# Chapter 6 Multiple Regression

In Chapter 5 we introduced ideas related to modeling for explanation, in particular that the goal of modeling is to make explicit the relationship between some outcome variable $$y$$ and some explanatory variable $$x$$. While there are many approaches to modeling, we focused on one particular technique: linear regression, one of the most commonly-used and easy-to-understand approaches to modeling. Furthermore to keep things simple we only considered models with one explanatory $$x$$ variable that was either numerical in Section 5.1 or categorical in Section 5.2.

In this chapter on multiple regression, we'll start considering models that include more than one explanatory variable $$x$$. You can imagine when trying to model a particular outcome variable, like teaching evaluation scores as in Section 5.1 or life expectancy as in Section 5.2, that it would be useful to include more than just one explanatory variable's worth of information.

Since our regression models will now consider more than one explanatory variable, the interpretation of the associated effect of any one explanatory variable must be made in conjunction with the other explanatory variables included in your model. Let's begin!

### Needed packages

Let's load all the packages needed for this chapter (this assumes you've already installed them). Recall from our discussion in Section 4.4.1 that loading the tidyverse package by running library(tidyverse) loads the following commonly used data science packages all at once:

• ggplot2 for data visualization
• dplyr for data wrangling
• tidyr for converting data to "tidy" format
• readr for importing spreadsheet data into R
• As well as the more advanced purrr, tibble, stringr, and forcats packages

If needed, read Section 1.3 for information on how to install and load R packages.

library(tidyverse)
library(moderndive)
library(skimr)
library(ISLR)

## 6.1 Two numerical explanatory variables

Let's first consider a multiple regression model with two numerical explanatory variables. The dataset we'll use is from "An Introduction to Statistical Learning with Applications in R (ISLR)", an intermediate-level textbook on statistical and machine learning. Its accompanying ISLR R package contains the datasets that the authors apply various machine learning methods to.

One frequently used dataset in this book is the Credit dataset, where the outcome variable of interest is the credit card debt of 400 individuals. Other variables like income, credit limit, credit rating, and age are included as well. Note that the Credit data is not based on real individuals' financial information, but rather is a simulated dataset used for educational purposes.

In this section, we'll fit a regression model where we have

1. A numerical outcome variable $$y$$, the cardholder's credit card debt
2. Two explanatory variables:
1. One numerical explanatory variable $$x_1$$, the cardholder's credit limit
2. Another numerical explanatory variable $$x_2$$, the cardholder's income (in thousands of dollars).

### 6.1.1 Exploratory data analysis

Let's load the Credit dataset, but 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 credit_ch6. Notice our slightly different use of the select() verb here than we introduced in Subsection 3.8.1. For example, we'll select the Balance variable from Credit but then save it with a new variable name debt. We do this because here the term "debt" is a little more interpretable than "balance."

library(ISLR)
credit_ch6 <- Credit %>%
as_tibble() %>%
select(debt = Balance, credit_limit = Limit,
income = Income, credit_rating = Rating, age = Age)

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 begin by looking at the raw values either in RStudio's spreadsheet viewer or by using the glimpse() function from the dplyr package. You can observe the effect of our use of select() to keep and rename the relevant variables.

