---
title: "Multiple Linear Regression - Confounding"
subtitle: "Notes and in-class exercises"
format: 
  html:
    embed-resources: true
    toc: true
---



You can download the .qmd file for this activity [here](../activity_templates/10-mlr-confounding.qmd) and open in R-studio. The rendered version is posted in the [course website](https://mutasim221b.github.io/Mac-STAT-155-Fall-25/) (Activities tab). I often experiment with the class activities (and see it in live!) and make updates, but I always post the final version before class starts. To be sure you have the most up-to-date copy, please download it once you’ve settled in before class begins.






```{r setup}
#| include: false
knitr::opts_chunk$set(
  collapse = TRUE, 
  warning = FALSE,
  message = FALSE,
  error = TRUE,
  fig.height = 2.75, 
  fig.width = 4.25,
  fig.env = 'figure',
  fig.pos = 'h',
  fig.align = 'center')
```


# Notes

- [Slides](https://docs.google.com/presentation/d/1ZCcWJaumyh_BohkXlsYZwHvPGpfQF8RrH5sc2sgwUf4/edit?usp=sharing) with comments on Quiz 1
- You can download a template file to work with [here](../activity_templates/10_mlr_confounding.qmd).
- **File organization:** Save this file in the "Activities" subfolder of your "STAT155" folder.

## Learning goals

By the end of this lesson, you should be familiar with:

- confounding variables
- how to control for confounding variables in our models
- how to represent the role of confounding variables using causal diagrams

## Readings and videos

Before class you should have read and watched:

- Sections 3.9.2 in the [STAT 155 Notes](https://mac-stat.github.io/Stat155Notes/)
- [Confounding (and other causal diagrams)](https://voicethread.com/myvoice/thread/15362352/96009456/96009456)
  - Watch from 0:00 - 6:54





# Exercises

## Exercise 1: Review

The `peaks` data includes information on hiking trails in the 46 "high peaks" in the Adirondack mountains of northern New York state:


```{r}
# Load useful packages and data
library(tidyverse)
peaks <- read_csv("https://mac-stat.github.io/data/high_peaks.csv") %>%
    mutate(ascent = ascent / 1000)

# Check it out 
head(peaks)
```


Below is a model of the `time` (in hours) that it takes to complete a hike by the hike's `length` (in miles), vertical `ascent`(in 1000s of feet), and `rating` (easy, moderate, or difficult):

```{r eval=FALSE}
peaks_model <- lm(time ~ length + ascent + rating, data = peaks)
coef(summary(peaks_model))
```

Interpret the `length` and `ratingeasy` coefficients in the model formula below by using our strategy:

> **Strategy:** When interpreting a coefficient for a variable x, compare two units whose values of x differ by 1 but who are identical for all other variables.

E[time | length, ascent, rating] = 6.511 + 0.459 length + 0.187 ascent - 3.169 ratingeasy - 2.477 ratingmoderate

---

Synthesis:

- Interpreting the coefficient $\beta_Q$ for a quantitative variable Q:
    - Holding all other variables constant, each unit increase in Q is associated with $\beta_Q$ change (note if it's an increase or decrease) in Y on average.

- Interpreting the coefficient $\beta_C$ for an indicator(or categorical) variable:
    - Holding all other variables constant, the average outcome for the group referenced by this indicator (group for whom indicator = 1), is $\beta_C$ higher/lower than that of the reference group.






## Live Note Taking!




## Exercise 2: Confounders

> **Research question:** Is there a wage gap, hence possibly discrimination, by marital status among 18-34 year olds?


To explore, we can revisit the `cps` data with employment information collected by the U.S. Current Population Survey (CPS) in 2018. View the codebook [here](https://github.com/Mac-STAT/data/blob/main/cps_2018_codebook.md).

```{r}
# Import data
cps <- read_csv("https://mac-stat.github.io/data/cps_2018.csv") %>% 
    filter(age >= 18, age <= 34) %>% 
    filter(wage < 250000)

# Check it out
head(cps)
```

Recall that a **simple linear regression model** of `wage` by `marital` suggests that single workers make $17,052 *less* than married workers:

```{r eval=FALSE}
wage_model_1 <- lm(wage ~ marital, data = cps)
coef(summary(wage_model_1))
```


That's a big gap!!

BUT this model ignores important **confounding variables** that might help explain this gap.

A confounding variable is a cause of both the predictor of interest (`marital`) and of the response variable (`wage`).

