Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications
Подождите немного. Документ загружается.
c. Of those with Master Card, .625 is the proportion
who also have a Visa Card.
d. Of those with Master Card, .375 is the proportion
who do not have a Visa Card.
e. Of those with at least one of the two cards, .769 is the
proportion who have a Visa card.
49. .217, .178
51. .436, .582
53. .0833
59. a. .067 b. .509
61. .287
63. a. 76.5% b. .235
65. .466, .288, .247
67. a. Because of independence, the conditional probability
is the same as the unconditional probability, .3.
b. .82 c. .146
71. .349, .651, (1 p)
n
,1 (1 p)
n
73. .99999969, .226
75. .9981
77. Yes, no
79. a. 2p p
2
b. 1 (1 p)
n
c. (1 p)
3
d. .9 + .1(1 p)
3
e. .0137
81. .8588, .9896
83. 2p(1 p)
85. a. 1/3, .444 b. .15 c. .291
87. .45, .32
89. a. 1/120 b. 1/5 c. 1/5
91. .9046
93. a. .904 b. .766
95. .008
97. .362, .348, .290
99. a. P(G | R
1
< R
2
< R
3
) ¼ 2/3, so classify as granite if
R
1
< R
2
< R
3
.
b. P(G | R
1
< R
3
< R
2
) ¼ .294, so classify as basalt if
R
1
< R
3
< R
2
.
P(G | R
3
< R
1
< R
2
) ¼ 1/15, so classify as basalt if
R
3
< R
1
< R
2
.
c. .175 d. p > 14/17
101. a. 1/24 b. 3/8
103. s ¼ 1
107. a. P(B
0
|survive) ¼ b
0
/[1 (b
1
+ b
2
)cd]
P(B
1
|survive) ¼ b
1
(1 cd)/[1 (b
1
+ b
2
)cd]
P(B
2
|survive) ¼ b
2
(1 cd)/[1 (b
1
+ b
2
)cd]
b. .712, .058, .231
Chapter 3
1.
S : FFF SFF FSF FFS FSS SFS SSF SSS
X:01112223
3. M ¼ the absolute value of the difference between the
outcomes, with possible values 0, 1, 2, 3, 4, 5 or 6;
W ¼ 1 if the sum of the two resulting numbers is even
and W ¼ 0 otherwise, a Bernoulli random variable.
5. No, X can be a Bernoulli random variable where a
success is an outcome in B, with B a particular subset
of the sample space.
7. a. Possible values are 0, 1, 2, ..., 12; discrete
b. With N ¼ # on the list, values are 0, 1, 2, ... , N;
discrete
c. Possible values are 1, 2, 3, 4, ...; discrete
d. { x:0 < x < 1 } if we assume that a rattlesnake can
be arbitrarily short or long; not discrete
e. With c ¼ amount earned per book sold, possible
values are 0, c,2c,3c, ... , 10,000c; discrete
f. { y:0 y 14} since 0 is the smallest possible pH
and 14 is the largest possible pH; not discrete
g. With m and M denoting the minimum and maximum
possible tension, respectively, possible values are {
x: m x M }; not discrete
h. Possible values are 3, 6, 9, 12, 15, ... — i.e., 3(1),
3(2), 3(3), 3(4), ...giving a first element, etc,; discrete
9. a. X is a discrete random variable with possible values
{2, 4, 6, 8, ...
}
b. X is a discrete random variable with possible values
{2, 3, 4, 5, ...}
11. a. p(4) ¼ .10 c. .45, .25
13. a. .70 b. .45 c. .55 d. .71 e. .65 f. .45
15. a. (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5)
b. p(0) ¼ .3, p(1) ¼ .6, p(2) ¼ .1, p(x) ¼ 0 otherwise
c. F(0) ¼ .30, F(1) ¼ .90, F(2) ¼ 1. The c.d.f. is
FðxÞ¼
0 x<0
:30 0 x<1
:90 1 x<2
12 x
8
>
>
<
>
>
:
17. a. .81 b. .162
c. The fifth battery must be an A, and one of the first
four must also be an A,so
p
(5) ¼ P(AUUUA or UAUUA or UUAUA or
UUUAA) ¼ .00324
d. P(Y ¼ y) ¼ (y 1)(.1)
y2
(.9)
2
, y ¼ 2,3,4,5,...
19. c. F(x) ¼ 0, x < 1, F(x) ¼ log
10
([x] + 1), 1 x 9,
F(x) ¼ 1, x > 9.
d. .602, .301
21. F(x) ¼ 0, x < 0; .10, 0 x < 1; .25, 1 x < 2; .45,
2 x < 3; .70, 3 x < 4; .90, 4 x < 5; .96,
5 x < 6; 1.00, 6 x
23. a. p(1) ¼ .30, p(3) ¼ .10, p(4) ¼ .05, p(6) ¼ .15,
p(12) ¼ .40
b. .30, .60
25. a. p(x) ¼ (1/3)(2/3)
x1
, x ¼ 1, 2, 3, ...
b. p(y) ¼ (1/3)(2/3)
y2
, y ¼ 2, 3, 4, ...
c. p(0) ¼ 1/6, p(z) ¼ (25/54)(4/9)
z1
, z ¼ 1, 2, 3, 4, ...
818 Chapter 3
29. a. .60 b. $110
31. a. 16.38, 272.298, 3.9936 b. 401 c. 2496 d. 13.66
33. Yes, because S(1/x
2
) is finite.
35. $700
37. E[h(X)] ¼ .408 > 1/3.5 ¼ .286, so you expect to win
more if you gamble.
39. V(X) ¼ V(X)
41. a. 32.5 b. 7.5
c. V(X) ¼ E[X(X–1)] + E(X) [E(X)]
2
43. a. 1/4, 1/9, 1/16, 1/25, 1/100
b. m ¼ 2.64, s ¼ 1.54, P(|X m| 2s) ¼ .04 < .25,
P(|X m| 3s) ¼ 0 < 1/9
The actual probability can be far below the Chebyshev
bound, so the bound is conservative.
c. 1/9, equal to the Chebyshev bound
d. P(1) ¼ .02, P(0) ¼ .96, P(1) ¼ .02
45. M
X
(t) ¼ .5e
t
/(1–.5e
t
), E(X) ¼ 2, V(X) ¼ 2
47. p
Y
(y) ¼ .75(.25)
y1
, y ¼ 1, 2, 3, ...
49. E(X) ¼ 5, V(X) ¼ 4
51. M
Y
ðtÞ¼e
t
2
=2
, E(X) ¼ 0, V(X) ¼ 1
53. E(X) ¼ 0, V(X) ¼ 2
59. a. .850 b. .200 c. .200 d. .701
e. .851 f. .000 g. .570
61. a. .354 b. .114 c. .919
63. a. .403 b. .787 c. .773
65. .1478
67. .4068, assuming independence
69. a. .0173 b. .8106, .4246 c. .0056, .9022, .5858
71. For p ¼ .9 the probability is higher for B (.9963 versus
.99 for A)
For p ¼ .5 the probability is higher for A (.75 versus
.6875 for B)
73. The tabulation for p > .5 is not needed.
75. a. 20, 16 (binomial, n ¼ 100, p ¼ .2) b. 70, 21
77. When p ¼ .5, the true probability for
k ¼ 2 is .0414,
compared to the bound of .25.
