Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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= 13 when
round up. Also, t = 7.95\u221210.26\u221a
1.102/10+0.572/10
= \u22125.90 with a P -value < 0.0005.
Decision: Reject H0; there is a significant difference in the steel rods.
10.113 n1 = n2 = 10, x¯1 = 21.5, s1 = 5.3177, x¯2 = 28.3 and s2 = 5.8699. Since f =
s21
s22
=
0.8207 and f0.05(9, 9) = 3.18, we conclude that the two variances are equal.
sp = 5.6001 and hence t =
21.5\u221228.3
(5.6001)
\u221a
1/10+1/10
= \u22122.71 with 0.005 < P -value < 0.0075.
Decision: RejectH0; the high income neighborhood produces significantly more wastew-
ater to be treated.
10.114 n1 = n2 = 16, x¯1 = 48.1875, s1 = 4.9962, x¯2 = 43.7500 and s2 = 4.6833. Since
f =
s21
s22
= 1.1381 and f0.05(15, 15) = 2.40, we conclude that the two variances are equal.
sp = 4.8423 and hence t =
48.1875\u221243.7500
(4.8423)
\u221a
1/16+1/16
= 2.59. This is a two-sided test. Since
0.005 < P (T > 2.59) < 0.0075, we have 0.01 < P -value < 0.015.
Decision: Reject H0; there is a significant difference in the number of defects.
10.115 The hypotheses are:
H0 : µ = 24× 10\u22124 gm,
H1 : µ < 24× 10\u22124 gm.
t = 22.8\u221224
4.8/
\u221a
50
= \u22121.77 with 0.025 < P -value < 0.05. Hence, at significance level of
\u3b1 = 0.05, the mean concentration of PCB in malignant breast tissue is less than
24× 10\u22124 gm.
Chapter 11
Simple Linear Regression and
Correlation
11.1 (a)
\u2211
i
xi = 778.7,
\u2211
i
yi = 2050.0,
\u2211
i
x2i = 26, 591.63,
\u2211
i
xiyi = 65, 164.04, n = 25.
Therefore,
b =
(25)(65, 164.04)\u2212 (778.7)(2050.0)
(25)(26, 591.63)\u2212 (778.7)2 = 0.5609,
a =
2050\u2212 (0.5609)(778.7)
25
= 64.53.
(b) Using the equation y\u2c6 = 64.53 + 0.5609x with x = 30, we find y\u2c6 = 64.53 +
(0.5609)(30) = 81.40.
(c) Residuals appear to be random as desired.
10 20 30 40 50 60
\u2212
30
\u2212
20
\u2212
10
0
10
20
30
Arm Strength
R
es
id
ua
l
11.2 (a)
\u2211
i
xi = 707,
\u2211
i
yi = 658,
\u2211
i
x2i = 57, 557,
\u2211
i
xiyi = 53, 258, n = 9.
b =
(9)(53, 258)\u2212 (707)(658)
(9)(57, 557)\u2212 (707)2 = 0.7771,
a =
658\u2212 (0.7771)(707)
9
= 12.0623.
149
150 Chapter 11 Simple Linear Regression and Correlation
Hence y\u2c6 = 12.0623 + 0.7771x.
(b) For x = 85, y\u2c6 = 12.0623 + (0.7771)(85) = 78.
11.3 (a)
\u2211
i
xi = 16.5,
\u2211
i
yi = 100.4,
\u2211
i
x2i = 25.85,
\u2211
i
xiyi = 152.59, n = 11. Therefore,
b =
(11)(152.59)\u2212 (16.5)(100.4)
(11)(25.85)\u2212 (16.5)2 = 1.8091,
a =
100.4\u2212 (1.8091)(16.5)
11
= 6.4136.
Hence y\u2c6 = 6.4136 + 1.8091x
(b) For x = 1.75, y\u2c6 = 6.4136 + (1.8091)(1.75) = 9.580.
(c) Residuals appear to be random as desired.
1.0 1.2 1.4 1.6 1.8 2.0
\u2212
1.
0
\u2212
0.
5
0.
0
0.
5
1.
