Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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\u3b1 = 0.05.
Critical region: \u3c72 > 5.991 with 2 degrees of freedom.
Computation:
Observed and expected frequencies
Years Lived Widow Widower Total
Less than 5 25 (32) 39 (32) 64
5 to 10 42 (41) 40 (41) 82
More than 10 33 (26) 21 (26) 54
Total 100 100 200
\u3c72 =
(25\u2212 32)2
32
+
(39\u2212 32)2
32
+ · · ·+ (21\u2212 26)
2
26
= 5.78.
Decision: Fail to reject H0; the proportions of widows and widowers are equal with
respect to the different time period.
10.97 The hypotheses are
H0 : Proportions of household within each standard of living category are equal,
H1 : Proportions of household within each standard of living category are not equal.
\u3b1 = 0.05.
Critical region: \u3c72 > 12.592 with 6 degrees of freedom.
Computation:
Observed and expected frequencies
Period Somewhat Better Same Not as Good Total
1980: Jan.
May.
Sept.
1981: Jan.
72 (66.6)
63 (66.6)
47 (44.4)
40 (44.4)
144 (145.2)
135 (145.2)
100 (96.8)
105 (96.8)
84 (88.2)
102 (88.2)
53 (58.8)
55 (58.8)
300
300
200
200
Total 222 484 294 1000
\u3c72 =
(72\u2212 66.6)2
66.6
+
(144\u2212 145.2)2
145.2
+ · · ·+ (55\u2212 58.8)
2
58.8
= 5.92.
Decision: Fail to reject H0; proportions of household within each standard of living
category are equal.
144 Chapter 10 One- and Two-Sample Tests of Hypotheses
10.98 The hypotheses are
H0 : Proportions of voters within each attitude category are the same for each of the
three states,
H1 : Proportions of voters within each attitude category are not the same for each of
the three states.
\u3b1 = 0.05.
Critical region: \u3c72 > 9.488 with 4 degrees of freedom.
Computation:
Observed and expected frequencies
Support Do not Support Undecided Total
Indiana
Kentucky
Ohio
82 (94)
107 (94)
93 (94)
97 (79)
66 (79)
74 (79)
21 (27)
27 (27)
33 (27)
200
200
200
Total 282 237 81 600
\u3c72 =
(82\u2212 94)2
94
+
(97\u2212 79)2
79
+ · · ·+ (33\u2212 27)
2
27
= 12.56.
Decision: Reject H0; the proportions of voters within each attitude category are not
the same for each of the three states.
10.99 The hypotheses are
H0 : Proportions of voters favoring candidate A, candidate B, or undecided are the
same for each city,
H1 : Proportions of voters favoring candidate A, candidate B, or undecided are not
the same for each city.
\u3b1 = 0.05.
Critical region: \u3c72 > 5.991 with 2 degrees of freedom.
Computation:
Observed and expected frequencies
Richmond Norfolk Total
Favor A
Favor B
Undecided
204 (214.5)
211 (204.5)
85 (81)
225 (214.5)
198 (204.5)
77 (81)
429
409
162
Total 500 500 1000
Solutions for Exercises in Chapter 10 145
\u3c72 =
(204\u2212 214.5)2
214.5
+
(225\u2212 214.5)2
214.5
+ · · ·+ (77\u2212 81)
2
81
= 1.84.
Decision: Fail to reject H0; the proportions of voters favoring candidate A, candidate
B, or undecided are not the same for each city.
10.100 The hypotheses are
H0 : p1 = p2 = p3,
H1 : p1, p2, and p3 are not all equal.
\u3b1 = 0.05.
Critical region: \u3c72 > 5.991 with 2 degrees of freedom.
Computation:
Observed and expected frequencies
Denver Phoenix Rochester Total
Watch Soap Operas
Do not Watch
52 (48)
148 (152)
31 (36)
119 (114)
37 (36)
113 (114)
120
380
Total 200 150 150 500
\u3c72 =
(52\u2212 48)2
48
+
(31\u2212 36)2
36
+ · · ·+ (113\u2212 114)
2
114
= 1.39.
Decision: Fail to reject H0; no difference among the proportions.
10.101 The hypotheses are
H0 : p1 = p2,
H1 : p1 > p2.
\u3b1 = 0.01.
Critical region: z > 2.33.
Computation: p\u2c61 = 0.31, p\u2c62 = 0.24, p\u2c6 = 0.275, and
z =
0.31\u2212 0.24\u221a
(0.275)(0.725)(1/100 + 1/100)
= 1.11.
Decision: Fail to reject H0; proportions are the same.
10.102 Using paired t-test, we observe that t = 1.55 with P -value > 0.05. Hence, the data
was not sufficient to show that the oxygen consumptions was higher when there was
little or not CO.
10.103 (a) H0 : µ = 21.8, H1 : µ 6= 21.8; critical region in both tails.
(b) H0 : p = 0.2, H1 : p > 0.2; critical region in right tail.
146 Chapter 10 One- and Two-Sample Tests of Hypotheses
(c) H0 : µ = 6.2, H1 : µ > 6.2; critical region in right tail.
