Linear regression is a method we can use to quantify the relationship between one or more predictor variables and a .

Often you may want to fit a regression model using one or more as predictor variables.

This tutorial provides a step-by-step example of how to perform linear regression with categorical variables in R.

Example: Linear Regression with Categorical Variables in R

Suppose we have the following data frame in R that contains information on three variables for 12 different basketball players:

• points scored
• hours spent practicing
• training program used

#create data frame
df <- data.frame(points=c(7, 7, 9, 10, 13, 14, 12, 10, 16, 19, 22, 18),
                 hours=c(1, 2, 2, 3, 2, 6, 4, 3, 4, 5, 8, 6),
                 program=c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3))

#view data frame
df

   points hours program
1       7     1       1
2       7     2       1
3       9     2       1
4      10     3       1
5      13     2       2
6      14     6       2
7      12     4       2
8      10     3       2
9      16     4       3
10     19     5       3
11     22     8       3
12     18     6       3

Suppose we would like to fit the following linear regression model:

points = β₀ + β₁hours + β₂program

In this example, hours is a continuous variable but program is a categorical variable that can take on three possible categories: program 1, program 2, or program 3. 

In order to fit this regression model and tell R that the variable “program” is a categorical variable, we must use as.factor() to convert it to a factor and then fit the model:

#convert 'program' to factor
df$program <- as.factor(df$program)

#fit linear regression model
fit <- lm(points ~ hours + program, data = df)

#view model summary
summary(fit)

Call:
lm(formula = points ~ hours + program, data = df)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.5192 -1.0064 -0.3590  0.8269  2.4551 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   6.3013     0.9462   6.660 0.000159 ***
hours         0.9744     0.3176   3.068 0.015401 *  
program2      2.2949     1.1369   2.019 0.078234 .  
program3      6.8462     1.5499   4.417 0.002235 ** 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.403 on 8 degrees of freedom
Multiple R-squared:  0.9392,	Adjusted R-squared:  0.9164 
F-statistic: 41.21 on 3 and 8 DF,  p-value: 3.276e-05

From the values in the Estimate column, we can write the fitted regression model:

points = 6.3013 + .9744(hours) + 2.2949(program 2) + 6.8462(program 3)

Here’s how to interpret the coefficient values in the output:

• hours: For each additional hour spent practicing, points scored increases by an average of 0.9744, assuming program is held constant.
  • The p-value is .015, which indicates that hours spent practicing is a statistically significant predictor of points scored at level α = .05.
• program2: Players who used program 2 scored an average of 2.2949 more points than players who used program 1, assuming hours is held constant.
  • The p-value is .078, which indicates that there is not a statistically significant difference in points scored by players who used program 2 compared to players who used program 1, at level α = .05.
• program3: Players who used program 3 scored an average of 2.2949 more points than players who used program 1, assuming hours is held constant.
  • The p-value is .002, which indicates that there is a statistically significant difference in points scored by players who used program 3 compared to players who used program 1, at level α = .05.

Using the fitted regression model, we can predict the number of points scored by a player based on their total hours spent practicing and the program they used.

#define new player
new <- data.frame(hours=c(5), program=as.factor(c(3)))

#use the fitted model to predict the points for the new player
predict(fit, newdata=new)

       1 
18.01923 

The model predicts that this new player will score 18.01923 points.

We can confirm this is correct by plugging in the values for the new player into the fitted regression equation:

• points = 6.3013 + .9744(hours) + 2.2949(program 2) + 6.8462(program 3)
• points = 6.3013 + .9744(5) + 2.2949(0) + 6.8462(1)
• points = 18.019

This matches the value we calculated using the predict() function in R.