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# Solucionario Walpole 8 ED

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Solutions for Exercises in Chapter 8 95
\u22122 \u22121 0 1 2
70
0
80
0
90
0
10
00
11
00
12
00
13
00
Normal Q\u2212Q Plot
Theoretical Quantiles
Sa
m
pl
e
Qu
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tile
s
8.32 (a) If the two population mean drying times are truly equal, the probability that the
difference of the two sample means is 1.0 is 0.0013, which is very small. This means
that the assumption of the equality of the population means are not reasonable.
(b) If the experiment was run 10,000 times, there would be (10000)(0.0013) = 13
experiments where X¯A \u2212 X¯B would be at least 1.0.
8.33 (a) n1 = n2 = 36 and z = 0.2/
\u221a
1/36 + 1/36 = 0.85. So,
P (X¯B \u2212 X¯A \u2265 0.2) = P (Z \u2265 0.85) = 0.1977.
(b) Since the probability in (a) is not negligible, the conjecture is not true.
8.34 The normal quantile-quantile plot is shown as
\u22122 \u22121 0 1 2
6.
65
6.
70
6.
75
6.
80
Normal Q\u2212Q Plot
Theoretical Quantiles
Sa
m
pl
e
Qu
an
tile
s
96 Chapter 8 Fundamental Sampling Distributions and Data Descriptions
8.35 (a) When the population equals the limit, the probability of a sample mean exceeding
the limit would be 1/2 due the symmetry of the approximated normal distribution.
(b) P (X¯ \u2265 7960 | µ = 7950) = P (Z \u2265 (7960\u2212 7950)/(100/\u221a25)) = P (Z \u2265 0.5) =
0.3085. No, this is not very strong evidence that the population mean of the
process exceeds the government limit.
8.36 (a) \u3c3X¯A\u2212X¯B =
\u221a
52
30
+ 5
2
30
= 1.29 and z = 4\u22120
1.29
= 3.10. So,
P (X¯A \u2212 X¯B > 4 | µA = µB) = P (Z > 3.10) = 0.0010.
Such a small probability means that the difference of 4 is not likely if the two
population means are equal.
(b) Yes, the data strongly support alloy A.
8.37 Since the probability that X¯ \u2264 775 is 0.0062, given that µ = 800 is true, it suggests
that this event is very rare and it is very likely that the claim of µ = 800 is not true.
On the other hand, if µ is truly, say, 760, the probability
P (X¯ \u2264 775 | µ = 760) = P (Z \u2264 (775\u2212 760)/(40/
\u221a
16)) = P (Z \u2264 1.5) = 0.9332,
which is very high.
8.38 Define Wi = lnXi for i = 1, 2, . . . . Using the central limit theorem, Z = (W¯ \u2212
µW1)/(\u3c3W1/
\u221a
n) \u223c n(z; 0, 1). Hence W¯ follows, approximately, a normal distribution
when n is large. Since
W¯ =
1
n
n\u2211
i=1
ln(Xi) =
1
n
ln
(
n\u220f
i=1
Xi
)
=
1
n
ln(Y ),
then it is easily seen that Y follows, approximately, a lognormal distribution.
8.39 (a) 27.488.
(b) 18.475.
(c) 36.415.
8.40 (a) 16.750.
(b) 30.144.
(c) 26.217.
8.41 (a) \u3c72\u3b1 = \u3c7
2
0.99 = 0.297.
(b) \u3c72\u3b1 = \u3c7
2
0.025 = 32.852.
(c) \u3c720.05 = 37.652. Therefore, \u3b1 = 0.05\u22120.045 = 0.005. Hence, \u3c72\u3b1 = \u3c720.005 = 46.928.
8.42 (a) \u3c72\u3b1 = \u3c7
2
0.01 = 38.932.
Solutions for Exercises in Chapter 8 97
(b) \u3c72\u3b1 = \u3c7
2
0.05 = 12.592.
