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Institute of Technology of Cambodia
2013-2014
Chapter II :
HYPOTHESIS TESTING
1. General
1.1. Introduction
Statistics develop techniques and methods for
analyzing data from observation, to identify the
characteristics of the population and identify a model
capable of generating these data. In this context, it is
necessary to make assumptions, that is to say to
make assertions about these features or this model.
Definition :
- A hypothesis is called parametric if it relates to
the parameters of the distribution. It is called
non-parametric in other cases.
- A parametric assumption is called simple if it is
associated with a unique value. It is said in other
cases multiple.
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Most often, the situation is summed up in two
alternative hypothesis made H0 and H1 , mutually
exclusive and are respectively called the null
hypothesis, or fundamental, and the alternative
hypothesis, or otherwise.
In general, the hypothesis H0 and H1 do not play
symmetric roles, and is chosen for the null
hypothesis H0 Assuming that we believe or we take,
or one that allows you to make calculations or one
whose rejection has serious consequences.
1.2. Test
Hypothesis to confront, H0 and H1 , are identified,
their validity is put to the test using a hypothesis test.
Definition : Hypothesis testing is a decision rule that
allows, on the basis of observed data and the risks of
error determined to accept or reject a statistical
hypothesis.
The decision rule is a test based on the observation
of a sample and not on the basis of full information,
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one is never sure of the accuracy of the conclusion:
there is always a risk error.
Definition : The first kind of error is to reject H0
wrongly: the risk of type I error is denoted  , is the
risk of error is taken into rejecting H0 while it is true.
It is also called the significance level of the test or,
more simply, the level of test.
P  type I error   
Definition : The second kind of error is to reject H1
wrongly : the risk of type II error is noted  , is the
risk of error is taken into rejecting H1 while it is true.
P  type II error   
  1–  is called the power of the test.
It tries to build tests that reduce risks to acceptable
levels. In general, we impose a threshold  not to
exceed (eg 1%, 5%, 10% by default), and given this
constraint, we seek to build the tests with the
highest possible power.
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Two types of decision error in a hypothesis test
Reject H0
Accept H0
H0 true
Type I error
good decision
H1 true
good decision
Type II error
1.3. Test Statistics
Definition:
A test based on a sample size n is determined by
region R of n called critical region, or region of
rejection of the hypothesis H0 .
The complementary A of R is called the acceptance
region of H0 .
The decision rule is to test the following : if
X   X 1 ,..., X n  is the vector of observed values, it
was decided to refuse H0 (and accept H1 ) if X  R ,
and decides to accept H0 if X  R .
In practice, we try to define a random variable D ,
called decision variable or test statistics, and the
distribution is known, at least under hypothesis H0 .
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The critical region will be the region where the
probability values of the test statistics tends to
increase when H0 is not true. This region is defined
using the risk  the first kind of test.
1.4. Critical Probability
Definition : Critical probability or critical level or Pvalue of hypothesis H0 ,  *, is the level of test at
which one rejects H0 given the results of
observations. The critical level  * depends on the
results of observations and test that uses. Knowing
the critical level  *, we can say what decision we
will take whatever level  chosen, in this case,
Reject H0 if and only if    *
The critical level is widely used in practice, because it
gives more information than a simple decision with a
pre-set level.
For example, if there is  the test statistic, a value of
the test statistics T , the critical probability  * the
hypothesis is given by :
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-  *  P T   | H0 
-  *  P T   | H0 
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(right tail test) ;
(left tail test) ;
-  *  P  T   | H0  (two tailed test).
The critical probability provides a measure of
credibility to the hypothesis H0 :
- a very low value of the critical probability means
that H0 is not valid,
- too high a value can doubt the randomness of
the experience and reliability of the data and
calculations.
2. Parametric Tests
Let X1 ,..., X n be an iid sample drawn from X , one
random variable depending on a parameter  , and
ˆn  T  X 1 ,..., X n  an estimator (a statistical test) of
 , and  0  . We consider the following simple
hypothesis testing :
(a)
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H0 :    0
H1 :    0
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(b)
(c)
H0 :    0
H1 :    0
H0 :    0
H1 :    0
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left tail test
two tail test
Example 1 : The manufacturer of a certain brand of
cigarettes says that nicotine does not exceed an
average 2.5 milligrams. Propose null and alternative
hypothesis to be used in testing this claim.
Example 2: Suppose that we are interested in the
mean compressive strength of a particular type of
concrete. Specifically, we are interested in deciding
whether or not that the mean compressive strength
is 2500 psi. Built the hypothesis for this problem.
Summary procedure for testing the hypothesis:
1. Propose the null hypothesis H0 where    0 .
2. Choose a suitable alternative hypothesis of one of
  0 ,   0 ,   0 .
3. Choose the level of test .
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4. Select the appropriate statistical test and establish
the critical region. (If the decision is based on the
critical probability or P-value, it is not necessary to
provide the critical region.)
5. Calculate the value of the test statistic of the
sample data.
6. Decision: Reject H0 if the test statistic has a value
in the critical region (or if the P-value is smaller
than or equal to the desired level of the test );
otherwise do not reject H0 .
2.1. Tests comparing a mean value of a
reference.
Theorem 1 : Let X1 ,..., X n be an iid sample drawn
from X , a normal random variable with expectation
 and variance  2 . We propose to test the
hypothesis :
H0 :   0
(a)
(b)
(c)
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H1 :   0 right-tailed test
H1 :   0 left-tailed test
H1 :   0 two-tailed test
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1. If the variance  2 is known, then
X 
- The test statistics Z  n

