## 6.1 One numerical and one categorical explanatory variable

Let’s revisit the instructor evaluation data from UT Austin we introduced in Section 5.1. We studied the relationship between teaching evaluation scores as given by students and “beauty” scores. The variable teaching score was the numerical outcome variable $$y$$, and the variable “beauty” score (bty_avg) was the numerical explanatory $$x$$ variable.

In this section, we are going to consider a different model. Our outcome variable will still be teaching score, but we’ll now include two different explanatory variables: age and (binary) gender. Could it be that instructors who are older receive better teaching evaluations from students? Or could it instead be that younger instructors receive better evaluations? Are there differences in evaluations given by students for instructors of different genders? We’ll answer these questions by modeling the relationship between these variables using multiple regression, where we have:

1. A numerical outcome variable $$y$$, the instructor’s teaching score, and
2. Two explanatory variables:
1. A numerical explanatory variable $$x_1$$, the instructor’s age.
2. A categorical explanatory variable $$x_2$$, the instructor’s (binary) gender.

It is important to note that at the time of this study due to then commonly held beliefs about gender, this variable was often recorded as a binary variable. While the results of a model that oversimplifies gender this way may be imperfect, we still found the results to be pertinent and relevant today.

### 6.1.1 Exploratory data analysis

Recall that data on the 463 courses at UT Austin can be found in the evals data frame included in the moderndive package. However, to keep things simple, let’s select() only the subset of the variables we’ll consider in this chapter, and save this data in a new data frame called evals_ch6. Note that these are different than the variables chosen in Chapter 5.

evals_ch6 <- evals %>%
select(ID, score, age, gender)

Recall the three common steps in an exploratory data analysis we saw in Subsection 5.1.1:

1. Looking at the raw data values.
2. Computing summary statistics.
3. Creating data visualizations.

Let’s first look at the raw data values by either looking at evals_ch6 using RStudio’s spreadsheet viewer or by using the glimpse() function from the dplyr package:

glimpse(evals_ch6)
Rows: 463
Columns: 4
$ID <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,…$ score  <dbl> 4.7, 4.1, 3.9, 4.8, 4.6, 4.3, 2.8, 4.1, 3.4, 4.5, 3.8, 4.5, 4.…
$age <int> 36, 36, 36, 36, 59, 59, 59, 51, 51, 40, 40, 40, 40, 40, 40, 40…$ gender <fct> female, female, female, female, male, male, male, male, male, …

Let’s also display a random sample of 5 rows of the 463 rows corresponding to different courses in Table 6.1. Remember due to the random nature of the sampling, you will likely end up with a different subset of 5 rows.

evals_ch6 %>% sample_n(size = 5)
TABLE 6.1: A random sample of 5 out of the 463 courses at UT Austin
ID score age gender
129 3.7 62 male
109 4.7 46 female
28 4.8 62 male
434 2.8 62 male
330 4.0 64 male

Now that we’ve looked at the raw values in our evals_ch6 data frame and got a sense of the data, let’s compute summary statistics. As we did in our exploratory data analyses in Sections 5.1.1 and 5.2.1 from the previous chapter, let’s use the skim() function from the skimr package, being sure to only select() the variables of interest in our model:

evals_ch6 %>% select(score, age, gender) %>% skim()
Skim summary statistics
n obs: 463
n variables: 3

── Variable type:factor
variable missing complete   n n_unique                top_counts ordered
gender       0      463 463        2 mal: 268, fem: 195, NA: 0   FALSE

── Variable type:integer
variable missing complete   n  mean  sd p0 p25 p50 p75 p100
age       0      463 463 48.37 9.8 29  42  48  57   73

── Variable type:numeric
variable missing complete   n mean   sd  p0 p25 p50 p75 p100
score       0      463 463 4.17 0.54 2.3 3.8 4.3 4.6    5

Observe that we have no missing data, that there are 268 courses taught by male instructors and 195 courses taught by female instructors, and that the average instructor age is 48.37. Recall that each row represents a particular course and that the same instructor often teaches more than one course. Therefore, the average age of the unique instructors may differ.

