Page 284 - DMTH404_STATISTICS

Statistics

Notes or

|t| > c
1
where c is so chosen that
1
Let c = t in accordance with the diskbution oft under Ho. Thus, the two sided test
1 m + n – 2, /2
obtained is

(X Y mn

 t m n 2, /2



S (m  n)
2
Example 2: Let X , . . . , X , be a random sample from N (,  ), p is known and  > 0, is

1 n
2
unknown. We wish to obtain a test statistic for testing H :  –  against an alternative H :  2
2
0 0 1
2
2
=  (   ).
1
0
We have
1  1 n 
P (X)  exp  2  (Xi   ) 2 

 n /2
1 (2 ) 2
1  1 1 
1  1 n 
P (X)  exp  2  (Xi   ) 2 

 n /2
0 (2 ) 2
0  0 1 
Using Neyman-Pearson Lemma, the test statistic is
P (X)
T(X) =  1  k
P (X)
 0
n /2
      1 1  n

  0  exp 1/2  2  2  (X   ) 2
i
  1      0  1  1
n
2
)

 (   2 0  (X   2 ) k, taking logarithms
1
i
1
n
2
  (X   2 ) k , since    2 0 ,under H 1

1
1
i
1
Here k is so determined that
1

P (T,(X) k) = 

0
 n 2 
 P  0  (X    k 1 = 
)
i
 1 
 n 2 
2
2
 P  0  (X   ) /  k / 0
1
0
i
 1 
276 LOVELY PROFESSIONAL UNIVERSITY