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?)

  • 4
    $\begingroup$ What did you do to conclude the interaction was significant? $\endgroup$
    – Glen_b
    Jun 5, 2014 at 2:36

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Jun 9, 2014 at 17:17
  • 1
    $\begingroup$ 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. $\endgroup$
    – Alexis
    Jun 9, 2014 at 20:37
  • $\begingroup$ What if there are multiple levels in the categorical variable i.e more than 2 even 7 or 8 $\endgroup$
    – CocoCrisp
    Nov 13, 2018 at 12:52
  • $\begingroup$ @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). $\endgroup$
    – Alexis
    Nov 14, 2018 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

$$\rm Output$$

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

                    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

$$\rm Plot$$



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.