Furthermore, let’s compute the correlation coefficient between our two numerical variables: score and age. Recall from Subsection 5.1.1 that correlation coefficients only exist between numerical variables. We observe that they are “weakly negatively” correlated.

evals_ch6 %>%
get_correlation(formula = score ~ age)
# A tibble: 1 x 1
cor
<dbl>
1 -0.107

Let’s now perform the last of the three common steps in an exploratory data analysis: creating data visualizations. Given that the outcome variable score and explanatory variable age are both numerical, we’ll use a scatterplot to display their relationship. How can we incorporate the categorical variable gender, however? By mapping the variable gender to the color aesthetic, thereby creating a colored scatterplot. The following code is similar to the code that created the scatterplot of teaching score over “beauty” score in Figure 5.2, but with color = gender added to the aes()thetic mapping.

ggplot(evals_ch6, aes(x = age, y = score, color = gender)) +
geom_point() +
labs(x = "Age", y = "Teaching Score", color = "Gender") +
geom_smooth(method = "lm", se = FALSE)

In the resulting Figure 6.1, observe that ggplot() assigns a default in red/blue color scheme to the points and to the lines associated with the two levels of gender: female and male. Furthermore, the geom_smooth(method = "lm", se = FALSE) layer automatically fits a different regression line for each group.

We notice some interesting trends. First, there are almost no women faculty over the age of 60 as evidenced by lack of red dots above $$x$$ = 60. Second, while both regression lines are negatively sloped with age (i.e., older instructors tend to have lower scores), the slope for age for the female instructors is more negative. In other words, female instructors are paying a harsher penalty for advanced age than the male instructors.

### 6.1.2 Interaction model

Let’s now quantify the relationship of our outcome variable $$y$$ and the two explanatory variables using one type of multiple regression model known as an interaction model. We’ll explain where the term “interaction” comes from at the end of this section.

In particular, we’ll write out the equation of the two regression lines in Figure 6.1 using the values from a regression table. Before we do this, however, let’s go over a brief refresher of regression when you have a categorical explanatory variable $$x$$.

Recall in Subsection 5.2.2 we fit a regression model for countries’ life expectancies as a function of which continent the country was in. In other words, we had a numerical outcome variable $$y$$ = lifeExp and a categorical explanatory variable $$x$$ = continent which had 5 levels: Africa, Americas, Asia, Europe, and Oceania. Let’s re-display the regression table you saw in Table 5.8:

TABLE 6.2: Regression table for life expectancy as a function of continent
term estimate std_error statistic p_value lower_ci upper_ci
intercept 54.8 1.02 53.45 0 52.8 56.8
continentAmericas 18.8 1.80 10.45 0 15.2 22.4
continentAsia 15.9 1.65 9.68 0 12.7 19.2
continentEurope 22.8 1.70 13.47 0 19.5 26.2
continentOceania 25.9 5.33 4.86 0 15.4 36.5

Recall our interpretation of the estimate column. Since Africa was the “baseline for comparison” group, the intercept term corresponds to the mean life expectancy for all countries in Africa of 54.8 years. The other four values of estimate correspond to “offsets” relative to the baseline group. So, for example, the “offset” corresponding to the Americas is +18.8 as compared to the baseline for comparison group Africa. In other words, the average life expectancy for countries in the Americas is 18.8 years higher. Thus the mean life expectancy for all countries in the Americas is 54.8 + 18.8 = 73.6. The same interpretation holds for Asia, Europe, and Oceania.