glimpse(credit_ch6)
Rows: 400
Columns: 5
$debt <int> 333, 903, 580, 964, 331, 1151, 203, 872, 279, 1350, 140…$ credit_limit  <int> 3606, 6645, 7075, 9504, 4897, 8047, 3388, 7114, 3300, 6…
$income <dbl> 14.9, 106.0, 104.6, 148.9, 55.9, 80.2, 21.0, 71.4, 15.1…$ credit_rating <int> 283, 483, 514, 681, 357, 569, 259, 512, 266, 491, 589, …
$age <int> 34, 82, 71, 36, 68, 77, 37, 87, 66, 41, 30, 64, 57, 49,… Furthermore, let's look at a random sample of five out of the 400 credit card holders in Table 6.1. Once again, note that due to the random nature of the sampling, you will likely end up with a different subset of five rows. set.seed(9) credit_ch6 %>% sample_n(size = 5) TABLE 6.1: Random sample of 5 credit card holders. debt credit_limit income credit_rating age 1259 8376 123.3 610 89 227 6033 108.0 449 64 467 4534 32.8 333 44 846 7576 94.2 527 44 436 4866 45.0 347 30 Now that we've looked at the raw values in our credit_ch6 data frame and have a sense of the data, let's move on to the next common step in an exploratory data analysis: computing 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: credit_ch6 %>% select(debt, credit_limit, income) %>% skim() Skim summary statistics n obs: 400 n variables: 3 ── Variable type:integer ─────────────────────────────────────────────────────── variable missing complete n mean sd p0 p25 p50 p75 p100 credit_limit 0 400 400 4735.6 2308.2 855 3088 4622.5 5872.75 13913 debt 0 400 400 520.01 459.76 0 68.75 459.5 863 1999 ── Variable type:numeric ─────────────────────────────────────────────────────── variable missing complete n mean sd p0 p25 p50 p75 p100 income 0 400 400 45.22 35.24 10.35 21.01 33.12 57.47 186.63 Observe the summary statistics for the outcome variable debt: the mean and median credit card debt are$520.01 and $459.50 respectively and that 25% of card holders had debts of$68.75 or less. Let's now look at one of the explanatory variables credit_limit: the mean and median credit card limit are $4735.6 and$4622.50 respectively while 75% of card holders had incomes of $57,470 or less. Since our outcome variable debt and the explanatory variables credit_limit and income are numerical, we can compute the correlation coefficient between the different possible pairs of these variables. First, we can run the cor() command as seen in Subsection 5.1.1 twice, once for each explanatory variable: cor(credit_ch6$debt, credit_ch6$credit_limit) cor(credit_ch6$debt, credit_ch6$income) Or we can simultaneously compute them by returning a correlation matrix which we display in Table 6.2. We can read off the correlation coefficient for any pair of variables by looking them up in the appropriate row/column combination. credit_ch6 %>% select(debt, credit_limit, income) %>% cor() TABLE 6.2: Correlation coefficients between credit card debt, credit limit, and income. debt credit_limit income debt 1.000 0.862 0.464 credit_limit 0.862 1.000 0.792 income 0.464 0.792 1.000 For example, the correlation coefficient of: 1. debt with itself is 1 as we would expect based on the definition of the correlation coefficient. 2. debt with credit_limit is 0.862. This indicates a strong positive linear relationship, which makes sense as only individuals with large credit limits can accrue large credit card debts. 3. debt with income is 0.464. This is suggestive of another positive linear relationship, although not as strong as the relationship between debt and credit_limit. 