When p ¼ .5, the true probability for k ¼ 3 is .0026,
compared to the bound of .1111.
When p ¼ .75, the true probability for k ¼ 2 is .0652,
compared to the bound of .25.
When p ¼ .75, the true probability for k ¼ 3 is .0039,
compared to the bound of .1111.
79. M
nX
(t) ¼ [p +(1 p)e
t
]
n
, E(n X) ¼ n(1 p),
V(n X) ¼ np(1 p)
Intuitively, the means of X and n X should add to n and
their variances should be the same.
81. a. .114 b. .879 c. .121 d. Use the binomial
distribution with n ¼ 15 and p ¼ .1
83. a. h(x; 15, 10, 20) b. .0325 c. .6966
85. a. h(x; 10, 10, 20) b. .0325 c. h(x; n, n,2n),
E(X) ¼ n/2, V(X) ¼ n
2
/[4(2n 1)]
87. a. nb(x; 2, .5) ¼ (x + 1).5
x+2
, x ¼ 0, 1, 2, 3, ...
b. 3/16 c. 11/16 d. 2, 4
89. nb(x; 6, .5), E(X) ¼ 6 ¼ 3(2)
93. a. .932 b. .065 c. .068 d. .491 e. .251
95. a. .011 b. .441 c. .554, .459 d. .944
97. a. .491 b. .133
99. a. .122, .808, .283 b. 12, 3.464 c. .530, .011
101. a. .099 b. .135 c. 2
103. a. 4 b. .215 c. 1.15 years
105. a. .221 b. 6,800,000 c. p(x; 1608.5)
111. b. 3.114, .405, .636
113. a. b(x; 15, .75) b. .6865 c. .313 d. 45/4, 45/16
e. .309
115. .9914
117. a. p(x; 2.5) b. .067 c. .109
119. 1.813, 3.05
121. p(2)
¼ p
2
, p(3) ¼ (1 p)p
2
, p(4) ¼ (1 p)p
2
,
p(x) ¼ [1 p(2)
...
p(x 3)](1 p)p
2
, x ¼ 5,
6, 7, ....
Alternatively, p(x) ¼ (1 p)p(x 1) +
p(1 p) p(x 2), x ¼ 5, 6, 7, ... ; 99950841
123. a. 0029 b. 0767, .9702
125. a. .135 b. .00144 c.
P
1
x¼0
½pðx; 2Þ
5
127. 3.590
129. a. No b. .0273
131. b. .6p(x; l)+.4p(x; m) c. (l + m)/2
d. (l + m)/2 + (l m)
2
/4
133. .5
137. X ~ b(x; 25, p), E(h(X)) ¼ 500p + 750,
s
h
ðXÞ
¼ 100
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pð1 pÞ
p
Independence and constant probability might not be
valid because of the effect that customers can have on
each other. Also, store employees might affect customer
decisions.
139.
X 01234
p(x) .07776 .10368 .19008 .20736 .17280
X 567 8
p(x) .13824 .06912 .03072 .01024
Chapter 3 819
Chapter 4
1. a. 25 b. .5 c. 7/16
3. b. .5 c. 11/16 d. .6328
5. a. 3/8 b. 1/8 c. .2969 d. .5781
7. a. f ðxÞ¼
1
10
for 25 x 35 and ¼ 0 otherwise
b. .2 c. .4 d. .2
9. a. .5618 b. .4382, .4382 c. .0709
11. a. 1/4 b. 3/16 c. 15/16
d.
ffiffiffi
2
p
e. f(x) ¼ x/2 for 0 x < 2, and f(x) ¼ 0
otherwise
13. a. 3 b. 0 for x 1, 1 1/x
3
for x > 1
c. 1/8, .088
15. a. F(x) ¼ 0 for x 0, F(x) ¼ x
3
/8 for 0 < x < 2,
F(x) ¼ 1 for x 2
b. 1/64 c. .0137, .0137 d. 1.817
17. b. 90th percentile of Y ¼ 1.8(90th percentile of X)+32
c. 100 pth percentile of Y ¼ a(100 pth percentile of
X)+b
19. a. 1.5, .866 b. .9245
21. a. .8182, .1113 b. .044
23. a. A +(B A)p
b. (A + B)/2, (B A)
2
/12, ðB AÞ=
ffiffiffiffiffi
12
p
c. (B
n+1
A
n+1
)/[(n + 1)(B A)]
25. 314.79
27. 248, 3.6
29. 1/(1 t/4), 1/4, 1/16
31. 100p,30p
33. f ðxÞ¼
1
10
for 5 x 5 and ¼ 0 otherwise
35. a. M(t) ¼ .15 e
.5t
/(.15 t), t < .15; E(X) ¼ 7.167,
V(X) ¼ 44.44
b. E(X) ¼ 7.167, V(X) ¼ 44.44
37. M(t) ¼ .15/(.15 t), E(X) ¼ 6.667, V(X) ¼ 44.44
This distribution is shifted left by .5, so the mean differs
by .5 but the variance is the same.
39. a. .4850 b. .3413 c. .4938 d. .9876
e. .9147 f. .9599 g. .9104 h. .0791
i. .0668 j. .9876
41. a. 1.34 b. 1.34 c. .674 d. .674
e. 1.555
43. a. .9772 b. .5 c. .9104 d. .8413
e. .2417 f. .6826
45. a.
.7977 b. .0004
c. The top 5% are the values above .3987.
47. The second machine
49. a. .2525 b. 39.96
51. .0510
53. a. .8664 b. .0124 c. .2718
55. a. .794 b. 5.88 c. 7.94 d. .265
57. No, because of symmetry.
59. a. approximate, .0391; binomial, .0437
b. approximate, .99993; binomial, .99976
61. a. .7287 b. .8643, .8159
63. a. approximate, .9933; binomial, .9905
b. approximate, .9874; binomial, .9837
c. approximate, .8051; binomial, .8066
67. a. .15866 b. .0013499 c. .999936658
Actual: .15866 .0013499 .999936658
d. .00000028665
69. a. 120 b. 1.329 c. .371 d. .735 e. 0
71. a. 5, 4 b. .715 c. .411
73. a. 1 b. 1 c. .982 d. .129
75. a. .449, .699, .148 b. .050, .018
77. a. \ A
i
b. Exponential with l ¼ .05
c. Exponential with parameter nl
83. a. .8257, .8257, .0636 b. .6637 c. 172.73
87. a. .9296 b. .2975 c. 98.18
89. a. 68.03, 122.09 b. .3196 c. .7257, skewness
91. a. 149.157, 223.595 b. .957 c. .0416
d. 148.41 e. 9.57 f. 125.90
93. a ¼ b
95. b. G(a + b) G(m + b)/[G(a + b + m ) G(b)], b/( a + b)
97. Yes, since the pattern in the plot is quite linear.
99. Yes
101. Yes
103. Form a new variable, the logarithms of the rainfall
values, and then construct a normal plot for the new
variable. Because of the linearity of this plot, normality
is plausible.