We can represent this idea with a **causal diagram**:

```{r fig.width = 6, fig.height = 2, echo = FALSE}
par(mar = rep(0,4))
plot(1, type = "n", xaxt = "n", yaxt = "n", bty = "n", xlab = "", ylab = "", xlim = c(0,6), ylim = c(1,4))
text(c("marital", "confounder", "wage"), x = c(1,3,5), y = c(1,4,1), cex = 1.1)
arrows(x0 = c(3,3), y0 = c(4,4)-0.2, x1 = c(1,5), y1 = c(1,1)+0.2, angle = 25, lwd = 4)
arrows(x0 = 1.7, y0 = 1, x1 = 4.5, y1 = 1, angle = 25, lwd = 4)
```

Another definition of a confounding variable is one that

- is a cause of the outcome (wage)
- is associated with the main variable of interest (marital status)
- NOT caused by the variable of interest

We can represent this on the causal diagram with a line from the confounder to the variable of interest (instead of an arrow):

```{r fig.width = 6, fig.height = 2, echo = FALSE}
par(mar = rep(0,4))
plot(1, type = "n", xaxt = "n", yaxt = "n", bty = "n", xlab = "", ylab = "", xlim = c(0,6), ylim = c(1,4))
text(c("marital", "confounder", "wage"), x = c(1,3,5), y = c(1,4,1), cex = 1.1)
arrows(x0 = 3, y0 = 4-0.2, x1 = 5, y1 = 1+0.2, angle = 25, lwd = 4)
lines(x = c(3, 1), y = c(3.8, 1.2), lwd = 4)
arrows(x0 = 1.7, y0 = 1, x1 = 4.5, y1 = 1, angle = 25, lwd = 4)
```

Name at least 2 potential confounders.


## Exercise 2b: How & why do confounders bias results?

Unaccounted-for confounders are often a source of **bias** in our models, meaning that when we ignore them, we often over- or under-estimate the true underlying relationship between a predictor and response variable. To explore why this is important, let's first look at how our focal predictor `marital` is associated with our response variable, `wage`:

```{r}
cps %>%
  ggplot(aes(x=marital, y=wage))+
  geom_boxplot()+
  theme_classic()
```

Now, let's consider `age` as a potential confounder. The following plot shows how age is associated with marital status:

```{r}
cps %>%
  ggplot(aes(x=age, y=marital))+
  geom_boxplot()+
  theme_classic()
```

...this should make sense, because the older a person is, the more likely they are to be married. Similarly, we can show how `age` is associated with `wage`:

```{r}
cps %>%
  ggplot(aes(x=age, y=wage))+
  geom_point()+
  geom_smooth(method="lm", se=F)+
  theme_classic()
```
Here we see that there is a positive correlation between age and wages (which again makes sense, because people who have been in the workforce longer typically earn more).

Let's revisit our initial plot showing the relationship between marital status and wages:

```{r}
cps %>%
  ggplot(aes(x=marital, y=wage))+
  geom_boxplot()+
  theme_classic()
```

Since we now know that age is associated with both being married *and* higher wages, this plot doesn't tell the full story--people who are married could simply be earning higher wages because they tend to be older, *not necessarily because they are married!* Age is therefore a **confounder** in the relationship between marital status and wages.




## Exercise 3: Controlling for confounders

The exercise above illustrates that it is important to *control* or *adjust* for confounding variables when trying to understand the *actual* causal relationship between a predictor (e.g. `marital`) and response (e.g. `wage`). 

a. Sometimes, we can control (adjust) for confounding variables through a carefully designed **experiment**. For example, in comparing the effectiveness (y) of 2 different cold remedies (x), we might want to control for the age, general health, and severity of symptoms among the participants. How might we do that?

b. BUT we're often working with **observational**, not experimental, data. Why? Well, explain what an experiment might look like if we wanted to explore the relationship between `wage` (y) and `marital` status (x) while controlling for `age`.







## Exercise 4: Age

We're in luck.

We can control (adjust) for confounding variables by *including them in our model*!

That's one of the superpowers of **multiple linear regression**.

Let's start simple, by *controlling for* age in our model of wages by marital status:

```{r eval=FALSE}
# Construct the model
wage_model_2 <- lm(wage ~ marital + age, cps)
coef(summary(wage_model_2))
```


a. Visualize this model by modifying the code below. 

