Three factor
multiple regression from Snedecor and Cochran (1967), table 13.10.1,
page 405.
Y = estimated plant available
phosphorus in the soil (20 C)
X1 = inorganic phosphorus
X2 = organic phosphorus
soluble in K2CO3 and hydrolized by hypobromite
X3= organic phosphorus
soluble in K2CO3 and NOT hydrolized by
hypobromite
All least squares
regression analyses start with the same three matrices.
|
X = |
|
1 |
0.4 |
53 |
158 |
|
Y = |
|
64 |
|
|
|
|
1 |
0.4 |
23 |
163 |
|
|
|
60 |
|
|
|
|
1 |
3.1 |
19 |
37 |
|
|
|
71 |
|
|
|
|
1 |
0.6 |
34 |
157 |
|
|
|
61 |
|
|
|
|
1 |
4.7 |
24 |
59 |
|
|
|
54 |
|
|
|
|
1 |
1.7 |
65 |
123 |
|
|
|
77 |
|
|
|
|
1 |
9.4 |
44 |
46 |
|
|
|
81 |
|
|
|
|
1 |
10.1 |
31 |
117 |
|
|
|
93 |
|
|
|
|
1 |
11.6 |
29 |
173 |
|
|
|
93 |
|
|
|
|
1 |
12.6 |
58 |
112 |
|
|
|
51 |
|
|
|
|
1 |
10.9 |
37 |
111 |
|
|
|
76 |
|
|
|
|
1 |
23.1 |
46 |
114 |
|
|
|
96 |
|
|
|
|
1 |
23.1 |
50 |
134 |
|
|
|
77 |
|
|
|
|
1 |
21.6 |
44 |
73 |
|
|
|
93 |
|
|
|
|
1 |
23.1 |
56 |
168 |
|
|
|
95 |
|
|
|
|
1 |
1.9 |
36 |
143 |
|
|
|
54 |
|
|
|
|
1 |
26.8 |
58 |
202 |
|
|
|
168 |
|
|
|
|
1 |
29.9 |
51 |
124 |
|
|
|
99 |
|
|
X´X = |
|
18 |
215 |
758 |
2214 |
|
X´Y = |
|
1463 |
|
|
|
|
215 |
4321.02 |
10139.5 |
27645 |
|
|
|
20706.2 |
|
|
|
|
758 |
10139.5 |
35076 |
96598 |
|
|
|
63825 |
|
|
|
|
2214 |
27645 |
96598 |
307894 |
|
|
|
187542 |
|
Y´Y
= [
131299 ]
Create a fully
augmented matrix of the form;
|
|
X´X |
X´Y |
I |
|
|
|
(X´Y)´ |
Y´Y |
0 |
|
The resulting matrix
contains;
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
n |
SX1 |
SX2 |
SX3 |
SY |
1 |
0 |
0 |
0 |
|
|
|
SX1 |
SX12 |
SX1X2 |
SX1X3 |
SX1Y |
0 |
1 |
0 |
0 |
|
|
|
SX2 |
SX1X2 |
SX22 |
SX2X3 |
SX2Y |
0 |
0 |
1 |
0 |
|
|
|
SX3 |
SX1X3 |
SX2X3 |
SX32 |
SX3Y |
0 |
0 |
0 |
1 |
|
|
|
SY |
SX1Y |
SX2Y |
SX3Y |
SY2 |
0 |
0 |
0 |
0 |
|
Numerically for this
problem given previously the matrix is;
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
18 |
215 |
758 |
2214 |
1463 |
1 |
0 |
0 |
0 |
|
|
|
215 |
4321.02 |
10139.5 |
27645 |
20706.2 |
0 |
1 |
0 |
0 |
|
|
|
758 |
10139.5 |
35076 |
96598 |
63825 |
0 |
0 |
1 |
0 |
|
|
|
2214 |
27645 |
96598 |
307894 |
187542 |
0 |
0 |
0 |
1 |
|
|
|
1463 |
20706.2 |
63825 |
187542 |
131299 |
0 |
0 |
0 |
0 |
|
The first step
(divide row 1 by value1,1) in the sweepout technique
produces,
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
11.944444 |
42.111111 |
123 |
81.277778 |
0.055556 |
0 |
0 |
0 |
|
|
|
215 |
4321.02 |
10139.5 |
27645 |
20706.2 |
0 |
1 |
0 |
0 |
|
|
|
758 |
10139.5 |
35076 |
96598 |
63825 |
0 |
0 |
1 |
0 |
|
|
|
2214 |
27645 |
96598 |
307894 |
187542 |
0 |
0 |
0 |
1 |
|
|
|
1463 |
20706.2 |
63825 |
187542 |
131299 |
0 |
0 |
0 |
0 |
|
And after sweeping
out the first column (subtracting a multiple of row 1 from all other
rows) we
have;
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
11.944444 |
42.111111 |
123 |
81.277778 |
0.055556 |
0 |
0 |
0 |
|
|
|
0 |
1752.964444 |
1085.611111 |
1200 |
3231.477778 |
-11.944444 |
1 |
0 |
0 |
|
|
|
0 |
1085.611111 |
3155.777778 |
3364 |
2216.444444 |
-42.111111 |
0 |
1 |
0 |