n
- Under the Hypothesis
X n  0
H0 : Z 
~ N  0,1

n
(1-1-a). Case (a) – right-tailed test:
The acceptance region of the null hypothesis H0
is an interval of the form: ,  z  where
z  0 and P  Z  z   
(1-1-b). Case (b) – left-tailed test:
The acceptance region of the null hypothesis H0 is
an interval of the form:   z ,  where z  0
and P  Z   z   
(1-1-c). Case (c) – two-tailed test:
The acceptance region of the null hypothesis H0
is an interval of the form:   z /2 ,  z /2  where
z /2  0 and P  Z  z /2   
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2. If the variance  2 is unknown, then
Xn  
- The test statistics T 
Sn
n
- Under the Hypothesis H0 :
X  0
T n
~ Student  n  1
Sn
n
(1-2-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form: , t  where t  0
and P T  t   
(1-2-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form:  t ,  where t  0
and P T  t   
(1-2-c). Case (c) – two-tailed test:
The acceptance region of the null hypothesis H0
is an interval of the form:  t /2 , t /2  where
t /2  0 and P  T  t /2   
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Example 3: The burning rate of a rocket propellant is
being studied. Specifications require that the mean
rate must be 40 cm/s. Furthermore, suppose that we
know that the standard deviation of the burning rate
is approximately 2 cm/s. The experimenter decides
to specify a type I error probability  0.05, and he
will base the test on a random sample of size n  25 .
Is the burning rate equal to 40 cm/s if the sample
mean burning rate obtained is 41.25 cm/s. Calculate
the P-Value.
Example 4: The Edison Electric Institute has
published figures on the annual number of kilowatthours spent by various appliances. It is claimed that a
vacuum spends an average of 46 kilowatt-hours per
year. If a random sample of 12 houses including a
planned study indicates that the cleaners spend an
average of 42 kilowatt-hours per year with a
standard deviation of 11.9 kilowatt-hours at the test
level of 5%, this suggests that the vacuum spend an
average of less than 46 kilowatt-hours per year?
Assume that the population of kilowatt-hours to be
normal.
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Theorem 2 : Let X1 ,..., X n (with n  30 ) be an iid
sample drawn from X , a real random variable with
expectation  and variance  2  0. We propose to
test the hypothesis:
H0 :   0
(a)
(b)
(c)
H1 :   0 right-tailed test
H1 :   0 left-tailed test
H1 :   0 two-tailed test
1.If the variance  2 is known, then
Xn  
- The test statistics Z 