4. As an added bonus, we can read off the correlation coefficient between the two explanatory variables of credit_limit and income as 0.792. We say there is a high degree of collinearity between the credit_limit and income explanatory variables. Collinearity (or multicollinearity) is a phenomenon where one explanatory variable in a multiple regression model is highly correlated with another. So in our case since credit_limit and income are highly correlated, if we knew someone's credit_limit, we could make pretty good guesses about their income as well. Thus, these two variables provide somewhat redundant information. However, we'll leave discussion on how to work with collinear explanatory variables to a more intermediate-level book on regression modeling. Let's visualize the relationship of the outcome variable with each of the two explanatory variables in two separate plots in Figure 6.1. ggplot(credit_ch6, aes(x = credit_limit, y = debt)) + geom_point() + labs(x = "Credit limit (in$)", y = "Credit card debt (in $)", title = "Debt and credit limit") + geom_smooth(method = "lm", se = FALSE) ggplot(credit_ch6, aes(x = income, y = debt)) + geom_point() + labs(x = "Income (in$1000)", y = "Credit card debt (in $)", title = "Debt and income") + geom_smooth(method = "lm", se = FALSE) geom_smooth() using formula 'y ~ x' geom_smooth() using formula 'y ~ x' Observe there is a positive relationship between credit limit and credit card debt: as credit limit increases so also does credit card debt. This is consistent with the strongly positive correlation coefficient of 0.862 we computed earlier. In the case of income, the positive relationship doesn't appear as strong, given the weakly positive correlation coefficient of 0.464. However, the two plots in Figure 6.1 only focus on the relationship of the outcome variable with each of the two explanatory variables separately. To visualize the joint relationship of all three variables simultaneously, we need a 3-dimensional (3D) scatterplot as seen in Figure 6.2. Each of the 400 observations in the credit_ch6 data frame are marked with a blue point where 1. The numerical outcome variable $$y$$ debt is on the vertical axis 2. The two numerical explanatory variables, $$x_1$$ income and $$x_2$$ credit_limit, are on the two axes that form the bottom plane. Furthermore, we also include the regression plane. Recall from Subsection 5.3.3 that regression lines are "best-fitting" in that of all possible lines we can draw through a cloud of points, the regression line minimizes the sum of squared residuals. This concept also extends to models with two numerical explanatory variables. The difference is instead of a "best-fitting" line, we now have a "best-fitting" plane that similarly minimizes the sum of squared residuals. Head to here to open an interactive version of this plot in your browser. Learning check (LC6.1) Conduct a new exploratory data analysis with the same outcome variable $$y$$ being debt but with credit_rating and age as the new explanatory variables $$x_1$$ and $$x_2$$. Remember, this involves three things: 1. Most crucially: Looking at the raw data values. 2. Computing summary statistics, such as means, medians, and interquartile ranges. 3. Creating data visualizations. What can you say about the relationship between a credit card holder's debt and their credit rating and age? ### 6.1.2 Regression plane Let's now fit a regression model and get the regression table corresponding to the regression plane in Figure 6.2. We'll consider a model fit with a formula of the form y ~ x1 + x2, where x1 and x2 represent our two explanatory variables credit_limit and income. Just like we did in Chapter 5, let's get the regression table for this model using our two-step process and display the results in Table 6.3. 1. We first "fit" the linear regression model using the lm(y ~ x1 + x2, data) function and save it in debt_model. 2. We get the regression table by applying the summary() function to debt_model. # Fit regression model: debt_model <- lm(debt ~ credit_limit + income, data = credit_ch6) # Get regression table: summary(debt_model)$coefficients
TABLE 6.3: Multiple regression table
Estimate Std. Error t value Pr(>|t|)
(Intercept) -385.179 19.465 -19.8 0
credit_limit 0.264 0.006 45.0 0
income -7.663 0.385 -19.9 0

