Introduction

In the previous lectures on Multiple Linear Regression, all predictors we used were either continuous (sleep hours, age, physical health days) or binary (sex, exercise). But many variables in epidemiology are categorical with more than two levels, including race/ethnicity, education, marital status, and disease staging.

When a categorical predictor has \(k\) levels, we cannot simply plug in the numeric codes (1, 2, 3, …) as if the variable were continuous. Doing so imposes an assumption that the categories are equally spaced and linearly related to the outcome, which is rarely appropriate for nominal variables and often inappropriate even for ordinal ones.

Dummy variables (also called indicator variables) provide the correct way to include categorical predictors in regression models. This lecture covers:

  • Why numeric coding of categories fails
  • How to construct dummy variables for dichotomous and multichotomous predictors
  • The concept of the reference group and how to change it
  • Interpreting dummy variable coefficients
  • Testing whether a categorical variable as a whole is significant (partial F-test)
  • Alternative coding schemes (effect coding, ordinal contrasts)

Setup and Data

library(tidyverse)
library(haven)
library(janitor)
library(knitr)
library(kableExtra)
library(broom)
library(gtsummary)
library(GGally)
library(car)
library(ggeffects)
library(plotly)

options(gtsummary.use_ftExtra = TRUE)
set_gtsummary_theme(theme_gtsummary_compact(set_theme = TRUE))

The BRFSS 2020 Dataset

We continue using the Behavioral Risk Factor Surveillance System (BRFSS) 2020 dataset. In this lecture, we focus on how categorical predictors, particularly education level, relate to mental health outcomes.

Research question for today:

How is educational attainment associated with the number of mentally unhealthy days in the past 30 days, after adjusting for age, sex, physical health, and sleep?

brfss_full <- read_xpt(
  "/Users/zoya_hayes/Desktop/EPI553/EPI553_Coding/data/brfss_2020"
) |>
  clean_names()
brfss_dv <- brfss_full |>
  mutate(
    # Outcome: mentally unhealthy days in past 30
    menthlth_days = case_when(
      menthlth == 88                    ~ 0,
      menthlth >= 1 & menthlth <= 30   ~ as.numeric(menthlth),
      TRUE                             ~ NA_real_
    ),
    # Physical health days
    physhlth_days = case_when(
      physhlth == 88                    ~ 0,
      physhlth >= 1 & physhlth <= 30   ~ as.numeric(physhlth),
      TRUE                             ~ NA_real_
    ),
    # Sleep hours
    sleep_hrs = case_when(
      sleptim1 >= 1 & sleptim1 <= 14   ~ as.numeric(sleptim1),
      TRUE                             ~ NA_real_
    ),
    # Age
    age = age80,
    # Sex
    sex = factor(sexvar, levels = c(1, 2), labels = c("Male", "Female")),
    # Education (6-level raw BRFSS variable EDUCA)
    # 1 = Never attended school or only kindergarten
    # 2 = Grades 1 through 8 (Elementary)
    # 3 = Grades 9 through 11 (Some high school)
    # 4 = Grade 12 or GED (High school graduate)
    # 5 = College 1 year to 3 years (Some college or technical school)
    # 6 = College 4 years or more (College graduate)
    # 9 = Refused
    education = factor(case_when(
      educa %in% c(1, 2, 3) ~ "Less than HS",
      educa == 4             ~ "HS graduate",
      educa == 5             ~ "Some college",
      educa == 6             ~ "College graduate",
      TRUE                   ~ NA_character_
    ), levels = c("Less than HS", "HS graduate", "Some college", "College graduate")),
    # General health status (5-level)
    gen_health = factor(case_when(
      genhlth == 1 ~ "Excellent",
      genhlth == 2 ~ "Very good",
      genhlth == 3 ~ "Good",
      genhlth == 4 ~ "Fair",
      genhlth == 5 ~ "Poor",
      TRUE         ~ NA_character_
    ), levels = c("Excellent", "Very good", "Good", "Fair", "Poor")),
    # Marital status
    marital_status = factor(case_when(
      marital == 1 ~ "Married",
      marital == 2 ~ "Divorced",
      marital == 3 ~ "Widowed",
      marital == 4 ~ "Separated",
      marital == 5 ~ "Never married",
      marital == 6 ~ "Unmarried couple",
      TRUE         ~ NA_character_
    ), levels = c("Married", "Divorced", "Widowed", "Separated",
                  "Never married", "Unmarried couple")),
    # Store the raw education numeric code for the "naive approach" demonstration
    educ_numeric = case_when(
      educa %in% c(1, 2, 3) ~ 1,
      educa == 4             ~ 2,
      educa == 5             ~ 3,
      educa == 6             ~ 4,
      TRUE                   ~ NA_real_
    )
  ) |>
  filter(
    !is.na(menthlth_days),
    !is.na(physhlth_days),
    !is.na(sleep_hrs),
    !is.na(age), age >= 18,
    !is.na(sex),
    !is.na(education),
    !is.na(gen_health),
    !is.na(marital_status)
  )

