Chapter 12
12.1 (a) simple; (b) composite (
β
not specified); (c) composite (parameter not specified);
(d) composite (parameter not specified).
12.4 ( 16; 0.90) ( 4; 0.10)
1 (0.1216 0.2702 0.2852 0.1901)
1 0.8671 0.1329
px px
αθ θ
( 16; 0.60) ( 4; 0.40)
0.000 0.0005 0.0031 0.0123 0.0159
px px
β
θθ
12.7
0
11 2 1.414
2
xz
n
zz
α
αα
σ
μ
0.5000 0.4207 0.8a
172 Mathematical Statistics, 8E
12.8
00
(1;)0px
β
β
0
00 0
00
11
(1;2)(1)
22
px
β
ββ β ββ
12.9
2
11
2112
3/4 3/4
14
x
x x dx dx
β
12.10 Proof same as in Example 12.4 except that the quantity
01
()n
μ
μ
is now positive and the
inequalities are
xK
inside c
xK
outside c
12.11
(1/ )
0
0
1
i
x
n
Le
θ
θ
1
(1/ )
1
1
1
i
x
n
Le
θ
θ
01
(1/ 1/ )
01
10
i
n
x
Lek
L
θθ
θ
θ
Chapter 12 173
12.12
000
(1 )
xnx
n
Lx
θθ
111
(1 )
xnx
n
Lx
θθ
12.13
100(0.40) 1.645
100(0.4)(0.6)
K
, K = 40
1.645(4.90) 31.94
12.14
1
() (1 )
x
fx
θθ
1, 2, 3,x
1
00 0
(1 )
x
L
θθ
1
11 1
(1 )
x
L
θθ
12.15
22
0
(1/2 )
0
0
1
(2)
x
nn
Le
σ
πσ
22
1
(1/2 )
1
1
1
2
x
nn
Le
σ
πσ
174 Mathematical Statistics, 8E
1
20
22
01
ln ln
11
nk
xK
σ
σ
σσ
12.16 The probabilities of making wrong decisions are
0.9
θ
0.6
θ
1
d
0.0114 0.1255 (a) (0.0114)(0.8) + (0.1255)(0.2)=0.034
12.17 (a)
07
20 0
7
2
16
20 0
7
2
25
20 1
721
2
12.18
0.95
θ
0.0022 0.0003 0.0025
α
0.90
θ
0.0319 0.0089 0.0020 0.0004 0.0001 0.0433
α
0.85
θ
1 1 (0.0388 0.1368 0.2293 0.2428 0.1821) 0.1702
β
β
Chapter 12 175
12.19
00
()( )
ii
xxxx
μ
μ
22 2
000
()()2()()()
iii
xxxxxxx
μ
μμ
μ
μ
12.20 (a)
(1 )
xnx
n
Lx
θθ
0
1
2
n
n
Lx
ln ln ln ( )ln(1 )
n
Lxnx
x
θθ
(b)
ln 2 ln ln ( ) ln( ) ln
ln()ln()
ln ( )ln( )
nnnxxnxnxk
xx nx nx k
xx nx nx K
176 Mathematical Statistics, 8E
12.21 (a)
(1/ )
1
x
n
Le
θ
θ
0
(1/ )
0
1
max
x
n
Le
θ
θ
(b)
0
(/)
n
nx
n
xk
ek
ne
θ
0
0
0
/
/
/
xn
xn
x
xek
n
xe n k K
xe K
θ
θ
θ
Chapter 12 177
t
12.24
22
0
(1/2 ) ( )
0
0
1
max (2)
xx
nn
Le
σ
πσ
22
ˆ
(1/2 ) ( )
0
1
max ˆ
(2)
xx
nn
Le
σ
πσ
22 2
0
22
0
/2
2
ˆ
(1/2) ( ) (1/ 1/ )
2
0
22 2 2
00
/2
2(1/2) ( ) /
2
0
()
11 1
ˆ()
()
n
xx
n
xx n
xx e
n
n
xx
xx e
n
σσ
σ
λσ
σσσ
λσ
178 Mathematical Statistics, 8E
12.26 Dividing numerator and denominator by
12
()/2
2
1
nn
s
yields
12.27
2
1
12
Lx
θ
β
0
(0) 1
dx
α
πσ
12.28 They would be committing a type I error if they erroneously reject the null hypothesis that 60%
of their passengers object to smoking inside the plane.
They would be committing a type I error if they erroneously accept this null hypothesis.
12.29 The doctor would commit a type I error if he/she erroneously rejects the null hypothesis that
12.30 (a) The manufacturer should use the alternative hypothesis
20
μ
and make the
12.31 (a)
12 1
:H
μ
μ
μ
μ
Chapter 12 179
12.32 With
9.6, 10.2,x
μ
and
80n
(a) Decision: reject
0
:H
since
0
H
is true, decision is in error.
μ
μ
μ
12.34 (a)
0
:H
the antipollution device is effective. A type I error would be made if the device is
12.35 (a) She will correctly reject the null hypothesis.
(b) She will erroneously reject the null hypothesis.
12.37 (a)
1.645 1.88n
3.525n
12.43n
13n
rounded up to nearest integer
12.38 (a) Yes; (b) Yes
12.39 (a)
12 12
/
10 /10 1.2 0.8
8
8
1
111
10
1 0.3012 0.4493 0.852
xx
edx e e e
180 Mathematical Statistics, 8E
12.40 Reject if
43.5x
65 2
64
x
σ
(a)
43.5 37 3.25, ( 43.5 37) ( 3.25) 0.00058
zPXPZ
μ
2
zPXPZ
μ
(b)
43.5 41 1.25, ( 43.5 41) ( 1.25) 0.8944
2
zPXPZ
μ
43.5 42 0.75, ( 43.5 42) ( 0.75) 0.7734
2
zPXPZ
μ
43.5 43 0.25, ( 43.5 43) ( 0.25) 0.5987
zPXPZ
μ
12.41 (a) Reject if
5x
Use Table II
11
λ
0.0375p
12
λ
0.0203p
(b)
10, 1 0.0671 0.9329, 7.5, 1 0.2415 0.7585
λλ
5, 1 0.6160 0.3840, 2.5, 1 0.9580 0.0420
λλ
Chapter 12 181
12.43
4534
16
4534
16
7 16 9 25 5 12 7 24
8106 8
(112 225 60 168) / 32
14 22.5 10 21
17.656
λ
12.44 From Exercise 12.21
0
(/)
0
n
nx n
xe
θ
λθ