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Table 5.15 Output of SAS for a general linear hypothesis testing

The SAS System

L Ginv(X'X) L' Lb-c

3.3414119E-6 1.4099083E-8 0.0013058159
1.4099083E-8 0.002613921 -0.027843496

Inv(L Ginv(X'X) L') Inv()(Lb-c)

299274.69475 -1.614241062 390.84258745
-1.614241062 382.56703747 -10.65411167

Dependent Variable: Y
Test: Numerator: 0.4035 DF: 2 F value: 1.1562
Denominator: 0.348994 DF: 26 Prob>F: 0.3303

The second hypothesis illustration a case m  0. Suppose prior
information suggested that the intercept 0 for a group of mean of this radiation

and wind should be 0.5 (0 = 0.5). We will construct a composite hypothesis by

adding constraint 0 = 0.5 to the two conditons in the first null hypothesis. The

null null hypothesis is H0 : K’  - m = 0 where

  
 1 0 0   0 0   0.5- 
     





K’ - m = 0 1 0 0  1  − 0 
    2   
 0 0 0 1     0 
  3 
For this hypotesis
 0.2949- - 0.5   0.7949- 




ˆ

K’ β - m =  0.0013  = 0.0013 

   
 - 0.0280   - 0.0280 

and
− 1
 2.85592259 51 6.173356e - 004 - 4.090973e - 002
 
-1
-1
(K’ ( X’X ) K) = 0.00061733 56 3.341412e - 006 1.409908e - 008 

 
 - 0.04090972 95 1.409908e - 008 2.613921e - 003 

~~* CHAPTER 5 THE MULTIPLE LINEAR REGRESSION MODEL *~~