105. The normal plot has a nonlinear pattern showing
positive skewness.
107. The plot deviates from linearity, especially at the low
end, where the smallest three observations are too small
relative to the others. The plot works for any l because l
is a scale parameter.
109. f
Y
( y) ¼ 2/y
3
, y > 1
111. f
Y
ðyÞ¼ye
y
2
=2
, y > 0
113. f
Y
( y) ¼ 1/16, 0 < y < 16
115. f
Y
( y) ¼ 1/[p(1 + y
2
)]
117. Y ¼ X
2
/16
119. f
Y
ðyÞ¼1=½2
ffiffiffi
y
p
,0< y < 1
121. f
Y
ðyÞ¼1=½4
ffiffiffi
y
p
,0< y < 1, f
Y
ðyÞ¼1=½8
ffiffiffi
y
p
,
1 < y < 9
125. p
Y
( y) ¼ (1 p)
y1
p, y ¼ 1, 2, 3, ...
820 Chapter 4
127. a. .4 b. .6 c. F(x) ¼ x/25, 0 x 25;
F(x) ¼0, x < 0; F(x) ¼ 1, x > 25 d. 12.5, 7.22
129. b. F(x) ¼ 1 16/(x +4)
2
, x 0; F(x) ¼ 0, x < 0
c. .247 d. 4 e. 16.67
131. a. .6563 b. 41.55 c. .3179
133. a. .00025, normal approximation; .000859, binomial
b. .0888, normal approximation; .0963, binomial
135. a. F(x) ¼1.5(1 1/x), 1 x 3; F(x) ¼0, x < 1;
F(x) ¼ 1, x > 3 b. .9, .4 c. 1.6479
d. .5333 e. .2662
137. a. 1.075, 1.075 b. .0614, .3331 c. 2.476
139. b. 95,693, 1/3
141. b. F(x) ¼ .5e
.2x
, x 0; F(x) ¼ 1 .5e
.2x
, x > 0
c. .5, .6648, .2555, .6703
143. a. k ¼ (a 1)5
a1
b. F(x) ¼ 0, x 5;
F(x) ¼ 1 (5/x)
a1
, x > 5 c. 5(a 1)/(a 2)
145. b. .4602, .3636 c. .5950 d. 140.178
147. a. Weibull b. .5422
149. a. l b. a x
a 1
/b
a
c. FðxÞ¼1 e
a xx
2
= 2bðÞ
ðÞ
,0 x b; F(x) ¼ 0,
x < 0; F(x) ¼ 1 e
ab/2
, x > b
f ðxÞ¼a 1 x=bðÞe
a xx
2
= 2bðÞ
ðÞ
,0 x b;
f(x) ¼ 0, x < 0, f(x) ¼ 0, x > b
This gives total probability less than 1, so some
probability is located at infinity (for items that last
forever).
151. m
R
v/20, s
R
v/800
155. F(q*) ¼ .818
Chapter 5
1. a. .20 b. .42 c. The probability of at least one
hose being in use at each pump is .70.
d.
x 012 y 012
p
X
(x) .16 .34 .50 p
Y
(y) .24 .38 .38
P(X 1) ¼ .50
e. dependent, .30 ¼ P(X ¼ 2 and Y ¼ 2) 6¼ P(X ¼ 2)
P(Y ¼ 2) ¼ (.50)(.38)
3. a. .15 b. .40 c. .22 ¼ P(A) ¼ P(|X
1
X
2
| 2)
d. .17, .46
e.
x
1
01234
p
1
(x
1
) .19 .30 .25 .14 .12
E(X
1
) ¼ 1.7
f.
x
2
0123
p
2
(x
2
) .19 .30 .28 .23
g. 0 ¼ p(4 , 0) 6¼ p
1
(4) p
2
(0) ¼ (.12)(.19) so the two
variables are not independent.
5. a. .54 b. .00018
7. a. .030 b. .120 c. .10, .30 d. .38
e. yes, p(x,y) ¼ p
X
(x) p
Y
(y)
9. a. .3/380,000 b. .3024 c. .3593
d. 10Kx
2
+ .05, 20 x 30 e. no
11. a. px; yðÞ¼e
l
l
x
=x!
e
y
y
y
=y!
for x ¼ 0, 1, 2, ...;
y ¼ 0, 1, 2, ... b. e
ly
ð1 þ l þ yÞ
c. e
ly
ðl þ yÞ
m
=m!, Poisson with parameter l + y
13. a. e
xy
, x 0, y 0 b. .3996 c. .5940
d. .3298
15. a. FðyÞ¼1 2e
2ly
þ e
3ly
for y 0, F(y) ¼ 0 for
y < 0; f ðyÞ¼4le
2ly
3e
3ly
for y 0, f(y) ¼ 0
for y < 0
b. 2/(3l)
17. a. .25 b. 1/p c. 2/p
d. f
X
ðxÞ¼2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R
2
x
2
p
pR
2
ðÞ
for R x R,
f
Y
ðyÞ¼2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R
2
y
2
p
pR
2
ðÞ
for R y R,no
19. .15
21. L
2
23. 1/4 h
25. 2/3
27. a. .1058 b. .0128
37. a. f
X
(x) ¼ 2x,0< x < 1, f
X
(x) ¼ 0 elsewhere
b. f
Y|X
(y| x) ¼ 1/x,0< y < x < 1 c. .6
d. no, the domain is not a rectangle
e. E(Y| X ¼ x) ¼ x/2, a linear function of x