0
Temperature
R
es
id
ua
l
11.4 (a)
\u2211
i
xi = 311.6,
\u2211
i
yi = 297.2,
\u2211
i
x2i = 8134.26,
\u2211
i
xiyi = 7687.76, n = 12.
b =
(12)(7687.26)\u2212 (311.6)(297.2)2
=
\u2212 0.6861,
a =
297.2\u2212 (\u22120.6861)(311.6)
12
= 42.582.
Hence y\u2c6 = 42.582\u2212 0.6861x.
(b) At x = 24.5, y\u2c6 = 42.582\u2212 (0.6861)(24.5) = 25.772.
11.5 (a)
\u2211
i
xi = 675,
\u2211
i
yi = 488,
\u2211
i
x2i = 37, 125,
\u2211
i
xiyi = 25, 005, n = 18. Therefore,
b =
(18)(25, 005)\u2212 (675)(488)
(18)(37, 125)\u2212 (675)2 = 0.5676,
a =
488\u2212 (0.5676)(675)
18
= 5.8254.
Hence y\u2c6 = 5.8254 + 0.5676x
Solutions for Exercises in Chapter 11 151
(b) The scatter plot and the regression line are shown below.
0 20 40 60
10
20
30
40
50
Temperature
G
ra
m
s
y^ = 5.8254 + 0.5676x
(c) For x = 50, y\u2c6 = 5.8254 + (0.5676)(50) = 34.205 grams.
11.6 (a) The scatter plot and the regression line are shown below.
40 50 60 70 80 90
20
40
60
80
Placement Test
Co
ur
se
 G
ra
de y^ = 32.5059 + 0.4711x
(b)
\u2211
i
xi = 1110,
\u2211
i
yi = 1173,
\u2211
i
x2i = 67, 100,
\u2211
i
xiyi = 67, 690, n = 20. Therefore,
b =
(20)(67, 690)\u2212 (1110)(1173)
(20)(67, 100)\u2212 (1110)2 = 0.4711,
a =
1173\u2212 (0.4711)(1110)
20
= 32.5059.
Hence y\u2c6 = 32.5059 + 0.4711x
(c) See part (a).
(d) For y\u2c6 = 60, we solve 60 = 32.5059 + 0.4711x to obtain x = 58.466. Therefore,
students scoring below 59 should be denied admission.
11.7 (a) The scatter plot and the regression line are shown here.
152 Chapter 11 Simple Linear Regression and Correlation
20 25 30 35 40 45 50
40
0
45
0
50
0
55
0
Advertising Costs
Sa
le
s
y^ = 343.706 + 3.221x
(b)
\u2211
i
xi = 410,
\u2211
i
yi = 5445,
\u2211
i
x2i = 15, 650,
\u2211
i
xiyi = 191, 325, n = 12. Therefore,
b =
(12)(191, 325)\u2212 (410)(5445)
(12)(15, 650)\u2212 (410)2 = 3.2208,
a =
5445\u2212 (3.2208)(410)
12
= 343.7056.
Hence y\u2c6 = 343.7056 + 3.2208x
(c) When x = $35, y\u2c6 = 343.7056 + (3.2208)(35) = $456.43.
(d) Residuals appear to be random as desired.
20 25 30 35 40 45 50
\u2212
10
0
\u2212
50
0
50
Advertising Costs
R
es
id
ua
l
11.8 (a) y\u2c6 = \u22121.70 + 1.81x.
(b) x\u2c6 = (54 + 1.71)/1.81 = 30.78.
11.9 (a)
\u2211
i
xi = 45,
\u2211
i
yi = 1094,
\u2211
i
x2i = 244.26,
\u2211
i
xiyi = 5348.2, n = 9.
b =
(9)(5348.2)\u2212 (45)(1094)
(9)(244.26)\u2212 (45)2 = \u22126.3240,
a =
1094\u2212 (\u22126.3240)(45)
9
= 153.1755.
Hence y\u2c6 = 153.1755\u2212 6.3240x.
Solutions for Exercises in Chapter 11 153
(b) For x = 4.8, y\u2c6 = 153.1755\u2212 (6.3240)(4.8) = 123.