Let's interpret the three values in the Estimate column. First, the Intercept value is -$385.179. This intercept represents the credit card debt for an individual who has credit_limit of$0 and income of $0. In our data however, the intercept has limited practical interpretation since no individuals had credit_limit or income values of$0. Rather, the intercept is used to situate the regression plane in 3D space.

Second, the credit_limit value is $0.264. Taking into account all the other explanatory variables in our model, for every increase of one dollar in credit_limit, there is an associated increase of on average$0.26 in credit card debt. Just as we did in Subsection 5.1.2, we are cautious to not imply causality as we saw in Subsection 5.3.2 that "correlation is not necessarily causation." We do this merely stating there was an associated increase.

Furthermore, we preface our interpretation with the statement "taking into account all the other explanatory variables in our model." Here, by all other explanatory variables we mean income. We do this to emphasize that we are now jointly interpreting the associated effect of multiple explanatory variables in the same model at the same time.

Third, income = -$7.663. Taking into account all the other explanatory variables in our model, for every increase of one unit in the variable income, in other words$1000 in actual income, there is an associated decrease of on average 7.663 in credit card debt. Putting these results together, the equation of the regression plane that gives us fitted values $$\widehat{y}$$ = $$\widehat{\text{debt}}$$ is: \begin{aligned} \widehat{y} &= b_0 + b_1 \cdot x_1 + b_2 \cdot x_2\\ \widehat{\text{debt}} &= b_0 + b_{\text{limit}} \cdot \text{limit} + b_{\text{income}} \cdot \text{income}\\ &= -385.179 + 0.264 \cdot\text{limit} - 7.663 \cdot\text{income} \end{aligned} Recall in the right-hand plot of Figure 6.1 that when plotting the relationship between debt and income in isolation, there appeared to be a positive relationship. In the last discussed multiple regression however, when jointly modeling the relationship between debt, credit_limit, and income, there appears to be a negative relationship of debt and income as evidenced by the negative slope for income of -7.663. What explains these contradictory results? A phenomenon known as Simpson's Paradox, whereby overall trends that exist in aggregate either disappear or reverse when the data are broken down into groups. In Subsection 6.3.3 we elaborate on this idea by looking at the relationship between credit_limit and credit card debt, but split along different income brackets.

Learning check

(LC6.2) Fit a new simple linear regression using lm(debt ~ credit_rating + age, data = credit_ch6) where credit_rating and age are the new numerical explanatory variables $$x_1$$ and $$x_2$$. Get information about the "best-fitting" regression plane from the regression table by applying the summary() function. How do the regression results match up with the results from your previous exploratory data analysis?

### 6.1.3 Observed/fitted values and residuals

Let's also compute all fitted values and residuals for our regression model using the code from Subsection 5.1.3 and present only the first 10 rows of output in Table 6.4. Remember that the coordinates of each of the blue points in our 3D scatterplot in Figure 6.2 can be found in the income, credit_limit, and debt columns. The fitted values on the regression plane are found in the debt_hat column and are computed using our equation for the regression plane in the previous section:

\begin{aligned} \widehat{y} = \widehat{\text{debt}} &= -385.179 + 0.264 \cdot \text{limit} - 7.663 \cdot \text{income} \end{aligned}

debt_model_data <- credit_ch6 %>%
select(debt, credit_limit, income) %>%
mutate(debt_hat = fitted(debt_model),
residual = residuals(debt_model)) %>%
rownames_to_column("ID")
TABLE 6.4: Regression points (First 10 credit card holders out of 400).
ID debt credit_limit income debt_hat residual
1 333 3606 14.9 454 -120.8
2 903 6645 106.0 559 344.3
3 580 7075 104.6 683 -103.4
4 964 9504 148.9 986 -21.7
5 331 4897 55.9 481 -150.0
6 1151 8047 80.2 1127 23.6
7 203 3388 21.0 349 -146.4
8 872 7114 71.4 948 -76.0
9 279 3300 15.1 371 -92.2
10 1350 6819 71.1 873 477.3

Let's interpret these results for the third card holder. Our regression model tells us that for a person with a credit_limit of $7,075 and an income of$104,600, we would expect them to have credit card debt of 683, on average. We calculated this number by plugging into our regression equation: \begin{aligned} \widehat{y} = \widehat{\text{debt}} &= -385.179 + 0.264 \cdot \text{limit} - 7.663 \cdot \text{income} \\ &= -385.179 + 0.264(7075) - 7.663(104.6) \\ &= 683 \end{aligned} However, this person had an actual credit card debt of580, so the residual for this observation is $$y - \widehat{y} = \580 - \683.37 = -\103.37$$. Note that Table 6.4 presents rounded values for debt_hat and residual.

## 6.2 One numerical & 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 now we'll now include two different explanatory variables: age and 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.

### 6.2.1 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)

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:

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 6.5. 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 6.5: 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 have a sense of the data, let's compute summary statistics.

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 for example 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 however that each row of our data 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 %>%
summarize(correlation = cor(score, age))
# A tibble: 1 x 1
correlation
<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)
geom_smooth() using formula 'y ~ x'

In the resulting Figure 6.3, 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 their age than the male instructors.

