How to Predict Values in R Using Multiple Regression Model


You can use the following basic syntax to predict values in R using a fitted multiple linear regression model:

#define new observation
new <- data.frame(x1=c(5), x2=c(10), x3=c(12.5))

#use fitted model to predict the response value for the new observation
predict(model, newdata=new)

The following example shows how to use this function in practice.

Example: Predict Values Using Fitted Multiple Linear Regression Model

Suppose we have the following dataset in R that contains information about basketball players:

#create data frame
df <- data.frame(rating=c(67, 75, 79, 85, 90, 96, 97),
                 points=c(8, 12, 16, 15, 22, 28, 24),
                 assists=c(4, 6, 6, 5, 3, 8, 7),
                 rebounds=c(1, 4, 3, 3, 2, 6, 7))

#view data frame
df

  rating points assists rebounds
1     67      8       4        1
2     75     12       6        4
3     79     16       6        3
4     85     15       5        3
5     90     22       3        2
6     96     28       8        6
7     97     24       7        7

Now suppose we fit a multiple linear regression model using points, assists, and rebounds as predictor variables and rating as the response variable:

#fit multiple linear regression model
model <- lm(rating ~ points + assists + rebounds, data=df)

#view model summary
summary(model)

Call:
lm(formula = rating ~ points + assists + rebounds, data = df)

Residuals:
      1       2       3       4       5       6       7 
-1.5902 -1.7181  0.2413  4.8597 -1.0201 -0.6082 -0.1644 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)  66.4355     6.6932   9.926  0.00218 **
points        1.2152     0.2788   4.359  0.02232 * 
assists      -2.5968     1.6263  -1.597  0.20860   
rebounds      2.8202     1.6118   1.750  0.17847   
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 3.193 on 3 degrees of freedom
Multiple R-squared:  0.9589,	Adjusted R-squared:  0.9179 
F-statistic: 23.35 on 3 and 3 DF,  p-value: 0.01396

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

Rating = 66.4355 + 1.2151(points) – 2.5968(assists) + 2.8202(rebounds)

We can use the following code to predict the rating of a new player who has 20 points, 5 assists, and 2 rebounds:

#define new player
new <- data.frame(points=c(20), assists=c(5), rebounds=c(2))

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

       1 
83.39607 

The model predicts that this new player will have a rating of 83.39607.

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

  • Rating = 66.4355 + 1.2151(points) – 2.5968(assists) + 2.8202(rebounds)
  • Rating = 66.4355 + 1.2151(20) – 2.5968(5) + 2.8202(2)
  • Rating = 83.39

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

Additional Resources

The following tutorials explain how to perform other common tasks in R:

How to Perform Simple Linear Regression in R
How to Perform Multiple Linear Regression in R
How to Create a Residual Plot in R

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