11.5 Tests about a Population Mean: Small Sample
Case
Example:
Objective: determine whether Heathrow airport provides superior service as
n  12, x  7.75, s  1.215 and   0.05 . That is, test the hypothesis
H 0 :   0  7 vs. H a :   0  7 .
Intuitively, if H 0 :   7 is true, for example, as
the sample mean
x
  7 , it is very likely that
should have value close to 7. Thus, a sensible statistical
procedure would be
where
c
reject H 0 :
x 7c
not reject H 0 :
x 7c
is some constant. Note that the above test is equivalent to
reject H 0 :
not reject H 0 :
where
c
c 

s
x 7
 c
s
n
x 7
 c
s
n
c
1.215
n
is some constant.
12
Similar to previous section, we can use level of significance to obtain the constant
c . As H 0
is true, for ease of explanation, suppose
  7,
X  0
X 7

~ T n  1  T 11 .
S
S
n
12
Then,
1
  0.05  the probabilit y that wro ngly reject H 0
 P( H 0 is true but is rejected)




X 7


P
 c  P T (11)  c 
 S





12 




t
 c  tn1,  t11, 0.05 and c  1.215

.
12  11,0.05

Thus, a sensible statistical procedure would be
x 7
reject H 0 : t 
1.215



12 

not reject H 0 : t  t n1,

x  0
 t n1,  t11, 0.05
 s



n

with


 X 7

  P  7; reject H 0   P
 t11, 0.05 ;   7   0.05
 S

n


Thus,
t
x 7
1.215



12 


7.75  7
1.215



12 

 2.14  t11,005  1.796 .
Therefore, we reject H 0 . The other approach is to use p-value. p-value
is the probability of making a type I error as reject H 0 at
x
, i.e.,


 X  0 x  0

p - value  P

,   7   PT (n  1)  t   PT (11)  2.14  0.028.
s
 S

n
n


Since p  value    0.05 , we reject H 0 .
2
General Case: as n  30 and level of significance
t

H 0 :   0
vs.
x  0
x  0

sX
 s
.


n

H a :   0
Then,
reject H 0 :
t  t n 1,
not reject H 0 :
t  t n 1,
In addition,
p - value  PT (n  1)  t 

H 0 :   0
vs.
H a :   0
Then,
reject H 0 :
t  t n 1,
not reject H 0 :
t  t n 1,
In addition,
p - value  PT (n  1)  t  .

H 0 :   0
vs.
H a :   0
Then,
reject H 0 :
t  t n 1,
not reject H 0 :
t  t n 1,
3
2
2

In addition,
p - value  PT (n  1)  t

Example 1:
Objective: determine if the mean filling weight is exactly 16 ounces.
That is to test
H 0 :   0  16
vs.
H a :   0  16
with the following data
16.02
16.22
15.82
15.92
16.22
16.32
16.12
15.92
Suppose the level of significance is 0.05.
[solution:]
n  8, x  16.07, s  0.18 .
Thus,
t
x  0
16.07  16

 1.10 .
 s

 0.18





n
8


Since
t  1.10  1.10  t n 1,  t 7,0.025  2.365 ,
2
we do not reject
H0 .
Example 2:
A sample of 6 observations, 18, 20, 16, 19, 17, 18, is taken. Suppose the population
is normally distributed. Then,
(a) for the hypothesis,
H 0 :   20 vs. H a :   20 ,
using   0.05 , test the hypothesis based on the classical hypothesis test?
(b) for the hypothesis,
H 0 :   18 vs. H a :   18
using   0.05 , test the hypothesis based on p-value?
(c) for the hypothesis,
4
H 0 :   17 vs. H a :   17
using   0.05 , test the hypothesis based on the confidence interval approach?
[solutions:]
6
n  6, x 
18  20  16  19  17  18
 18, s 
6
 x  x 
i 1
2
i
6 1
 2.
(a)  0  20,
t
x   0 18  20

 3.464  t n1,  t 5, 0.05  2.015  reject H 0
s
2
n
6
(b)  0  18,
x   0 18  18

 0  p - value  PT (n  1)  t   PT (5)  0  0.5    0.05
s
2
n
6
 not reject H 0
t
(c) A 95% confidence interval for  is
x  t n1,
2
s
2
 18  t 5,0.025
 18  2.571  0.577   16.516,19.483
n
6
Since 17  16.516,19.483 , we do not reject H 0 .
Example 3:
Consider the following hypothesis test,
H 0 :   50 vs. H a :   50. .
Assume a sample of 16 items provides the following test statistics. Use   0.05 , test
the hypothesis based on p-value as
(i)
t  1.055
(ii)
t  3.261 .
[solution:]
(i)
n  16,  p - value  PT (n  1)  t   PT (15)  1.055  PT (15)  1.753    0.05
 do not reject H 0
5
(ii)
n  16,  p - value  PT (n  1)  t   PT (15)  3.261  PT (15)  1.753    0.05
 reject H 0
Online Exercise:
Exercise 11.5.1
Exercise 11.5.2
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