Orthogonality
Orthogonality exists when the condition Sficicj=0
and Sfici=0
for each equation.
| fi |
9/16 |
3/16 |
3/16 |
1/16 |
| c1 |
1 |
1 |
-3 |
-3 |
| c2 |
1 |
-3 |
1 |
-3 |
| c3 |
1 |
-3 |
-3 |
9 |
First to show that Sfici
= 0:
Sfici
when i = 1 
9/16(1) + 3/16(1) + 3/16(-3) + 1/16(-3) = 0
Sfici when i
= 2 9/16(1)
+ 3/16(-3) + 3/16(1) + 1/16(-3) = 0
Sfici when i
= 3 9/16(1)
+ 3/16(-3) + 3/16(-3) + 1/16(9) = 0
Now to show that SSficicj
= 0
i=1, j=2
(9/16)(1)(1) + 3/16(1)(-3) + 3/16(-3)(1)
+
1/16(-3)(-3) = 0
i=1, j=3 (9/16)(1)(1)
+ 3/16(1)(-3) + 3/16(-3)(-3)
+
1/16(-3)(9) = 0
i=2, j=3 (9/16)(1)(1)
+ 3/16(-3)(-3) + 3/16(1)(-3)
+
1/16(-3)(9) = 0
** The concept of orthogonal tests of hypothesis is that each test
'stands alone' and the results of one test does not influence another
test.
X
= (a
+ a
- 3a
- 3a
)
3n
X
= (a
- 3a
+ a
- 3a)
3n
XL
= (a
- 3a
-3a
+ 9a)
9n
