12.2 Hypothesis Test about the Difference
Large Sample Case ( n1  30, n2  30 ):
I.
Motivating Example :
Objective:
we want to test if the mean scores in two training center are different
1 : the mean score in the first training center.
 2 : the mean score in the second training center.
We want to test
H 0 : 1   2 vs. H a : 1   2
( H 0 : 1   2  0 vs. H a : 1   2  0 )
with
  0.05 . In addition,
n1  30, n2  40, x1  82.5, x2  78,  12  82 ,  22  102
A sensible statistical procedure would be
where
c
reject H 0 :
x1  x2  c
not reject H 0 :
x1  x2  c
is some constant.
Next Question: how to determine the value of c ?
Answer: control  (the probability of making type I error) to
determine the value of c .
As H 0 is true,

1   2  X1  X 2  N 0,  X2
Then,
1
1X2
.
  the probabilit y that wro ngly reject H 0

 P ( H 0 is true but is rejected)  P 1   2 ; X 1  X 2  c
 X X
c
2
 P 1

  X X
 X1  X 2
1
2


c
X
 z
1X2





c
P Z 

 X1  X 2





 12
 c   X 1  X 2 z 
n1
2
2


 22
n2
z
2.
Thus,
reject H 0 :
x1  x2   X 1  X 2 z
not reject H 0 :
x1  x2   X 1  X 2 z
2
2
is a sensible statistical procedure. Furthermore, denote
z
x1  x2
X
1

X2
x1  x2
 12
n1
Thus, by dividing
X
1X2

 22 .
n2
on the both sides, the above sensible
statistical procedure can be simplified to
reject H 0 :
z  z
not reject H 0 :
z  z
2
2
In addition,
p - value  the probabilit y of making type I error by rejecting H 0 at x1  x2 as 1   2
 X X

x  x2
2
 P X 1  X 2  x1  x2 , 1   2  P 1
 1
, 1   2 
 

 X1  X 2
 X1  X 2

 P Z  z 


Therefore, in this example,
2
z 
x1  x 2
X
Thus, we reject

1X2
82.5  78
 2.09  2.09  1.96  z 0.025  z
8 2 10 2

30 40
2
H 0 . Also,
p  value  P Z  z   P Z  2.09  0.0366  0.05   .
we reject
H0
General
based on p-value.
Case:
significance
n1  30, n2  30 and level of
as

 As  1 ,  2 are known, let
z
x1  x2  0
X
1

x1  x2  0
X2
 12
n1
(I): H 0 : 1  2  0
vs.

 22 .
n2
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): H 0 : 1  2  0
vs.
H a : 1  2  0
Then,
reject H 0 :
z  z
not reject H 0 :
z  z
3
In addition,
p - value  PZ  z 
(III): H 0 : 1  2  0
H a : 1  2  0
vs.
Then,
In addition,
reject H 0 :
z  z
not reject H 0 :
z  z
2
2
p - value  P Z  z 
 As  1 ,  2 are unknown, let
z
x1  x2  0
x  x2  0
 1
sX1  X 2
s12
s22 .

n1
n2
(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): H 0 : 1  2  0
vs.
H a : 1   2   0
Then,
reject H 0 :
z  z
not reject H 0 :
z  z
In addition,
4
p - value  PZ  z 
(III): H 0 : 1  2  0
H a : 1   2   0
vs.
Then,
reject H 0 :
z  z
not reject H 0 :
z  z
In addition,
2
2
p - value  P Z  z 
Example:
Consider the following results for two samples randomly taken from two populations.
Sample 1
Sample 2
Sample size
64
49
Mean
1150
921
Standard deviation
90
65
Let
1
and
2
be the population means.
(a) For   0.05 , test
H 0 : 1   2  200
(b) For   0.01, please use p-value to test
using the classical hypothesis test.
H 0 : 1   2  200 .
(c) For   0.05 , please use the confidence interval method to test the hypothesis
H 0 : 1   2  200 .
[solution:]
(a)
x1  1150, x2  921, s1  90, s2  65, n1  64, n2  49, 0  200,  0.05 .
Then,
z
x1  x2   0
s12 s22

n1 n2

1150  921  200
90 2 652

64 49
5
 1.99  z  z0.05  1.645
Therefore, we reject H 0 .
(b)
p  value  PZ  z   PZ  1.99  0.0233    0.01 .
Therefore, we do not reject H 0 .
(c)
A 95% confidence interval for
x1  x 2   z

1   2
is
s12 s 22
90 2 65 2

 1150  921  z 0.025

n1 n 2
64 49
2
 229  1.96 14.587   200.41,257.59
Since
200  200.41,257.59 , we reject
.
H0 .
II. Small Sample Case ( n1  30, n2  30 ):
Similar to 11.1, two assumptions are made:
1. Both populations have normal distribution.
2. The variance of the populations are equal (  1
2
  22   2 )
Motivating Example :
Objective:
we want to test if the mean project-completion time using the new
software package is shorter than using current technology
1 : the mean project-completion time using the current technology
 2 : the mean project-completion time using the new software package.
We want to test
H 0 : 1   2 vs. H a : 1   2
( H 0 : 1   2  0 vs. H a : 1   2  0 )
with
  0.05 . In addition,
6
n1  12, n2  12, x1  325, x2  288, s1  40, s2  44 .
Thus,
s 
2
p
n1  1s12  n2  1s22
n1  n2  2
11  40 2  11  44 2

