Simple 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.

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 xi column shows scores on the aptitude test. Similarly, the yi column shows statistics grades. The last two rows show sums and mean scores that we will use to conduct the regression analysis.

Student xi yi (xi - x) (yi - y) (xi - x)2 (yi - y)2 (xi - x)(yi - y)
1 95 85 17 8 289 64 136
2 85 95 7 18 49 324 126
3 80 70 2 -7 4 49 -14
4 70 65 -8 -12 64 144 96
5 60 70 -18 -7 324 49 126
Sum 390 385 730 630 470
Mean 78 77

The regression equation is a linear equation of the form: ŷ = b0 + b1x . To conduct a regression analysis, we need to solve for b0 and b1. Computations are shown below.

b1 = Σ [ (xi - x)(yi - y) ] / Σ [ (xi - x)2]
b1 = 470/730 = 0.644
  b0 = y - b1 * x
b0 = 77 - (0.644)(78) = 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 would be:

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

R2 = { ( 1 / N ) * Σ [ (xi - x) * (yi - y) ] / (σx * σy ) }2

where N is the number of observations used to fit the model, Σ is the summation symbol, xi is the x value for observation i, x is the mean x value, yi is the y value for observation i, y is the mean y value, σx is the standard deviation of x, and σy is the standard deviation of y. Computations for the sample problem of this lesson are shown below.

σx = sqrt [ Σ ( xi - x )2 / N ]
σx = sqrt( 730/5 ) = sqrt(146) = 12.083

 

σy = sqrt [ Σ ( yi - y )2 / N ]
σy = sqrt( 630/5 ) = sqrt(126) = 11.225

R2 = { ( 1 / N ) * Σ [ (xi - x) * (yi - y) ] / (σx * σy ) }2
R2 = [ ( 1/5 ) * 470 / ( 12.083 * 11.225 ) ]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.