# 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).