### 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# | #1# |

#3# | #8# |

#4# | #8# |

#5# | #6# |

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+1+8+8+6}{5}=\dfrac{29}{5}=5.8

\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# | #0.2# | #4# | #-0.4# |

#2# | #1# | #-1# | #-4.8# | #1# | #4.8# |

#3# | #8# | #0# | #2.2# | #0# | #0# |

#4# | #8# | #1# | #2.2# | #1# | #2.2# |

#5# | #6# | #2# | #0.2# | #4# | #0.4# |

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{-0.4+4.8+0+2.2+0.4}{4+1+0+1+4}\\\\

&=& 0.7\\\\

b_0 &=& \bar{Y} - b_1 \cdot \bar{X}\\\\

&=& 5.8 - (0.7)\cdot3\\\\

&=& 3.7

\end{array}\]

So the regression equation is:

\[\begin{array}{rcl}

\hat{Y} &=& b_0 + b_1X\\\\

&=& 3.7+ 0.7X

\end{array}\]

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

In this case, #(\bar{X},\bar{Y})= (3,5.8)# and entering #X=3# into the equation gives:

\[\begin{array}{rcl}

\hat{Y} &=& 3.7 + 0.7\cdot 3\\\\

&=& 5.8

\end{array}\]

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