Hypt. tests

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STATISTICAL INFERENCE
PART VI
HYPOTHESIS TESTING
1
TESTS OF HYPOTHESIS
• A hypothesis is a statement about a
population parameter.
• The goal of a hypothesis test is to decide
which of two complementary hypothesis is
true, based on a sample from a population.
2
TESTS OF HYPOTHESIS
• STATISTICAL TEST: The statistical procedure
to draw an appropriate conclusion from
sample data about a population parameter.
• HYPOTHESIS: Any statement concerning an
unknown population parameter.
• Aim of a statistical test: test a hypothesis
concerning the values of one or more
population parameters.
3
NULL AND ALTERNATIVE HYPOTHESIS
• NULL HYPOTHESIS=H0
– E.g., a treatment has no effect or there is no
change compared with the previous situation.
• ALTERNATIVE HYPOTHESIS=HA
– E.g., a treatment has a significant effect or there is
development compared with the previous
situation.
4
TESTS OF HYPOTHESIS
• Sample Space, A: Set of all possible values of sample
values x1,x2,…,xn.
(x1,x2,…,xn) A
• Parameter Space, : Set of all possible values of the
parameters.
 =Parameter Space of Null Hypothesis Parameter
Space of Alternative Hypothesis
 = 0   1
H0:0
H1: 1
5
TESTS OF HYPOTHESIS
• Critical Region, C, is a subset of A which leads
to rejection region of H0.
Reject H0 if (x1,x2,…,xn)C
Not Reject H0 if (x1,x2,…,xn)C΄
• A test defines the critical region.
• A test is a rule which leads to a decision to fail
to reject or reject H0 on the basis of the
sample information.
6
TEST STATISTIC AND REJECTION
REGION
• TEST STATISTIC: The sample statistic on which
we base our decision to reject or not reject
the null hypothesis.
• REJECTION REGION: Range of values such
that, if the test statistic falls in that range, we
will decide to reject the null hypothesis,
otherwise, we will not reject the null
hypothesis.
7
TESTS OF HYPOTHESIS
• If the hypothesis completely specify the
distribution, then it is called a simple
hypothesis. Otherwise, it is composite
hypothesis.
• =(1, 2)
H0:1=3f(x;3, 2)
Composite Hypothesis
H1:1=5f(x;5, 2)
If 2 is known, simple hypothesis.
8
TESTS OF HYPOTHESIS
H0 is True
Reject H0
Type I error
P(Type I error) = 
H0 is False
Correct Decision
1-
Do not reject H0
Correct Decision
1-
Type II error
P(Type II error) = 
Tests are based on the following principle:
Fix , minimize .
()=Power function of the test for all .
= P(Reject H0)=P((x1,x2,…,xn)C)
9
TESTS OF HYPOTHESIS
    P  Reject H 0 H 0 is true

  0
 P Type I error
    
Type I error=Rejecting H0 when H0 is true
max
  0
         max. prob. of Type I error
    P  Reject H 0 H 1 is true

  1
 1  P  Not Reject H 0 H 1 is true
  1    
max
  1
         max. prob. of Type II error
10
PROCEDURE OF STATISTICAL TEST
1. Determining H0 and HA.
2. Choosing the best test statistic.
3. Deciding the rejection region (Decision
Rule).
4. Conclusion.
11
HOW TO DERIVE AN APPROPRIATE
TEST
Definition: A test which minimizes the
Type II error (β) for fixed Type I error
(α) is called a most powerful test or
best test of size α.
12
MOST POWERFUL TEST (MPT)
H0:=0 Simple Hypothesis
H1:=1  Simple Hypothesis
Reject H0 if (x1,x2,…,xn)C
The Neyman-Pearson Lemma:
Reject


L  0 
C    x1 , x 2 ,  , x n  :
 k
L  1 


L  0 
k
H0 if L 
L  1 
  P L  k    0   k
  1  P  L  k   1 
Proof: Available in
text books (e.g.
Bain &
Engelhardt, 1992,
p.g.408)
13
EXAMPLES
• X~N(, 2) where 2 is known.
H0:  =  0
H1:  =  1
where
 0 >  1.
Find the most powerful test of size .
14
Solution
1
L( | X )  (
L ( 1 | X )
L(0 | X )
 exp{ 
2 
 exp{ 
1
2
1
2
2
2
 (X i   )
2
}
for
 0
i 1
n
[( X i   1 )  ( X i   0 ) ]}
2 
2
2
i 1
n
[ 2 X i (  0   1 )  (  1   0 )]}  k
2 
2
2
i 1
 2 X  ( 1   0 ) 
2
n
) exp{ 
n
1
2
2
ln k
n (  0  1 )
sin ce
 0  1
(  0  1 )
X 

c
n ( 1   0 )
2

ln k
15
Solution, cont.
• What is c?: It is a constant that satisfies
  P(X  c | 0 )  P(
c   0  Z
X~N(, 2).
X  0
c  0

