Reading: Chapter 4.3
I. Uniformly Most Powerful Tests
When the alternative hypothesis is composite,
H
1
:
 
1
, then the power can be different for different alternatives. For each particular alternative

1
, a test is the most powerful level
 test for the alternative

1 alternative
H
1
:
 
1 if the test is most powerful for the simple
.
If a particular test function

* x is the most powerful level
 test for all alternatives
 
1
, then we say that uniformly most powerful (UMP) level

* x is a

test and we should clearly use

* x as our test function under the Neyman-
Pearson paradigm.
Example of uniformly most powerful test:
Let
X
1
, , X n be iid N

and suppose we want to test
H
0
:
 
0 versus
H
1 powerful level
:
 test of
 
0
H
0
:
. For each
 
0

1
 versus
H
1
:

0
, the most
 
1
rejects the null hypothesis for
 n i

1
( X n i
 
0
)
  
1
(1
 
)
. Since this same test function is most powerful for each

1
 
0
, this test function is UMP.
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But suppose we consider the alternative hypothesis H
1
:
 
0
.
Then there is no UMP test. The most powerful test for each
H
1
:
 
1
, where

1
 
0
, rejects the null hypothesis for
 n i

1
( X i
 
0
)
  
1
(1
 
)
, but the most powerful test for each n
H
1
:
 
1
, where

1
 
0
, rejects for
 n i

1
( X i n
 
0
)
  
1
.
Note that the test that rejects the null hypothesis for
 i n

1
( X n i
 
0
)
  
1
(1
 
) alternative

1
 
0
cannot be most powerful for an
by part (c) (necessity) of the Neyman-
Pearson Lemma since it is not a likelihood ratio test for

1
 
0
.
It is rare for UMP tests to exist. However, for one-sided alternative, they exist in some problems.
A condition under which UMP tests exists is when the family of distributions being considered possesses a property called monotone likelihood ratio .
Definition: The one-parameter family of distributions
  
} is said to be a monotone likelihood ratio family in the one-dimensional statistic ( ) if for

0
 
1
(a) p x

0
)
 p x

1
) for all x (identifiability)
2

(b) p p (
( x |

1
) x |

)
(where is an increasing function of ( ) p (
0 p ( x |

1
) x |

0
)
 
if p ( x |

0
)
 p x |

1
)

0 p p (
( x |

1
) x |

0
)
 
if p ( x |

0
)
 p x |

1
)

0
).
and
The basic property of a family with monotone likelihood ratio is: for every pair of parameter values

0
 
1
, the sample points have the same ordering in terms of the likelihood ratio. This means that the most powerful test of
H
0
:
 
0
vs. H
1
:
 
1 will have the same critical region for every
 
1

0
.
Examples of families with monotone likelihood ratio:
(a) For
X
1
, , X n iid Exponential (

), p p (
( x |

1
) x |

0
)
 i n 

1

1 e
 
1
X i i n 

1

0 e
 
0
X i





1



0
 n exp


(
 
1 0
) i n 

1
X i


For
 
1

0
, p ( x | p ( x |

1

0
)
) is an increasing function of
 i n 

1
X i so the family has monotone likelihood ratio in
T
  i n 

1
X i
.
(b) Consider the one-parameter exponential family model
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p ( x |

)
 h x
 
T x

B

If ( ) is strictly increasing in
 
, then the family is
. If ( ) monotone likelihood ratio in ( ) decreasing in ratio in

T ( )
 
. is strictly
, then the family is monotone likelihood
Example 2: Let X

1 .
   x
(1
 
)


1
 
 n log(1
 
)]
This is a one-parameter exponential family with
   log

1
 
   n log(1
 
), ( )

, ( )
Since ( ) is strictly increasing in monotone likelihood ratio in T ( )


 x .
, the family is
.
Note: Not all one-parameter families have monotone likelihood ratio. For example, the Cauchy distribution is not monotone likelihood ratio in x :
 
1

1
  
)
2
For
 
0 ,

( | 0)

1
1 (
 x
2

)
2
,
. which is not increasing in x .
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Theorem (4.3.1 and Corollary): If the one-parameter family of distributions { ( | ) :
 
} has monotone likelihood ratio in
, then there exists a UMP level
 test for testing
H
0
:
0
versus
H
1
:
0
and it is given by

*
 



 x x

 c c x
 c where c and
 are determined so that
Proof: Fix an alternative
 
0
E

0

*  
.
. Because the family is monotone likelihood ratio in ( )
L x
 
0 1
 p x
1
) p x


0
)
, the likelihood ratio statistic is an increasing function of ( ) gives that

* x
. The Neyman-Pearson lemma
is a most powerful test for testing H
0
:
 
0 versus H
1
:
 
1
. Since this holds for all
 
1

0
, we have that

* x is UMP.
Now consider
H
0

* x still UMP?
:
0 versus
H
1
:
0
. Is the above test
Fact: The test

* x
0 is level
 for
H
0
:
0
, the probability of rejection is at most
(i.e., for every

).
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Before proving this fact, we prove Corollary 4.2.1.
Corollary 4.2.1: If

* is a most powerful level
 test of
H
 
0 than or equal to p
0 x
:

0
)
 p vs.
H
1 x


1
: with equality if and only if
)
 
1
, then the power of

* at

1 with probability one (under both is greater
H
0
and
H
1
).
Proof: The test

( )
 regardless of
  for all x (i.e., reject with probability
 part of the Neyman-Pearson lemma, if powerful, then x ) has level p ( X |