Going back to our multiple regression model for teaching score using age and gender in Figure 6.1, we generate the regression table using the same two-step approach from Chapter 5: we first “fit” the model using the lm() “linear model” function and then we apply the get_regression_table() function. This time, however, our model formula won’t be of the form y ~ x, but rather of the form y ~ x1 * x2. In other words, our two explanatory variables x1 and x2 are separated by a * sign:

# Fit regression model:
score_model_interaction <- lm(score ~ age * gender, data = evals_ch6)

# Get regression table:
get_regression_table(score_model_interaction)
TABLE 6.3: Regression table for interaction model
term estimate std_error statistic p_value lower_ci upper_ci
intercept 4.883 0.205 23.80 0.000 4.480 5.286
age -0.018 0.004 -3.92 0.000 -0.026 -0.009
gendermale -0.446 0.265 -1.68 0.094 -0.968 0.076
age:gendermale 0.014 0.006 2.45 0.015 0.003 0.024

Looking at the regression table output in Table 6.3, there are four rows of values in the estimate column. While it is not immediately apparent, using these four values we can write out the equations of both lines in Figure 6.1. First, since the word female comes alphabetically before male, female instructors are the “baseline for comparison” group. Thus, intercept is the intercept for only the female instructors.

This holds similarly for age. It is the slope for age for only the female instructors. Thus, the red regression line in Figure 6.1 has an intercept of 4.883 and slope for age of -0.018. Remember that for this data, while the intercept has a mathematical interpretation, it has no practical interpretation since instructors can’t have zero age.

What about the intercept and slope for age of the male instructors in the blue line in Figure 6.1? This is where our notion of “offsets” comes into play once again.

The value for gendermale of -0.446 is not the intercept for the male instructors, but rather the offset in intercept for male instructors relative to female instructors. The intercept for the male instructors is intercept + gendermale = 4.883 + (-0.446) = 4.883 - 0.446 = 4.437.

Similarly, age:gendermale = 0.014 is not the slope for age for the male instructors, but rather the offset in slope for the male instructors. Therefore, the slope for age for the male instructors is age + age:gendermale $$= -0.018 + 0.014 = -0.004$$. Thus, the blue regression line in Figure 6.1 has intercept 4.437 and slope for age of -0.004. Let’s summarize these values in Table 6.4 and focus on the two slopes for age:

TABLE 6.4: Comparison of intercepts and slopes for interaction model
Gender Intercept Slope for age
Female instructors 4.883 -0.018
Male instructors 4.437 -0.004

Since the slope for age for the female instructors was -0.018, it means that on average, a female instructor who is a year older would have a teaching score that is 0.018 units lower. For the male instructors, however, the corresponding associated decrease was on average only 0.004 units. While both slopes for age were negative, the slope for age for the female instructors is more negative. This is consistent with our observation from Figure 6.1, that this model is suggesting that age impacts teaching scores for female instructors more than for male instructors.

Let’s now write the equation for our regression lines, which we can use to compute our fitted values $$\widehat{y} = \widehat{\text{score}}$$.

\begin{aligned} \widehat{y} = \widehat{\text{score}} &= b_0 + b_{\text{age}} \cdot \text{age} + b_{\text{male}} \cdot \mathbb{1}_{\text{is male}}(x) + b_{\text{age,male}} \cdot \text{age} \cdot \mathbb{1}_{\text{is male}}(x)\\ &= 4.883 -0.018 \cdot \text{age} - 0.446 \cdot \mathbb{1}_{\text{is male}}(x) + 0.014 \cdot \text{age} \cdot \mathbb{1}_{\text{is male}}(x) \end{aligned}

Whoa! That’s even more daunting than the equation you saw for the life expectancy as a function of continent in Subsection 5.2.2! However, if you recall what an “indicator function” does, the equation simplifies greatly. In the previous equation, we have one indicator function of interest:

$\mathbb{1}_{\text{is male}}(x) = \left\{ \begin{array}{ll} 1 & \text{if } \text{instructor } x \text{ is male} \\ 0 & \text{otherwise}\end{array} \right.$

Second, let’s match coefficients in the previous equation with values in the estimate column in our regression table in Table 6.3:

1. $$b_0$$ is the intercept = 4.883 for the female instructors
2. $$b_{\text{age}}$$ is the slope for age = -0.018 for the female instructors
3. $$b_{\text{male}}$$ is the offset in intercept = -0.446 for the male instructors
4. $$b_{\text{age,male}}$$ is the offset in slope for age = 0.014 for the male instructors