(d) H0 : p = 0.7, H1 : p < 0.7; critical region in left tail.
(e) H0 : p = 0.58, H1 : p 6= 0.58; critical region in both tails.
(f) H0 : µ = 340, H1 : µ < 340; critical region in left tail.
10.104 The hypotheses are
H0 : p1 = p2,
H1 : p1 > p2.
\u3b1 = 0.05.
Critical region: z > 1.645.
Computation: p\u2c61 = 0.24, p\u2c62 = 0.175, p\u2c6 = 0.203, and
z =
0.24\u2212 0.175\u221a
(0.203)(0.797)(1/300 + 1/400)
= 2.12.
Decision: Reject H0; there is statistical evidence to conclude that more Italians prefer
white champagne at weddings.
10.105 n1 = n2 = 5, x¯1 = 165.0, s1 = 6.442, x¯2 = 139.8, s2 = 12.617, and sp = 10.02. Hence
t =
165\u2212 139.8
(10.02)
\u221a
1/5 + 1/5
= 3.98.
This is a one-sided test. Therefore, 0.0025 < P -value < 0.005 with 8 degrees of
freedom. Reject H0; the speed is increased by using the facilitation tools.
10.106 (a) H0 : p = 0.2, H1 : p > 0.2; critical region in right tail.
(b) H0 : µ = 3, H1 : µ 6= 3; critical region in both tails.
(c) H0 : p = 0.15, H1 : p < 0.15; critical region in left tail.
(d) H0 : µ = $10, H1 : µ > $10; critical region in right tail.
(e) H0 : µ = 9, H1 : µ 6= 9; critical region in both tails.
10.107 The hypotheses are
H0 : p1 = p2 = p3,
H1 : p1, p2, and p3 are not all equal.
\u3b1 = 0.01.
Critical region: \u3c72 > 9.210 with 2 degrees of freedom.
Computation:
Solutions for Exercises in Chapter 10 147
Observed and expected frequencies
Distributor
Nuts 1 2 3 Total
Peanuts
Other
345 (339)
155 (161)
313 (339)
187 (161)
359 (339)
141 (161)
1017
483
Total 500 500 500 1500
\u3c72 =
(345\u2212 339)2
339
+
(313\u2212 339)2
339
+ · · ·+ (141\u2212 161)
2
161
= 10.19.
Decision: Reject H0; the proportions of peanuts for the three distributors are not equal.
10.108 The hypotheses are
H0 : p1 \u2212 p2 = 0.03,
H1 : p1 \u2212 p2 > 0.03.
Computation: p\u2c61 = 0.60 and p\u2c62 = 0.48.
z =
(0.60\u2212 0.48)\u2212 0.03\u221a
(0.60)(0.40)/200 + (0.48)(0.52)/500
= 2.18.
P -value = P (Z > 2.18) = 0.0146.
Decision: Reject H0 at level higher than 0.0146; the difference in votes favoring the
proposal exceeds 3%.
10.109 The hypotheses are
H0 : p1 = p2 = p3 = p4,
H1 : p1, p2, p3, and p4 are not all equal.
\u3b1 = 0.01.
Critical region: \u3c72 > 11.345 with 3 degrees of freedom.
Computation:
Observed and expected frequencies
Preference Maryland Virginia Georgia Alabama Total
Yes
No
65 (74)
35 (26)
71 (74)
29 (26)
78 (74)
22 (26)
82 (74)
18 (26)
296
104
Total 100 100 100 100 400
\u3c72 =
(65\u2212 74)2
74
+
(71\u2212 74)2
74
+ · · ·+ (18\u2212 26)
2
26
= 8.84.
Decision: Fail to reject H0; the proportions of parents favoring Bibles in elementary
schools are the same across states.
148 Chapter 10 One- and Two-Sample Tests of Hypotheses
10.110 d¯ = \u22122.905, sd = 3.3557, and t = d¯sd/\u221an = \u22122.12. Since 0.025 < P (T > 2.12) < 0.05
with 5 degrees of freedom, we have 0.05 < P -value < 0.10. There is no significant
change in WBC leukograms.
10.111 n1 = 15, x¯1 = 156.33, s1 = 33.09, n2 = 18, x¯2 = 170.00 and s2 = 30.79. First we do the
f -test to test equality of the variances. Since f =
s21
s22
= 1.16 and f0.05(15, 18) = 2.27,
we conclude that the two variances are equal.
To test the difference of the means, we first calculate sp = 31.85. Therefore, t =
156.33\u2212170.00
(31.85)
\u221a
1/15+1/18
= \u22121.23 with a P -value > 0.10.
Decision: H0 cannot be rejected at 0.05 level of significance.
10.112 n1 = n2 = 10, x¯1 = 7.95, s1 = 1.10, x¯2 = 10.26 and s2 = 0.57. First we do the f -test to
test equality of the variances. Since f =
s21
s22
= 3.72 and f0.05(9, 9) = 3.18, we conclude
that the two variances are not equal at level 0.10.
To test the difference of the means, we first find the degrees of freedom v