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 |
|
|
|
0 |
1200 |
3364 |
35572 |
7593 |
-123 |
0 |
0 |
1 |
|
|
|
0 |
3231.477778 |
2216.444444 |
7593 |
12389.61111 |
-81.277778 |
0 |
0 |
0 |
|
We start the second
column sweep by dividing row 2 by value2,2,
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
11.944444 |
42.111111 |
123 |
81.277778 |
0.055556 |
0 |
0 |
0 |
|
|
|
0 |
1 |
0.6193 |
0.684555 |
1.843436 |
-0.006814 |
0.00057 |
0 |
0 |
|
|
|
0 |
1085.611111 |
3155.777778 |
3364 |
2216.444444 |
-42.111111 |
0 |
1 |
0 |
|
|
|
0 |
1200 |
3364 |
35572 |
7593 |
-123 |
0 |
0 |
1 |
|
|
|
0 |
3231.477778 |
2216.444444 |
7593 |
12389.61111 |
-81.277778 |
0 |
0 |
0 |
|
and finish sweeping the second column to obtain;
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
0 |
34.713915 |
114.823375 |
59.258959 |
0.136943 |
-0.006814 |
0 |
0 |
|
|
|
0 |
1 |
0.6193 |
0.684555 |
1.843436 |
-0.006814 |
0.00057 |
0 |
0 |
|
|
|
0 |
0 |
2483.458674 |
2620.839842 |
215.189831 |
-34.713915 |
-0.6193 |
1 |
0 |
|
|
|
0 |
0 |
2620.839842 |
34750.53439 |
5380.87679 |
-114.823375 |
-0.684555 |
0 |
1 |
|
|
|
0 |
0 |
215.189831 |
5380.87679 |
6432.588616 |
-59.258959 |
-1.843436 |
0 |
0 |
|
The third column
starts with,
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
0 |
34.713915 |
114.823375 |
59.258959 |
0.136943 |
-0.006814 |
0 |
0 |
|
|
|
0 |
1 |
0.6193 |
0.684555 |
1.843436 |
-0.006814 |
0.00057 |
0 |
0 |
|
|
|
0 |
0 |
1 |
1.055318 |
0.086649 |
-0.013978 |
-0.000249 |
0.000403 |
0 |
|
|
|
0 |
0 |
2620.839842 |
34750.53439 |
5380.87679 |
-114.823375 |
-0.684555 |
0 |
1 |
|
|
|
0 |
0 |
215.189831 |
5380.87679 |
6432.588616 |
-59.258959 |
-1.843436 |
0 |
0 |
|
and after being
swept out produces,
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
0 |
0 |
78.189139 |
56.251024 |
0.622176 |
0.001843 |
-0.013978 |
0 |
|
|
|
0 |
1 |
0 |
0.030996 |
1.789774 |
0.001843 |
0.000725 |
-0.000249 |
0 |
|
|
|
0 |
0 |
1 |
1.055318 |
0.086649 |
-0.013978 |
-0.000249 |
0.000403 |
0 |
|
|
|
0 |
0 |
0 |
31984.71367 |
5153.782984 |
-78.189139 |
-0.030996 |
-1.055318 |
1 |
|
|
|
0 |
0 |
0 |
5153.782984 |
6413.942579 |
-56.251024 |
-1.789774 |
-0.086649 |
0 |
|
Finally the fourth
column in the X´X matrix is started and swept out,
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
0 |
0 |
78.189139 |
56.251024 |
0.622176 |
0.0018428
|
-0.0139781
|
0 |
|
|
|
0 |
1 |
0 |
0.030996 |
1.789774 |
0.001843 |
0.0007249
|
-0.0002494
|
0 |
|
|
|
0 |
0 |
1 |
1.055318 |
0.086649 |
-0.013978 |
-0.0002494
|
0.0004027
|
0 |
|
|
|
0 |
0 |
0 |
1 |
0.161133 |
-0.002445 |
-0.0000010
|
-0.0000330
|
0.000031 |
|
|
|
0 |
0 |
0 |
5153.782984 |
6413.942579 |
-56.251024 |
-1.7897741
|
-0.0866492
|
0 |
|
and the final result
is;
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
0 |
0 |
0 |
43.652198 |
0.813316 |
0.0019185 |
-0.0113982 |
-0.002445 |
|
|
|
0 |
1 |
0 |
0 |
1.78478 |
0.001919 |
0.0007249 |
-0.0002483 |
-0.000001 |
|
|
|
0 |
0 |
1 |
0 |
-0.083397 |
-0.011398 |
-0.0002483 |
0.0004375 |
-0.000033 |
|
|
|
0 |
0 |
0 |
1 |
0.161133 |
-0.002445 |
-0.0000010 |
-0.0000330 |
0.000031 |
|
|
|
0 |
0 |
0 |
0 |
5583.499658 |
-43.652198 |
-1.7847797 |
0.0833971 |
-0.161133 |
|
The resulting
matrix is of the form;
|
|
I |
B |
(X´X)-1 |
|
|
|
0 |
SSE |
-B´ |
|
and contains the
values
|
|
X0 |
X1 |
X2 |
X3 |
X´Y |
c0 |
c1 |
c2 |
c3 |
|
|
|
1 |
0 |
0 |
0 |
b0 |
c00 |
c01 |
c02 |
c03 |
|
|
|
0 |
1 |
0 |
0 |
b1 |
c10 |
c11 |
c12 |
c13 |
|
|
|
0 |
0 |
1 |
0 |
b2 |
c20 |
c21 |
c22 |
c23 |
|
|
|
0 |
0 |
0 |
1 |
b3 |
c30 |
c31 |
c32 |
c33 |
|
|
|
0 |
0 |
0 |
0 |
SSE |
-b0 |
-b1 |
-b2 |
-b3 |
|
The solution to the
regression equation is then,
Yi =
43.652 + 1.785X1i - 0.083X2i + 0.161X3i
+ e
The sums of squares
are given by the sequential reduction in the YY matrix
|
MATRIX |
Y´Y VALUE |
INTERPRETATION of the
REPLACEMENT VALUE |
DIFFERENCE from PREVIOUS VALUE |
INTERPRETATION of the DIFFERENCE |
|
Original |
131299 |
SY2
(uncorrected) |
|
|
|
|
12389.6111 |
SY2-(SY)2/n =
SSY|X0 |
118909.3840 |
(Y)2/n = C.F. |
|
|
6432.5886 |
SSY|X0,X1 |
5957.0225 |
SeqSSX1 |
|
|
6413.9426 |
SSY|X0,X1,X2 |
18.6460 |
SeqSSX2 |
|
|
5583.4997 |
SSY|X0,X1,X2,X3 =
SSE |
830.4429 |
SeqSSX3 |
Partial sums of
squares, or fully adjusted sums of squares, are given by
PARTIAL SS = bk2/ckk
Partial SSX1
= b12/c11 =
1.78482 / 0.0007249
= 4394.1523
Partial SSX2
= b22/c22 =
0.083402 / 0.0004375
= 15.8979
Partial SSX3
= b32/c33 =
0.16112 / 0.00003127
= 830.4429
Recall that I number
the positions in the X´X matrix differently, from k = 0, 1, ... , p
(where p is
the number of parameters excluding the intercept) instead of starting
at 1 as
other matrices. This is done in order to
be able to associate the matrix position with the regression
coefficient
subscript.