Solution for HW5 - University of Minnesota · Solution for HW5 . IE 4521 Fall 2011 . 7.2.3 (a)
3
Click here to load reader
Transcript of Solution for HW5 - University of Minnesota · Solution for HW5 . IE 4521 Fall 2011 . 7.2.3 (a)
Solution for HW5
IE 4521 Fall 2011
7.2.3
(a) 1 2ˆ2 2X Xμ = + , 1 2 1 2
1 1ˆ( ) ( ) ( ) ( ) ( ) 2.52 2 4 4X XVar Var Var Var X Var Xμ = + = + =
(b)
1 2ˆ (1 )pX p Xμ = + −
2
ˆ( ) ( )
10 12 6
Var p Var X
p p
μ = +
= − +
,
This is minimized when
2 2 2 2 2 21 2 1 2(1 ) ( ) ( ) (1 2 ) ( ) 4 6 12 6p Var X p Var X p p Var X p p p− = + − + = + − +
2min
12 0.62 10
ˆ( ) 10 0.6 12 0.6 6 2.4
p
Var μ
−= − =
×= × − × + =
(c) The relative efficiency is minˆ( ) 2.4 0.96ˆ( ) 2.5
VarVar
μμ
= =
7.2.5
(a) 1 1 11ˆ[ ] ( ) ( ) ( )n n nnE a E x a E x a a a aμ μ μ μ= + + = + + = +L L L
The point estimate for a parameter is said to be unbiased if
ˆ[ ]E μ μ=
So 1 1na a+ =L
(b) Minimize ˆ( )Var μ = 2 21( na a 2)σ+L subject to the constraints
1 1na a+ =L2 2 2
1 1 1 2 1 3 1
2 2 2 2 2 2 21 1 2 1 3
2 21
( ) 2 2 2
( )
n n n n
n n
n
a a a a a a a a a a
a a a a a a a a
n a a
−
−
+ + = + + + + +
≤ + + + + + + + + +
= + +
L L L
L L
L
21 n
So
22 2 11
( ) 1( ) nn
a aa an n
+ ++ + ≥ =
LL
And
22 2 11
(( ) nn
a aa an
+ ++ + =
LL
) will hold if and only if 1 na a= =L
So 1 1 1na a na+ + = =L 11
na an
= = =L
7.2.6 2 2
12 2
22 2
3
( ) 0.02 (0.13 ) 0.0369
( ) 0.07 (0.05 ) 0.0725
( ) 0.005 (0.24 ) 0.0625
MSE
MSE
MSE
2
2
2
θ θ θ θ
θ θ θ θ
θ θ θ
= + =
= + =
= + = θ
1θ has the smallest MSE.
7.3.7
(a)
1
20
0.025,20
( ) / ~
( ( ) / ) ( / 21 ) 0.95
(1 0.95) / 2 0.025
0.455221
nn X s t
P X s c P t c
tc
μ
μ−−
− ≤ = ≤ =
− =
= =
(b)
20
0.005,20
( / 21 ) 0.9(1 0.99) / 2 0.005
0.620921
P t c
tc
≤ =
− =
= =
9
7.3.19
If a sample size n=100 is used, then the probability is (0.24 0.05 0.24 0.05) (19 (100,0.24) 29)P p P B− ≤ ≤ + = ≤ ≤
Using a normal approximation this can be estimated as
29 0.5 100 0.24 19 0.5 100 0.24( ) ( ) (1.288) ( 1.288) 0.8022100 0.24 0.76 100 0.24 0.76+ − × − − ×
Φ −Φ = Φ −Φ −× × × ×
=
If a sample if size n=200 is used, then the probability is
(38 (200,0.24) 58)P B≤ ≤
Using a normal approximation this can be estimated as
58 0.5 200 0.24 38 0.5 200 0.24( ) ( ) (1.738) ( 1.738) 0.9178200 0.24 0.76 200 0.24 0.76+ − × − − ×
Φ −Φ = Φ −Φ −× × × ×
=
7.3.20
174, 2.8, 302.8 2.8ˆ ~ (174, )
30ˆ(173 175) (1.956) ( 1.956) 0.9496
n
N
P
μ σ
μ
μ
= = =×
≤ ≤ = Φ −Φ − =