f. V(Y| X ¼ x) ¼ x
2
/12
39. a. f
X
(x) ¼ 2e
2x
,0< x < 1, f
X
(x) ¼ 0, x 0
b. f
Y|X
(y| x) ¼ e
y+x
,0< x < y < 1
c. P(Y > 2| x ¼ 1) ¼ 1/e
d. no, the domain is not rectangular
e. E(Y| X ¼ x) ¼ x + 1, a linear function of x
f. V(Y| X ¼ x) ¼ 1
41. a. E(Y| X ¼ x) ¼ x/2, a linear function of x; V(Y|
X ¼ x) ¼ x
2
/12
b. f(x, y) ¼ 1/x,0< y < x < 1
c. f
Y
(y) ¼ln(y), 0 < y < 1
d. E(Y) ¼ 1/4, V(Y) ¼ 7/144
e. E(Y) ¼ 1/4, V(Y) ¼ 7/144
43. a. p
Y|X
(0
|
1) ¼ 4/17, p
Y|X
(1
|
1) ¼ 10/17, p
Y|X
(2
|
1) ¼ 3/17
b. p
Y|X
(0
|
2) ¼ .12, p
Y|X
(1
|
2) ¼ .28, p
Y|X
(2
|
2) ¼ .60
c. .40
d. p
X|Y
(0
|
2) ¼ 1/19, p
X|Y
(1
|
2) ¼ 3/19, p
X|Y
(2
|
2) ¼ 15/19
45. a. E(Y| X ¼ x) ¼ x
2
/2 b. V(Y| X ¼ x) ¼ x
4
/12
c. f
Y
(y) ¼ y
–.5
1, 0 < y < 1
47. a. p(1,1) ¼ p(2,2) ¼ p(3,3) ¼ 1/9, p(2,1) ¼ p(3,1)
¼ p(3,2) ¼ 2/9
b. p
X
(1) ¼ 1/9, p
X
(2) ¼ 3/9, p
X
(3) ¼ 5/9
c. p
Y|X
(1
|
1) ¼ 1, p
Y|X
(1
|
2) ¼ 2/3, p
Y|X
(2
|
2) ¼ 1/3,
p
Y|X
(1
|
3) ¼ .4, p
Y|X
(2
|
3) ¼ .4, p
Y|X
(3
|
3) ¼ .2
d. E(Y| X ¼ 1) ¼ 1, E(Y| X ¼ 2) ¼ 4/3,
E(Y| X ¼ 3) ¼ 1.8, no
e. V(Y| X ¼ 1) ¼ 0, V(Y| X ¼ 2) ¼ 2/9,
V(Y| X ¼ 3) ¼ .56
49. a. p
X|Y
(1
|
1) ¼ .2, p
X|Y
(2
|
1) ¼ .4, p
X|Y
(3
|
1) ¼ .4,
p
X|Y
(2
|
2) ¼ 1/3, p
X|Y
(3
|
2) ¼ 2/3, p
X|Y
(3
|
3) ¼ 1
b. E(X| Y ¼ 1) ¼ 2.2, E(X| Y ¼ 2) ¼ 8/3,
E(X| Y ¼ 3) ¼ 3, no
c. V(X| Y ¼ 1) ¼ .56, V(X| Y ¼ 2) ¼ 2/9,
V(X| Y ¼ 3) ¼ 0
Chapter 5
821
51. a. 2x –10 b. 9 c. 3 d. .0228
53. a. p
X
(x) ¼ .1, x ¼ 0, 1, 2, ...,9;p
Y|X
(y| x) ¼ 1/9, y ¼ 0,
1, 2, ...,9,y 6¼ x;
p
X,Y
(x, y) ¼ 1/90, x, y ¼ 0, 1, 2, ...,9,y 6¼ x
b. E(Y| X ¼ x) ¼ 5 x/9, x ¼ 0, 1, 2, ..., 9, a linear
function of x
55. a. .6x, .24x b. 60 c. 60
57. a. .1410 b. .1165
With positive correlation, the deviations from their means
of X and Y are likely to have the same sign.
59. a. If U ¼ X
1
+ X
2
, f
U
(u) ¼ u
2
,0< u < 1, f
U
(u) ¼
2u – u
2
,1< u < 2, f
U
(u) ¼ 0, elsewhere
b. If V ¼ X
2
X
1
, f
V
(v) ¼ 2 2v,0< v < 1,
f
V
(v) ¼ 0, elsewhere
61. 4y
3
[(ln(y
3
)]
2
,0< y
3
< 1
65. a. g
5
(y) ¼ 5y
4
/10
5
, 25/3 b. 20/3 c. 5
d. 1.409
67. g
Y
5
jY
1
y
5
j4ðÞ¼½2=3½ðy
5
4Þ=6
3
,4< y
5
< 10; 8.8
69. 1/(n + 1), 2/(n + 1), 3/(n + 1), ..., n/(n +1)
71.
Gðnþ1ÞGðiþ1=yÞ
GðiÞGðnþ1þ1=yÞ
,
Gðnþ1ÞGðiþ2=yÞ
GðiÞGðnþ1þ2=yÞ
Gðnþ1ÞGðiþ1=yÞ
GðiÞGðnþ1þ1=yÞ
hi
2
73. a. .0238 b. $2025
75.
g
i;j
y
i
; y
j
¼
n!
ði1Þ!ðji1Þ!ðnjÞ!
Fðy
i
Þ
i1
ðFðy
j
ÞFðy
i
ÞÞ
ji1
ð1 Fðy
j
ÞÞ
nj
f ðy
i
Þf ðy
j
Þ,
1 < y
i
< y
j
< 1
77. a. f
W
2
w
2
ðÞ¼nðn 1Þ
Ð
1
1
ðFðw
1
þ w
2
ÞFðw
1
ÞÞ
n2
f ðw
1
Þf ðw
1
þ w
2
Þdw
1
b. f
W
2
w
2
ðÞ¼nðn 1Þw
2
n2
ð1 w
2
Þ,0< w
2
< 1
79. f(x) ¼ e
x/2
e
x
, x 0; f(x) ¼ 0, x < 0.
81. a. 3/81,250
b.
f
X
ðxÞ¼
ð
30x
20x
kxydy ¼ kð250x 10x
2
Þ; 0 x 20
ð
30x
0
kxydy ¼ kð450x 30x
2
þ
1
2
x
3
Þ; 20<x 30
8
>
>
<
>
>
:
f
Y
(y) ¼ f
X
(y) dependent
c. .3548 d. 25.969 e. 32.19, .894
f. 7.651
83. 7/6
87. c. If p(0 ) ¼ .3, p(1) ¼ .5, p(2) ¼ .2, then 1 is the smaller of
the two roots, so extinction is certain in this case with m < 1.
If p(0) ¼ .2, p(1) ¼ .5, p(2) ¼ .3, then 2/3 is the smaller of
the two roots, so extinction is not certain with m > 1.
