11.4 Two-Tailed Tests about a Population Mean:
Large Sample Case
Objective:
Find a sensible statistical procedure to test
H 0 :   0 vs. H a :   0 .
Derivation of a sensible test:
   2 

  . Then, it is
is true, X  N   0 , 

  n 
Intuitively, if H 0 :    0
very likely that the sample mean
On the other hand, if
x
x
should have value close to
is much larger or smaller than
0 .
 0 , it seems
to indicate that H a :    0 is a sensible hypothesis. Thus, a sensible
statistical procedure would be
where
c
reject H 0 :
x  0  c
not reject H 0 :
x  0  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 .
Since
2





  0 , X  N  0 , 
 .

 n  

Then,
1
  the probabilit y that wro ngly reject H 0

 P( H 0 is true but is rejected)  P    0 ; X   0  c







 X  0
c 
c

 P

  P Z 










n
n

n  



c





n
 z  c 
z
2
Thus,
x  0 
reject H 0 :
x  0 
not reject H 0 :
2
n
.
z
2
n
z
2
n
is a sensible statistical procedure. Furthermore, denote
z

Thus, by dividing
n
x  0

X
x  0

.


n

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 x as    0

 P X  0  x  0 ,   0

 P Z  z 
2


 X  0

x  0
 P

,   0 

 

n
n


Example:
Objective: check if the produced golf balls have an average distance in carry and
roll of 280 yards as n  36, x  278.5,   12 and   0.05 ,
[solutions:]
H 0 :   0  280 vs. H a :   0  280 .
Then,
z
x  0
278.5  280

 0.75 .


12





n
36 


Since
z   0.75  0.75  z  z 0.025  1.96 ,
2
we thus do not reject H 0 . In addition,
p - value  P Z  z   P Z   0.75   P Z  0.75  0.4532    0.05
Therefore, we can not reject H 0
General Case: as n  30 and level of significance
H 0 :   0
vs
H a :   0
 As  is known, let
z
x  0

X
x  0

.


n

Then,
reject H 0 :
z  z
not reject H 0 :
z  z
In addition,
3
2
2

p - value  P Z  z

 As  is unknown,
z
x  0
x  0

sX
 s
.


n

Then,
In addition,
reject H 0 :
z  z
not reject H 0 :
z  z
2
2
p - value  P Z  z

Interval estimation and hypothesis testing
For
H 0 :   0
vs.
H a :   0
with

known, we do not
reject H 0 as
z 
x  0
 z

2
  z

2
0  x
 z

2
n
  z
n

n
2
 x  z
  0  x  z

2
n

n
2
  0  x  z

2
n

 

  0 falls into  x  z
, x  z

2
2
n
n

  0 falls into 1     100% confidence interval for 
4
On the other hand, we reject H 0 as


0   x  z
, x  z
 
2
n
n 
 0 do not falls into 1     100% confidence interval for 

2
Note:
Since for any
H 0 :   0
H a :   0 ,
vs.
  P Z  z ,    0   1    P Z  z ,    0  .
2
2




Thus,








X




X
0
1    P Z  z ,    0   P
 z   P  0
 z 
2
2
2


 

 

n
n








0  X


 
 P   z 
 z   P  z
  0  X  z


2
2
2
2
n
n




n




 
 P X  z
  0  X  z

2
2
n
n


Example (continue):
In the previous example, n  36, x  278.5,   12 ,   0.05
H 0 :   0  280 vs. H a :   0  280 .
We can also use confidence interval approach to hypothesis testing. A 95% confidence
interval for  is
x  z 0.05
Since

2
n
 278.5  1.96
12
36
 278.5  3.92  274.58, 282.42
 0  280  274.58, 282.42 , we do not reject
5
H0 .
A confidence Interval approach to hypothesis testing
n  30 , level of significance
H 0 :   0
Step 1: Construct a
vs.
1   100%

and
H a :   0
confidence interval
 As  is known,
x  z

2

 

  x  z
, x  z

2
2
n
n
n

 As  is unknown,
x  z
Step 2: If
2
0
s

  x  z
2
n

s
, x  z
2
n
s 

n
falls into the above confidence intervals, then
do not reject H 0 . Otherwise, reject H 0 .
Example 1:
The average starting salary of a college graduate is $19000 according to government’s
report. The average salary of a random sample of 100 graduates is $18800. The
standard error is 800.
(a) Is the government’s report reliable as the level of significance is 0.05.
(b) Find the p-value and test the hypothesis in (a) with the level of significance
  0.01.
(c) The other report by some institute indicates that the average salary is $18900.
Construct a 95% confidence interval and test if this report is reliable.
[solutions:]
(a)
H 0 :    0  19000 vs. H a :    0  19000,
n  100, x  18800, s  800,   0.05
Then,
6
z 
x  0
18800  19000

  2.5  2.5  z  z 0.025  1.96 .
s
800
2
n
100
Therefore, reject H 0 .
(b)
p - value  P Z  z   P Z  2.5  2  PZ  2.5  0.0124  0.01
Therefore, not reject H 0 .
(c)
H 0 :    0  18900 vs H a :    0  18900,
A 95% confidence interval is
x  z
Since
s
n
2
 18800  z 0.025
800
100
 18800  1.96  80  18643.2, 18956.8 .
 0  18900  18643.2, 18956.8 , Therefore, not reject H 0 .
Example 2:
A sample of 49 provides a sample mean of 38 and a sample standard deviation of 7.
Let
  0.05 . Please test the hypothesis
H 0 : u  40 vs. H a : u  40 .
based on
(a) classical hypothesis test
(b) p-value
(c) confidence interval.
[solution:]
x  38, s  7, u 0  40, n  49, z 
x  u 0 38  40

 2 .
s
7
n
49
(a)
z  2  1.96  z 0.025
we reject
H0 .
7
(b)
p  value  P Z  z   P Z  2  2 * (1  0.9772)  0.0456  0.05  
we reject
H0 .
(c)
100  (1   )%  95% confidence interval is
x  z
Since
s
2
n
 38  z 0.025
7
49
 38  1.96  36.04, 39.96.
40  36.04, 39.96, we reject H 0 .
Online Exercise:
Exercise 11.4.1
Exercise 11.4.2
8