(c) \u3c720.01 = 23.209 and \u3c7
2
0.025 = 20.483 with \u3b1 = 0.01 + 0.015 = 0.025.
8.43 (a) P (S2 > 9.1) = P
(
(n\u22121)S2
\u3c32
> (24)(9.1)
6
)
= P (\u3c72 > 36.4) = 0.05.
(b) P (3.462 < S2 < 10.745) = P
(
(24)(3.462)
6
< (n\u22121)S
2
\u3c32
< (24)(10.745)
6
)
= P (13.848 < \u3c72 < 42.980) = 0.95\u2212 0.01 = 0.94.
8.44 \u3c72 = (19)(20)
8
= 47.5 while \u3c720.01 = 36.191. Conclusion values are not valid.
8.45 Since (n\u22121)S
2
\u3c32
is a chi-square statistic, it follows that
\u3c32(n\u22121)S2/\u3c32 =
(n\u2212 1)2
\u3c34
\u3c32S2 = 2(n\u2212 1).
Hence, \u3c32S2 =
2\u3c34
n\u22121 , which decreases as n increases.
8.46 (a) 2.145.
(b) \u22121.372.
(c) \u22123.499.
8.47 (a) P (T < 2.365) = 1\u2212 0.025 = 0.975.
(b) P (T > 1.318) = 0.10.
(c) P (T < 2.179) = 1 \u2212 0.025 = 0.975, P (T < \u22121.356) = P (T > 1.356) = 0.10.
Therefore, P (\u22121.356 < T < 2.179) = 0.975\u2212 0.010 = 0.875.
(d) P (T > \u22122.567) = 1\u2212 P (T > 2.567) = 1\u2212 0.01 = 0.99.
8.48 (a) Since t0.01 leaves an area of 0.01 to the right, and \u2212t0.005 an area of 0.005 to the
left, we find the total area to be 1 \u2212 0.01 \u2212 0.005 = 0.985 between \u2212t0.005 and
t0.01. Hence, P (\u2212t0.005 < T < t0.01) = 0.985.
(b) Since \u2212t0.025 leaves an area of 0.025 to the left, the desired area is 1\u22120.025 = 0.975.
That is, P (T > \u2212t0.025) = 0.975.
8.49 (a) From Table A.4 we note that 2.069 corresponds to t0.025 when v = 23. Therefore,
\u2212t0.025 = \u22122.069 which means that the total area under the curve to the left of
t = k is 0.025 + 0.965 = 0.990. Hence, k = t0.01 = 2.500.
(b) From Table A.4 we note that 2.807 corresponds to t0.005 when v = 23. Therefore
the total area under the curve to the right of t = k is 0.095+0.005 = 0.10. Hence,
k = t0.10 = 1.319.
(c) t0.05 = 1.714 for 23 degrees of freedom.
98 Chapter 8 Fundamental Sampling Distributions and Data Descriptions
8.50 From Table A.4 we find t0.025 = 2.131 for v = 15 degrees of freedom. Since the value
t =
27.5\u2212 30
5/4
= \u22122.00
falls between \u22122.131 and 2.131, the claim is valid.
8.51 t = (24 \u2212 20)/(4.1/3) = 2.927, t0.01 = 2.896 with 8 degrees of freedom. Conclusion:
no, µ > 20.
8.52 x¯ = 0.475, s2 = 0.0336 and t = (0.475\u2212 0.5)/0.0648 = \u22120.39. Hence
P (X¯ < 0.475) = P (T < \u22120.39) \u2248 0.35.
So, the result is inconclusive.
8.53 (a) 2.71.
(b) 3.51.
(c) 2.92.
(d) 1/2.11 = 0.47.
(e) 1/2.90 = 0.34.
