Example Method of Least Squares
The given example explains how to find the equation of a straight line or a least square line by using the method of least square, which is very useful in statistics as well as in mathematics.
Example:
Fit a least square line for the following data. Also find the trend values and show that $$\sum \left( {Y – \widehat Y} \right) = 0$$.
$$X$$

1

2

3

4

5

$$Y$$

2

5

3

8

7

Solution:
$$X$$

$$Y$$

$$XY$$

$${X^2}$$

$$\widehat Y = 1.1 + 1.3X$$

$$Y – \widehat Y$$

1

2

2

1

2.4

0.4

2

5

10

4

3.7

+1.3

3

3

9

9

5.0

2

4

8

32

16

6.3

1.7

5

7

35

25

7.6

0.6

$$\sum X = 15$$

$$\sum Y = 25$$

$$\sum XY = 88$$

$$\sum {X^2} = 55$$

Trend Values

$$\sum \left( {Y – \widehat Y} \right) = 0$$

The equation of least square line $$Y = a + bX$$
Normal equation for ‘a’ $$\sum Y = na + b\sum X{\text{ }}25 = 5a + 15b$$ — (1)
Normal equation for ‘b’ $$\sum XY = a\sum X + b\sum {X^2}{\text{ }}88 = 15a + 55b$$ —(2)
Eliminate $$a$$ from equation (1) and (2), multiply equation (2) by 3 and subtract from equation (2). Thus we get the values of $$a$$ and $$b$$.
Here $$a = 1.1$$ and $$b = 1.3$$, the equation of least square line becomes $$Y = 1.1 + 1.3X$$.
For the trends values, put the values of $$X$$ in the above equation (see column 4 in the table above).