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# Solucionario Walpole 8 ED

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```9.57 n = 1600, p\u2c6 = 2
3
, q\u2c6 = 1
3
, and z0.025 = 1.96.
(a) 2
3
± (1.96)
\u221a
(2/3)(1/3)
1600
= 2
3
± 0.023, which yields 0.644 < p < 0.690.
(b) Error \u2264 (1.96)
\u221a
(2/3)(1/3)
1600
= 0.023.
9.58 n = (1.96)
2(0.32)(0.68)
(0.02)2
= 2090 when round up.
9.59 n = (2.05)
2(0.57)(0.43)
(0.02)2
= 2576 when round up.
9.60 n = (2.575)
2(0.228)(0.772)
(0.05)2
= 467 when round up.
9.61 n = (2.33)
2(0.08)(0.92)
(0.05)2
= 160 when round up.
9.62 n = (1.96)
2
(4)(0.01)2
= 9604 when round up.
9.63 n = (2.575)
2
(4)(0.01)2
= 16577 when round up.
9.64 n = (1.96)
2
(4)(0.04)2
= 601 when round up.
9.65 nM = nF = 1000, p\u2c6M = 0.250, q\u2c6M = 0.750, p\u2c6F = 0.275, q\u2c6F = 0.725, and z0.025 = 1.96.
So
(0.275\u2212 0.250)± (1.96)
\u221a
(0.250)(0.750)
1000
+
(0.275)(0.725)
1000
= 0.025± 0.039,
which yields \u22120.0136 < pF \u2212 pM < 0.0636.
Solutions for Exercises in Chapter 9 111
9.66 n1 = 250, n2 = 175, p\u2c61 =
80
250
= 0.32, p\u2c62 =
40
175
= 0.2286, and z0.05 = 1.645. So,
(0.32\u2212 0.2286)± (1.645)
\u221a
(0.32)(0.68)
250
+
(0.2286)(0.7714)
175
= 0.0914± 0.0713,
which yields 0.0201 < p1 \u2212 p2 < 0.1627. From this study we conclude that there is
a significantly higher proportion of women in electrical engineering than there is in
chemical engineering.
9.67 n1 = n2 = 500, p\u2c61 =
120
500
= 0.24, p\u2c62 =
98
500
= 0.196, and z0.05 = 1.645. So,
(0.24\u2212 0.196)± (1.645)
\u221a
(0.24)(0.76)
500
+
(0.196)(0.804)
500
= 0.044± 0.0429,
which yields 0.0011 < p1 \u2212 p2 < 0.0869. Since 0 is not in this confidence interval, we
conclude, at the level of 90% confidence, that inoculation has an effect on the incidence
of the disease.
9.68 n5\u25e6C = n15\u25e6C = 20, p\u2c65\u25e6C = 0.50, p\u2c615\u25e6C = 0.75, and z0.025 = 1.96. So,
(0.5\u2212 0.75)± (1.96)
\u221a
(0.50)(0.50)
20
+
(0.75)(0.25)
20
= \u22120.25± 0.2899,
which yields \u22120.5399 < p5\u25e6C \u2212 p15\u25e6C < 0.0399. Since this interval includes 0, the
significance of the difference cannot be shown at the confidence level of 95%.
9.69 nnow = 1000, p\u2c6now = 0.2740, n91 = 760, p\u2c691 = 0.3158, and z0.025 = 1.96. So,
(0.2740\u2212 0.3158)± (1.96)
\u221a
(0.2740)(0.7260)
1000
+
(0.3158)(0.6842)
760
= \u22120.0418± 0.0431,
which yields \u22120.0849 < pnow \u2212 p91 < 0.0013. Hence, at the confidence level of 95%,
the significance cannot be shown.
9.70 n90 = n94 = 20, p\u2c690 = 0.337, and 0\u2c694 = 0.362
(a) n90p\u2c690 = (20)(0.337) \u2248 7 and n94p\u2c694 = (20)(0.362) \u2248 7.
(b) Since z0.025 = 1.96, (0.337\u22120.362)± (1.96)
\u221a
(0.337)(0.663)
20
+ (0.362)(0.638)
20
= \u22120.025±
0.295, which yields \u22120.320 < p90 \u2212 p94 < 0.270. Hence there is no evidence, at
the confidence level of 95%, that there is a change in the proportions.
9.71 s2 = 0.815 with v = 4 degrees of freedom. Also, \u3c720.025 = 11.143 and \u3c7
2
0.975 = 0.484.
So,
(4)(0.815)
11.143
< \u3c32 <
(4)(0.815)
0.484
, which yields 0.293 < \u3c32 < 6.736.