8.54 s21 = 10.441 and s
2
2 = 1.846 which gives f = 5.66. Since, from Table A.6, f0.05(9, 7) =
3.68 and f0.01(9, 7) = 6.72, the probability of P (F > 5.66) should be between 0.01 and
0.05, which is quite small. Hence the variances may not be equal. Furthermore, if a
computer software can be used, the exact probability of F > 5.66 can be found 0.0162,
or if two sides are considered, P (F < 1/5.66) + P (F > 5.66) = 0.026.
8.55 s21 = 15750 and s
2
2 = 10920 which gives f = 1.44. Since, from Table A.6, f0.05(4, 5) =
5.19, the probability of F > 1.44 is much bigger than 0.05, which means that the two
variances may be considered equal. The actual probability of F > 1.44 is 0.3436 and
P (F < 1/1.44) + P (F > 1.44) = 0.7158.
8.56 The box-and-whisker plot is shown below.
5
10
15
20
Box\u2212and\u2212Whisker Plot
Solutions for Exercises in Chapter 8 99
The sample mean = 12.32 and the sample standard deviation = 6.08.
8.57 The moment-generating function for the gamma distribution is given by
MX(t) = E(e
tX) =
1
\u3b2\u3b1\u393(\u3b1)
\u222b \u221e
0
etxx\u3b1\u22121e\u2212x/\u3b2 dx
=
1
\u3b2\u3b1(1/\u3b2 \u2212 t)\u3b1
1
(1/\u3b2 \u2212 t)\u2212\u3b1\u393(\u3b1)
\u222b \u221e
0
x\u3b1\u22121e\u2212x(
1
\u3b2
\u2212t) dx
=
1
(1\u2212 \u3b2t)\u3b1
\u222b \u221e
0
x\u3b1\u22121e\u2212x/(1/\u3b2\u2212t)
\u22121
[(1/\u3b2 \u2212 t)\u22121]\u3b1\u393(\u3b1) dx =
1
(1\u2212 \u3b2t)\u3b1 ,
for t < 1/\u3b2, since the last integral is one due to the integrand being a gamma density
function. Therefore, the moment-generating function of an exponential distribution,
by substituting \u3b1 to 1, is given by MX(t) = (1\u2212 \u3b8t)\u22121. Hence, the moment-generating
function of Y can be expressed as
MY (t) =MX1(t)MX2(t) · · ·MXn(t) =
n\u220f
i=1
(1\u2212 \u3b8t)\u22121 = (1\u2212 \u3b8t)\u2212n,
which is seen to be the moment-generating function of a gamma distribution with
\u3b1 = n and \u3b2 = \u3b8.
8.58 The variance of the carbon monoxide contents is the same as the variance of the coded
measurements. That is, s2 = (15)(199.94)\u221239
2
(15)(14)
= 7.039, which results in s = 2.653.
8.59 P
(
S21
S22
< 4.89
)
= P
(
S21/\u3c3
2
S22/\u3c3
2 < 4.89
)
= P (F < 4.89) = 0.99, where F has 7 and 11
degrees of freedom.
8.60 s2 = 114, 700, 000.
8.61 LetX1 andX2 be Poisson variables with parameters \u3bb1 = 6 and \u3bb2 = 6 representing the
number of hurricanes during the first and second years, respectively. Then Y = X1+X2
has a Poisson distribution with parameter \u3bb = \u3bb1 + \u3bb2 = 12.
(a) P (Y = 15) = e
\u2212121215
15!
= 0.0724.
(b) P (Y \u2264 9) =
9\u2211
y=0
e\u22121212y
y!
= 0.2424.
8.62 Dividing each observation by 1000 and then subtracting 55 yields the following data:
\u22127, \u22122, \u221210, 6, 4, 1, 8, \u22126, \u22122, and \u22121. The variance of this coded data is
(10)(311)\u2212 (\u22129)2
(10)(9)
= 33.656.
Hence, with c = 1000, we have
s2 = (1000)2(33.656) = 33.656× 106,
and then s = 5801 kilometers.
100 Chapter 8```