n
- Under the Hypothesis
X n  0
H0 : Z 
Î N  0,1

n
(2-1-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :  ,  z  where z  0
and P  Z  z   
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(2-1-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :   z ,   where z  0
and P  Z   z   
(2-1-c). Case (c) – two-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :   z /2 ,  z/2  where
z /2  0 and P  Z  z /2   
2. If the variance  2 is unknown, then
Xn  
- The test statistics Z 
Sn
n
- Under the Hypothesis
X n  0
H0 : Z 
Î N  0,1
Sn
n
(2-2-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :  ,  z  where z  0
and P  Z  z   
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(2-2-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :   z ,   where z  0
and P  Z   z   
(2-2-c). Case (c) – two-tailed test :
The acceptance region of the null hypothesis is
an interval of the form :   z /2 ,  z/2  where
z /2  0 and P  Z  z /2   
Example 4 : A researcher claims that the average
wind speed in a certain city is 8 miles per hour. A
sample of 32 days has average wind speed of 8.2
miles per hour. The standard deviation of the sample
is 0.6 mile per hour. Test level of 0.05, is there
enough evidence to reject the claim? Use the P-value
method.
2.2. Test medium: the case of two samples
Theorem 3 : Let X 1 and X 2 be two random
variables
independent
normal
expectancy
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respectively 1 and 2 and variance respectively  12
and  22 , and X 1, 1 ,..., X 1, n1 be an iid sample drawn
from X 1, X 2,1 ,..., X2, n2 be an iid sample drawn from
X 2 . We propose to test the hypothesis :
H0 : 1  2  d0
(a)
(b)
(c)
H1 : 1  2  d0
H1 : 1  2  d0
H1 : 1  2  d0
right-tailed test
left-tailed test
two-tailed test
1. If the variance  12 and  22 are known, then
- The test statistics
X 1  X 2    1  2 

Z
 12  22

n1 n2
- Under the Hypothesis H0 :
X1  X 2   d0

Z
~ N  0,1
2
2
1  2

n1 n2
(3-1-a). Case (a) – right-tailed test :
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The acceptance region of the null hypothesis H0 is
an interval of the form :  ,  z  where z  0
and P  Z  z   
(3-1-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :   z ,   where z  0
and P  Z   z   
(3-1-c). Case (c) – two-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :   z /2 ,  z/2  where
z /2  0 and P  Z  z /2   
2. If  12   22   2 where  2 is unknown, then
- The test statistics
X 1  X 2    1  2 

T
1 1
Sp

n1 n2
where S p 
 n1  1 S12   n2  1 S22
n1  n2  2
- Under the Hypothesis H0 :
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X

T
1
 X 2   d0
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~ Student  n1  n2  2 
1 1
Sp

n1 n2
(3-2-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :  , t  where t  0
and P T  t   
(3-2-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :  t ,   where t  0
and P T  t   
(3-2-c). Case (c) – two-tailed test:
The acceptance region of the null hypothesis H0
is an interval of the form :  t /2 , t/2  where
t /2  0 and P  T  t /2   
3. If  12   22 and both are unknown, then
- The test statistics
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X

