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 The method of least squares is a standard 
approach to the approximate solution of over 
determined systems, i.e., sets of equations in which 
there are more equations than unknowns. 
 "Least squares" means that the overall solution 
minimizes the sum of the squares of the errors 
made in the results of every single equation. 
 The least-squares method is usually credited to Carl 
Friedrich Gauss (1795), but it was first published 
by Adrien-Marie Legendre.
 A linear regression is a statistical analysis assessing 
the association between two variables. It is used to 
find the relationship between two variables. 
Regression Equation(y) = a + bx Slope(b) = 
(NΣXY - (ΣX)(ΣY)) / (NΣX2 - (ΣX)2) 
Intercept(a) = (ΣY - b(ΣX)) / N
 To fit the straight line y=a+bx: 
 Substitute the observed set of n values in this 
equation. 
 Form the normal equations for each constant i.e. 
Σy=na+bΣx, Σxy=aΣx+bΣx2 . 
 Solve these normal equations as simultaneous 
equations of a and b. 
 Substitute the values of a and b in y=a+bx, which is 
the required line of best fit.
Fit a straight line to the x and y values in the following 
Table: 
xi yi xiyi xi 
2 
1 0.5 0.5 1 
2 2.5 5 4 
3 2 6 9 
4 4 16 16 
5 3.5 17.5 25 
6 6 36 36 
7 5.5 38.5 49 
28 24 119.5 140 
28  i x   24.0 i y 
140 2  i  x  i i 119.5 x y 
3.428571 
24 
x   y   
7 
4 
28 
7 
3.428571 
24 
x   y   
7 
4 
28 
7
0.8392857 
   
n x y  
x y 
i i i i 
2 2 
  
n x x 
( ) 
 
7  119.5  28  
24 
2 
7 140 28 
 
 
a  y  
a x 
  
0 1 
 
3.428571 0.8392857 4 0.07142857 
1 
    
a 
i i 
* 
* Y  a  a X 
0 1 
Y = 0.07142857 + 0.8392857x
Suppose if we want to know the approximate y value 
for the variable x = 4. Then we can substitute the 
value in the above equation. 
Y *  a  a X 
* 
0 1 
Y = 0.07142857 + 0.8392857x 
Y = 0.07142857 + 0.8392857(4) 
Y = 0.07142857 + 3.3571428 
Y = 3.42857137 Ans
Fit the following Equation: 
0 . 24  i y 
to the data in the following table: 
xi yi X*=log xi Y*=logyi 
1 0.5 0 -0.301 
2 1.7 0.301 0.266 
3 3.4 0.477 0.534 
4 5.7 0.602 0.753 
5 8.4 0.699 0.922 
15 19.7 2.079 2.141 
b y  a x 
log log( 2 ) 
2 
log y log a b log x 2 2   
* * 
Y y, X x, 
  
let  log  
log 
a log 
a , a b 
0 2 1 2 
* 
* Y  a  a X 
0 1
METHOD OF LEAST SQURE
METHOD OF LEAST SQURE
METHOD OF LEAST SQURE
METHOD OF LEAST SQURE

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METHOD OF LEAST SQURE

  • 1.
  • 2.
  • 3.  The method of least squares is a standard approach to the approximate solution of over determined systems, i.e., sets of equations in which there are more equations than unknowns.  "Least squares" means that the overall solution minimizes the sum of the squares of the errors made in the results of every single equation.  The least-squares method is usually credited to Carl Friedrich Gauss (1795), but it was first published by Adrien-Marie Legendre.
  • 4.  A linear regression is a statistical analysis assessing the association between two variables. It is used to find the relationship between two variables. Regression Equation(y) = a + bx Slope(b) = (NΣXY - (ΣX)(ΣY)) / (NΣX2 - (ΣX)2) Intercept(a) = (ΣY - b(ΣX)) / N
  • 5.  To fit the straight line y=a+bx:  Substitute the observed set of n values in this equation.  Form the normal equations for each constant i.e. Σy=na+bΣx, Σxy=aΣx+bΣx2 .  Solve these normal equations as simultaneous equations of a and b.  Substitute the values of a and b in y=a+bx, which is the required line of best fit.
  • 6. Fit a straight line to the x and y values in the following Table: xi yi xiyi xi 2 1 0.5 0.5 1 2 2.5 5 4 3 2 6 9 4 4 16 16 5 3.5 17.5 25 6 6 36 36 7 5.5 38.5 49 28 24 119.5 140 28  i x   24.0 i y 140 2  i  x  i i 119.5 x y 3.428571 24 x   y   7 4 28 7 3.428571 24 x   y   7 4 28 7
  • 7. 0.8392857    n x y  x y i i i i 2 2   n x x ( )  7  119.5  28  24 2 7 140 28   a  y  a x   0 1  3.428571 0.8392857 4 0.07142857 1     a i i * * Y  a  a X 0 1 Y = 0.07142857 + 0.8392857x
  • 8. Suppose if we want to know the approximate y value for the variable x = 4. Then we can substitute the value in the above equation. Y *  a  a X * 0 1 Y = 0.07142857 + 0.8392857x Y = 0.07142857 + 0.8392857(4) Y = 0.07142857 + 3.3571428 Y = 3.42857137 Ans
  • 9.
  • 10. Fit the following Equation: 0 . 24  i y to the data in the following table: xi yi X*=log xi Y*=logyi 1 0.5 0 -0.301 2 1.7 0.301 0.266 3 3.4 0.477 0.534 4 5.7 0.602 0.753 5 8.4 0.699 0.922 15 19.7 2.079 2.141 b y  a x log log( 2 ) 2 log y log a b log x 2 2   * * Y y, X x,   let  log  log a log a , a b 0 2 1 2 * * Y  a  a X 0 1