1. No category

Two Sample
Elementary Multivariate Statistical Inference
Contents
1
Notation
II.G-1
2
General Algorithm
II.G-1
3
Paired Data
3.1 Example . . . . . . . . . . . . . .
3.2 Idea . . . . . . . . . . . . . . . .
3.3 Assumptions . . . . . . . . . . .
3.4 Formulas . . . . . . . . . . . . .
3.5 Repeated Measure Design (RMD)
.
.
.
.
.
II.G-1
. . II.G-1
. . II.G-2
. . II.G-2
. . II.G-2
. . II.G-3
.
.
.
.
II.G-5
. . II.G-5
. . II.G-5
. . II.G-5
. . II.G-7
4
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Comparing Mean Vectors From Two Populations—Unpaired Data
4.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Example: Profile Analysis . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
II.G. Two-Sample Data
1
Notation
both p
dimensional
2
(
Sample
Population
Summary Characteristics
X1 , S1
µ1 , Σ1
X2 , S2
µ2 , Σ2
sample 1 X1,1 , . . . , X1,n1
sample 2 X2,1 , . . . , X2,n2
General Algorithm
1. Can data be paired? If so, use paired data approach.
2. For unpaired data, are samples large? — large sample approach.
3. If unpaired data are small samples, check to make sure that the data is normal and Σ1 = Σ2 .
If not, recode data.
3
3.1
Paired Data
Example
Sales before and after a major advertising campaign. In genearl, comparison of two treatment
responses.
sales per day
sales per day
before treatment
after treatment
store
product A product B
product A product B
16
15
25
10
1
28
9
13
18
2
27
12
22
14
3
..
..
..
..
..
.
.
.
.
.
.
..
.
.
..
..
..
11
.
|
"
X11 =
25
10
{z
1st sample
#
"
, X12 =
13
18
}
#
"
, ... X21 =
II.G-1
{z
|
}
2nd sample
16
15
#
"
, X22 =
28
9
#
, ...
3.2
Idea
Di = X1,i − X2,i . Treat D1 , . . . , Dn as one sample data and analyze it using the methods in II.A
through II.F.
"
#
"
#
µ11 − µ21
9
Note: µD = µ1 − µ2 =
. Eg. D1 =
.
µ12 − µ22
−5
3.3
Assumptions
D1 , . . . , Dn are i.i.d. If n − p is small then we need the normality assumption.
3.4
Formulas
1. Data transformation:
"
Let Xi =
X1i
X2i
#
and set C = (Ip | − Ip )p×2p .
Then D = CX = X1 − X2 , SD = CSC0 , Di = CXi .
2. Hypothesis testing:
H0 : µD = 0 (no treatment effect) versus H1 : µD 6= 0.
Test statistic:
T 2 = nD 0 S−1
DD ∼
(n − 1)p
Fp,n−p if µD = 0. Reject H0 if T 2 > Tα2
n−p
2
Eg. T 2 = 13.6, T.05
= 9.47, so reject H0 . Evidently there is treatment effect.
3. Confidence ellipsoids:
2
(a) For µD = µ1 − µ2 , n(D − µD )0 S−1
D (D − µD ) ≤ Tα , use eigenvalues and eigenvectors
of SD . Half axis length is
q
λi Tα2 /n
(b) Simultaneous confidence intervals: for all compounds ` = a 0 µD , the simultaneous
confidences:
s
a 0 SD a
0
a D ± Tα
.
n
Eg. 95% simultaneous confidence interval for the change in sales per day for
product A:
"
a=
1
0
#
"
, D=
13.27
−9.36
s
13.27 ± T.05
#
"
, SD =
418.61 88.38
88.38 199.26
418.61
= (−5.7, 32.3).
11
II.G-2
#
"
For product B : a =
0
1
s
#
− 9.36 ± Tα
199.26
= (−22.5, 3.7).
11
Gain in competetive edge:
i.e., change in market difference:
product A − product B
After
µ11 − µ12
−Before
µ21 − µ22
Change (µ11 − µ21 ) − (µ12 − µ22 )
"
a=
3.5
1
−1
s
#
13.27 − (−9.36) ± Tα
418.61 + 199.26 − 2 · 88.38
≡ (3.1, 42.1).
11
Repeated Measure Design (RMD)
1. Definition: In an RMD each individual (experimental unit) receives each of q treatments
over successive periods of time. The order in which one receives treatment is randomized.
2. Notation:

Xi =





Xi1
Xi2
..
.






Xiq
← first treatment
response
qth treatment
← response






responses


on the
mean = 


ith




individual
µ1
µ2
..
.

