### Linear Regression

#### Introduction

#### Regression

#### Regression Analysis

#### Statistical Inference

### Simple Linear Regression Lessons

# Linear Regression Example

In this lesson, we apply regression analysis to some fictitious data, and we show how to interpret the results of our analysis.

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(e.g., a recent version of Edge, Chrome, Firefox, or Opera), you can watch a video treatment of this lesson.

**Note:** Regression computations are usually handled by a software package or a
graphing calculator. For this
example, however, we will do the computations "manually", since
the gory details have educational value.

## Problem Statement

Last year, five randomly selected students took a math aptitude test before they began their statistics course. The Statistics Department has three questions.

- What linear regression equation best predicts statistics performance, based on math aptitude scores?
- If a student made an 80 on the aptitude test, what grade would we expect her to make in statistics?
- How well does the regression equation fit the data?

## How to Find the Regression Equation

In the table below, the x_{i} column shows scores on the
aptitude test. Similarly, the y_{i} column shows statistics
grades. The last two columns show deviations scores - the difference between the
student's score and the average score on each test. The last two rows show sums and mean
scores that we will use to conduct the regression analysis.

Student | x_{i} |
y_{i} |
(x_{i}-x) |
(y_{i}-y) |
---|---|---|---|---|

1 | 95 | 85 | 17 | 8 |

2 | 85 | 95 | 7 | 18 |

3 | 80 | 70 | 2 | -7 |

4 | 70 | 65 | -8 | -12 |

5 | 60 | 70 | -18 | -7 |

Sum |
390 | 385 | ||

Mean |
78 | 77 |

And for each student, we also need to compute the squares of the deviation scores (the last two columns in the table below).

Student | x_{i} |
y_{i} |
(x_{i}-x)^{2} |
(y_{i}-y)^{2} |
---|---|---|---|---|

1 | 95 | 85 | 289 | 64 |

2 | 85 | 95 | 49 | 324 |

3 | 80 | 70 | 4 | 49 |

4 | 70 | 65 | 64 | 144 |

5 | 60 | 70 | 324 | 49 |

Sum |
390 | 385 | 730 | 630 |

Mean |
78 | 77 |

And finally, for each student, we need to compute the product of the deviation scores.

Student | x_{i} |
y_{i} |
(x_{i}-x)(y_{i}-y) |
---|---|---|---|

1 | 95 | 85 | 136 |

2 | 85 | 95 | 126 |

3 | 80 | 70 | -14 |

4 | 70 | 65 | 96 |

5 | 60 | 70 | 126 |

Sum |
390 | 385 | 470 |

Mean |
78 | 77 |

The regression equation is a linear equation of the form:
ŷ = b_{0} + b_{1}x . To conduct a regression
analysis, we need to solve for b_{0} and b_{1}.
Computations are shown below. Notice that all of our inputs for the
regression analysis come from the above three tables.

First, we solve for the regression coefficient (b_{1}):

b_{1} = Σ [ (x_{i} - x)(y_{i} - y) ] / Σ [ (x_{i} - x)^{2}]

b_{1} = 470/730

b_{1} = 0.644

Once we know the value of the regression coefficient (b_{1}), we can solve for the regression slope (b_{0}):

b_{0} = y - b_{1} * x

b_{0} = 77 - (0.644)(78)

b_{0} = 26.768

Therefore, the regression equation is: ŷ = 26.768 + 0.644x .

## How to Use the Regression Equation

Once you have the regression equation, using it is a snap. Choose
a value for the independent variable (*x*), perform the
computation, and you have an estimated value (ŷ)
for the dependent variable.

In our example, the independent variable is the student's score on the aptitude test. The dependent variable is the student's statistics grade. If a student made an 80 on the aptitude test, the estimated statistics grade (ŷ) would be:

ŷ = b_{0} + b_{1}x

ŷ = 26.768 + 0.644x = 26.768 + 0.644 * 80

ŷ = 26.768 + 51.52 = 78.288

**Warning:** When you use a regression equation,
do not use values for the independent variable that are outside
the range of values used to create the equation. That is called
**extrapolation**, and it can produce unreasonable
estimates.

In this example, the aptitude test scores used to create the regression equation ranged from 60 to 95. Therefore, only use values inside that range to estimate statistics grades. Using values outside that range (less than 60 or greater than 95) is problematic.

## How to Find the Coefficient of Determination

Whenever you use a regression equation, you should ask how well the equation fits the data. One way to assess fit is to check the coefficient of determination, which can be computed from the following formula.

R^{2} = { ( 1 / N ) * Σ [ (x_{i} - x) * (y_{i} - y) ] / (σ_{x} * σ_{y} ) }^{2}

where N is the number of
observations used to fit the model, Σ is the summation symbol,
x_{i} is the x value for observation i,
x is the mean x value,
y_{i} is the y value for observation i,
y is the mean y value,
σ_{x} is the standard deviation of x, and
σ_{y} is the standard deviation of y.

Computations for the sample problem of this lesson are shown below. We begin by computing the standard deviation of x (σ_{x}):

σ_{x} = sqrt [ Σ ( x_{i} - x )^{2} / N ]

σ_{x} = sqrt( 730/5 ) = sqrt(146) = 12.083

Next, we find the standard deviation of y, (σ_{y}):

σ_{y} = sqrt [ Σ ( y_{i} - y )^{2} / N ]

σ_{y} = sqrt( 630/5 ) = sqrt(126) = 11.225

^{2}):

R^{2} = { ( 1 / N ) * Σ [ (x_{i} - x) * (y_{i} - y) ] / (σ_{x} * σ_{y} ) }^{2}

R^{2} = [ ( 1/5 ) * 470 / ( 12.083 * 11.225 ) ]^{2}

R^{2} = ( 94 / 135.632 )^{2} = ( 0.693 )^{2} = 0.48

A coefficient of determination equal to 0.48 indicates that about 48% of the variation in statistics grades (the dependent variable) can be explained by the relationship to math aptitude scores (the independent variable). This would be considered a good fit to the data, in the sense that it would substantially improve an educator's ability to predict student performance in statistics class.

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