Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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for associate professors in research
institutions is $2000 higher than for those in other institutions.
10.33 The hypotheses are
H0 : µ1 \u2212 µ2 = 0.5 micromoles per 30 minutes,
H1 : µ1 \u2212 µ2 > 0.5 micromoles per 30 minutes.
\u3b1 = 0.01.
Critical region: t > 2.485 with 25 degrees of freedom.
Computation: s2p =
(14)(1.5)2+(11)(1.2)2
25
= 1.8936, and t = (8.8\u22127.5)\u22120.5\u221a
1.8936
\u221a
1/15+1/12
= 1.50. Do
not reject H0.
10.34 The hypotheses are
H0 : µ1 \u2212 µ2 = 8,
H1 : µ1 \u2212 µ2 < 8.
Computation: s2p =
(10)(4.7)2+(16)(6.1)2
26
= 31.395, and t = (85\u221279)\u22128\u221a
31.395
\u221a
1/11+1/17
= \u22120.92.
Using 28 degrees of freedom and Table A.4, we obtain that 0.15 < P -value < 0.20.
Decision: Do not reject H0.
10.35 The hypotheses are
H0 : µ1 \u2212 µ2 = 0,
H1 : µ1 \u2212 µ2 < 0.
\u3b1 = 0.05
Critical region: t < \u22121.895 with 7 degrees of freedom.
Computation: sp =
\u221a
(3)(1.363)+(4)(3.883)
7
= 1.674, and t = 2.075\u22122.860
1.674
\u221a
1/4+1/5
= \u22120.70.
Decision: Do not reject H0.
10.36 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 6= µ2.
Computation: sp =
\u221a
51002+59002
2
= 5515, and t = 37,900\u221239,800
5515
\u221a
1/12+1/12
= \u22120.84.
Using 22 degrees of freedom and since 0.20 < P (T < \u22120.84) < 0.3, we obtain 0.4 <
P -value < 0.6. Decision: Do not reject H0.
128 Chapter 10 One- and Two-Sample Tests of Hypotheses
10.37 The hypotheses are
H0 : µ1 \u2212 µ2 = 4 kilometers,
H1 : µ1 \u2212 µ2 6= 4 kilometers.
\u3b1 = 0.10 and the critical regions are t < \u22121.725 or t > 1.725 with 20 degrees of
freedom.
Computation: t = 5\u22124
(0.915)
\u221a
1/12+1/10
= 2.55.
Decision: Reject H0.
10.38 The hypotheses are
H0 : µ1 \u2212 µ2 = 8,
H1 : µ1 \u2212 µ2 < 8.
\u3b1 = 0.05 and the critical region is t < \u22121.714 with 23 degrees of freedom.
Computation: sp =
\u221a
(9)(3.2)2+(14)(2.8)2
23
= 2.963, and t = 5.5\u22128
2.963
\u221a
1/10+1/15
= \u22122.07.
Decision: Reject H0 and conclude that µ1 \u2212 µ2 < 8 months.
10.39 The hypotheses are
H0 : µII \u2212 µI = 10,
H1 : µII \u2212 µI > 10.
\u3b1 = 0.1.
Degrees of freedom is calculated as
v =
(78.8/5 + 913.333/7)2
(78.8/5)2/4 + (913/333/7)2/6
= 7.38,
hence we use 7 degrees of freedom with the critical region t > 2.998.
Computation: t = (110\u221297.4)\u221210\u221a
78.800/5+913.333/7
= 0.22.
Decision: Fail to reject H0.
10.40 The hypotheses are
H0 : µS = µN ,
H1 : µS 6= µN .
Degrees of freedom is calculated as
v =
(0.3914782/8 + 0.2144142/24)2
(0.3914782/8)2/7 + (0.2144142/24)2/23
= 8.
Computation: t = 0.97625\u22120.91583\u221a
0.3914782/8+0.2144142/24
= \u22120.42. Since 0.3 < P (T < \u22120.42) < 0.4,
we obtain 0.6 < P -value < 0.8.
