Multivariate Statistical Analysis Mid Term Solution
November 9, 2012
4 pages, 14 problems, 100 points
1. Eigenvalues: 9, 1
1
1
Corresponding normalized eigenvectors: [√2
]′
√2
and
1
[√2
−1 ′
√2
]
. (7%)
---------------------------------------------------------------------------------------------1
1
5 4
2. [
] = 9 [√2
] [√2
1
4 5
1
]+
√2
1
1
√2
[−1] [√2
√2
−1
√2
]=
√2
9
9
1
[29
2
9]
2
2
2
+[
1
1
−2
−2
1 ].
(7%)
2
---------------------------------------------------------------------------------------------3. max
𝐱′𝐒𝐱
𝐱≠0 𝐱′𝐱
= 9, min
𝐱′𝐒𝐱
𝐱≠0 𝐱′𝐱
= 1. (7%)
---------------------------------------------------------------------------------------------4. (1)When at least one deviation vector lies in the (hyper) plane formed by any one
linear combinations of the others
In other words, the columns of the matrix of deviations are linearly dependent
(2) When the sample size is not larger than the number of variables.
(7%)
---------------------------------------------------------------------------------------------𝟒
5. 𝝁
̂ = 𝐱̅ = [ ],
𝟔
[
𝟓−𝟒 [
] 𝟓−𝟒
𝟕−𝟔
̂ = 𝟏 {[𝟑 − 𝟒] [𝟑 − 𝟒
𝚺
𝟒
𝟔−𝟔
𝟕 − 𝟔] + [
𝟒−𝟒 [
] 𝟒−𝟒
𝟕−𝟔
𝟒−𝟒 [
] 𝟒−𝟒
𝟔 − 𝟔] + [
𝟒−𝟔
𝟏
𝟕 − 𝟔]} = 𝟒 [
𝟒 − 𝟔] +
𝟐 𝟏
]. (7%)
𝟏 𝟔
----------------------------------------------------------------------------------------------
Based on Eq. (5 - 6) or (5 - 7),
1  5  4   0.5
T 2  n x  μ 0 ' S 1 x  μ 0   30 0.5 1.5 
9   4 5   1.5 
  8.5 10
10
  0.5 1.5
6.
. (7%)
   18.5  61.6667
3
 9.5  3
( n  1) p
29  2
58
Critical value 
Fp , n  p (0.05) 
F2, 28 (0.05) 
 3.34  6.92
n p
28
28
T 2  critical value.  reject H 0
----------------------------------------------------------------------------------------------
From Eq. (5 - 18),
p( n  1)
Fp ,n  p (0.1)  5.179
n p
x  μ ' S 1 x  μ   5.179  0.17263
. (7%)
30
p( n  1)
By Eq. (5 - 19), major axes i
Fp ,n  p (0.1) and directions e i
n( n  p )
n x  μ ' S 1 x  μ  
7.
namely, major axes 3 0.172633  1.2464 and 0.172633  0.4155.
directions
1 1 0.7071
1  1   0.7071 
1  0.7071 and
 

2 
2   1   0.7071

----------------------------------------------------------------------------------------------
Use Eq. (5.24),
8.
x1 
p( n  1)
s
Fp , n  p (0.1) 11  1  x1 
n p
n
x2 
p( n  1)
s
Fp , n  p (0.1) 22  2  x2 
n p
n
p( n  1)
s
Fp , n  p (0.1) 11
n p
n
p( n  1)
s
Fp , n  p (0.1) 22
n p
n
p( n  1)
Fp , n  p (0.1)  5.17857  2.27565
n p
.
0.5 - 5.17857
5
5
 1  0.5  5.17857
, 1  - 0.429, 1.429 
30
30
0.5  5.17857
5
5
 2  0.5  5.17857
, 2  - 0.429, 1.429 
30
30
(7%)
---------------------------------------------------------------------------------------------From Eq. (5 - 29),
x1  tn 1 (
0.1 s11
0.1 s
)
 1  x1  tn 1 ( ) 11
2p n
2p n
0.1 s22
0.1 s
)
 2  x2  tn 1 ( ) 22
2p
n
2p
n
9.
t29 (0.025)  2.045
x 2  t n 1 (
0.5  2.045
. (7%)
5
5
 1  0.5  2.045
, 1  - 0.3349, 1.3349 
30
30
5
5
 2  0.5  2.045
, 2  - 0.3349, 1.3349 
30
30
---------------------------------------------------------------------------------------------10. one-at-a-time confidence intervals: each interval consider only one dimension at a
time, don’t care the other dimensions.
0.5  2.045
T 2 simultaneous confidence intervals: all linear combinations of the unknown
parameters.
2
Thus T simultaneous confidence intervals are with more strict constraints, and
need longer intervals to satisfy them . (7%)
----------------------------------------------------------------------------------------------
11. Initial
estimate
~1 
42
3
2
,
~2 
6 48
6
3
(3  3)2  (4  3)2  (2  3)2 2 ~
(6  6) 2  (4  6) 2  (8  6) 2 8
~
 11 
 ,  22 
 ,
3
3
3
3
(3  3)( 6  6)  ( 4  3)( 4  6)  ( 2  3)(8  6)
4
~12 
 .
3
3
 1 
~11 | ~12 