T 
1
 X 2    1  2 
S12 S 22

n1 n2
- Under the Hypothesis H0 :
X1  X 2   d0

T 
~ Student  
S12 S22

n1 n2
where  
S
2
1
S
2
1
/ n1  S / n2
/ n1
2
2
  S
2
2
2

2
/ n2

2
n1  1
n2  1
(3-3-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :  , t  where t  0
and P T   t   
(3-3-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :  t ,   where t  0
and P T   t   
(3-3-c). Case (c) – two-tailed test :
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The acceptance region of the null hypothesis H0
is an interval of the form :  t /2 , t/2  where
t /2  0 and P  T   t /2   
Example 5 : A random sample of size n1  25 taken
from a normal population with a standard deviation
 1  5.2 has an average X1  81. A second random
sample of size n1  36 taken from a different
population normal with a standard deviation
 2  3.4 has an average X 2  76. Test the
hypothesis that  1   2 against the alternative
 1   2 . Quote of the P-value in your conclusion.
Example 6: A large automobile manufacturing
company is trying to decide whether to buy brand A
or brand B tires for its new models. To help reach a
decision, an experiment is conducted using 12 of
each brand. The tires are run until they wear out. The
results are
X A  37900 Kilometres,
Brand A:
S A  5100 Kilometres
X B  39800 Kilometres,
Brand B:
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S B  5900 Kilometres.
Test of the hypothesis test level of 0.05 there is no
difference in the two tire brands. Assume that the
populations are approximately normally distributed.
Theorem 4 : Let X A and X P be two normal random
variables dependent expectancy respectively  A and
 P ( X A and X P may be the same variable in a
population of processing and a reference population)
and X A, 1 ,..., X A, n is an iid sample drawn from X A ,
X P ,1 ,..., X P, n is an iid sample drawn from X P . We
1 n
put
and
D  X A  XP,
D   Di
n i 1
2
1 n
SD 
Di  D  .


n  1 i1
We propose to test the hypothesis:
H0 :  D :  A   P  d0
(a)
(b)
(c)
H1 : D  d0
H1 :  D  d0
H1 : D  d0
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left-tailed test
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Then,
D  D
- The test statistics T 
SD
n
- Under Hypothesis H0 :
D  d0
T
~ Student  n  1
SD
n
(4-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form: , t  where t  0
and P T  t   
(4-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0
is an interval of the form:  t ,  where t  0
and P T  t   
(4-c). Case (c) – two-tailed test:
The acceptance region of the null hypothesis H0
is an interval of the form:  t /2 , t /2  where
t /2  0 and P  T  t /2   
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Example 7: A physical education director says that
taking a particular vitamin haltéro one pill can
increase the strength. Eight athletes are selected and
tested for strength data by using the standard bench
press. After two weeks of regular training,
supplemented by vitamins, they are tested again.
Test the effectiveness of the dosage of the vitamin
  0.05. Each value of these data is the maximum
number of pounds the athlete can bench press.
Assume the variable is approximately normally
distributed.
Athlete
1
2
3
4
5
6
7
8
Before
210
230
182
205
262
253
219 216
After
219
236
179
204
270
250
222 216
2.3. Testing the value of a proportion
Theoreme 5 : Let X ~ B  p  and X 1 ,..., X n (with n
very large) an iid sample drawn from X . We
propose to test the hypothesis :
H0 : p  p0
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(a) H1 : p  p0
(b) H1 : p  p0
(c) H1 : p  p0
Then,
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right-tailed test
left-tailed test
two-tailed test
- The test statistics Z 
Xn  p
p 1  p 
n
- Under Hypothesis H0 :
X n  p0
Z
Î N  0,1
p0 1  p0 
n
(5-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0
is an interval of the form :  , z  where
z  0 and P  Z  z   
(5-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form :   z ,   where z  0
and P  Z   z   
(5-c). Case (c) – two-tailed test:
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The acceptance region of the null hypothesis H0 is
an interval of the form :   z /2 ,  z/2  where
z /2  0 and P  Z  z /2   
Example 8 : An oil company claims that a fifth of
homes in a certain city are heated by oil. We have
reason to doubt this statement though, in a random
sample of 1000 houses in this city, it is found that
236 are heated by oil? Use a significance level of
0.01.
2.4. Tests of proportions: if two samples
Theorem 6 : Let X 1 ~ B  p1  and X 2 ~ B  p2  be
two independent random variables ; X 1,1 ,..., X 1, n1 is
an iid sample drawn from X 1 and X 2, 1 ,..., X 2, n 2 is
an iid sample drawn from X 2 .
We propose to test the hypothesis:
H0 : p1  p2
(a)
(b)
H1 : p1  p2
H1 : p  p0
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right-tailed test
left-tailed test
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2013-2014
(c) H1 : p1  p2
two-tailed test
Then,
- The test statistics
X 1  X 2    p1  p2 