 1768.
12  12  2
A sensible statistical procedure would be
reject H 0 :
x1  x2  c
not reject H 0 :
x1  x2  c
where c is some constant. The above statistical procedure is equivalent to the
following statistical test:
reject H 0 : t 
not reject H 0 : t 
x1  x2
c

 c


s X1  X 2
s X1  X 2
c
s X 1  X 2
 c
As H 0 is true,
X1  X 2
1   2 
1
1 

S  
n
n
2 
 1
 T n1  n2  2
2
p
where
S p2
,
is the sample statistic with possible values s 2p .
Then,
  0.05  the probabilit y that wro ngly reject H 0
 P ( H 0 is true but is rejected)



X1  X 2
 P




1
2 1
 S p   
 n1 n2 


 P T n1  n2  2   c











c
X1  X 2

c 
  P







1
1
1
2 1


s 2p    
S



p

 n1 n2  
 n1 n2 


7
 c  tn1 n2 2,  tn1 n2 2, 0.05 .
Thus,
reject H 0 : t 
x1  x2

s X1  X 2
not reject H 0 : t  t n1 n2 2,0.05
x1  x2
1 1
s 2p   
 n1 n2 
 t n1 n2 2,
 t n1 n2 2,0.05  t n1 n2 2,
is a sensible statistical procedure. In addition,
p - value  the probabilit y of making type I error by rejecting H 0 at x1  x 2 as 1   2






X1  X 2
x1  x 2
 P

, 1   2 


1
1
1
2 1
s 2p   
 S p   

 n1 n2 
 n1 n2 


 PT n1  n2  2  t 
Therefore, in this example,
t
x1  x2
1
1
s 2p   
 n1 n2 
Thus, we reject

325  288
1
1
1768  
 12 12 
 2.16  1.717  t 22,0.05  t n1 n2 2,
H0 .
Also,
p  value  PT n1  n2  2  t   PT 22  2.16  0.0209  0.05  
we reject
H0
based on p-value.
8
General
Case:
significance
as
n1  30, n2  30 and level of

t
x1  x2  0

sX 1  X 2
x1  x2  0
1
1  .
s 
n  n 

2 
 1
2
p
(I):
H 0 : 1  2  0
vs.
H a : 1  2  0
Then,
reject H 0 :
t  t n1  n2  2,
not reject H 0 :
t  t n1  n2  2,
In addition,
p - value  PT n1  n2  2  t 
(II):
H 0 : 1  2  0
vs.
H a : 1   2   0
Then,
reject H 0 :
t  t n1  n2  2,
not reject H 0 :
t  t n1  n2  2,
In addition,
p - value  PT n1  n2  2  t 
(III):
H 0 : 1  2  0
vs.
Then,
9
H a : 1  2  0
t  t n1  n2  2 ,
reject H 0 :
t  t n1  n2  2 ,
not reject H 0 :
2
2
In addition,
p - value  PT n1  n2  2  t 
Example:
Consider the following results for two samples randomly taken from two normal
populations with equal variance
Sample 1
Sample 2
Sample size
10
12
Mean
48
44
Standard deviation
9
8
(a) Test H 0 : 1   2  3 vs. H a : 1   2  3 at   0.1 using the classical
hypothesis test.
(b) Test H 0 : 1   2  4 vs. H a : 1   2  4 at   0.05 using p-value.
(c) Test H 0 : 1   2  3 vs. H a : 1   2  3 at   0.05 using the confidence
interval method.
(d) At 95% confidence, how many data would have to be taken to provide an interval
with length 6 given equal sample sizes in two populations?
[solution:]
(a)
n1  10, n2  12, x1  48, x2  44, s1  9, s2  8, 0  3 .
Then,
s 
2
p
n1  1s12  n2  1s22
n1  n2  2
9  9 2  11  8 2

 71.65
10  12  2
Thus,
t 
x1  x2   0
1 1
s 2p   
 n1 n2 

48  44  3
1 1
71.65  
 10 12 
Therefore, we reject H 0 .
(b)  0  4
10
 1.93  t n  n 2,  t 20,0.05  1.7247
1
2
2



p  value  P T n1  n2  2  t   P T n1  n2  2 




x1  x2   0
1 1
s 2p   
 n1 n2 














48  44  4
 P T 20 

1 1

71.65  

 12 10  

 PT 20  2.207   P T 20  2.086  0.05
Therefore, we reject H 0 .
(c) A 95% confidence interval for 1   2 is
1 1
1
1
s 2p     48  44  t 20, 0.025 71.65    
1
2
2
 12 10 
 n1 n2 
 4  2.086  3.6243   3.56,11.56
x1  x2   t n n 2,
Since 3   3.56,11.56 , we do not reject H 0 .
(d)
As sample sizes are large and equal sample sizes ( n1  n2  n ) in two populations, the
1   100% confidence interval for 1   2 is
 x1  x2   z
The length of the confidence interval is 2 z
z
2
s12
s2
 2  z0.025
n
n
s12
s 22

.
n
n
2
2
s12 s 22

. Therefore,
n
n
92
82

 1.96
n
n


92
82

3
n
n
92
82
92  82  1.96 2
 3 



 61.89
 n
n
n
32
 1.96 
2




s12  s22 z2
n 

2
,
E
:
the
marginal
error
2


E




Therefore,
n  62
and total 124 data need to be taken.
11
Online Exercise:
Exercise 12.2.1
Exercise 12.2.2
12