Let’s put this all together and compute the fitted value $$\widehat{y} = \widehat{\text{score}}$$ for female instructors. Since for female instructors $$\mathbb{1}_{\text{is male}}(x)$$ = 0, the previous equation becomes

\begin{aligned} \widehat{y} = \widehat{\text{score}} &= 4.883 - 0.018 \cdot \text{age} - 0.446 \cdot 0 + 0.014 \cdot \text{age} \cdot 0\\ &= 4.883 - 0.018 \cdot \text{age} - 0 + 0\\ &= 4.883 - 0.018 \cdot \text{age}\\ \end{aligned}

which is the equation of the red regression line in Figure 6.1 corresponding to the female instructors in Table 6.4. Correspondingly, since for male instructors $$\mathbb{1}_{\text{is male}}(x)$$ = 1, the previous equation becomes

\begin{aligned} \widehat{y} = \widehat{\text{score}} &= 4.883 - 0.018 \cdot \text{age} - 0.446 + 0.014 \cdot \text{age}\\ &= (4.883 - 0.446) + (- 0.018 + 0.014) * \text{age}\\ &= 4.437 - 0.004 \cdot \text{age}\\ \end{aligned}

which is the equation of the blue regression line in Figure 6.1 corresponding to the male instructors in Table 6.4.

Phew! That was a lot of arithmetic! Don’t fret, however, this is as hard as modeling will get in this book. If you’re still a little unsure about using indicator functions and using categorical explanatory variables in a regression model, we highly suggest you re-read Subsection 5.2.2. This involves only a single categorical explanatory variable and thus is much simpler.

Before we end this section, we explain why we refer to this type of model as an “interaction model.” The $$b_{\text{age,male}}$$ term in the equation for the fitted value $$\widehat{y}$$ = $$\widehat{\text{score}}$$ is what’s known in statistical modeling as an “interaction effect.” The interaction term corresponds to the age:gendermale = 0.014 in the final row of the regression table in Table 6.3.

We say there is an interaction effect if the associated effect of one variable depends on the value of another variable. That is to say, the two variables are “interacting” with each other. Here, the associated effect of the variable age depends on the value of the other variable gender. The difference in slopes for age of +0.014 of male instructors relative to female instructors shows this.

Another way of thinking about interaction effects on teaching scores is as follows. For a given instructor at UT Austin, there might be an associated effect of their age by itself, there might be an associated effect of their gender by itself, but when age and gender are considered together there might be an additional effect above and beyond the two individual effects.

### 6.1.3 Parallel slopes model

When creating regression models with one numerical and one categorical explanatory variable, we are not just limited to interaction models as we just saw. Another type of model we can use is known as a parallel slopes model. Unlike interaction models where the regression lines can have different intercepts and different slopes, parallel slopes models still allow for different intercepts but force all lines to have the same slope. The resulting regression lines are thus parallel. Let’s visualize the best-fitting parallel slopes model to evals_ch6.

Unfortunately, the geom_smooth() function in the ggplot2 package does not have a convenient way to plot parallel slopes models. Evgeni Chasnovski thus created a special purpose function called geom_parallel_slopes() that is included in the moderndive package. You won’t find geom_parallel_slopes() in the ggplot2 package, but rather the moderndive package. Thus, if you want to be able to use it, you will need to load both the ggplot2 and moderndive packages. Using this function, let’s now plot the parallel slopes model for teaching score. Notice how the code is identical to the code that produced the visualization of the interaction model in Figure 6.1, but now the geom_smooth(method = "lm", se = FALSE) layer is replaced with geom_parallel_slopes(se = FALSE).

ggplot(evals_ch6, aes(x = age, y = score, color = gender)) +
geom_point() +
labs(x = "Age", y = "Teaching Score", color = "Gender") +
geom_parallel_slopes(se = FALSE)

Observe in Figure 6.2 that we now have parallel lines corresponding to the female and male instructors, respectively: here they have the same negative slope. This is telling us that instructors who are older will tend to receive lower teaching scores than instructors who are younger. Furthermore, since the lines are parallel, the associated penalty for being older is assumed to be the same for both female and male instructors.