(Note: The last line where we add a `geom_line` layer adds in trend lines similar to what we might obtain using `geom_smooth`, but it uses the exact fitted values from our model. `geom_smooth`, on the other hand, adds in trend lines based on fitting two separate models to the `married` and `single` subsets of the data. Try adding `geom_smooth(method="lm", se=F, linetype="dashed")` to the plot to see how they compare).

```{r eval=FALSE}
ggplot(cps, aes(y = wage, x = age, color = marital)) +
    geom(size = 0.1, alpha = 0.5) +
    geom_line(aes(y = wage_model_2$fitted.values), linewidth = 0.5)
```

b. Suppose 2 workers are the *same age*, but one is married and one is single. By how much do we expect the single worker's wage to differ from the married worker's wage? (How does this compare to the $17,052 marital gap among all workers?)

c. How can we interpret the `maritalsingle` coefficient?
    

    



## Exercise 5: More confounders

Let's control for even more potential confounders!

Model wages by marital status while controlling for `age` *and* years of `education`:   

```{r eval=FALSE}
wage_model_3 <- lm(wage ~ marital + age + education, cps)
coef(summary(wage_model_3))
```

a. With so many variables, this is a tough model to visualize. If you *had* to draw it, how would the model trend appear: 1 point, 2 points, 2 lines, 1 plane, or 2 planes? Explain your rationale. Hint: pay attention to whether your predictors are quantitative or categorical.

b. Given our research question, which coefficient is of primary interest? Interpret this coefficient.

c. Interpret the two other coefficients in this model.




## Exercise 6: Even more

Let's control for *another* potential confounder, the job `industry` in which one works (categorical):

```{r eval=FALSE}
wage_model_4 <- lm(wage ~ marital + age + education + industry, cps)
coef(summary(wage_model_4))
```
    
If we *had* to draw it, this model would appear as 12 planes.

The original plane explains the relationship between wage and the 2 quantitative predictors, age and education.

Then this plane is split into 12 (2*6) individual planes, 1 for each possible combination of marital status (2 possibilities) and industry (6 possibilities).

a. Interpret the main coefficient of interest for our research question.

b. When controlling for a worker's age, marital status, and education level, which industry tends to have the highest wages? The lowest? Note: the following table shows the 6 industries:

```{r eval=FALSE}
cps %>% count(industry)
```








## Exercise 7: Biggest model yet

Build a model that helps us explore `wage` by `marital` status while controlling for: `age`, `education`, job `industry`, typical number of work `hours`, and `health` status.

Store this model as `wage_model_5`.


```{r eval=FALSE}
wage_model_5 <- lm(wage ~ marital + age + education + industry + hours + health, cps)
coef(summary(wage_model_5))
```







## Exercise 8: Reflection

Take two workers -- one is married and the other is single.

The models above provided the following insights into the typical difference in wages for these two groups:    
    
Model            Assume the two people have the same...    Wage difference
---------------- --------------------------------------- -----------------
`wage_model_1`   NA                                               -$17,052
`wage_model_2`   age                                               -$7,500
`wage_model_3`   age, education                                    -$6,478
`wage_model_4`   age, education, industry                          -$5,893
`wage_model_5`   age, education, industry, hours, health           -$4,993



a. Though not the case in every analysis, the `marital` coefficient got closer and closer to 0 as we controlled for more confounders. Explain the significance of this phenomenon, in context - what does it *mean*?

b. Do you still find the wage gap for single vs married people to be meaningfully "large"? Can you think of any remaining factors that might explain part of this remaining gap? Or do you think we've found evidence of inequitable wage practices for single vs married workers?













## Exercise 9: A new (extreme) example

For a more extreme example of why it's important to control for confounding variables, let's return to the `diamonds` data:

```{r}
# Import and wrangle the data
data(diamonds)
diamonds <- diamonds %>% 
    mutate(
        cut = factor(cut, ordered = FALSE),
        color = factor(color, ordered = FALSE),
        clarity = factor(clarity, ordered = FALSE)
    ) %>% 
    select(price, clarity, cut, color, carat)
head(diamonds)
```

Our goal is to explore how the `price` of a diamond depends upon its `clarity` (a measure of quality).

Clarity is classified as follows, in order from best to worst:

clarity  description
-------- ------------------------------------------------
IF       flawless (no internal imperfections) 
VVS1     very very slightly imperfect
VVS2     " "
VS1      very slightly imperfect
VS2      " "
SI1      slightly imperfect
SI2      " "
I1       imperfect


a. Check out a model of `price` by `clarity`. What clarity has the highest average price? The lowest? (This is surprising!)       

```{r eval=FALSE}
diamond_model_1 <- lm(price ~ clarity, data = diamonds)

# Get a model summary
coef(summary(diamond_model_1))
```

b. What *confounding variable* might explain these results? What's your rationale?
    