11.10 (a) z\u2c6 = cdw, ln z\u2c6 = ln c + (ln d)w; setting y\u2c6 = ln z, a = ln c, b = ln d, and y\u2c6 = a + bx,
we have
x = w 1 2 2 3 5 5
y = ln z 8.7562 8.6473 8.6570 8.5932 8.5142 8.4960\u2211
i
xi = 18,
\u2211
i
yi = 51.6639,
\u2211
i
x2i = 68,
\u2211
i
xiyi = 154.1954, n = 6.
b = ln d =
(6)(154.1954)\u2212 (18)(51.6639)
(6)(68)\u2212 (18)2 = \u22120.0569,
a = ln c =
51.6639\u2212 (\u22120.0569)(18)
6
= 8.7813.
Now c = e8.7813 = 6511.3364, d = e\u22120.0569 = 0.9447, and z\u2c6 = 6511.3364× 0.9447w.
(b) For w = 4, z\u2c6 = 6511.3364× 0.94474 = $5186.16.
11.11 (a) The scatter plot and the regression line are shown here.
1300 1400 1500 1600 1700 1800
30
00
35
00
40
00
45
00
50
00
Temperature
Th
ru
st
y^ = \u2212 1847.633 + 3.653x
(b)
\u2211
i
xi = 14, 292,
\u2211
i
yi = 35, 578,
\u2211
i
x2i = 22, 954, 054,
\u2211
i
xiyi = 57, 441, 610, n = 9.
Therefore,
b =
(9)(57, 441, 610)\u2212 (14, 292)(35, 578)
(9)(22, 954, 054)\u2212 (14, 292)2 = 3.6529,
a =
35, 578\u2212 (3.6529)(14, 292)
9
= \u22121847.69.
Hence y\u2c6 = \u22121847.69 + 3.6529x.
11.12 (a) The scatter plot and the regression line are shown here.
154 Chapter 11 Simple Linear Regression and Correlation
30 40 50 60 70
25
0
26
0
27
0
28
0
29
0
30
0
31
0
32
0
Temperature
Po
w
er
 C
on
su
m
ed
y^ = 218.255 + 1.384x
(b)
\u2211
i
xi = 401,
\u2211
i
yi = 2301,
\u2211
i
x2i = 22, 495,
\u2211
i
xiyi = 118, 652, n = 8. Therefore,
b =
(8)(118, 652)\u2212 (401)(2301))
(8)(22, 495)\u2212 (401)2 = 1.3839,
a =
2301\u2212 (1.3839)(401)
8
= 218.26.
Hence y\u2c6 = 218.26 + 1.3839x.
(c) For x = 65\u25e6F, y\u2c6 = 218.26 + (1.3839)(65) = 308.21.
11.13 (a) The scatter plot and the regression line are shown here. A simple linear model
seems suitable for the data.
30 40 50 60 70
25
0
26
0
27
0
28
0
29
0
30
0
31
0
32
0
Temperature
Po
w
er
 C
on
su
m
ed
y^ = 218.255 + 1.384x
(b)
\u2211
i
xi = 999,
\u2211
i
yi = 670,
\u2211
i
x2i = 119, 969,
\u2211
i
xiyi = 74, 058, n = 10. Therefore,
b =
(10)(74, 058)\u2212 (999)(670)
(10)(119, 969)\u2212 (999)2 = 0.3533,
a =
670\u2212 (0.3533)(999)
10
= 31.71.
Hence y\u2c6 = 31.71 + 0.3533x.
(c) See (a).
Solutions for Exercises in Chapter 11 155
11.14 From the data summary, we obtain
b =
(12)(318)\u2212 [(4)(12)][(12)(12)]
(12)(232)\u2212 [(4)(12)]2 = \u22126.45,
a = 12\u2212 (\u22126.45)(4) = 37.8.
Hence, y\u2c6 = 37.8 \u2212 6.45x. It appears that attending professional meetings would not
result in publishing more papers.
11.15 The least squares estimator A of \u3b1 is a linear combination of normally distributed
random variables and is thus normal as well.
E(A)