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I dug through previous simple slope/effect questions and couldn't find what I was looking for, but happy to be pointed to it if it exists.

On p. 17 of Jaccard and Turrisi (2003), they offer the following basic example of a regression equation with a single interaction term:

$Y= a + β_1X+β_2Z+β_3XZ+ε$

They explain how to interpret the coefficients for X and Z thus (p. 24):

The coefficient for X estimates the effect of X on Y when Z is at a specific value, namely, when Z = 0. The coefficient for Z estimates the effect of Z on Y when X is at a specific value, namely, when X = 0.

These are simple slopes (or "simple effects" as J & T call them). This makes sense to me. However, I don't understand how to interpret the non-interaction coefficients in the following case, which adds W and XW as new predictors:

$Y= a + β_1X+β_2Z+β_3W+β_4XZ+β_5XW+ε$

When only one interaction term is present, it is straightforward to interpret $β_1$ and $β_2$. But in a model where X is present in two (or more) interaction terms, how is $β_1$ to be interpreted? Is it now the effect of X on Y when both Z and W = 0, or something else? I'm also guessing the interpretation of $β_2$ remains unchanged as Z only appears in one interaction.

Thanks in advance.

Reference

Jaccard, J., & Turrisi, R. (2003). Interaction effects in multiple regression (No. 72). Sage.

intermediary
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1 Answers1

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Algebraic Expression

We can just as easily reshape the linear equation you have below (I changed $a$ to $\beta_0$ for consistency in notation):

$$ y = \beta_0 + \beta_1{X} + \beta_2{Z} + \beta_3{W} + \beta_4{XZ} + \beta_5{XW} + \epsilon $$

Into this congruent equation, re-expressed with $X$ as the conditional effect (see Aiken & West, 1991, p.13):

$$ y = (\beta_1 + \beta_4Z + \beta_5{W})X + \beta_0 + \beta_2Z + \beta_3W + \epsilon $$

You can then see that the intercept and individual main effects of $\beta_2Z$ and $\beta_3W$ are unchanged (and algebraically, neither is $\beta_1X$). The only parts that are not additive here are the products formed from $\beta_4XZ$ and $\beta_5XW$. We are only estimating the products of $X$ with $W$ and $Z$ explicitly for interactions. So to be explicit, the $XW$ here is an interaction, albeit a simple two-way interaction between $X$ and $W$. I assume here that three-way interactions have not been modeled in this equation, else it would be included as another coefficient.

The interpretation for this model is still the same. You set all the $\beta$ coefficients to zero, thereafter a change in one coefficient is conditional on all the rest being equal to zero. To find out what the effect of $X$ is on the outcome, we can just plug this into our earlier re-expressed equation like so, leaving only our $\beta_0$ and $\beta_1$ alone:

$$ y = (\beta_1 + 0 + 0)X + \beta_0 + 0 + 0 + \epsilon $$

Which simplifies to:

$$ y = \beta_0 + \beta_1{X} + \epsilon $$

Simulated Example

Using your example above, we can simulate this directly in R, setting all of the mean and standard deviations to their defaults with $n = 1000$.

#### Simulate Data ####
set.seed(123)
n <- 1000
x <- rnorm(n)
y <- rnorm(n)
z <- rnorm(n)
w <- rnorm(n)

Then we can fit the data into a regression which explicitly models the main effects of $X$,$W$,and $Z$, along with the interactions with $X$.

#### Fit Data ####
fit <- lm(
  y ~ x + z + w + x:z + x:w
)
summary(fit)

The results can be shown below:

Call:
lm(formula = y ~ x + z + w + x:z + x:w)

Residuals:
    Min      1Q  Median      3Q     Max 
-2.7916 -0.6280 -0.0261  0.6446  3.4334


Coefficients:
            Estimate Std. Error t value Pr(>|t|)

(Intercept) -0.01657    0.03105  -0.534   0.5937

x            0.02560    0.03073   0.833   0.4050

z            0.05043    0.03112   1.620   0.1055

w           -0.01367    0.03104  -0.440   0.6598

x:z         -0.05401    0.03003  -1.798   0.0724 .
x:w         -0.05447    0.03035  -1.795   0.0730 .




Signif. codes:  0 ‘*’ 0.001 ‘’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


Residual standard error: 0.9761 on 994 degrees of freedom
Multiple R-squared:  0.009623,  Adjusted R-squared:  0.004641 
F-statistic: 1.932 on 5 and 994 DF,  p-value: 0.08655

Using the example I gave earlier, we can simply set $W$ and $Z$ to zero and $X$ to an arbitrary value of $2$. By creating new data, we simply fill this into our linear equation and use predict to see what it should come out as:

#### Create New Data ####
new.data <- data.frame(
  z = 0,
  w = 0,
  x = 2
)

Predict New Data


pred <- predict(fit,
        newdata = new.data)

The estimated value of the prediction is $0.03462692$. If we calculate our coefficients the same way we spoke about earlier, we should get the same result as the pred object (I have commented which of our $\beta$ coefficients go where in hash comments next to the R code, as by default R codes the coefficients by their sequential order):

#### Check if Predicted Outcome Matches Coefficients ####
pred == (
  fit$coefficients[1] # beta 0
  + (fit$coefficients[2] * 2) # beta 1
  + (fit$coefficients[3] * 0) # beta 2
  + (fit$coefficients[4] * 0) # beta 3
  + (fit$coefficients[5] * 0) # beta 4
  + (fit$coefficients[6] * 0) # beta 5
  )

By doing so, our prediction comes true:

   1 
TRUE

Citation

Aiken, L. S., & West, S. G. (1991). Multiple regression: Testing and interpreting interactions. Sage Publications.

Shawn Hemelstrand
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