Decision: Fail to reject H0.
Solutions for Exercises in Chapter 10 129
10.41 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 6= µ2.
\u3b1 = 0.05.
Degrees of freedom is calculated as
v =
(7874.3292/16 + 2479/5032/12)2
(7874.3292/16)2/15 + (2479.5032/12)2/11
= 19 degrees of freedom.
Critical regions t < \u22122.093 or t > 2.093.
Computation: t = 9897.500\u22124120.833\u221a
7874.3292/16+2479.5032/12
= 2.76.
Decision: Reject H0 and conclude that µ1 > µ2.
10.42 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 6= µ2.
\u3b1 = 0.05.
Critical regions t < \u22122.776 or t > 2.776, with 4 degrees of freedom.
Computation: d¯ = \u22120.1, sd = 0.1414, so t = \u22120.10.1414/\u221a5 = \u22121.58.
Decision: Do not reject H0 and conclude that the two methods are not significantly
different.
10.43 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 > µ2.
Computation: d¯ = 0.1417, sd = 0.198, t =
0.1417
0.198/
\u221a
12
= 2.48 and 0.015 < P -value < 0.02
with 11 degrees of freedom.
Decision: Reject H0 when a significance level is above 0.02.
10.44 The hypotheses are
H0 : µ1 \u2212 µ2 = 4.5 kilograms,
H1 : µ1 \u2212 µ2 < 4.5 kilograms.
Computation: d¯ = 3.557, sd = 2.776, t =
3.557\u22124.5
2.778/
\u221a
7
= \u22120.896, and 0.2 < P -value < 0.3
with 6 degrees of freedom.
Decision: Do not reject H0.
130 Chapter 10 One- and Two-Sample Tests of Hypotheses
10.45 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 < µ2.
Computation: d¯ = \u221254.13, sd = 83.002, t = \u221254.1383.002/\u221a15 = \u22122.53, and 0.01 < P -value <
0.015 with 14 degrees of freedom.
Decision: Reject H0.
10.46 The hypotheses are
H0 : µ1 = µ2,
H1 : µ1 6= µ2.
\u3b1 = 0.05.
Critical regions are t < \u22122.365 or t > 2.365 with 7 degrees of freedom.
Computation: d¯ = 198.625, sd = 210.165, t =
198.625
210.165/
\u221a
8
= 2.67.
Decision: Reject H0; length of storage influences sorbic acid residual concentrations.
10.47 n = (1.645+1.282)
2(0.24)2
0.32
= 5.48. The sample size needed is 6.
10.48 \u3b2 = 0.1, \u3c3 = 5.8, \u3b4 = 35.9 \u2212 40 = \u22124.1. Assume \u3b1 = 0.05 then z0.05 = 1.645,
z0.10 = 1.28. Therefore,
n =
(1.645 + 1.28)2(5.8)2
(\u22124.1)2 = 17.12 \u2248 18 due to round up.
10.49 1\u2212 \u3b2 = 0.95 so \u3b2 = 0.05, \u3b4 = 3.1 and z0.01 = 2.33. Therefore,
n =
(1.645 + 2.33)2(6.9)2
3.12
= 78.28 \u2248 79 due to round up.
10.50 \u3b2 = 0.05, \u3b4 = 8, \u3b1 = 0.05, z0.05 = 1.645, \u3c31 = 6.28 and \u3c32 = 5.61. Therefore,
n =
(1.645 + 1.645)2(6.282 + 5.612)
82
= 11.99 \u2248 12 due to round up.
10.51 n = 1.645+0.842)
2(2.25)2
[(1.2)(2.25)]2
= 4.29. The sample size would be 5.
10.52 \u3c3 = 1.25, \u3b1 = 0.05, \u3b2 = 0.1, \u3b4 = 0.5, so \u2206 = 0.5
1.25
= 0.4. Using Table A.8 we find
n = 68.
