Page 126 - Basic Statistics

121

Worked Solution:

The hypothesis tested is H0: 1 = 3 = 0. General linear hypothesis is
defined as follows.

.

  0  

 0 1 0   0  1    0
K’  =      =  



 0 0 0 1    2   0 
  
 3 
-1
By matrix multiplication of K’ ( X’X ) K, obtained a (2x2) submatrix extracted
from the matrix ( X’X ) ,
-1

[,1] [,2] [,3] [,4]
[1,]
[2,] 3.341412e-006 1.409908e-008
[3,]
[4,] 1.409908e-008 2.613921e-003

ˆ
From the results of previous work example 5.3, has obtained β = 0.0013 and
1
ˆ
β = -0.0280. Furthermore, the sum of squares Q can be calculated as follows
3

ˆ
ˆ
-1
-1
Q = (K’ β )’ (K’ ( X’X ) K) (K’ β )
 0.0013  T  3.341412e - 006 1.409908e - 008  -1  0.0013 
=       





 - 0.0280   1.409908e - 008 2.613921e - 003   - 0.0280 
= 0.807

Q k / . 0 807 2 /
F = = = . 1 1555
S 2 . 0 349195

Which is much less than the critical value of F for  = 0.05 and 2 and 26 degrees

of freedom, F0.05;2;26 = 3.37 , There is no reason to reject H0 that 1 = 3 = 0.

Output using Statistical Analysis System (SAS) software presented of the

results which showing in Table 5.15.

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