Since this interval contains 1, the claim that \u3c32 seems valid.
112 Chapter 9 One- and Two-Sample Estimation Problems
9.72 s2 = 16 with v = 19 degrees of freedom. It is known \u3c720.01 = 36.191 and \u3c7
2
0.99 = 7.633.
Hence
(19)(16)
36.191
< \u3c32 <
(19)(16)
7.633
, or 8.400 < \u3c32 < 39.827.
9.73 s2 = 6.0025 with v = 19 degrees of freedom. Also, \u3c720.025 = 32.852 and \u3c7
2
0.975 = 8.907.
Hence,
(19)(6.0025)
32.852
< \u3c32 <
(19)(6.0025)
8.907
, or 3.472 < \u3c32 < 12.804,
9.74 s2 = 0.0006 with v = 8 degrees of freedom. Also, \u3c720.005 = 21.955 and \u3c7
2
0.995 = 1.344.
Hence,
(8)(0.0006)
21.955
< \u3c32 <
(8)(0.0006)
1.344
, or 0.00022 < \u3c32 < 0.00357.
9.75 s2 = 225 with v = 9 degrees of freedom. Also, \u3c720.005 = 23.589 and \u3c7
2
0.995 = 1.735.
Hence,
(9)(225)
23.589
< \u3c32 <
(9)(225)
1.735
, or 85.845 < \u3c32 < 1167.147,
which yields 9.27 < \u3c3 < 34.16.
9.76 s2 = 2.25 with v = 11 degrees of freedom. Also, \u3c720.05 = 19.675 and \u3c7
2
0.95 = 4.575.
Hence,
(11)(2.25)
19.675
< \u3c32 <
(11)(2.25)
4.575
, or 1.258 < \u3c32 < 5.410.
9.77 s21 = 1.00, s
2
2 = 0.64, f0.01(11, 9) = 5.19, and f0.01(9, 11) = 4.63. So,
1.00/0.64
5.19
<
\u3c321
\u3c322
< (1.00/0.64)(4.63), or 0.301 <
\u3c321
\u3c322
< 7.234,
which yields 0.549 < \u3c31
\u3c32
< 2.690.
9.78 s21 = 5000
2, s22 = 6100
2, and f0.05(11, 11) = 2.82. (Note: this value can be found by
using \u201c=finv(0.05,11,11)\u201d in Microsoft Excel.) So,(
5000
6100
)2
1
2.82
<
\u3c321
\u3c322
<
(
5000
6100
)2
(2.82), or 0.238 <
\u3c321
\u3c322
< 1.895.
Since the interval contains 1, it is reasonable to assume that \u3c321 = \u3c3
2
2 .
9.79 s2I = 76.3, s
2
II = 1035.905, f0.05(4, 6) = 4.53, and f0.05(6, 4) = 6.16. So,(
76.3
1035.905
)(
1
4.53
)
<
\u3c32I
\u3c32II
<
(
76.3
1035.905
)
(6.16), or 0.016 <
\u3c32I
\u3c32II
< 0.454.
Hence, we may assume that \u3c32I 6= \u3c32II .
Solutions for Exercises in Chapter 9 113
9.80 sA = 0.7794, sB = 0.7538, and f0.025(14, 14) = 2.98 (Note: this value can be found by
using \u201c=finv(0.025,14,14)\u201d in Microsoft Excel.) So,(
0.7794
0.7538
)2(
1
2.98
)
<
\u3c32A
\u3c32B
<
(
0.7794
0.7538
)2
(2.98), or 0.623 <
\u3c32A
\u3c32B
< 3.186.
Hence, it is reasonable to assume the equality of the variances.
9.81 The likelihood function is
L(x1, . . . , xn) =
n\u220f
i=1
f(xi; p) =
n\u220f
i=1
pxi(1\u2212 p)1\u2212xi = pnx¯(1\u2212 p)n(1\u2212x¯).
Hence, lnL = n[x¯ ln(p) + (1 \u2212 x¯) ln(1 \u2212 p)]. Taking derivative with respect to p
and setting the derivative to zero, we obtain \u2202 ln(L)
\u2202p
= n
(
x¯
p
\u2212 1\u2212x¯
1\u2212p
)
= 0, which yields
x¯
p
\u2212 1\u2212x¯
1\u2212p = 0. Therefore, p\u2c6 = x¯.
