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