Page 285 - DMTH404_STATISTICS

Unit 21: Confidence Intervals

Notes
n
2
2
2
2
Under the null hypothesis, since  =  0  (X   ) / has a  distribution (chi-square
,
}
2
i
n
0
1
distribution with n degrees of freedom). Let  2 n,  be the upper - probability point of  . The
2
n
test statistic is thus
n
2
 (X   )  k1 and hence
i
1
 n 2 
2
2
 
C0 = X| (X   ) /  c n,
0
i
 1 
2
2
2
On the other hand, if the alternative hypothesis is H :  =  (  <  ), then the test statistic is
2
1 1 1 0
and hence
n
2
 (X    k 2
)
i
1
where  2 is the lower  -probability point of the  distribution with n degrees of freedom.
2
n,1
2
Example 3: Let X , . . . , X and Y , . . . , Y be independent random samples from N( ,  )
1 m 1 n l 1
2
2
2
2
2
and N( ,  ). We wish to obtain a test statistic for testing H :  =  against H :    .
2 2 0 1 2 1 1 2
2
2
,
,
Here  = {( 1 , 2 ,  2 2 ) : –       0,i  1,2}
i
1
1
2
2
2
,i
and  = {( , , 2 , 2 ) : –      1,2,      0}
0 1 2 1 2 i 1 2
We shall use   ( 1 ,  , 2 1 , 2 2 ).
Also L (8 | X, Y)
m n m /2 n /2

 1  2  1   1   1 n 2 1 n 2 
)

=    2   2  exp  2  (X   1 )  2  (Y   2 
i
i
 2    1    1   2 1 1 2 2 1 
2
The maximum likelihood estimates of  ,  ,  2 , are respectively
l 2 1 2
1 m 1 n
ˆ    X  X, ˆ   Y  Y
1 i 2 i
m 1 n 1
2  1 m 2  1 n
2
   (X  X) ,    (Y  Y) 2
1 i 2 i
m 1 n 1
2
2
Further, if      2 , the maximum likelihood estimate of  is
2
2
1
1  m n 2 
2
2
ˆ    (X  X)   (Y  Y) 
i
i

(m n)  1 1 
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