1
) p ( X |

0
)
 k and power


( )
. By the necessity
  were most
with probability one. Therefore, p ( X |

1
)
 kp ( X |

0
) with probability one which implies that k =1 and consequently that p ( X |

1
)
 p ( X |

0
) with probability one.
Proof of fact: Let
  
*
)

E

[

* x
. We want to show
  
*
)
  for
0
. Fix

2
 
0 powerful test for testing H
0
:
2
. Then
vs. H
1
:

* x
 
0 is the most
at level
  
*
( , )
2
. But from Corollary 4.2.1,
   
*    
*
( , ) ( , )
0 2
.
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Back to the question of is
H
0
:
 
0 versus H
1
:

* x
0
?
UMP level
 for testing
Yes, it is UMP. Proof: Consider a level
 must also be a level
 test for
H
0

:
 
0 test
 for
H
0
:
 
0
. Then, because
.

* x is UMP level
 for testing
  
*
)
     
0
.
H
0
:
0 versus
H
1
:
0
,
Summary statement of Theorem 4.3.1 and Corollary:
If the one-parameter family of distributions
  
} has monotone likelihood ratio in ( ) level
 test for testing
H
0
:
0
, then there exists a UMP
versus
H
1
:
0
and it is given by

*
 



 where c and
 x
 c x
 c x
 c are determined so that
E

0

*  
.
Example 2 continued: For X

H
0
:
0
vs.
H
1
:
0
, the UMP level
and testing
 test is
1 if x
 c

*  


if x
 c
0 if x
 c where c is the smallest integer such that
P

0
( X
 c )
  and
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

0


[
 
P

0
( X
 c )] / P

0
( X
 c )
. For example, for n

10 ,
0.5
and
 
0.05
, the UMP test is

*
( )




1 if x

8
0.893 if x

8
0 if x

8
[Note: In practice, randomized tests are generally not used and we would face the choice of either being conservative and using the test that rejects for x

8 or allowing for a higher than 0.05 level and using the test that rejects for x

7 ]
II. Power and Sample Size
In the Neyman-Pearson paradigm, we choose a level test and then seek to find the most powerful level
If the power of the most powerful level


 for the test. test is unacceptably small, the only way to increase it is by increasing the sample size.
If a study is being planned in advance, the sample size can be chosen.
In a power (sample size) calculation, we calculate the smallest sample size needed to attain a certain power
 for the level
 test that we will use.
Example: There is a theory that the anticipation of a birthday can prolong a person’s life. We plan to set up a study to see if there is evidence for this theory by randomly sampling the obituaries of n people and counting how many people died in the
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three-month period preceding their birthday. We would like to use a level 0.05 test. How large should n be?
Let p denote the probability that a randomly chosen person dies in the three-month period preceding their birthday. The theory that anticipation of a birthday can prolong a person’s life corresponds to p

0.25
.
Since the goal of the study is to see if there is evidence for the theory that anticipation of a birthday can prolong a person’s life, the alternative hypothesis should correspond to the theory being true and the null hypothesis should correspond to the theory being false, i.e.,
H
0
: p

0.25, H
1
: p

0.25
.
The data will be X , the number of people who die in the threemonth period preceding their birthday. X will have a binomial
( , ) distribution. Because the family of binomial distributions has monotone likelihood ratio in X , the UMP level 0.05 test of
H
0
: p

0.25, H
1
: p

0.25
is

*  



1 if X
 c

if X

0 if X
 c c where c and
 are chosen so that
E p

0.25
 x

0.05
.
Let

( , ) denote the probability of rejecting the null hypothesis for the test

* x for a sample size of n when the
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true probability of a death in the three months preceding a birthday is p .
For a large enough sample size n , the central limit theorem approximation to the binomial gives that

( , )
  p
(

| )
 p
(
 c )
 p
(
 c )

P p


X
 np
 c
 np

 np (1
 p ) np (1
 p ) c
 np

 np (1
 p ) where
 is the standard normal CDF.
Note that

( , ) is a continuous function of p . This continuity of the power shows that high power cannot be attained for p sufficiently close to 0.25, since

(0.25, )

0.05
(since for a fixed n , the c and
 in

* x are chosen so that the test has size
0.05).
For many problems, the points in the alternative are not of equal importance and we care more about having high power at some points than other points. In particular, we may be willing to have an indifference region , a subset of the alternative hypothesis on which we are willing to tolerate low power. For example, we might be willing to tolerate low power for
0.20
0.25
.
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When there is an indifference region, the power calculation chooses the smallest sample size to have a certain power
 for all points in the alternative that are outside of the indifference region.
 is often chosen to be 0.8.
For our problem with
 
0.05
,
 
0.8
and an indifference region of 0.20
0.25
, the approximate sample size needed is the n that solves that the following two equations in ( , )




 
0.05
(1) n c

0.25* n
(1) sets the size of the test to 0.05
(2)


 n c

0.20 * n

 
0.80
(2) sets the power of the test to 0.80 or more outside the indifference region (note that

( , ) is a decreasing function of p so that to have power 0.80 outside the indifference region, it is sufficient to set the power to 0.80 at the edge of the indifference region).
Applying
 
1
to both sides of equations (1) and (2), we obtain
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c

0.25* n
 
1.645
n c

0.20* n

0.842
n
Solving these equations for n gives n

215 . Thus, a sample size of approximately 215 is needed to have power 0.8 to detect p

0.2
with a level 0.05 test.
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