Hypothesis Testing
• Tests on a single sample:
•
If we have the experiment X1,
X2,………, Xn representing a random
sample from a distribution with
mean   and variance  2  0
Then the hypotheses are:
H 0 :   0
H a :   0
Hypothesis Testing
• Standardization of X
Z
X 


n X   

n
If
  0
and
n X   

follow an N(x;0,1)
Hypothesis Testing
• Then
P(  z  
2
n X   

 z )  1  
2
Hypothesis Testing
• From diagram we reject H0 if mean
less than a or mean greater than b
where:
a  0  z
2

n
or b   0  z 
2

n
Hypothesis Testing
• Example 1:
A random sample of 100 recorded death
in USA the past year, showed an
average life span of 71.8 years.
Assuming a population standard
deviation of 8.9 years, does this seem
to indicate that the mean of life span
today is greater than 70 years? Use
0.05 level of significance.
Hypothesis Testing
• Solution:
1. Hypothesis:
H 0 :   70
H a :   70
at   0.05
Hypothesis Testing
•
Solution:
Hypothesis Testing
• Solution:
2. Critical region
1    0.95
From table z = 1.645
Calculate z as: z  n X     10  (71.8  70)  2.02

8.9
Hypothesis Testing
• Solution:
 2.02 is greater th an 1.645
Then we reject H0 so we conclude that the life
span today is greater than 70 years.
P-value
While z is at rejection area then
P-value = (1-P( z < 2.02))
From table
P-value = 1- 0.9783 = 0.0217 < alpha
so that we reject null hypothesis.
Hypothesis Testing
• Tests on single sample [variance
unknown]
2

is
known
and
S
is unknown
• When
then we should use student t-distribution under
the following assumptions:
1.The random variables (X1, X2, ………., Xn)
represent random sample from a normal
Distribution with
 is known and  is unknown
2
Hypothesis Testing
Then the random variable
n (X  )
S
has a t.distribution with (n-1) degree
of freedom.
Hypothesis Testing
2. The structure of the test is identical
that for sigma is known with the
exception that the value sigma in the
test is replaced by computed estimate
S.
3.The standard normal distribution is
replaced by t.distribution as:
t ,( n 1) 
n (x  )
S
Hypothesis Testing
•
•
•
Example 2:
The Edison Electric Institute has published
figures on the number of kilowatt hours
used annually by various houses. It is
claimed that a vacuum cleaner uses an
average of 46 kilowatt hours per year. If a
random sample of 12 homes they found
vacuum cleaner 42 kilowatt hours per year
with standard deviation 11.9 kw/hr.
Does this suggest at the 0.05 level of
significant that vacuum cleaner use less
than 46 kw/hr?
Hypothesis Testing
• Solution:
1. Hypothesis
H 0 :   46 kw/hr.
H a :   46 kw/hr.
  0.05 , and df  (12 - 1)  11
Critical region: 1- .05 = 0.95
Hypothesis Testing
• From table then
• T < -1.796
• 4. Computations: mean = 42, S =11.9,
n=12.
12 (42  46)
t
 1.164
11.9
Hypothesis Testing
• While calculated t is less than tabulated
t 1.164 < 1.796
Then
We cannot reject H0 and we conclude
that the average number of kilowatt
hours used is not less than 46.
Hypothesis Testing
• Example 3:
• Test the hypothesis that the average
content of containers of a particular
lubricant is 10 litters if the contents of
random sample of 10 containers are:
10.2, 9.7, 10.1, 10.3, 10.1,
9.8,9.9,10.4,10.3,and 9.8 litters use 0.01
level of significant and assume that the
distribution of content is normal.
Hypothesis Testing
• Solution:
• Hypothesis
H 0 :   10 litter.
H a :   10 litters.
  0.01 with two tails


2
 0.005 and degree of freedom  9
Hypothesis Testing
3.Then the critical region is t < -3.25 and t
> 3.25 , these t are from table
4. Calculated t:
t
n (x  )


10 (10.06  10)
 0.77
0.246
Hypothesis Testing
Solution:
10.2  9.7  .....  9.8
x
 10.06
10
2
2
(
10
.
2

10
.
06
)

....

(
9
.
8

10
.
06
)
2 
 0.060516
9
   0.246
Calculated t is less than tabulated t then
we don’t reject H0 so that the mean is 10