9.82 (a) The likelihood function is
L(x1, . . . , xn) =
n\u220f
i=1
f(xi;\u3b1, \u3b2) = (\u3b1\u3b2)
n
n\u220f
i=1
x\u3b2\u22121i e
\u2212\u3b1x\u3b2i
= (\u3b1\u3b2)ne
\u2212\u3b1
nP
i=1
x\u3b2i
(
n\u220f
i=1
xi
)\u3b2\u22121
.
(b) So, the log-likelihood can be expressed as
lnL = n[ln(\u3b1) + ln(\u3b2)]\u2212 \u3b1
n\u2211
i=1
x\u3b2i + (\u3b2 \u2212 1)
n\u2211
i=1
ln(xi).
To solve for the maximum likelihood estimate, we need to solve the following two
equations
\u2202 lnL
\u2202\u3b1
= 0, and
\u2202 lnL
\u2202\u3b2
= 0.
9.83 (a) The likelihood function is
L(x1, . . . , xn) =
n\u220f
i=1
f(xi;µ, \u3c3) =
n\u220f
i=1
{
1\u221a
2\u3c0\u3c3xi
e\u2212
[ln(xi)\u2212µ]
2
2\u3c32
}
=
1
(2\u3c0)n/2\u3c3n
n\u220f
i=1
xi
exp
{
\u2212 1
2\u3c32
n\u2211
i=1
[ln(xi)\u2212 µ]2
}
.
114 Chapter 9 One- and Two-Sample Estimation Problems
(b) It is easy to obtain
lnL = \u2212n
2
ln(2\u3c0)\u2212 n
2
ln(\u3c32)\u2212
n\u2211
i=1
ln(xi)\u2212 1
2\u3c32
n\u2211
i=1
[ln(xi)\u2212 µ]2.
So, setting 0 = \u2202 lnL
\u2202µ
= 1
\u3c32
n\u2211
i=1
[ln(xi) \u2212 µ], we obtain µ\u2c6 = 1n
n\u2211
i=1
ln(xi), and setting
0 = \u2202 lnL
\u2202\u3c32
= \u2212 n
2\u3c32
+ 1
2\u3c34
n\u2211
i=1
[ln(xi)\u2212 µ]2, we get \u3c3\u2c62 = 1n
n\u2211
i=1
[ln(xi)\u2212 µ\u2c6]2.
9.84 (a) The likelihood function is
L(x1, . . . , xn) =
1
\u3b2n\u3b1\u393(\u3b1)n
n\u220f
i=1
(
x\u3b1\u22121i e
\u2212xi/\u3b2) = 1
\u3b2n\u3b1\u393(\u3b1)n
(
n\u220f
i=1
xi
)\u3b1\u22121
e
\u2212
nP
i=1
(xi/\u3b2)
.
(b) Hence
lnL = \u2212n\u3b1 ln(\u3b2)\u2212 n ln(\u393(\u3b1)) + (\u3b1\u2212 1)
n\u2211
i=1
ln(xi)\u2212 1
\u3b2
n\u2211
i=1
xi.
Taking derivatives of lnL with respect to \u3b1 and \u3b2, respectively and setting both
as zeros. Then solve them to obtain the maximum likelihood estimates.
9.85 L(x) = px(1\u2212 p)1\u2212x, and lnL = x ln(p) + (1\u2212 x) ln(1\u2212 p), with \u2202 lnL
\u2202p
= x
p
\u2212 1\u2212x
1\u2212p = 0,
we obtain p\u2c6 = x = 1.
9.86 From the density function b\u2217(x; p) =
(
x\u22121
k\u22121
)
pk(1\u2212 p)x\u2212k, we obtain
lnL = ln
(
x\u2212 1
k \u2212 1
)
+ k ln p+ (n\u2212 k) ln(1\u2212 p).
Setting \u2202 lnL
\u2202p
= k
p
\u2212 n\u2212k
1\u2212p = 0, we obtain p\u2c6 =
k
n
.
9.87 For the estimator S2,
V ar(S2) =
1
(n\u2212 1)2V ar
[
n\u2211
i=1
(xi \u2212 x¯)2
]
=
1
(n\u2212 1)2V ar(\u3c3
2\u3c72n\u22121)
=
1
(n\u2212 1)2\u3c3
4[2(n\u2212 1)] = 2\u3c3
4
n\u2212 1 .
For the estimator \u3c3\u3022, we have
V ar(\u3c3\u3022) =
2\u3c34(n\u2212 1)
n2
.
Solutions for Exercises in Chapter 9 115
9.88 n = 7, d¯ = 3.557, sd = 2.776, and t0.025 = 2.447 with 6 degrees of freedom. So,
3.557± (2.447)2.776\u221a
7
= 3.557± 2.567,
which```