Z
p1 1  p1  p2 1  p2 

n1
n2
- Under Hypothesis H0 :
X1  X 2 

Z
Î N  0,1
1 1
ˆ
ˆ
X 1 X   
 n1 n2 
n1 X 1  n2 X 2
ˆ
where X 
.
n1  n2
(6-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis H0 is
an interval of the form: ,  z  where z  0
and P  Z  z   
(6-b). Case (b) – left-tailed test:

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
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The acceptance region of the null hypothesis H0 is
an interval of the form:   z ,  where z  0
and P  Z   z   
(6-c). Case (c) – two-tailed test:
The acceptance region of the null hypothesis H0 is
an interval of the form:   z /2 ,  z /2  where
z /2  0 and P  Z  z /2   
Example 9: A cigarette manufacturing company
distributes two brands of cigarettes. It is found that
56 of 200 smokers prefer brand A and 29 of 150
smokers prefer brand B, can we conclude at the 0.06
test that brand A is sold more that brand B?
2.5. Tests comparing
reference value
a
variance
to
a
Theorem 7 : Let X ~ N   , 2  and X 1 ,..., X n is
an iid sample drawn from X .
We propose to test the hypothesis :
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H0 :  2   02
(a) H1 :  2   02
(b) H1 :  2   02
(c) H1 :  2   02
Then,
right-tailed test
left-tailed test
two-tailed test
- The test statistics
2
n

1
S


n
2 
2
- Under Hypothesis H0 :
2
n

1
S


n
2
2
 
~

 n  1 .
2
0
(7-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis
H0 is an interval of the form : 0, 2  where
2  0 and P   2  2   
(7-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis
H0 is an interval of the form :  12 ,  
where 12  0 and P   2  12   1  
(7-c).
Case (c) – two-tailed test:
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The acceptance region of the null hypothesis
H0 is an interval of the form :  12 /2 , 2 /2 


where 12 /2 , 2 /2  0, P  2  2 /2   / 2
and P   2  12 /2   1   / 2 .
Example 10 : Manufacture of car batteries claims
that its battery life is approximately normally
distributed with a standard deviation equal to 0.9
years. If a random sample of size ten of these
batteries has a standard deviation of 1.2 years, do
you think   0.9 ? Use a level test 0.05.
2.6. Tests of variances: if two samples
Theorem 8 : Let X 1 ~ N   1 ,  12  and X 1, 1 ,..., X 1, n1
is an iid sample drawn from X 1 ; X 2 ~ N
2

,

 2 2
and X 2, 1 ,..., X 2, n 2 is an iid sample drawn from X 2 .
We propose to test the hypothesis:
H0 :  12   22
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(a) H1 :  12   22
(b) H1 :  12   22
(c) H1 :  12   22
Then,
2013-2014
right-tailed test
left-tailed test
two-tailed test
S12 /  12
- The test statistics F  2 2
S2 /  2
- Under Hypothesis H0 :
S12
F  2 ~ F  n1  1, n2  1 .
S2
(8-a). Case (a) – right-tailed test :
The acceptance region of the null hypothesis
H0 is an interval of the form :  0, f  where
f  0 and P  F  f   
(8-b). Case (b) – left-tailed test :
The acceptance region of the null hypothesis
H0 is an interval of the form :  f1 , 
where f1  0 and P  F  f1   1  
(8-c). Case (c) – two-tailed test:
The acceptance region of the null hypothesis
H0 is an interval of the form :  f1 /2 , f /2 
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where f1 /2 , f /2  0, P  F  f /2    / 2
and P  F  f1 /2   1   / 2 .
Example 11 : A medical researcher wants to see if the
variance of the heart rate (beats per minute) of
smokers is different from the variance in heart people
who do not smoke. Two samples are selected, and
the data are shown. Use   0.05, Is there enough
evidence to support the claim?
Smoking
NonSmoking
n1 = 26
n2 = 18
S12  36
S 22  10
To be continued
…
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