11.2 One-Tailed Tests about a Population Mean:
Large Sample Case
Example (continue):
Suppose the population variance is
 2  0.18 2 . Also, suppose a random sample
of 36 cans of coffee is selected. Intuitively, if H 0 :   3 is true,
have
large
value
   2 
( X  N   , 
  ).
  n  


For
example,
as
x
should
 7
  0.18  2 
X  N  7, 
 
  36   . Then, it is very likely that the sample mean x


,
should
have value close to 7 and therefore should be much larger than 3. On the other hand,
if
x
is much larger than 3, it seems to indicate that H 0 :   3 is a sensible
hypothesis. That is, not rejection of H 0 :   3 might be a sensible choice.
Thus, a sensible statistical procedure would be
where
c
reject H 0 :
x 3 c
not reject H 0 :
x 3 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 .
Suppose
H0
  0.01
is the predetermined Type I error (the level of significance). As
is true, for ease of explanation, suppose
  0.18  2 
  3, X  N  3, 
  .
  36  


1
Then,
  0.01  the probabilit y that wro ngly reject H 0
 P( H 0 is true but is rejected)  P  3; X  3  c 




X

3
c

 P

  0.18
  0.18






36  
36  


c
 0.18



36 





c

P Z 

 0.18




36  


  z 0.01  2.33  c  2.33  0.03  0.0699 .
Thus, a sensible statistical procedure would be
reject H 0 :
x  3  0.0699
not reject H 0 :
x  3  0.0699
with
  P  3; reject H 0   PX  3  0.0699;   3  0.01 .
Based on the above statistical procedure, we can use sample mean
x
to test the
hypothesis. For example,
if x  2  x  3  2  3  1  0.0699  reject H 0
if x  2.92  x  3  2.92  3  0.08  0.0699  reject H 0
if x  2.97  x  3  2.97  3  0.03  0.0699  not reject H 0
if x  4  x  3  4  3  1  0.0699  not reject H 0 .
Note:
Even as the true mean
  3 , not   3 , the Type I error is still
smaller than 0.01, i.e.,
  PX  3  0.0699;   3  0.01 .
2
That is, the above statistical procedure can control the Type I error
u  3.
less than or equal to 0.01 for any
Note:
The hypothesis in this example is
H 0 :   0  3
vs.
H a :   0  3 .
The resulting statistical procedure is
reject H 0 :
x  3  0.0699
not reject H 0 :
x  3  0.0699

reject H 0 :
not reject H 0 :
x 3

x  0
  z   z0.01 




n

 0.18



36 

x  0
  z0.01




n

 0.0699
 0.18



36 

General Case: as n  30 and level of significance
 As  is known, let
z
x  0

X
x  0

.


n

(I):
H 0 :   0
vs.
H a :   0
Then,
reject H 0 :
z   z
not reject H 0 :
z   z
(II):
H 0 :   0
vs.
3
H a :   0

Then,
reject H 0 :
z  z
not reject H 0 :
z  z
 As  is unknown,
z
x  0
x  0

sX
 s
.


n

(I):
H 0 :   0
vs.
H a :   0
Then,
reject H 0 :
z   z
not reject H 0 :
z   z
(II):
H 0 :   0
vs.
H a :   0
Then,
reject H 0 :
z  z
not reject H 0 :
z  z
Example 1:
H 0 :   600
Suppose
vs.
H a :   600 .
n  36, x  605, s  12 . What is the conclusion as   0.05 ?
[solution:]
z
x  0
605  600 5

  2.5  z  z0.05  1.64
.
 s  12
 2

 

n 
36 

 reject
H0
4
Example 2:
n  50
A sample with
provides a sample mean of 36 and sample standard
deviation of 12. Consider the following hypothesis test
H 0 :   30 vs. H a :   30 .
Using a critical value, test the hypothesis at the 10% level of significance (classical
approach).
[solution:]
n  50,   0.1, x  36, s  12, 0  30,
z
Therefore, we reject
x   0 36  30

 3.535  z 0.1  1.28
s
12
50
n
H0 .
Note:
 z
in (I) or
z
in (II) are called critical value.
Online Exercise:
Exercise 11.2.1
5