# interaction of categorical and continuous variables

I have a dependent variable that is continuous and I have two independent variables: one continuous and one categorical (with 2 categories)

The interaction between the independent variables is significant. Which statistical analysis should I use (in R) to proceed with the analysis and document the interaction?

(Should I simply analyze each of the two categories separately using simple linear regression?)

• What did you do to conclude the interaction was significant? – Glen_b Jun 5 '14 at 2:36

## 2 Answers

In the scenario you describe least squares regression will allow you to tell a very straightforward story:

First of all, imagine that you have no dichotomous independent variable. So:

(1) $$y_{i} = \beta_{0} + \beta_{1}x_{1i} + \varepsilon_{i}$$

Your regression describes the relationship between your dependent variable $$y$$ and your continuous independent variable $$x_{1}$$ as a straight line, with intercept $$\beta_{0}$$ and slope $$\beta_{1}$$. Cool? Cool.

Now add both the dichotomous independent variable $$x_{2}$$ and the interaction between $$x_{1}$$ and $$x_{2}$$ to the model:

(2) $$y_{i} = \beta_{0} + \beta_{1}x_{1i} + \beta_{2}x_{2i} + \beta_{3}x_{1i}x_{2i} + \varepsilon_{i}$$

So now what is your model telling you? Well, (assuming $$x_{2}$$ is coded 0/1) when $$x_{2} = 0$$, then the model reduces to equation (1) because $$\beta_{2} \times 0 = 0$$ and $$\beta_{3} \times x_{1} \times 0 = 0$$. So that is easy-peasy puddin' pie.

What about when $$x_{2} =1$$? Well now the $$y$$-intercept is $$\beta_{0} + \beta_{2}$$ (Right? Because $$\beta_{2} \times 1 = \beta_{2}$$).

And the slope of the line relating $$y$$ to $$x_{1}$$ is now $$\beta_{1} + \beta_{3}$$ (Right? Because $$\beta_{1}\times x_{1} + \beta_{3} \times x_{1} \times 1 = \beta_{1}\times x_{1} + \beta_{3} \times x_{1} = (\beta_{1} + \beta_{3})\times x_{1}$$).

So when $$x_{2}=1$$ you simply have a second regression line relating $$y$$ to $$x_{1}$$, with a different intercept (if $$\beta_{2} \ne 0$$) and a different slope (if $$\beta_{3} \ne 0$$ which will be true if you tested a significant interaction term in, say, ANOVA).

How do you communicate this? A single graph with two regression lines overlaying your data (possibly with different colored/shaped/sized markers when $$x_{2}=1$$), and a label indicating which line corresponds to $$x_{2}=0$$ and $$x_{2}=1$$. Also providing your audience with the values of the $$\beta$$s and their standard errors and/or confidence intervals is good (like, in a table of multiple regression results).

Cool? Cool.

Finally, while all the above tells you about trend relationships between $$y$$ and $$x_{1}$$ given $$x_{2}$$, least squares regression also tells you about strength of association. If you had a single independent variable, you'd probably want to use something like $$R^{2}$$ to describe this strength of association, but when you add variables $$R^{2}$$ doesn't quite mean what it did before. So you might use generalized $$R^{2}$$, or Pseudo-$$R^{2}$$ or some such.

• How do I follow up such a significant interaction between categorical and continuous variable? That is, how can I breakdown the interaction effect for each sub-category of the factor variable? – user2564011 Jun 9 '14 at 17:17
• Well. You can do the same thing, except instead of two lines, you would have $k$ lines corresponding to $k$ groups. Assume you have 4 groups, A, B, C, and D. Then you could use three 0/1 indicator variables for groups B, C and D. When B, C and D = zero, the estimates reduce to equation (1). When only C and D = 0, the estimates reduce to something like equation (2), same when B and D = 0 and when B and C = 0. – Alexis Jun 9 '14 at 20:37
• What if there are multiple levels in the categorical variable i.e more than 2 even 7 or 8 – Liger Nov 13 '18 at 12:52
• @Liger I already answered this with an example in the comment directly above the one you just wrote (I used a categorical variable with 4 groups, but categorical variables with others numbers work the same way, just need to be effect coded, and you are good to go). – Alexis Nov 14 '18 at 1:41

@Alexis seems to cover the equations pretty well. Here's some example code in :

set.seed(8);d8a=data.frame(x=rnorm(99),z=rbinom(99,1,.5))    #Data sim'd to fit the scenario
d8a$y=(d8a$x+rnorm(99,0,3))*(2*d8a\$z-1)                      #Guarantees an interaction
summary(lm(y~scale(x)*factor(z),d8a))       #Fits a GLM with OLS – this is the part you need


$$\rm Output$$

Residuals:
Min      1Q  Median      3Q     Max
-6.1575 -2.1416 -0.2051  1.8558  6.5765

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)         -0.17374    0.43557  -0.399 0.690867
scale(x)            -1.11354    0.45224  -2.462 0.015608 *
factor(z)1           0.01546    0.58976   0.026 0.979144
scale(x):factor(z)1  2.24831    0.59689   3.767 0.000287 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.922 on 95 degrees of freedom
Multiple R-squared: 0.1328, Adjusted R-squared: 0.1054
F-statistic:  4.85 on 3 and 95 DF,  p-value: 0.003492


$$\rm Plot$$

require(ggplot2);ggplot(d8a,aes(x,y,color=factor(z)))+stat_smooth(method=lm)+geom_point()