10.53 (a) The hypotheses are
H0 : Mhot \u2212Mcold = 0,
H1 : Mhot \u2212Mcold 6= 0.
Solutions for Exercises in Chapter 10 131
(b) Use paired T -test and find out t = 0.99 with 0.3 < P -value < 0.4. Hence, fail to
reject H0.
10.54 Using paired T -test, we find out t = 2.4 with 8 degrees of freedom. So, 0.02 <
P -value < 0.025. Reject H0; breathing frequency significantly higher in the presence
of CO.
10.55 The hypotheses are
H0 : p = 0.40,
H1 : p > 0.40.
Denote by X for those who choose lasagna.
P -value = P (X \u2265 9 | p = 0.40) = 0.4044.
The claim that p = 0.40 is not refuted.
10.56 The hypotheses are
H0 : p = 0.40,
H1 : p > 0.40.
\u3b1 = 0.05.
Test statistic: binomial variable X with p = 0.4 and n = 15.
Computation: x = 8 and np0 = (15)(0.4) = 6. Therefore, from Table A.1,
P -value = P (X \u2265 8 | p = 0.4) = 1\u2212 P (X \u2264 7 | p = 0.4) = 0.2131,
which is larger than 0.05.
Decision: Do not reject H0.
10.57 The hypotheses are
H0 : p = 0.5,
H1 : p < 0.5.
P -value = P (X \u2264 5 | p = 0.05) = 0.0207.
Decision: Reject H0.
10.58 The hypotheses are
H0 : p = 0.6,
H1 : p < 0.6.
So
P -value \u2248 P
(
Z <
110\u2212 (200)(0.6)\u221a
(200)(0.6)(0.4)
)
= P (Z < \u22121.44) = 0.0749.
Decision: Fail to reject H0.
132 Chapter 10 One- and Two-Sample Tests of Hypotheses
10.59 The hypotheses are
H0 : p = 0.2,
H1 : p < 0.2.
Then
P -value \u2248 P
(
Z <
136\u2212 (1000)(0.2)\u221a
(1000)(0.2)(0.8)
)
= P (Z < \u22125.06) \u2248 0.
Decision: Reject H0; less than 1/5 of the homes in the city are heated by oil.
10.60 The hypotheses are
H0 : p = 0.25,
H1 : p > 0.25.
\u3b1 = 0.05.
Computation:
P -value \u2248 P
(
Z >
28\u2212 (90)(0.25)\u221a
(90)(0.25)(0.75)
)
= P (Z > 1, 34) = 0.091.
Decision: Fail to reject H0; No sufficient evidence to conclude that p > 0.25.
10.61 The hypotheses are
H0 : p = 0.8,
H1 : p > 0.8.
\u3b1 = 0.04.
Critical region: z > 1.75.
Computation: z = 250\u2212(300)(0.8)\u221a
(300)(0.8)(0.2)
= 1.44.
Decision: Fail to reject H0; it cannot conclude that the new missile system is more
accurate.
10.62 The hypotheses are
H0 : p = 0.25,
H1 : p > 0.25.
\u3b1 = 0.05.
Critical region: z > 1.645.
Computation: z = 16\u2212(48)(0.25)\u221a
(48)(0.25)(0.75)
= 1.333.
Decision: Fail to reject H0. On the other hand, we can calculate
P -value = P (Z > 1.33) = 0.0918.
Solutions for Exercises in Chapter 10 133
10.63 The hypotheses are
H0 : p1 = p2,
H1 : p1 6= p2.
Computation: p\u2c6 = 63+59
100+125
= 0.5422, z = (63/100)\u2212(59/125)\u221a
(0.5422)(0.4578)(1/100+1/125)
= 2.36, with
P -value = 2P (Z > 2.36) = 0.0182.
Decision: Reject H0 at level 0.0182. The proportion of urban residents who favor