)
 / n
 / n

n
since
For a pre-specified α, most powerful test says,

Reject Ho if
X  c    Z
0

n
X  0

  Z

n
16
Examples
• Example2: See Bain & Engelhardt, 1992,
p.g.410 Find MPT of
Ho: p=p0 vs H1: p=p1 > p0
• Example 3: See Bain & Engelhardt, 1992,
p.g.411 Find MPT of
Ho: X~Unif(0,1) vs H1: X~Exp(1)
17
UNIFORMLY MOST POWERFUL (UMP) TEST
• If a test is most powerful against every possible value in a
composite alternative, then it will be a UMP test.
• One way of finding UMPT is to find MPT by NeymanPearson Lemma for a particular alternative value, and
then show that test does not depend on the specific
alternative value.
• Example: X~N(, 2), we reject Ho if
Note that this does not depend on
X  0
  Z

particular value of μ1, but only on the
n
fact that  0 >  1. So this is a UMPT of H0:  = 0 vs H1:  <  0.
18
UNIFORMLY MOST POWERFUL (UMP)
TEST
• To find UMPT, we can also use Monotone
Likelihood Ratio (MLR).
• If L=L(0)/L(1) depends on (x1,x2,…,xn) only
through the statistic y=u(x1,x2,…,xn) and L is an
increasing function of y for every given 0>1,
then we have a monotone likelihood ratio (MLR)
in statistic y.
• If L is a decreasing function of y for every given
0>1, then we have a monotone likelihood ratio
(MLR) in statistic −y.
19
UNIFORMLY MOST POWERFUL (UMP)
TEST
• Theorem: If a joint pdf f(x1,x2,…,xn;) has MLR
in the statistic Y, then a UMP test of size 
• for H0:0 vs H1:>0 is to reject H0 if Yc
where P(Y c0)=.
• for H0:0 vs H1:<0 is to reject H0 if Yc
where P(Y  c0)=.
20
EXAMPLE
• X~Exp()
H0:0
H1:>0
Find UMPT of size .
21
EXAMPLE
• Xi~Poi(), i=1,2,…,n
Determine whether (X1,…,Xn) has MLR property.
Find a UMP level α test for testing H0:=0
versus H1:<0.
22
GENERALIZED LIKELIHOOD RATIO TEST
(GLRT)
• GLRT is the generalization of MPT and
provides a desirable test in many applications
but it is not necessarily a UMP test.
23
GENERALIZED LIKELIHOOD RATIO TEST
(GLRT)
H0:0
H1: 1
L    f  x1 , x 2 ,  , x n ; 
r .s .
  f  x1 ;  , f  x 2 ;  ,  , f  x n ;  
L    L ˆ  and
Let L ˆ   max
 
MLE of 
ˆ   max L    L ˆ 
L 
0
0
  0
MLE of  under H0
24
GENERALIZED LIKELIHOOD RATIO TEST
(GLRT)
 
 
ˆ
L 
0
 
 The Generalize d Likelihood
ˆ
L 
Ratio
GLRT: Reject H0 if 0
25
EXAMPLE
• X~N(, 2)
H0:  = 0
H1:   0
Derive GLRT of size .
26
EXAMPLE
• Let X1,…,Xn be independent r.v.s, each with
shifted exponential p.d.f.:
f (x | ,) 
1

exp{  ( x   ) /  } I [ ,  ] ( x )
where λ is known.
Find the LRT to test H0:=0 versus H1:>0.
ASYMPTOTIC DISTRIBUTION OF −2ln
• GLRT: Reject H0 if 0
• GLRT: Reject H0 if -2ln>-2ln0=c
 2 ln 
under H 0
~
asympt .
k
2
where k is the number of parameters to be
tested.
Reject H0 if -2ln>  ,k
2
28
TWO SAMPLE TESTS
X ~ Bin  n 1 , p 1 , r .s .
Y ~ Bin  n 2 , p 2 , r .s .
H 0 : p1  p 2  p 0
H 1 : p1  p 2
Derive GLRT of size , where X and Y are
independent; p0, p1 and p2 are unknown.
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