    





## Exercise 10: Size

It turns out that `carat`, the *size* of a diamond, is an important confounding variable.

Let's explore what happens when we control for this in our model:

```{r eval=FALSE}
diamond_model_2 <- lm(price ~ clarity + carat, data = diamonds)

# Get a model summary
coef(summary(diamond_model_2))

# Plot the model
diamonds %>% 
    ggplot(aes(y = price, x = carat, color = clarity)) + 
    geom_line(aes(y = diamond_model_2$fitted.values))
```

What do you think now?

Which clarity has the highest expected price?

The lowest?

Provide numerical evidence from the model.









## Exercise 11: Simpson's Paradox

Controlling for `carat` didn't just *change* the `clarity` coefficients, hence our understanding of the relationship between `price` and `clarity`...
It flipped the *signs* of many of these coefficients.

This extreme scenario has a name: **Simpson's paradox**.

CHALLENGE: Explain *why* this happened and support your argument with *graphical* evidence.

HINTS: Think about the causal diagram below. *How* do you think `carat` influences `clarity`? *How* do you think `carat` influences `price`? Make 2 `ggplot()` that support your answers.


```{r fig.width = 6, fig.height = 2, echo = FALSE}
par(mar = rep(0,4))
plot(1, type = "n", xaxt = "n", yaxt = "n", bty = "n", xlab = "", ylab = "", xlim = c(0,6), ylim = c(1,4))
text(c("clarity", "carat", "price"), x = c(1,3,5), y = c(1,4,1), cex = 1.1)
arrows(x0 = 3, y0 = 4-0.2, x1 = 5, y1 = 1+0.2, angle = 25, lwd = 4)
lines(x = c(3, 1), y = c(3.8, 1.2), lwd = 4)
arrows(x0 = 1.7, y0 = 1, x1 = 4.5, y1 = 1, angle = 25, lwd = 4)
```



```{r eval=FALSE}
diamonds %>% 
    ggplot(aes(y = price, x = carat)) + 
    geom_point()

```

The bigger the diamond the bigger the price


```{r eval=FALSE}
diamonds %>% 
    ggplot(aes(y = carat, x = clarity)) + 
    geom_boxplot()

```

BUT the bigger the diamond, the more flawed it tends to be. Thus flawed diamonds looked more expensive, but only because they also tend to be bigger (and size is a bigger driver of price).




## Exercise 12: Final conclusion

What's your final conclusion about diamond prices?    

Which diamonds are more expensive: flawed ones or flawless ones?
    










\
\
\
\






# Solutions

```{r eval = TRUE, echo = FALSE}
# Load useful packages and data
library(tidyverse)
peaks <- read_csv("https://mac-stat.github.io/data/high_peaks.csv") %>%
    mutate(ascent = ascent / 1000)

# Import data
cps <- read_csv("https://mac-stat.github.io/data/cps_2018.csv") %>% 
    filter(age >= 18, age <= 34) %>% 
    filter(wage < 250000)
```


## Exercise 1: Review

```{r eval = TRUE}
peaks_model <- lm(time ~ length + ascent + rating, data = peaks)
coef(summary(peaks_model))
```

- `length` coefficient:
    - Among hikes with the same vertical ascent and challenge rating, each additional mile of the hike is associated with a 0.46 hour increase in completion time on average.
    - Holding vertical ascent and challenge rating constant (fixed), each additional mile of the hike is associated with a 0.46 hour increase in completion time on average.

- `ratingeasy` coefficient:
    - Among hikes with the same length and vertical ascent, the average completion time of easy hikes is 3.2 hours less than that of difficult hikes (reference category).
    - Holding constant hike length and vertical ascent, the average completion time of easy hikes is 3.2 hours less than that of difficult hikes.




## Exercise 2: Confounders

age, education, job industry, ...

`marital` vs `wage`:

```{r eval = TRUE}
cps %>%
  ggplot(aes(x=marital, y=wage))+
  geom_boxplot()+
  theme_classic()
```

`age` vs `marital`:

```{r eval = TRUE}
cps %>%
  ggplot(aes(x=age, y=marital))+
  geom_boxplot()+
  theme_classic()
```

`age` vs `wage`:

```{r eval = TRUE}
cps %>%
  ggplot(aes(x=age, y=wage))+
  geom_point()+
  geom_smooth(method="lm", se=F)+
  theme_classic()
```


## Exercise 3: Controlling for confounders

a. create 2 separate groups that are as similar as possible with respect to these variables. give the groups different remedies.

b. we'd have to get 2 groups that are similar with respect to age, and assign 1 group to get married and 1 group to be single. that would be weird (and unethical).



