11.3 Modern Approach: P-value
Example (continue):
In the coffee example,
H0 :   3
Suppose
vs
H a :   3 ,  2  0.18 2 ,   0.01
x  2.92 , the p-value in this case is
p - value  the probabilit y of making type I error by rejecting H 0 at x  2.92 as   3
 PX  x ,   3  PX  2.92,   3


 X 3
2.92  3 
8

 P

  0.0038
  P Z 
0
.
18
0
.
18
3




36
36 

 p - value  0.0038  level of significan ce    0.01
(the error we might make < (the allowable error)
as reject H 0 at x  2.92 )
The error we might make by rejecting H 0 at x  2.92 is smaller than the
allowable error. Therefore, we reject H 0 .
Suppose
x  2.97 ,
p - value  the probabilit y of making type I error by rejecting H 0 at x  2.97 as   3
 PX  x ,   3  PX  2.97,   3


 X 3
2.97  3 
 P

  PZ  1  0.1587
0
.
18
0
.
18


36
36 

 p - value  0.1587  level of significan ce    0.01
Since the error we might make by rejecting H 0 at x  2.97 is larger than the
allowable error. Therefore, we can not reject H 0 .
1
Note:
The conclusion based on the p-value is the same as the one based on
the previous hypothesis testing procedure.
Note:
In the above example, let
x
be the sample mean and  0  3 . Then,








X


x


x


0
0
0
  P Z 

p - value  PX  x ,    0   P

 


 


.
 





n 
n  
n  



For example, as



2.92  3 
x  2.92  p - value  P Z 
  0.0038
0
.
18


36 




2.97  3 
x  2.97  p - value  P Z 
  0.1587
0
.
18


36 

General Case: as n  30 and level of significance
(a) As  is known, let
z
x  0

X
x  0

.


n

(I):
H 0 :   0
Then,
vs.
H a :   0
p - value  PZ  z 
2

(II):
H 0 :   0
Then,
vs.
H a :   0
p - value  PZ  z 
(b) As  is unknown,
z
x  0
x  0

sX
 s
.


n

(I):
H 0 :   0
Then,
vs.
H a :   0
p - value  PZ  z 
(II):
H 0 :   0
Then,
vs.
H a :   0
p - value  PZ  z 
In (a) and (b), we then compare p-value and the predetermined level
of significance

.
If p - value   , then reject H 0 .
Otherwise, we do not reject H 0 .
3
Example 1 (continue):
H 0 :   600
Suppose
vs.
H a :   600 .
n  36, x  605, s  12 . What is the conclusion based on p-value as
  0.05 .
[solutions:]
 0  600 . Then,




x  0 

p - value  P Z 


 s




n  






605  600 

P Z
 PZ  2.5  0.0062  0.05  

12




36  


H0 .
Therefore, we reject
Example 2 (continue):
n  50
A sample with
deviation
of
12.
provides a sample mean of 36 and sample standard
Consider
H 0 :   30 vs. H a :   30 .
the
following
hypothesis
Find the p-value. What is your conclusion as
  0.0001?
[solution:]
p  value  PZ  z   PZ  3.535  0.0002 .
Since
p  value  0.0002    0.0001 ,
we do not reject
test
H0 .
Online Exercise:
Exercise 11.3.1
Exercise 11.3.2
4