Finding the Regression Equation

Chapter 11: Regression Analysis: Simple Linear Regression

Finding the Regression Equation

A regression line is the best-fitting straight line through a set of data points. What makes a line best-fitting is that it minimizes the differences between the predicted and observed values of #Y#.

This section introduces a method by which the slope and the intercept of the regression line can directly be calculated.
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Calculation Regression Coefficient and Intercept

Performing a simple linear regression analysis results in a regression equation of the form:

\[\hat{Y}=b_0 + b_1 \cdot X\]

To calculate the slope #b_1# of the regression line, use of the following formula:

\[b_1 =\cfrac{\sum\limits_{i=1}^n{(X_i-\bar{X})(Y_i-\bar{Y})}}{\sum\limits_{i=1}^n{(X_i-\bar{X})^2}}\]

Once the slope is known, it is possible to calculate the intercept #b_0# of the regression with the following formula:

\[b_0 = \bar{Y} - b_1 \cdot \bar{X}\]

Consider the following #5# pairs of data points:

#X_i#	#Y_i#
#1#	#6#
#2#	#6#
#3#	#1#
#4#	#4#
#5#	#7#

Find the regression line corresponding to these points.

The first step in determining the regression line is to calculate the mean values of #X# and #Y#.
\[\begin{array}{rcl}
\bar{X}&=&\displaystyle\cfrac{\sum\limits_{i=1}^n{X_i}}{n} = \dfrac{1+2+3+4+5}{5}=\dfrac{15}{5}=3\\\\
\bar{Y}&=&\displaystyle\cfrac{\sum\limits_{i=1}^n{Y_i}}{n} = \dfrac{6+6+1+4+7}{5}=\dfrac{24}{5}=4.8
\end{array}\]
Next, find the values of #(X_i-\bar{X}), (Y_i-\bar{Y}), (X_i-\bar{X})^2# and #(X_i-\bar{X})(Y_i-\bar{Y})# for each pair of data points:

#X#	#Y#	#X_i - \bar{X}#	#Y_i - \bar{Y}#	#(X_i - \bar{X})^2#	#(X_i - \bar{X})(Y_i - \bar{Y})#
#1#	#6#	#-2#	#1.2#	#4#	#-2.4#
#2#	#6#	#-1#	#1.2#	#1#	#-1.2#
#3#	#1#	#0#	#-3.8#	#0#	#0#
#4#	#4#	#1#	#-0.8#	#1#	#-0.8#
#5#	#7#	#2#	#2.2#	#4#	#4.4#

With this information the slope #b_1# and intercept #b_0# can be calculated:
\[\begin{array}{rcl}
b_1 &=& \displaystyle\cfrac{\sum\limits_{i=1}^n{(X_i-\bar{X})(Y_i-\bar{Y})}}{\sum\limits_{i=1}^n{(X_i-\bar{X})^2}}\\\\
&=& \cfrac{-2.4-1.2+0-0.8+4.4}{4+1+0+1+4}\\\\
&=& 0.0\\\\
b_0 &=& \bar{Y} - b_1 \cdot \bar{X}\\\\
&=& 4.8 - (0.0)\cdot3\\\\
&=& 4.8
\end{array}\]
So the regression equation is:
\[\begin{array}{rcl}
\hat{Y} &=& b_0 + b_1X\\\\
&=& 4.8+ 0.0X
\end{array}\]

Note that the regression line always passes through the mean point #(\bar{X},\bar{Y})#.

In this case, #(\bar{X},\bar{Y})= (3,4.8)# and entering #X=3# into the equation gives:
\[\begin{array}{rcl}
\hat{Y} &=& 4.8 + 0.0\cdot 3\\\\
&=& 4.8
\end{array}\]

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