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 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.

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

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

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. Notice that all of our inputs for the regression analysis come from the above three tables.

First, we solve for the regression coefficient (b₁):

b₁ = Σ [ (x_i - x)(y_i - y) ] / Σ [ (x_i - x)²]

b₁ = 470/730

b₁ = 0.644

Once we know the value of the regression coefficient (b₁), we can solve for the regression slope (b₀):

b₀ = y - b₁ * x

b₀ = 77 - (0.644)(78)

b₀ = 26.768

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