89. a. P((X,Y) ∈ A) ¼ F(b, d) F(b, c) F(a, d)+F(a, b)
b. P((X,Y) ∈ A) ¼ F(10, 6)
F(10, 1) F(4, 6) + F(4, 1)
P((X,Y) ∈ A) ¼ F(b, d) F(b, c–1) – F(a–1, d)+
F(a–1, b–1)
c. At each (x*, y*), F(x*, y*) is the sum of the
probabilities at points (x, y) such that x x* and
y y*
F(x, y) x
100 250
200 .50 1
y 100 .30 .50
0 .20 .25
d. F(x, y) ¼ .6x
2
y +.4xy
3
,0 x 1; 0 y 1;
F(x, y) ¼ 0, x 0; F(x, y) ¼ 0, y 0;
F(x, y) ¼ .6x
2
+.4x,0 x 1, y > 1;
F(x, y) ¼ .6y +.4y
3
, x > 1, 0 y 1; F(x, y) ¼ 1,
x > 1, y > 1
P(.25 X .75, .25 Y .75) ¼ .23125
e. F(x, y) ¼ 6x
2
y
2
, x+y 1, 0 x 1; 0 y 1,
x 0, y 0
F(x, y) ¼ 3x
4
8x
3
+6x
2
+3y
4
8y
3
+6y
2
1,
x+y> 1, x 1, y 1
F(x, y) ¼ 0, x 0; F(x, y) ¼ 0, y 0;
F(x, y) ¼ 3x
4
–8x
3
+6x
2
,0 x 1, y > 1
F(x, y) ¼ 3y
4
–8y
3
+6y
2
,0 y 1, x > 1
F(x, y) ¼ 1, x > 1, y > 1
91. a. 2x, x b. 40 c. .100
93. M
W
(t) ¼ 2/[(1–1000t)(2–1000t)], 1500
Chapter 6
1. a.
x 25 32.5 40 45 52.5 65
pð
xÞ .04 .20 .25 .12 .30 .09
Eð
XÞ¼44:5 ¼ m
b.
s
2
0 112.5 312.5 800
p(s
2
) .38 .20 .30 .12
E(S
2
) ¼ 212.25 ¼ s
2
3. x/n 0 .1.2.3.4
p(x/n) 0.0000 0.0000 0.0001 0.0008 0.0055
.5 .6 .7 .8 .9 1.0
0.0264 0.0881 0.2013 0.3020 0.2684 0.1074
5. a.
x 1 1.5 2 2.5 3 3.5 4
pð
xÞ .16 .24 .25 .20 .10 .04 .01
b. Pð
X 2:5Þ¼:85
c.
r 0123
p(r) .30 .40 .22 .08
d. .24
7.
xpð
xÞ
2.8 0.052077
3.0 0.034718
3.2 0.021699
3.4 0.012764
3.6 0.007091
3.8 0.003732
4.0 0.001866
xpð
xÞ
1.4 0.090079
1.6 0.112599
1.8 0.125110
2.0 0.125110
2.2 0.113736
2.4 0.094780
2.6 0.072908
xpð
xÞ
0.0 0.000045
0.2 0.000454
0.4 0.002270
0.6 0.007567
0.8 0.018917
1.0 0.037833
1.2 0.063055
822 Chapter 6
11. a. 12, .01
b. 12, .005
c. With less variability, the second sample is more
closely concentrated near 12.
13. a. No, the distribution is clearly not symmetric.
A positively skewed distribution —perhaps Weibull,
lognormal, or gamma.
b. .0746
c. .00000092. No, 82 is not a reasonable value for m.
15. a. .8366 b. no
17. 43.29
19. a. .9802, .4802 b. 32
21. a. .9839 b. .8932
27. a. 87,850, 19,100,116
b. In case of dependence, the mean calculation is still
valid, but not the variance calculation.
c. .9973
29. a. .2871 b. .3695
31. .0317; Because each piece is played by the same
musicians, there could easily be some dependence. If
they perform the first piece slowly, then they might
perform the second piece slowly, too,
33. a. 45 b. 68.33 c. 1, 13.67 d. 5, 68.33
35. a. 50, 10.308 b. .0076 c. 50 d. 111.56
e. 131.25
37. a. .9615 b. .0617
39. a. .5, n(n + 1)/4 b. .25, n(n + 1)(2n + 1)/24
41. 10:52.74
43. .48
45. b. M
Y
(t) ¼ 1/[1 t
2
/(2n)]
n
47. Because w
2
n
is the sum of n independent random variables,
each distributed as w
2
1
, the Central Limit Theorem applies.
53. a. 3.2 b. 10.04, the square of the answer to (a)
57. a. n
2
/(n
2
2), n
2
> 2
b. 2n
2
2
ðn
1
þ n
2
2Þ=½n
1
ðn
2
2Þ
2
ðn
2
4Þ, n
2
> 4
61. a. 4.32
65. a. The approximate value, .0228, is smaller because of
skewness in the chi-squared distribution
b. This approximation gives the answer .03237, agreeing
with the software answer to this number of decimals.
67. No, the sum of the percentiles is not the same as the
percentile of the sum, except that they are the same for
the 50th percentile. For all other percentiles, the
percentile of the sum is closer to the 50th percentile
than is the sum of the percentiles
69. a. 2360, 73.70 b. .9713
71. .9685
73. .9093 Independence is questionable because con-
sumption one day might be related to consumption the
next day.
75. .8340
77. a. r ¼ s
2
W
=ðs
2
W
þ s
2
E
Þ
b. r ¼ .9999
79. 26, 1.64
81. If Z
1
and Z
2
are independent standard normal
observations, then let
X ¼ 5Z
1
+ 100, Y ¼ 2ð:5Z
1
þð
ffiffiffi
3
p
=2ÞZ
2
Þþ50
Chapter 7
1. a. 113.73, X b. 113,
e
X
c. 12.74, S, an estimator for the population standard
deviation
d. The sample proportion of students exceeding 100 in
IQ is 30/33 ¼ .91
e. .112, S=
X
3. a. 1.3481,
X b. 1.3481, X
c. 1.78,
X þ 1:282S
d. .67 e. .0846
5. a. 1,703,000 b. 1,599,730 c. 1,601,438
7. a. 120.6 b. 1,206,000, 10,000
X c. .8
d. 120,
e
X
9. a.
X, 2.113 b.
ffiffiffiffiffiffiffiffi
l=n
p
, .119
11. b.
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p
1
ð1 p
1
Þ=n
1
þ p
2
ð1 p
2
Þ=n
2
p
c. In part (b) replace p
1
with X
1
/n
1
and replace p
2
with
X
2
/n
2
d. .245 e. .0411
13. a. .9876 b. .6915
15. a.
^
y ¼
P
X
2
i
=ð2nÞ b. 74.505
17. b. 4/9
19. a.
^
p ¼ 2
^
l :30 ¼ :20 c.
^
p ¼ð100
^
l 9Þ=70
21. a. .15 b. yes c. .4437
23. a.
^
y ¼ð2x 1Þ=ð1 xÞ¼3
b.
^
y ¼½n=S lnðx
i
Þ 1 ¼ 3:12
25.
^
p ¼ r=ðr þ xÞ¼:15 This is the number of successes
over the number of trials, the same as the result in
Exercise 21. It is not the same as the estimate of
Exercise 17.
27. a. ^s
2
¼
1
n
P
X
2
i
b. ^s
2
¼
1
n
P
X
2
i
29. a.
^
y ¼
P
X
2
i
=ð2nÞ¼74:505, the same as in Exercise 15
b.
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
^
y lnð2Þ
q
¼ 10:16
31.
^
l ¼lnð
^
pÞ=24 ¼ :0120
33. No, statistician A does not have more information.
35.