### 6.2.2 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 6.3 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 5.10:

TABLE 6.6: Linear regression table
Estimate Std. Error t value Pr(>|t|)
(Intercept) 54.8 1.02 53.45 0
continentAmericas 18.8 1.80 10.45 0
continentAsia 15.9 1.65 9.68 0
continentEurope 22.8 1.70 13.47 0
continentOceania 25.9 5.33 4.86 0

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 6.3, 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 summary() function. This time however, our model formula won't be of the form y ~ x, but rather of the form y ~ x1 + x2 + x1 * x2. In other words, we include a main effect for each of our two explanatory variables x1 and x2, as well as an interaction term x1 * x2. In terms of the general mathematical equation, an interaction model with two explanatory variables is of the form:

$\widehat{y} = b_0 + b_1 \cdot x_1 + b_2 \cdot x_2 + b_{1,2} \cdot x_1 \cdot x_2$

# Fit regression model:
score_model_interaction <- lm(score ~ age + gender + age * gender, data = evals_ch6)
# Get regression table:
summary(score_model_interaction)coefficients TABLE 6.7: Regression table for interaction model. Estimate Std. Error t value Pr(>|t|) (Intercept) 4.883 0.205 23.80 0.000 age -0.018 0.004 -3.92 0.000 gendermale -0.446 0.265 -1.68 0.094 age:gendermale 0.014 0.006 2.45 0.015 Looking at the regression table output in Table 6.7, we see there are four rows of values in the Estimate column that correspond to the 4 estimated components of the model: $$b_0, \ b_1, \ b_2,$$ and $$b_{1,2}$$. Note that we chose to use the notation $$b_{1,2}$$ to make it clear this is the coefficient for the interaction term between $$x_1$$ and $$x_2$$, but we could have easily decided to denote this as $$b_3$$ instead. While it is not immediately apparent, using these four values we can write out the equations of both lines in Figure 6.3. First, since the word female comes alphabetically before male, female instructors are the "baseline for comparison" group. Therefore 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 6.3 has an intercept of 4.883 and slope for age of -0.018. Remember that for this particular data, while the intercept has a mathematical interpretation, it has no practical interpretation since there can't be any instructors with age zero. What about the intercept and slope for age of the male instructors (i.e. the blue line in Figure 6.3)? 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. Therefore, 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. Therefore the blue regression line in Figure 6.3 has intercept 4.437 and slope for age of -0.004. Let's summarize these values in Table 6.8 and focus on the two slopes for age: TABLE 6.8: 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 6.3, 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_{\mbox{age}} \cdot \mbox{age} + b_{\mbox{male}} \cdot \mathbb{1}_{\mbox{is male}}(x) + b_{\mbox{age,male}} \cdot \mbox{age} \cdot \mathbb{1}_{\mbox{is male}}\\ &= 4.883 -0.018 \cdot \mbox{age} - 0.446 \cdot \mathbb{1}_{\mbox{is male}}(x) + 0.014 \cdot \mbox{age} \cdot \mathbb{1}_{\mbox{is male}} \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" AKA "dummy variable" does, the equation simplifies greatly. In the previous equation, we have one indicator function of interest: $\mathbb{1}_{\mbox{is male}}(x) = \left\{ \begin{array}{ll} 1 & \text{if } \text{instructor } x \text{ is male} \\ 0 & \text{otherwise}\end{array} \right.