Chapter 11
11.2 (a)
12
12
12
[0 ( )] 1
() 1
()
pkxx
px x k
px x k
θα
θα
θα
  








2
22
11
2k
θα
θ

2
1
2k
α
2
1
2
k
α
1
2
k
α
Chapter 11 163
11.3
()1
1
pR cR
pR
c
θα
θθα
 

 


11.4 By inspection
/
3/2 /22/3
/2 /3 2 /3
2
zz zz
zzz
αα α α
αα α


length of first confidence interval is less than that of 2nd confidence interval
11.5 Length of confidence interval:
Since the normal density function
()
fx
is decreasing for
21
0, x
δδ

, thus
/2
2
k
Lz n
α
σ

164 Mathematical Statistics, 8E
11.6
/2
1px z n
α
σ
μ
α

  


11.7 Substitute
/2, 1 /2
for
n
s
tz
nn
αα
σ

If
x
, the mean of a random sample of size n from a formal population with the mean
, is
used as an estimate of
, we can assert with
(1 )100%
α
confidence that the error is less than
/2, 1
n
s
tn
α
.
μ
μ
12
11.9
22222
1212
12 12 12
11 2
() 22 2
p
nnnn
ES nn nn nn
σσ σσ


  
therefore unbiased
11.10
12 2 1
2
12
12 1 2 12
()( ) 1
11 (2)
22
p
Zxx
TYnn S
nn n n nn
μ
μ
σ

 

 
Chapter 11 165
11.13
/2
;
(1 )
xn
zn
α
θ
θθ

 
/
2
(1 )
xn z
n
α
θ
θθ

  
Let
*
θ
= value of
θ
with
θθθ
  
closest to
1
2
. By Theorem 11.7,
/2
*(1 *)
ez n
α
θθ
and
2
/
2
2
*(1 *) z
ne
α
θθ

11.16 If
12
nnn
, then
112 2
/2
ˆˆˆˆ
(1 ) (1 )
Ez n
α
θθθθ
 
The right-hand side of this inequality is maximized when
12
1.
2
θθ

Thus,
/2
1
2
Ez n
α
,
2
2
/
2
,
2
z
En
α
and
2
/
2
2
2
z
nE
α
.
/
/
166 Mathematical Statistics, 8E
11.20 n = 150 9.4
σ
9.4 1.96(9.4)
1.96 1.50
12.247
150
E
11.21 9.4
61.8 2.575 61.8 1.98, 59.82 63.78
150
μ

11.24
0.005
;
s
xz n
45
52.80 2.575 ,
64
or (51.35, 54.25).
11.25
0.025
2.68
1.96 0.83 min.
40
s
ez n
 
11.26
0.025
9.4 900 150
1.96 1.37.
1 900 1
150
Nn
ez N
n
σ

 

Chapter 11 167
11.33
22
12
12 0.05
12
() ;xx z nn
σσ

22
4.8 3.5
5.2 1.645 16 25
 , or ( 7.49, 2.91).
11.34
22
12
12 0.05
12
() ;
ss
xx z nn
 
22
19.4 18.8
7.4 2.575 61
 , or ( 16.31,1.51).
11.36
11 2 2
8260, 251.89, 7930, s 206.52xs x  
22
2
4(251.89) 4(206.52) 53,049.54
8
p
s
 230.32
p
s
11
8260 7930 3.355(230.32) 55
 
12
330 488.75
158.75 818.75 million calorie per ton
μμ

168 Mathematical Statistics, 8E
11.41 (0.76)(0.24)
1.96 0.053
250
e

11.42 (0.18)(0.82)
0.18 2.575 100
0.18 0.099
0.081 0.279
θ

11.45
2
2
(1.96) 2401
4(0.02)
n

11.46
2
1.96
(0.03)(0.70) (0.21)(9604) 2017
0.02
n




Chapter 11 169
11.49 84 0.336
250 156 0.624
250
(0.336)(0.664) (0.624)(0.376)
(0.336 0.624) 1.96 250 250

0.288 0.084
12
0.372 0.204
θθ

11.51 (0.096)(0.904) (0.170)(0.830)
2.33 500 400
2.33(0.022939) 0.053
e


11.54
22
2
11(0.625) 11(0.625)
19.675 4.575
σ

2
0.2184 0.939
σ
 0.47 0.97
σ

11.55 4.5 4.5
2.575 2.575
11
128 128
σ


3.67 5.83
σ

170 Mathematical Statistics, 8E
2
6.39
σ
2
σ
11.60 Using MINITAB we enter the data into C1 and we give the command
MTB> Tinterval 95.0 C1
Obtaining
N MEAN STDEV SEMEAN 95.0 PERCENT C.I.
20 6.145 1.467 0.328 (5.458, 6.832)