Q
n
i¼1
x
i
;
P
n
i¼1
x
i
37. I(.5 max(x
1
, x
2
, ..., x
n
) y min(x
1
, x
2
, ..., x
n
))
39. a. 2X(n X)/[n(n 1)]
41. a.
X b. FððX cÞ=
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 1=n
p
Þ
Chapter 7
823
43. a. Vð
~
yÞ¼y
2
= nnþ2ðÞ½b. y
2
/n
c. The variance in (a) is below the bound of (b), but the
theorem does not apply because the domain is a
function of the parameter.
45. a.
x b. N(m, s
2
/n)
c. Yes, the variance is equal to the Crame
´
r-Rao bound
d. The answer in (b) shows that the asymptotic
distribution of the theorem is actually exact here.
47. a. 2/s
2
b. The answer in (a) is different from the answer,
1/(2s
4
), to 46(a), so the information does depend on
the parameterization.
49.
^
l ¼ 6=ð6t
6
t
1
... t
5
Þ¼6=ðx
1
þ 2x
2
þ ...þ 6x
6
Þ¼:0436,
where x
1
¼ t
1
, x
2
¼ t
2
t
1
, ..., x
6
¼ t
6
t
5
53. 1.275, s ¼ 1.462
55. b. no, Eð
^
s
2
Þ¼s
2
=2, so 2
^
s
2
is unbiased
59. .416, .448
61. d(X) ¼ (1)
X
, d(200) ¼ 1, d(199) ¼1
63. b.
^
b ¼
P
x
i
y
i
P
x
2
i
¼ 30:040, the estimated minutes
per item;
^
s
2
¼
1
n
P
ðy
i
^
bx
i
Þ
2
¼ 16:912;
25
^
b ¼ 751
Chapter 8
1. a. 99.5% b. 85% c. 2.97 d. 1.15
3. a. A narrower interval has a lower probability b. No,
m is not random
c. No, the interval refers to m, not individual observations
d. No, a probability of .95 does not guarantee 95
successes in 100 trials
5. a. (4.52, 5.18) b. (4.12, 5.00) c. 55 d. 94
7. Increase n by a factor of 4. Decrease the width by a factor
of 5.
9. a. ð
x 1:645 s=
ffiffiffi
n
p
; 1Þ; (4.57, 1)
b. ð
x z
a
s=
ffiffiffi
n
p
; 1Þ
c. ð1;
x þ z
a
s=
ffiffiffi
n
p
Þ;(1, 59.7)
11. 950; .8724 (normal approximation), .8731 (binomial)
13. a. (.99, 1.07) b. 158
15. a. 80% b. 98% c. 75%
17. .06, which is positive, suggesting that the population
mean change is positive
19. (.513, .615)
21. .218
23. (.439, .814)
25. a. 381 b. 339
29. a. 1.341 b. 1.753 c. 1.708 d. 1.684
e. 2.704
31. a. 2.228 b. 2.131 c. 2.947 d. 4.604
e. 2.492 f. 2.715
33. a. (38.081, 38.439) b. (100.55, 101.19), yes
35. a. Assuming normality, a 95% lower confidence bound
is 8.11. When the bound is calculated from repeated
independent samples, roughly 95% of such bounds
should be below the population mean.
b. A 95% lower prediction bound is 7.03. When the
bound is calculated from repeated independent
samples, roughly 95% of such bounds should be
below the value of an independent observation.
37. a. 378.85 b. 413.09 c. (333.88, 407.50)
39. 95% prediction interval: (.0498, .0772)
41. a. (169.36, .179.37)
b. (134.30, 214.43), which includes 152
c. The second interval is much wider, because it allows
for the variability of a single observation.
d. The normal probability plot gives no reason to doubt
normality. This is especially important for part (b), but
the large sample size implies that normality is not so
critical for (a).
45. a. 18.307 b. 3.940 c. .95 d. .10
47. b. (2.34, 5.60)
49. a. (7.91, 12.00)
b. Because of an outlier, normality is questionable for
this data set.
c. In MINITAB, put the data in C1 and execute the
following macro 999 times
Let k3 ¼ N(c1)
sample k3 c1 c3;
replace.
let k1 ¼ mean(c3)
stack k1 c5 c5
end
51. a. (26.61, 32.94)
b. Because of outliers, the weight gains do not seem
normally distributed.
c. In MINITAB, see Exercise 49(c).
53. a. (38.46, 38.84)
b. Although the normal probability plot is not perfectly
straight, there is not enough deviation to reject
normality.
c. In MINITAB, see Exercise 49(c).
55. a. (169.13, 205.43)
b. Because of an outlier, normality is questionable for
this data set.
c. In MINITAB, see Exercise 49(c).
57. a. In MINITAB, put the data in C1 and execute the
following macro 999 times
Let k3 ¼ N(c1)
sample k3 c1 c3;
replace.
let k1 ¼ stdev(c3)
stack k1 c5 c5
end
b. Assuming normality, a 95% confidence interval for
s is (3.541, 6.578), but the interval is inappropriate
because the normality assumption is clearly not
satisfied.
824 Chapter 8
59. a. (.198, .230) b. .048
c. A 90% prediction interval is (.149, .279)
61. 246
63. a. A 95% confidence interval for the mean is (.163,
.174). Yes, this interval is below the interval for 59(a).
b. (.089, .326)
65. (0.1263, 0.3018)
67. a. yes b. (196.88, 222.62)
69. c. Vð
^
bÞ¼s
2
=Sx
2
i
, s
^
b
¼ s=
ffiffiffiffiffiffiffiffi
Sx
2
i
p
d. Put the x
i
’s far from 0 to minimize s
^
b
e.
^
b t
a=2;n1
s=
ffiffiffiffiffiffiffiffi
Sx
2
i
p
, (29.93, 30.15)
73. a. .00985 b. .0578
75. a. ðx ðs=
ffiffiffi
n
p
Þt
:025;n1;d
; x ðs=
ffiffiffi
n
p
Þt
:975;n1;d
Þ
b. (3.01, 4.46)
77. a. 1/2
n
b. n/2
n
c. (n + 1)/2
n
,1 (n + 1)/2
n1
,
(29.9, 39.3) with confidence level .9785
79. a. P(A
1
\A
2
) ¼ .95
2
b. P(A
1
\A
2
) .90
c. P(A
1
\A
2
) 1 a
1
a
2
; P(A
1
\A
2
\ ...\ A
k
)
1 a
1
a
2
...
a
k
Chapter 9
1. a. yes b. no c. no d. yes e. no f. yes
5. H
0
: s ¼ .05 vs. H
a
: s < .05. Type I error: Conclude that
the standard deviation is less than .05 mm when it is
really equal to .05 mm. Type II error: Conclude that the
standard deviation is .05 mm when it is really less than
.05.
7. A type I error here involves saying that the plant is not in
compliance when in fact it is. A type II error occurs when
we conclude that the plant is in compliance when in fact it
isn’t. A government regulator might regard the type II
error as being more serious.
