29)(10.2)2
58
= 10.35, then
P
[
T >
34.0
10.35
\u221a
1/30 + 1/30
]
= P (Z > 12.72) \u2248 0.
Hence, the conclusion is that running increases the mean RMR in older women.
126 Chapter 10 One- and Two-Sample Tests of Hypotheses
10.28 The hypotheses are
H0 : µC = µA,
H1 : µC > µA,
with sp =
\u221a
(24)(1.5)2+(24)(1.25)2
48
= 1.3807. We obtain t = 20.0\u221212.0
1.3807
\u221a
2/25
= 20.48. Since
P (T > 20.48) \u2248 0, we conclude that the mean percent absorbency for the cotton fiber
is significantly higher than the mean percent absorbency for acetate.
10.29 The hypotheses are
H0 : µ = 35 minutes,
H1 : µ < 35 minutes.
\u3b1 = 0.05 and df = 19.
Critical region: t < \u22121.729.
Computation: t = 33.1\u221235
4.3/
\u221a
20
= \u22121.98.
Decision: Reject H0 and conclude that it takes less than 35 minutes, on the average,
to take the test.
10.30 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 6= µ2.
Since the variances are known, we obtain z = 81\u221276\u221a
5.22/25+3.52/36
= 4.22. So, P -value\u2248 0
and we conclude that µ1 > µ2.
10.31 The hypotheses are
H0 : µA \u2212 µB = 12 kilograms,
H1 : µA \u2212 µB > 12 kilograms.
\u3b1 = 0.05.
Critical region: z > 1.645.
Computation: z = (86.7\u221277.8)\u221212\u221a
(6.28)2/50+(5.61)2/50
= \u22122.60. So, fail to reject H0 and conclude that
the average tensile strength of thread A does not exceed the average tensile strength
of thread B by 12 kilograms.
10.32 The hypotheses are
H0 : µ1 \u2212 µ2 = $2, 000,
H1 : µ1 \u2212 µ2 > $2, 000.
Solutions for Exercises in Chapter 10 127
\u3b1 = 0.01.
Critical region: z > 2.33.
Computation: z = (70750\u221265200)\u22122000\u221a
(6000)2/200+(5000)2/200
= 6.43, with a P -value= P (Z > 6.43) \u2248
0. Reject H0 and conclude that the mean salary