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).