9. a. R
1
b. A type I error involves saying that the two companies
are not equally favored when they are. A type II error
involves saying that the two companies are equally
favored when they are not.
c. binomial, n ¼ 25, p ¼ .5; .0433
d. .3, .4881; .4, .8452; .6, .8452; .7, .4881
e. If only 6 favor the first company, then reject the null
hypothesis and conclude that the first company is not
preferred.
11. a. H
0
: m ¼ 10 vs. H
a
: m 6¼ 10 b. .0099
c. .5319. .0076 d. c ¼ 2.58 e. c ¼ 1.96
f.
x ¼ 10:02, so do not reject H
0
g. Recalibrate if z 2.58 or z 2.58
13. b. .00043, .0000075, less than .01
15. a. .0301 b. .0030 c. .0040
17. a. Because z ¼ 2.56 > 2.33, reject H
0
b. .84
c. 142 d. .0052
19. a. Because z ¼2.27 > 2.58, do not reject H
0
b. .22 c. 22
21. Test H
0
: m ¼ .5 vs. H
a
: m 6¼ .5
a. Do not reject H
0
because t
.025,12
¼ 2.179 >
|
1.6
|
b. Do not reject H
0
because t
.025,12
¼ 2.179 >
|
1.6
|
c. Do not reject H
0
because t
.005,24
¼ 2.797 >
|
2.6
|
d. Reject H
0
because t
.005,24
¼ 2.797 <
|
3.9
|
23. Because t ¼ 2.24 1.708 ¼ t
.05,25
, reject H
0
: m ¼ 360.
Yes, this suggests contradiction of prior belief.
25. Because
|
z
|
¼ 3.37 1.96, reject the null hypothesis.
It appears that this population exceeds the national
average in IQ.
27. a. no, t ¼.02 b. 58
c. n ¼ 20 total observations
29. a. Because t ¼ .50 < 1.895 ¼ t
.05,7
do not reject H
0
.
b. .73
31. Because t ¼1.24 > 1.397 ¼t
.10,8
, we do not have
evidence to question the prior belief.
35. a. The distribution is fairly symmetric, without outliers.
b. Because t ¼ 4.25 3.499 ¼ t
.005,7
, there is strong
evidence to say that the amount poured differs from
the industry standard, and indeed bartenders tend to
exceed the standard.
c. Yes, the test in (b) depends on normality, and a normal
probability plot gives no reason to doubt the
assumption.
d. .643, .185, .016
37. a. Do not reject H
0
: p ¼ .10 in favor of H
a
: p > .10
because z ¼ 1.33 < 1.645. Because the null
hypothesis is not rejected, there could be a type II
error.
b. .49, .27. c. 362
39. a. Do not reject H
0
: p ¼ .02 in favor of H
a
: p < .02
because z ¼1.1 > 1.645. There is no strong
evidence suggesting that the inventory be postponed.
b. .195. c. <.0000001.
41. a. Reject H
0
because z ¼ 3.08 2.58. b. .03
43. Using n ¼ 25, the probability of 5 or more leaky faucets
is .0980 if p ¼ .10, and the probability of 4 or fewer leaky
faucets is .0905 if p ¼ .3. Thus, the rejection region is
5 or more, a ¼ .0980, and b ¼ .0905.
45. a. reject b. reject c. do not reject d. reject
e. do not reject
47. a. .0778 b. .1841 c. .0250 d. .0066 e. .5438
49. a. P ¼ .0403 b. P ¼ .0176 c. P ¼ .1304
d. P ¼ .6532 e. P ¼ .0021 f. P ¼ .000022
51. Based on the given data, there is no reason to believe
that pregnant women differ from others in terms of true
average serum receptor concentration.
53. a. Because the P-value is .17, no modification is
indicated. b. 997
55. Because t ¼1.759 and the P-value ¼ .089, which is
less than .10, reject H
0
: m ¼ 3.0 against a two-tailed
alternative at the 10% level. However, the P-value
exceeds .05, so do not reject H
0
at the 5% level. There
Chapter 9
825
is just a weak indication that the percentage is not equal to
3% (lower than 3%).
57. a. Test H
0
: m ¼ 10 vs. H
a
: m < 10
b. Because the P-value is .017 < .05, reject H
0
,
suggesting that the pens do not meet specifications.
c. Because the P-value is .045 > .01, do not reject H
0
,
suggesting there is no reason to say the lifetime is
inadequate.
d. Because the P-value is .0011, reject H
0
. There is good
evidence showing that the pens do not meet
specifications.
61. a. 98, .85, .43, .004, .0000002
b. .40, .11, .0062, .0000003
c. Because the null hypothesis will be rejected with high
probability, even with only slight departure from the
null hypothesis, it is not very useful to do a .01 level
test.
63. b. 36.61 c. yes
65. a. Sx
i
c b. yes
67. Yes, the test is UMP for the alternative H
a
: y > .5
because the tests for H
0
: y ¼ .5 vs. H
a
: y ¼ p
0
all
have the same form for any p
0
> .5.
69. b. .05
c. .04345, .05826; Because .04345 < .05, the test is not
unbiased.
d. .05114; not most powerful
71. b. The value of the test statistic is 3.041, so the P-value is
.081, compared to .089 for Exercise 55.
73. A sample size of 32 should suffice.
75. a. Test H
0
: m ¼ 2150 vs. H
a
: m > 2150
b. t ¼ð
x 2150Þ=ðs=
ffiffiffi
n
p
Þ c. 1.33 d. .101
e. Do not reject H
0
at the .05 level.
77. Because t ¼ .77 and the P-value is .23, there is no
evidence suggesting that coal increases the mean heat
flux.
79. Conclude that activation time is too slow at the .05 level,
but not at the .01 level.
81. A normal probability plot gives no reason to doubt the
normality assumption. Because the sample mean is 9.815,
giving t ¼ 4.75 and a (upper tail) P-value of .00007,
reject the null hypothesis at any reasonable level. The
true average flame time is too high.
83. Assuming normality, calculate t ¼ 1.70, which gives a
two tailed P-value of .102. Do not reject the null
hypothesis H
0
: m ¼ 1.75.
85. The P-value for a lower tail test is .0014 (normal
approximation, .0005), so it is reasonable to reject the
idea that p ¼ .75 and conclude that fewer than 75% of
mechanics can identify the problem.
87. Because t ¼ 6.43, giving an upper tail P-value of
.0000002, conclude that the population mean time
exceeds 15 minutes.
89. Because the P-value is .013 > .01, do not reject the null
hypothesis at the .01 level.