## Exercise 4: Age

```{r eval = TRUE}
# Construct the model
wage_model_2 <- lm(wage ~ marital + age, cps)
coef(summary(wage_model_2))
```

a. .

```{r eval = TRUE}
cps %>%
ggplot(aes(y = wage, x = age, color = marital)) +
    geom_point(size = 0.1, alpha = 0.5) +
    geom_line(aes(y = wage_model_2$fitted.values), size = 0.5)
```    

bonus! adding in the `geom_smooth` layer:

```{r eval = TRUE}
cps %>%
ggplot(aes(y = wage, x = age, color = marital)) +
  geom_point(size = 0.1, alpha = 0.5) +
  geom_line(aes(y = wage_model_2$fitted.values), size = 0.5)+
  geom_smooth(method="lm", se=F, linetype="dashed")
```


b. -$7500


c. 
    - When controlling for ("holding constant") age, single workers make $7500 less than married workers on average.
    - Among workers of the same age, single workers make $7500 less than married workers on average.
    

    
    



## Exercise 5: More confounders

```{r eval = TRUE}
wage_model_3 <- lm(wage ~ marital + age + education, cps)
coef(summary(wage_model_3))
```

a. 2 planes: There are 2 quantitative predictors which form the dimensions of a regression plane (including response, it’s a 3D!). Because marital status is a categorical predictor with two levels, the model produces two such planes—one for each group.

b. The `maritalsingle` coefficient is of main interest:
    - Among workers of the same age and years of education, single workers earn $6478 less than married workers.

c.
    - `age` coefficient: Among workers of the same marital status and years of education, each additional year of age is associated with a $1677 increase in salary on average.
    - `education` coefficient: Among workers of the same marital status and age, each additional year of education is associated with a $4285 increase in salary on average.





## Exercise 6: Even more

```{r eval = TRUE}
wage_model_4 <- lm(wage ~ marital + age + education + industry, cps)
coef(summary(wage_model_4))
```

a. Among workers of the same job industry, education, and age, single workers make $5893 less than a married worker on average.

b. highest = construction (because it has the highest positive coefficient), lowest = service (because it has the most negative coefficient)





## Exercise 7: Biggest model yet

```{r eval = TRUE}
wage_model_5 <- lm(wage ~ marital + age + education + industry + hours + health, cps)
coef(summary(wage_model_5))
```





## Exercise 8: Reflection

a. These confounders explained away more and more of the wage gap between single and married workers.
b. Answers will vary. A potential factor that we haven't considered is a worker's role within a given industry.













## Exercise 9: A new (extreme) example

clarity  description
-------- ------------------------------------------------
IF       flawless (no internal imperfections) 
VVS1     very very slightly imperfect
VVS2     " "
VS1      very slightly imperfect
VS2      " "
SI1      slightly imperfect
SI2      " "
I1       imperfect

```{r eval = TRUE}
diamond_model_1 <- lm(price ~ clarity, data = diamonds)

# Get a model summary
coef(summary(diamond_model_1))
```

a. highest = SI2, lowest = VVS1

b. will vary.
    
    





## Exercise 10: Size

```{r eval = TRUE}
diamond_model_2 <- lm(price ~ clarity + carat, data = diamonds)

# Get a model summary
coef(summary(diamond_model_2))

# Plot the model
diamonds %>% 
    ggplot(aes(y = price, x = carat, color = clarity)) + 
    geom_line(aes(y = diamond_model_2$fitted.values))
```

highest = IF, lowest = I1 (reference category)

This is what we would have expected!








## Exercise 11: Simpson's Paradox

The bigger the diamond the bigger the price:

```{r eval = TRUE}
diamonds %>% 
    ggplot(aes(y = price, x = carat)) + 
    geom_point()
```


BUT the bigger the diamond, the more flawed it tends to be:

```{r eval = TRUE}
diamonds %>% 
    ggplot(aes(y = carat, x = clarity)) + 
    geom_boxplot()
```


Thus flawed diamonds looked more expensive, but only because they also tend to be bigger (and size is a bigger driver of price).




## Exercise 12: Final conclusion

Flawless diamonds are more expensive.
    



