# How to do a test of joint significance (in R)

## Task

If we have a multivariate linear model, how do we test the joint significance of all the variables in the model? In other words, how do we test the overall significance of the regression model?

## Solution

Let’s assume that you already made your multiple regression model, similar to the one shown below. You can visit this task, , to see how to construct a multivariate linear model.

Let’s assume that you already made your multivariate linear model, similar to the one shown below. If you still need to create one, first see how to fit a multivariate linear model.

We use example data here, but you would use your own data instead.

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x1 <- c( 2, 7, 4, 3, 11, 18, 6, 15, 9, 12)
x2 <- c( 4, 6, 10, 1, 18, 11, 8, 20, 16, 13)
x3 <- c(11, 16, 20, 6, 14, 8, 5, 23, 13, 10)
y <- c(24, 60, 32, 29, 90, 45, 130, 76, 100, 120)
model <- lm(y ~ x1 + x2 + x3)

Now we want to test whether the model is significant. We will use a null hypothesis that states that all of the model’s coefficients are equal to zero, that is, they are not jointly significant in predicting $y$. We can write $H_0: \beta_0 = \beta_1 = \beta2 = \beta_3 = 0$.

We also choose a value $0 \le \alpha \le 1$ as our Type 1 error rate. Herer we’ll use $\alpha=0.05$.

The summary output for the model will give us both the F-statistic and the p-value.

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summary(model)

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Call:
lm(formula = y ~ x1 + x2 + x3)
Residuals:
Min 1Q Median 3Q Max
-25.031 -20.218 -8.373 22.937 35.640
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 77.244 27.366 2.823 0.0302 *
x1 -2.701 2.855 -0.946 0.3806
x2 7.299 2.875 2.539 0.0441 *
x3 -4.861 2.187 -2.223 0.0679 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 30.13 on 6 degrees of freedom
Multiple R-squared: 0.5936, Adjusted R-squared: 0.3904
F-statistic: 2.921 on 3 and 6 DF, p-value: 0.1222

In the final line of the output, we can see that the F-statistic is 2.921. The corresponding $p$-value in the same line is 0.1222, which is greater than $\alpha$, so we do not have sufficient evidence to reject the null hypothesis.

We cannot conclude that the independent variables in our model are jointly significant in predicting the response variable.

Content last modified on 24 July 2023.

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Contributed by Elizabeth Czarniak (CZARNIA_ELIZ@bentley.edu)