However, observe also in Figure 6.2 that these two lines have different intercepts as evidenced by the fact that the blue line corresponding to the male instructors is higher than the red line corresponding to the female instructors. This is telling us that irrespective of age, female instructors tended to receive lower teaching scores than male instructors.

In order to obtain the precise numerical values of the two intercepts and the single common slope, we once again “fit” the model using the lm() “linear model” function and then apply the get_regression_table() function. However, unlike the interaction model which had a model formula of the form y ~ x1 * x2, our model formula is now of the form y ~ x1 + x2. In other words, our two explanatory variables x1 and x2 are separated by a + sign:

# Fit regression model:
score_model_parallel_slopes <- lm(score ~ age + gender, data = evals_ch6)
# Get regression table:
get_regression_table(score_model_parallel_slopes)
TABLE 6.5: Regression table for parallel slopes model
term estimate std_error statistic p_value lower_ci upper_ci
intercept 4.484 0.125 35.79 0.000 4.238 4.730
age -0.009 0.003 -3.28 0.001 -0.014 -0.003
gendermale 0.191 0.052 3.63 0.000 0.087 0.294

Similarly to the regression table for the interaction model from Table 6.3, we have an intercept term corresponding to the intercept for the “baseline for comparison” female instructor group and a gendermale term corresponding to the offset in intercept for the male instructors relative to female instructors. In other words, in Figure 6.2 the red regression line corresponding to the female instructors has an intercept of 4.484 while the blue regression line corresponding to the male instructors has an intercept of 4.484 + 0.191 = 4.675. Once again, since there aren’t any instructors of age 0, the intercepts only have a mathematical interpretation but no practical one.

Unlike in Table 6.3, however, we now only have a single slope for age of -0.009. This is because the model dictates that both the female and male instructors have a common slope for age. This is telling us that an instructor who is a year older than another instructor received a teaching score that is on average 0.009 units lower. This penalty for being of advanced age applies equally to both female and male instructors.

Let’s summarize these values in Table 6.6, noting the different intercepts but common slopes:

TABLE 6.6: Comparison of intercepts and slope for parallel slopes model
Gender Intercept Slope for age
Female instructors 4.484 -0.009
Male instructors 4.675 -0.009

Let’s now write the equation for our regression lines, which we can use to compute our fitted values $$\widehat{y} = \widehat{\text{score}}$$.

\begin{aligned} \widehat{y} = \widehat{\text{score}} &= b_0 + b_{\text{age}} \cdot \text{age} + b_{\text{male}} \cdot \mathbb{1}_{\text{is male}}(x)\\ &= 4.484 -0.009 \cdot \text{age} + 0.191 \cdot \mathbb{1}_{\text{is male}}(x) \end{aligned}

Let’s put this all together and compute the fitted value $$\widehat{y} = \widehat{\text{score}}$$ for female instructors. Since for female instructors the indicator function $$\mathbb{1}_{\text{is male}}(x)$$ = 0, the previous equation becomes

\begin{aligned} \widehat{y} = \widehat{\text{score}} &= 4.484 -0.009 \cdot \text{age} + 0.191 \cdot 0\\ &= 4.484 -0.009 \cdot \text{age} \end{aligned}

which is the equation of the red regression line in Figure 6.2 corresponding to the female instructors. Correspondingly, since for male instructors the indicator function $$\mathbb{1}_{\text{is male}}(x)$$ = 1, the previous equation becomes

\begin{aligned} \widehat{y} = \widehat{\text{score}} &= 4.484 -0.009 \cdot \text{age} + 0.191 \cdot 1\\ &= (4.484 + 0.191) - 0.009 \cdot \text{age}\\ &= 4.675 -0.009 \cdot \text{age} \end{aligned}

which is the equation of the blue regression line in Figure 6.2 corresponding to the male instructors.