← average
treatment 1
response





µq
3. Example of contrasts:
(a) Control or standard (i.e., µ1 ) contrasted with other treatments.









µ1 − µ2
µ1 − µ3
..
.
..
.
µ1 − µq



















=
(q−1)×1
1 −1
1
..
.
..
.
1
0
..
.
..
.
0
0
−1
..
.
···
··· ··· 0
..
..
.
.
..
.. ..
.
.
.
.. ..
.
.
0
· · · 0 −1










µ1
µ2
..
.
..
.
µq


















=
c1 0
c2 0
..
.
..
.
cq−1 0





 mean.



Yi = CXi .
Contrast matrix: rows are linearly independent, and each row sums to 0.
(b) Successive treatment






µ1 − µ2
µ2 − µ3
..
.
µq−1 − µq













=
1 −1
..
.
0
.. . .
.
.
0 ···
0 ··· 0
..
.. ..
.
.
.
.. ..
.
.
0
0 1 −1
II.G-3







µ1
µ2
..
.
µq







=




c1 0
c2 0
..
.
cq−1 0



 mean.


(c) Eg. Anesthetic testing
Response: millisecond between heart beats
4
present
3
Halothene
Factor 1
Design:
absent
2
low
CO2
1
high
Factor 2
sleeping dog data (Table 6.2, page 282)
Treatments
dog
1
2
3
4
1 426 609 556 600
x1 0
..
..
..
..
..
..
.
.
.
.
.
.
x19 0
19
Contrasts:
Halothene (µ3 + µ4 ) − (µ1 + µ2 ) = present − absent
CO2
(µ1 + µ3 ) − (µ2 + µ4 ) = high − low
interaction (µ1 + µ4 ) − (µ2 + µ3 )


−1 −1
1
1
1 −1 
C=
 1 −1
 , Yi = CXi .
1 −1 −1
1
0
4. Analysis: H0 : µY = 0, i.e., Cµ = 0 (all contrasts are zero). T 2 = nY S−1
Y Y = 116.
Recall that Y = CX, SY = CSC0 ,
Tα2 =
(n − 1)(q − 1)
Fq−1,n−(q−1) (α) = 10.94 (note that it’s q − 1 dimensional).
n − (q − 1)
Simultaneous 95% confidence interval is
s
0
ci X ± Tα
II.G-4
ci 0 Sci
.
n
4
Comparing Mean Vectors From Two Populations—Unpaired
Data
4.1
Assumptions
1. X11 , . . . , X1n1 are i.i.d. random vectors from a population with characteristics µ1 and Σ1 .
2. X12 , . . . , X2n2 are i.i.d. random vectors from a population with characteristics µ2 and Σ2 .
3. The samples are independent of each other and both populations are p dimensional.
4. If n1 − p or n2 − p is small, then assume that each population is normally distributed with
the same covariance matrix Σ = Σ1 = Σ2 .
Note: If Σ1 ≈ Σ2 , procedure works OK.
Σ1 = (σ1,ij ), Σ2 = (σ2,ij ). Trouble if σ1,ii > 4σ2,ii or vice versa.
4.2
Example
Electricity consumption for houses with and without air conditioning (AC) during July in Wisconsin. (Eg 6.4, page 289).
Population 1: Consumption for homes with AC, n1 = 45 homes
Population 2: Consumption for homes without AC, n2 = 55 homes
Xi1 = total consumption during peak periods
Xi2 = total consumption during off peak periods
Data summary:
"
#
"
#
204.4
13825 23823
X1 =
S1 =
with AC
556.6 #
23823 73107 #
"
"
130.0
8632 19616
without AC X2 =
S2 =
355.0
19616 55964
4.3
Analysis
1. Estimate of µD = µ1 −µ2 is D = X1 −X2 and (standard error)2 is cov(D) = Σ1 /n1 +Σ2 /n2
which describes the variability in our estimate.
Eg. D = (74.4, 201.6)0 , Σ1 /n1 + Σ2 /n2 must be estimated. If n1 − p and n2 − p are both
large then this covariance is estimated by S1 /n1 + S2 /n2 .
Eg. (above)
S1 S2
+
=
n1 n2
"
464.17 886.08
886.08 2642.15
II.G-5
#
.
So the estimate for the peak hours has standard error
√
464.17 = 21.5.
For small samples then the assumption Σ = Σ1 = Σ2 is required and so Σ1 /n1 + Σ2 /n2 =
(1/n1 + 1/n2 )Σ which is estimated by (1/n1 + 1/n2 )Spooled . Here
Spooled =
SS1 + SS2
total SS
(n1 − 1)S1 + (n2 − 1)S2
=
=
.
(n1 − 1) + (n2 − 1)
total d.f.’s
n1 + n2 − 2
Eg. (above)
Spooled
44S1 + 54S2
=
=
98
Hence,
"
10963.7 21505.5
21505.5 63661.3
1
1
+
Spooled =
n1 n2
"
#
442.98 868.91
868.91 2572.17
.
#
.
2.
Definition: S =