91. a. For the test of H
0
: m ¼ m
0
vs. H
a
: m > m
0
at level a,
reject H
0
if 2Sx
i
/m
0
w
2
a;2n
For the test of H
0
: m ¼ m
0
vs. H
a
: m < m
0
at level a,
reject H
0
if 2Sx
i
/m
0
w
2
1a;2n
For the test of H
0
: m ¼ m
0
vs. H
a
: m 6¼ m
0
at level a,
reject H
0
if 2Sx
i
/m
0
w
2
a=2;2n
or
if 2Sx
i
/m
0
w
2
1a=2;2n
b. Because Sx
i
¼ 737, the test statistic is 2Sx
i
/m
0
¼ 19.65, which gives a P-value of .52. There is no
reason to reject the null hypothesis.
93. a. yes
Chapter 10
1. a. .4; it doesn’t b. .0724, .269
c. Although the CLT implies that the distribution will be
approximately normal when the sample sizes are each
100, the distribution will not necessarily be normal
when the sample sizes are each 10.
3. Do not reject H
0
because z ¼ 1.76 < 2.33
5. a. H
a
says that the average calorie output for sufferers is
more than 1 cal/cm
2
/min below that for non-sufferers.
Reject H
0
in favor of H
a
because z ¼2.90 2.33
b. .0019 c. .819 d. .66
7. Yes, because z ¼ 1.83 1.645.
9. a.
x
y ¼ 6:2
b. z ¼ 1.14, two-tailed P-value ¼ .25, so do not reject
the null hypothesis that the population means are
equal.
c. No, the values are positive and the standard deviation
exceeds the mean.
d. 95% CI: (10.0, 29.8)
11. a. A 95% CI for the true difference, fast food mean – not
fast food mean is (219.6, 538.4)
b. The one-tailed P-value is .014, so reject the null
hypothesis of a 200-calorie difference at the .05
level, and conclude that yes, there is strong evidence.
13. 22. No.
15. b. It increases.
17. Because z ¼ 1.36, there is no reason to reject the
hypothesis of equal population means ( p ¼ .17).
19. Because z ¼ .59, there is no reason to conclude that the
population mean is higher for the no-involvement group
(p ¼ .28).
21. Because t ¼3.35 3.30 ¼ t
.001,42
, yes, there is
evidence that experts do hit harder.
23. b. No c. Because
|
t
|
¼
|
.38
|
< 2.228 ¼ t
.025,10
, no,
there is no evidence of a difference.
25. Because the one-tailed P-value is .005 .01, conclude at
the .01 level that the difference is as stated.
This could result in a type I error.
27. Yes, because t ¼ 2.08 with P-value ¼ .046.
29 b. (127.6, 202.0) c. 131.8
826 Chapter 10
31. Because t ¼ 1.82 with P-value .046 .05, conclude at
the .05 level that the difference exceeds 1.
33. a. x
yðÞt
a=2;mþn2
s
p
ffiffiffiffiffiffiffiffiffiffi
1
m
þ
1
n
q
b. (.24, 3.64)
c. (.34, 3.74), which is wider because of the loss of a
degree of freedom
35. a. The slender distribution appears to have a lower mean
and lower variance.
b. With t ¼ 1.88 and a P-value of .097, there is no
significant difference at the .05 level.
37. With t ¼ 2.19 and a two-tailed P-value of .031, there is a
significant difference at the .05 level but not the .01 level.
39. With t ¼ 3.89 and one-tailed P-value ¼ .006, conclude
at the 1% level that true average movement is less for the
TightRope treatment. Normality is important, but the
normal probability plot does not indicate a problem.
41. a. The 95% confidence interval for the difference of
means is (.000046, .000446), which has only positive
values. This omits 0 as a possibility, and says that the
conventional mean is higher.
b. With t ¼ 2.68 and P-value ¼ .010, reject at the .05
level the hypothesis of equal means in favor of the
conventional mean being higher.
43. With t ¼ 1.87 and a P-value of .049, the difference is
(barely) significantly greater than 5 at the .05 level.
45. a. No b. 49.1 c. 49.1
47. 1 2 3 4
x 10 20 30 40
y 11 21 31 41
49. a. Because |z| ¼ |4.84| 1.96, conclude that there is a
difference. Rural residents are more favorable to the
increase.
b. .9967
51. (.016, .171)
53. Because z ¼ 4.27 with P-value .000010, conclude that
the radiation is beneficial.
55. a. H
0
: p
3
¼ p
2
, H
a
: p
3
> p
2
b. (X
3
X
2
)/n
c. ðX
3
X
2
Þ=
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X
2
þ X
3
p
d. With z ¼ 2.67, P ¼ .004, reject H
0
at the .01 level.
57. 769
59. Because z ¼ 3.14 with P ¼ .002, reject H
0
at the .01
level. Conclude that lefties are more accident-prone.
61. a. .0175 b. .1642 c. .0200 d. .0448
e. .0035
63. No, because f ¼ 1.814 < 6.72 ¼ F
.01,9,7
.
65. Because f ¼ 1.2219 with P ¼ .505, there is no reason to
question the equality of population variances.
67. 8.10
69. a. (.158, .735)
b. Here is a macro that can be executed 999 times in
MINITAB:
# start with X in C1, Y in C2
let k3 ¼ N(c1)
let k4 ¼ N(c2)
sample k3 c1 c3;
replace.
sample k4 c2 c4;
replace.
let k1 ¼ mean(c3)-mean(c4)
stack k1 c5 c5
end
71. a. Here is a macro that can be executed 999 times in
MINITAB:
# start with X in C1, Y in C2
let k3 ¼ N(c1)
let k4 ¼ N(c2)
sample k3 c1 c3;
replace.
sample k4 c2 c4;
replace.
let k2 ¼ medi(c3)-medi(c4)
stack k2 c6 c6
end
73. a. (.593, 1.246)
b. Here is a macro that can be executed 999 times in
MINITAB:
# start with X in C1, Y in C2
let k3 ¼ N(c1)
let k4 ¼ N(c2)
sample k3 c1 c3;
replace.
sample k4 c2 c4;
replace.
let k5 ¼ stdev(c3)/stdev(c4)
stack k5 c12 c12
end
75. a. Because t ¼2.62 with a P-value of .018, conclude
that the population means differ. At the 5% level,
blueberries are significantly better.
b. Here is a macro that can be executed repeatedly in
MINITAB:
# start with data in C1, group var in C2
let k3 ¼ N(c1)
Sample k3 c1 c3.
unstack c3 c4 c5;
subs c2.
let k9 ¼ mean(c4)-mean(c5)
stack k9 c6 c6
end
77. a. Because f ¼ 4.46 with a two-tailed P-value of .122,
there is no evidence of unequal population variances.
b. Here is a macro that can be executed repeatedly in
MINITAB:
let k1 ¼ n(C1)
Sample K1 c1 c3.
unstack c3 c4 c5;
subs c2.
let k6 ¼ stdev(c4)/stdev(c5)
stack k6 c6 c6
end
79. a. A MINITAB macro is given in #75(b).
81. a. (11.85, 6.40)
b. See Exercise 57(a) in Chapter 8.
Chapter 10
827