Great! We’ve considered both an interaction model and a parallel slopes model for our data. Let’s compare the visualizations for both models side-by-side in Figure 6.3.

At this point, you might be asking yourself: “Why would we ever use a parallel slopes model?”. Looking at the left-hand plot in Figure 6.3, the two lines definitely do not appear to be parallel, so why would we force them to be parallel? For this data, we agree! It can easily be argued that the interaction model on the left is more appropriate. However, in the upcoming Subsection 6.3.1 on model selection, we’ll present an example where it can be argued that the case for a parallel slopes model might be stronger.

### 6.1.4 Observed/fitted values and residuals

For brevity’s sake, in this section we’ll only compute the observed values, fitted values, and residuals for the interaction model which we saved in score_model_interaction. You’ll have an opportunity to study the corresponding values for the parallel slopes model in the upcoming Learning check.

Say, you have an instructor who identifies as female and is 36 years old. What fitted value $$\widehat{y}$$ = $$\widehat{\text{score}}$$ would our model yield? Say, you have another instructor who identifies as male and is 59 years old. What would their fitted value $$\widehat{y}$$ be?

We answer this question visually first for the female instructor by finding the intersection of the red regression line and the vertical line at $$x$$ = age = 36. We mark this value with a large red dot in Figure 6.4. Similarly, we can identify the fitted value $$\widehat{y}$$ = $$\widehat{\text{score}}$$ for the male instructor by finding the intersection of the blue regression line and the vertical line at $$x$$ = age = 59. We mark this value with a large blue dot in Figure 6.4.

What are these two values of $$\widehat{y}$$ = $$\widehat{\text{score}}$$ precisely? We can use the equations of the two regression lines we computed in Subsection 6.1.2, which in turn were based on values from the regression table in Table 6.3:

• For all female instructors: $$\widehat{y} = \widehat{\text{score}} = 4.883 - 0.018 \cdot \text{age}$$
• For all male instructors: $$\widehat{y} = \widehat{\text{score}} = 4.437 - 0.004 \cdot \text{age}$$

So our fitted values would be: $$4.883 - 0.018 \cdot 36 = 4.24$$ and $$4.437 - 0.004 \cdot 59 = 4.20$$, respectively.

Now what if we want the fitted values not just for these two instructors, but for the instructors of all 463 courses included in the evals_ch6 data frame? Doing this by hand would be long and tedious! This is where the get_regression_points() function from the moderndive package can help: it will quickly automate the above calculations for all 463 courses. We present a preview of just the first 10 rows out of 463 in Table 6.7.

regression_points <- get_regression_points(score_model_interaction)
regression_points
TABLE 6.7: Regression points (First 10 out of 463 courses)
ID score age gender score_hat residual
1 4.7 36 female 4.25 0.448
2 4.1 36 female 4.25 -0.152
3 3.9 36 female 4.25 -0.352
4 4.8 36 female 4.25 0.548
5 4.6 59 male 4.20 0.399
6 4.3 59 male 4.20 0.099
7 2.8 59 male 4.20 -1.401
8 4.1 51 male 4.23 -0.133
9 3.4 51 male 4.23 -0.833
10 4.5 40 female 4.18 0.318

It turns out that the female instructor of age 36 taught the first four courses, while the male instructor taught the next 3. The resulting $$\widehat{y}$$ = $$\widehat{\text{score}}$$ fitted values are in the score_hat column. Furthermore, the get_regression_points() function also returns the residuals $$y-\widehat{y}$$. Notice, for example, the first and fourth courses the female instructor of age 36 taught had positive residuals, indicating that the actual teaching scores they received from students were greater than their fitted score of 4.25. On the other hand, the second and third courses this instructor taught had negative residuals, indicating that the actual teaching scores they received from students were less than 4.25.

Learning check

(LC6.1) Compute the observed values, fitted values, and residuals not for the interaction model as we just did, but rather for the parallel slopes model we saved in score_model_parallel_slopes.