$ Second, let's match coefficients in the previous equation with values in the Estimate column in our regression table in Table 6.7: 1. $$b_0$$ is the Intercept = 4.883 for the female instructors 2. $$b_{\mbox{age}}$$ is the slope for age = -0.018 for the female instructors 3. $$b_{\mbox{male}}$$ is the offset in intercept for the male instructors 4. $$b_{\mbox{age,male}}$$ is the offset in slope for age 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 $$\mathbb{1}_{\mbox{is male}}(x)$$ = 0, the previous equation becomes \begin{aligned} \widehat{y} = \widehat{\text{score}} &= 4.883 - 0.018 \cdot \mbox{age} - 0.446 \cdot 0 + 0.014 \cdot \mbox{age} \cdot 0\\ &= 4.883 - 0.018 \cdot \mbox{age} - 0 + 0\\ &= 4.883 - 0.018 \cdot \mbox{age}\\ \end{aligned} which is the equation of the red regression line in Figure 6.3 corresponding to the female instructors in Table 6.8. Correspondingly, since for male instructors $$\mathbb{1}_{\mbox{is male}}(x)$$ = 1, the previous equation becomes \begin{aligned} \widehat{y} = \widehat{\text{score}} &= 4.883 - 0.018 \cdot \mbox{age} - 0.446 + 0.014 \cdot \mbox{age}\\ &= (4.883 - 0.446) + (- 0.018 + 0.014) * \mbox{age}\\ &= 4.437 - 0.004 \cdot \mbox{age}\\ \end{aligned} which is the equation of the blue regression line in Figure 6.3 corresponding to the male instructors in Table 6.8. 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 $$b_{\mbox{age,male}}$$ 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 6.7. We say there is an interaction effect if the associated effect of one variable depends on the value of another variable. In other words, the two variables are "interacting" with each other. In our case, the associated effect of the variable age depends on the value of the other variable gender. This was evidenced by the difference in slopes for age of +0.014 of male instructors relative to female instructors. 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 an additional effect above and beyond the two individual effects. ### 6.2.3 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. We can think of a parallel slopes model as a restricted case of the interaction model where we've forced $$b_{1,2}$$, the coefficient of the interaction term $$x_{1} \cdot x_2$$, to be zero. Therefore, the mathematical equation for a parallel slopes model with two explanatory variables is of the form: \begin{aligned} \widehat{y} = \widehat{\text{score}} &= b_0 + b_{1} \cdot x_1 + b_{2} \cdot x_2 + b_{1,2} \cdot x_1 \cdot x_2\\ &= b_0 + b_{1} \cdot x_1 + b_{2} \cdot x_2 + 0 \cdot x_1 \cdot x_2\\ &= b_0 + b_{1} \cdot x_1 + b_{2} \cdot x_2 \end{aligned} Unfortunately, the ggplot2 package does not have a convenient way to plot a parallel slopes model, so we just display it for you in Figure 6.4 but leave the code for a more advanced data visualization class. Plotting parallel slopes models is now much easier using the geom_parallel_slopes() function. We suggest you use geom_parallel_slopes() instead of gg_parallel_slopes(); read the help file by running ?geom_parallel_slopes to learn how. Observe in Figure 6.4 that we have parallel lines corresponding to the female and male instructors respectively: here they have the same negative slope. This is different from the interaction model displayed in Figure 6.3, which allowed male and female to have different slopes. Figure 6.4 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 aging is assumed to be the same for both female and male instructors. However, observe also in Figure 6.4 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 summary() function. Our model formula this time is of the form y ~ x1 + x2, where x1 and x2 represent the two predictor variables, age and gender, but we do not include the extra interaction term x1 * x2. # Fit regression model: score_model_parallel_slopes <- lm(score ~ age + gender, data = evals_ch6) # Get regression table: summary(score_model_parallel_slopes)coefficients
TABLE 6.9: Regression table for parallel slopes model.
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.484 0.125 35.79 0.000
age -0.009 0.003 -3.28 0.001
gendermale 0.191 0.052 3.63 0.000

Looking at the regression table output in Table 6.9, we see there are three rows of values in the Estimate column. Similar to what we did in Subsection 6.2.2, using these three values we can write out the equations of both lines in Figure 6.4.

Again, since the word female comes alphabetically before male, female instructors are the "baseline for comparison" group. Therefore, Intercept is the intercept for only the female instructors. Thus the red regression line in Figure 6.4 has an intercept of 4.484.

