12.3 Inference About the Difference: Matched
Samples
Motivating Example:
Objective:
we want to compare two production methods. The original data are
1
2
3
4
5
6
Method
1
6.0
5.0
7.0
6.2
6.0
6.4
Method
2
5.4
5.2
6.5
5.9
6.0
5.8
Worker
1 : the mean completion time for production method 1
 2 : the mean completion time for production method 2
We want to test
H 0 : 1   2  0 vs. H a : 1   2  0
with
  0.05 .
Two different designs can be considered:
1. A simple random sample of worker using method 1 is selected. A
second simple random sample of worker using method 2 is selected.
The methods in 10.1 and 10.2 can be used.
Disadvantage:
the variation between workers is not considered. The effect of the
different production methods might not be distinguished with the effect of
the capability of different workers, especially in small sample.
2. (Matched sample design): only one simple random sample of
workers is selected. Each worker use both methods. Each worker
provides a pair of data values, one value for method 1 and the other
for method 2.
Advantage:
1
Two production methods are tested under similar conditions (i.e., with the
same workers); Variation between workers is eliminated.
x
Let
1,1
, x2,1 , x1, 2 , x2, 2 ,, x1, n , x2, n  the matched sample from
two populations, where
x1,1 , x1, 2 ,, x1, n are from population 1 and
x2,1 , x2, 2 ,, x2, n are from population 2. In this example,
Let
n
d i  x1,i  x2,i , d 
d
i 1
 d
n
i
n
,s 
2
d
i 1
d
2
i
n 1
General Case:
 n  30 and level of significance
z 
sd
n

d  0
sd
,
n
is the standard error of d
(I):
H 0 : 1  2  0
vs.
H a : 1  2  0
Then,
In addition,
reject H 0 :
z   z
not reject H 0 :
z   z
p - value  PZ  z 
(II):
2
.
H 0 : 1  2  0
vs.
H a : 1   2   0
Then,
In addition,
reject H 0 :
z  z
not reject H 0 :
z  z
p - value  PZ  z 
(III):
H 0 : 1  2  0
vs.
H a : 1  2  0
Then,
In addition,
z  z
not reject H 0 :
z  z
2
2
p - value  P Z  z  .
1001   %
d  z
reject H 0 :
2
confidence interval for
1  2
is
sd
sd
sd 

 d  z
, d  z
2
2
n
n
n


 n  30 and level of significance
t 

d  0
sd
,
n
(I):
3
H 0 : 1  2  0
vs.
H a : 1  2  0
Then,
In addition,
reject H 0 :
t  t n 1,
not reject H 0 :
t  t n 1,
p - value  PT n 1  t 
(II):
H 0 : 1  2  0
vs.
H a : 1   2   0
Then,
In addition,
reject H 0 :
t  t n 1,
not reject H 0 :
t  t n 1,
p - value  PT n 1  t 
(III):
H 0 : 1  2  0
vs.
H a : 1  2  0
Then,
reject H 0 :
t  t n 1,
2
not reject H 0 :
t  t n 1,
2
In addition,
p - value  PT n  1  t  .
1001   %
confidence interval for
4
1  2
is
sd
sd
sd 

 d  t n 1,
, d  t n 1,
2
2
n 
n
n 
d  t n 1,
2
Motivating Example (continue):
di
0.6
-0.2
n
d
0.5
6
d d
i
i 1
n

i 1
6
0.3
 d
n
i
 0.3, s d 
i 1
d
n 1
 d
6
2
i
0.0

i 1
0.6
d
2
i
 0.335 .
5
Then,
t
d

sd
n
0.3
0.335
 2.19  t  2.19  2.571  t 5,0.025  t n 1,
2.
6
Thus, we do not reject H 0 . In addition,
p  value  PT n  1  t   PT 5  2.19  0.08  0.05   .
Therefore, we also do not reject H 0 based on p-value.
A 95% confidence interval for is
s
0.335
0.335
d  t n 1, d  0.3  t 5, 0.025 
 0.3  2.571 
 0.3  0.35   0.05, 0.65 .
2
n
6
6
Since 0   0.05, 0.65 , we do not reject H 0 .
Example:
To determine the effectiveness of a new weight control diet, ten randomly selected
students observed the diet for 4 weeks with the results shown below.
Dieter
Weight (before)
Weight (after)
A
138
135
B
151
147
C
129
132
D
125
127
E
168
155
F
139
131
G
152
144
H
140
142
5
I
137
137
J
180
180
We like to test the hypothesis
H 0 : 1   2 ,
where
1
and
2
are the
mean weights of the students before and after taking the weight control diet,
respectively.
(a) For   0.1, test the above hypothesis using the classical hypothesis test.
(b) For   0.05 , please use the confidence interval method to test the above
hypothesis.
(c) For   0.2 , please use p-value to test the above hypothesis.
[solution:]
(a)
d1
d2
d3
d4
d5
d6
d7
d8
d9
d 10
3
4
-3
-2
13
8
8
-2
0
0
Therefore,
t
d  2.9, s d  5.322 . Thus,
d
 sd



n


We do not reject
2.9
 5.322



10 

 1.723  t  1.723  1.833  t 9,0.05  t n 1, .
2
H0 .
(b) A 95% confidence interval for
d  t n1,
Since
sd
2
n
 2.9  t9,0.025 
1   2
is
5.322
5.322
 2.9  2.262 
 2.9  3.807   0.907, 6.707 .
10
10
0   0.907, 6.707 , we do not reject H 0 .
(c)
p  value  PT n  1  t   PT 9  1.723  0.2  PT 9  1.383 .
Therefore, we reject
H0 .
Online Exercise:
6
Exercise 12.3.1
Exercise 12.3.2
7