Download February 26, 2002

February 26, 2002
Rice Ch 9: #12, 13, 16, 18 (Group assignment)
simple and composite hypotheses
power
power curve
Power
Power is the probability of correctly rejecting the null hypothesis if it is false. For composite
alternatives, the power will depend on the particular alternative under which the probability is
computed.
Suppose we are testing
H 0 :   750
versus
H 1 :   750
Using the t-statistic
x  750
s/ n
, for   .05 , c  1.645 . The rejection region is (,1.645] .
If n  30 ,   20 , and   740 , the power is
x  750
x  740
10
10
P{
 1.645}  P{
 1.645 
}  (1.645 
)  (1.09)  .86
s / 30
s / 30
s / 30
20 / 30
Comparison of simple hypotheses
H 0 : f 0 ( x) joint density of data under the null hypothesis
H A : f A (x) joint density of data under the alternative hypothesis
f 0 ( x)
p 0 ( x)
(for discrete data, the likelihood ratio is
, which is the ratio of
f A ( x)
p A ( x)
a probability under the null hypothesis and a probability under the alternative.
likelihood ratio:
Clearly, the rejection region should be of the form {x : f 0 ( x) / f A ( x)  c} . The critical value
would be determined by the equation
f 0 ( x)dx  

{ f 0 ( x ) / f A ( x )  c }
The power is

{ f 0 ( x ) / f A ( x )c }
f A ( x)dx  
{ f 0 ( x ) / f A ( x )c }
f A ( x)
f 0 ( x)dx
f 0 ( x)
Test for composite hypotheses (generalized likelihood ratio test)
f (x |  )
parameter
H0 :
 0
HA:
  1
   0  1
Compare the best choice of  in  0 to the best choice of  over all.

max   0 f ( x |  )
max   f ( x |  )
Theorem Under smoothness conditions on the probability density of probability mass function
involved, the null distribution of  2 log  tends to a chi-square distribution with degrees of
freedom dim   dim  0 .
Definition If Z1 ,, Z are independent standard normal random variables, then
V  Z12    Z2
is chi-square distributed with  degrees of freedom
Example Let X 1 ,, X n be iid random variables with normal distribution
f ( x | , ) 
Test
H0 :   0
versus
HA :   0
1
2 

e
( x )2
2 2