Remember, the value for gendermale of 0.191 is not the intercept for the male instructors, but rather the offset in intercept for male instructors relative to female instructors. Therefore, the intercept for male instructors is Intercept + gendermale = $$4.484 + 0.191 = 4.675$$. In other words, in Figure 6.4 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 4.675. 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 6.7 however, we now only have a single slope for age of -0.009. This is because the model specifies 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 0.018 units lower. This penalty for aging applies equally for both female and male instructors.

Let's summarize these values in Table 6.10, noting the different intercepts but common slopes:

TABLE 6.10: 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

Recall that the common slope occurs because we chose not to include the interaction term $$\mbox{age} \cdot \mathbb{1}_{\mbox{is male}}$$ in our model. This is equivalent to assuming $$b_{age,male} = 0$$ and therefore not allowing there to be an "offset" in slope for males.

Let's now write the equation for our parallel slopes 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_{\mbox{age}} \cdot \mbox{age} + b_{\mbox{male}} \cdot \mathbb{1}_{\mbox{is male}}(x)\\ &= 4.484 -0.009 \cdot \mbox{age} + 0.191 \cdot \mathbb{1}_{\mbox{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 $$\mathbb{1}_{\mbox{is male}}(x)$$ = 0, the previous equation becomes

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

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

\begin{aligned} \widehat{y} = \widehat{\text{score}} &= 4.484 -0.009 \cdot \mbox{age} + 0.191 \cdot 1\\ &= (4.484 + 0.191) - 0.009 \cdot \mbox{age}\\ &= 4.67 -0.009 \cdot \mbox{age} \end{aligned}

which is the equation of the blue regression line in Figure 6.4 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 6.5.

Plotting parallel slopes models is now much easier using the
geom_parallel_slopes() function. We suggest you use
geom_parallel_slopes() instead of gg_parallel_slopes(); read the help
file by running ?geom_parallel_slopes to learn how.
geom_smooth() using formula 'y ~ x'

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 6.5, 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 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.

### 6.2.4 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 these values for the parallel slopes model in the upcoming Learning Check.

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

We answer this question visually first 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 6.6. 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 6.6.

geom_smooth() using formula 'y ~ x'

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 6.2.2, which in turn were based on values from the regression table in Table 6.7:

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

So our fitted values would be: 4.883 - 0.018 $$\cdot$$ 36 = 4.25 and 4.437 - 0.004 $$\cdot$$ 59 = 4.20 respectively.

Now what if we want the fitted values not just for the instructors of these two courses, 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 our data wrangling code from Subsection 5.1.3 can help: it will quickly automate this for all 463 courses. We present a preview of just the first 10 rows out of 463 in Table 6.11.

score_model_interaction_data <- evals_ch6 %>%
select(score, age, gender) %>%
mutate(score_hat = fitted(score_model_interaction),
residual = residuals(score_model_interaction)) %>%
rownames_to_column("ID")
score_model_interaction_data
TABLE 6.11: 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

In fact, 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. The residuals $$y-\widehat{y}$$ are displayed in the residuals column. Notice for example the first and fourth courses the female instructor of age 36 taught had positive residuals, indicating that the actual teaching score they received from students was more than their fitted score of 4.25. On the other hand, the second and third course this instructor taught had negative residuals, indicating that the actual teaching score they received from students was less than their fitted score of 4.25.

Learning check

(LC6.3) 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.

## 6.4 Conclusion

### 6.4.1 What's to come?

Congratulations! We've completed the "Data modeling" portion of this book. We're ready to proceed to the next part of the book: "Statistical Theory." These chapters will lay the foundation for key ideas in statistics such as randomization (Chapter 7), populations and samples (Chapter 8), and sampling distributions (Chapter 9).

In Parts I and II of the book, we've been focusing only on exploratory data analysis and exploring relationships that exist in our observed dataset. Once we've established some of the statistical theory in Part III, we will be able to move beyond exploratory data analysis and into "statistical inference" in Part IV, where we will learn how (and when it is appropirate) to make inferences about statistical relationships in a population beyond our dataset.