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