S1 + S2 ,
if large samples
n2  1 + 1 Spooled , if small samples.
n1
n2
n1
3. Hypotheses Tests
H0 : µ1 = µ2 (i.e., µD = µ1 − µ2 = 0) versus H1 : µ1 6= µ2 . D = X1 − X2 . Test statistic
is
(df)p
H
F
T 2 = (D − 0)0 S−1 (D − 0) = D0 S−1 D ∼0
df − p + 1 p,df−p+1
|
≈χ2p
{z
}
for large samples
For small samples, df = n1 + n2 − 2. Reject H0 if T 2 > Tα2 .
Eg. (above, using large sample approach)
T 2 = 15.66, α = .05, Tα2 ≈ χ22 (.05) = 5.99, reject H0 . If small sample approach was used,
T 2 = 16.17, Tα2 = (98)(2)
F2,97 (.05) = 6.26.
97
Fact: The compound most responsible for rejecting H0 : µ1 = µ2 is ` = a 0 µD , where the
direction of a is estimated by S−1 D.
Eg. (above)
"
−1
S D=
.041
.063
#
.
i.e., the difference in off peak electricity consumption between those with AC and those
without AC contributes more (.063 versus .041) than the difference in on peak electricity
consumption to the rejection of H0 : µ1 = µ2 .
4. Confidence ellipsoid for µD = µ1 − µ2 .
general formula: (estimate − µD )0 (cov)−1 (estimate − µD ) ≤ Tα2 .
Here, (D − µD )0 S−1 (D − µD ) ≤ Tα2 .
Use eigenvalues-eigenvectors for the description of the ellipsoid.
II.G-6
√
5. Simultaneous confidence interval for a 0 µD : a 0 D ± Tα a 0 Sa
peak
diff
off peak
diff
small samples (22, 127) µ11 − µ21 (75, 329) µ12 − µ22
large samples (22, 127) a = (1, 0)0 (76, 327) a = (0, 1)0
4.4
Example: Profile Analysis
1. Idea: Compare two or more groups that are each given p treatments (tests, questions, etc.)
All the responses for different groups are independent.
2. Example:
Q1 : Describe your contribution to marriage
Q2 : Describe your outcomes to marriage
The answers to each question are in scale from 1 (extremely negative) to 5 (extremely positive).
Q3 : Level of passionate love for your spouse
Q4 : Level of compassionate love for your spouse
The answers are in scale from 1 (none) to 5 (tremendous)
4.6
population 1: married men, n1 = 30
population 2: married women, n2 = 30.
Data Profile:
Husband
Wife
●
4.2
●
4.0
mean ratings
4.4
●
●
●
Q1
Q3
Q2
question
II.G-7
Q4
3. Analysis
P1: Are profile parallel?
H0 : µ1i − µ1,i−1 = µ2i − µ2,i−1 , i = 2, · · · , p
Cµ1 = Cµ2 or C(µ1 − µ2 ) = 0
−1
1
0 0

1 0 
C =  0 −1
.
0
0 −1 1
P2: Are profile identical? (if parallel)
H0 : µ1i = µ2i for i = 1, · · · , p.
If parallel, this is equivalent to 10 µ1 = 10 µ2
or 10 (µ1 − µ2 ) = 0.
P3: Are profile level? (if identical, and all questions are of same scale).
H0 : µ11 = µ12 = · · · = µ1p = µ21 = · · · = µ2p .
Test statistic: T 2 = (estimate)0 (cov)−1 (estimate).
P1 :
P2 :
P3 :
T 2 = [C(X1 − X2 )]0 (CSC0 )−1 [C(X1 − X2 )]; reject H0 if T 2 > Tα2
T 2 = [10 (X1 − X2 )]0 (10 S1)−1 [10 (X1 − X2 )]; reject H0 if T 2 > F1,n−2 (α)
T 2 = n(CX)0 (CSpooled C0 )−1 (CX); reject H0 if T 2 > Fp−1,n−p (α),
1
1
where X = grand average, n = n1 + n2 , S =
+
Spooled
n1 n2
(n − 2)(p − 1)
Tα2 =
Fp−1,n−p (α).
n−p
II.G-8