~ 
ˆ   , Σ
Partition μ
        , and predict



~12 | ~22 
 2 

4 3
1
~
x11  ~1  ~12~22
( x12  ~2 )  3  (  ) (6  6)  3
3 8
~
2  4 3 4
2
1
x11
 ~11  ~12~22
 12  ~
x112          32  9
3  3 8 3
~
x11x12  ~
x11x12  3  6  18
~
~  x11  x21  x31  3  6  9 
T1  

 
 x12  x22  x32   18  18
~

2
2
2
~ 
x11
 x21
 x31
T2 
 x ~x  x x  x x
 11 12 21 22 31 32
~
~  T1  3 ,
μ
n 6
 29
~
~ T2 ~ ~  3
Σ
 μμ  
50
n

3
~

x11x12  x21x22  x31x32  29 50 

50 116
2
2
2


x12  x22  x32

50 
 2
3



3  3 6   3
 4
116  6


3 
 3
 4
3 
8 

3 
. (7%)
---------------------------------------------------------------------------------------------12. Eq. (6-21), p. 285, Textbook
n1  1
n2  1
60 5 4 60 7 2 6 3
S pooled 
S1 
S2 


.
n1  n2  2
n1  n2  2
120 4 5 120 2 7 3 6
Eq. (6-24), p. 286, Textbook
,
 1 1 

T  x1  x2 '   S pooled 
 n1 n2 

1
x1  x2   47.4444 ,
2
larger than the critical value
( n1  n2  2) p
120  2
240
Fp , n1  n2  p 1 ( ) 
F2,120 (0.1) 
 2.35  4.74 .
( n1  n2  p  1)
119
119
Thus the
hypothesis H 0 : μ1  μ 2 is rejected at 10% significance level. (7%)
---------------------------------------------------------------------------------------------Follow the equation in p. 291,
1i  2i : ( x1i  x2i )  tn
1  n2
2
13. 11  21 :  2  t120 (0.025)
12  22 : 1  t120 (0.025)
(

1 1
)    sii, pooled
2 p  n1 n2 
2
 6  2  1.98  0.4435, i.e., (-2.88, - 1.12)
61
.
2
 6  1  1.98  0.4435, i.e., (0.12, 1.88)
61
(7%)
---------------------------------------------------------------------------------------------14. Assume that the pair of paired samples are with n data, mean difference 𝑑̅ , and
standard deviation of difference sd. The confidence interval by the paired
𝛼 𝑠
comparison is 𝑑̅ ± 𝑡𝑛−1 ( 2 ) 𝑑𝑛. While if we use the unpaired method to the same
√
𝛼
1
1
paired samples, the confidence interval will be 𝑑̅ ± 𝑡𝑛+𝑛−2 ( 2 ) √(𝑛 + 𝑛) 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
Since
𝑛−1
𝑛−1
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑛+𝑛−2 𝑠𝑑 + 𝑛+𝑛−2 𝑠𝑑 = 𝑠𝑑 ,
the
confidence
interval
using
unpaired method can be reduced to
𝛼
1 1
𝛼 2
𝑑̅ ± 𝑡𝑛+𝑛−2 ( ) √( + ) 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑑̅ ± 𝑡2𝑛−2 ( ) √ 𝑠𝑑
2
𝑛 𝑛
2 𝑛
The lengths of the confidence intervals for the paired and the unpaired methods
𝛼 𝑠𝑑
will be 2𝑡𝑛−1 ( 2 )
√𝑛
𝛼 √2𝑠𝑑
.
√𝑛
and 2𝑡2𝑛−2 ( 2 )
𝛼
Because for n not too small, 𝑡𝑛−1 ( 2 )
𝛼 𝑠𝑑
does not larger than √2𝑡2𝑛−2 (𝛼), 2𝑡𝑛−1 ( 2 )
√𝑛
Thus the paired method is more accurate. (9%)
𝛼 √2𝑠𝑑
.
√𝑛
is smaller than 2𝑡2𝑛−2 ( 2 )