# Reproducible random sample
set.seed(1220)
brfss_dv <- brfss_dv |>
  select(menthlth_days, physhlth_days, sleep_hrs, age, sex,
         education, gen_health, marital_status, educ_numeric) |>
  slice_sample(n = 5000)

# Save for lab activity
saveRDS(brfss_dv,
  "/Users/zoya_hayes/Desktop/EPI553/EPI553_Coding/data/brfss_dv_2020.rds")

tibble(Metric = c("Observations", "Variables"),
       Value  = c(nrow(brfss_dv), ncol(brfss_dv))) |>
  kable(caption = "Analytic Dataset Dimensions") |>
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Analytic Dataset Dimensions
Metric Value
Observations 5000
Variables 9

Descriptive Statistics

brfss_dv |>
  select(menthlth_days, physhlth_days, sleep_hrs, age, sex,
         education, gen_health) |>
  tbl_summary(
    label = list(
      menthlth_days  ~ "Mentally unhealthy days (past 30)",
      physhlth_days  ~ "Physically unhealthy days (past 30)",
      sleep_hrs      ~ "Sleep (hours/night)",
      age            ~ "Age (years)",
      sex            ~ "Sex",
      education      ~ "Education level",
      gen_health     ~ "General health status"
    ),
    statistic = list(
      all_continuous()  ~ "{mean} ({sd})",
      all_categorical() ~ "{n} ({p}%)"
    ),
    digits = all_continuous() ~ 1,
    missing = "no"
  ) |>
  add_n() |>
  bold_labels() |>
  italicize_levels() |>
  modify_caption("**Table 1. Descriptive Statistics — BRFSS 2020 Analytic Sample (n = 5,000)**") |>
  as_flex_table()
**Table 1. Descriptive Statistics — BRFSS 2020 Analytic Sample (n = 5,000)**

Characteristic

N

N = 5,0001

Mentally unhealthy days (past 30)

5,000

3.8 (7.9)

Physically unhealthy days (past 30)

5,000

3.3 (7.9)

Sleep (hours/night)

5,000

7.0 (1.4)

Age (years)

5,000

54.9 (17.5)

Sex

5,000

Male

2,303 (46%)

Female

2,697 (54%)

Education level

5,000

Less than HS

290 (5.8%)

HS graduate

1,348 (27%)

Some college

1,340 (27%)

College graduate

2,022 (40%)

General health status

5,000

Excellent

1,065 (21%)

Very good

1,803 (36%)

Good

1,426 (29%)

Fair

523 (10%)

Poor

183 (3.7%)

1Mean (SD); n (%)

ggplot(brfss_dv, aes(x = education, fill = education)) +
  geom_bar(alpha = 0.85) +
  geom_text(stat = "count", aes(label = after_stat(count)), vjust = -0.3) +
  scale_fill_brewer(palette = "Blues") +
  labs(
    title = "Distribution of Education Level",
    subtitle = "BRFSS 2020 Analytic Sample (n = 5,000)",
    x = "Education Level",
    y = "Count"
  ) +
  theme_minimal(base_size = 13) +
  theme(legend.position = "none")
Distribution of Education Level in Analytic Sample

Distribution of Education Level in Analytic Sample 

ggplot(brfss_dv, aes(x = education, y = menthlth_days, fill = education)) +
  geom_boxplot(alpha = 0.7, outlier.alpha = 0.2) +
  scale_fill_brewer(palette = "Blues") +
  labs(
    title = "Mentally Unhealthy Days by Education Level",
    subtitle = "BRFSS 2020 (n = 5,000)",
    x = "Education Level",
    y = "Mentally Unhealthy Days (Past 30)"
  ) +
  theme_minimal(base_size = 13) +
  theme(legend.position = "none")
Mental Health Days by Education Level

Mental Health Days by Education Level

Part 1: Guided Practice — Dummy Variables

1. Categorical Variables: The Problem

1.1 Types of Categorical Variables

Categorical predictor variables come in two forms:

Type Definition Examples
Nominal Categories with no natural ordering Sex, race/ethnicity, marital status, blood type
Ordinal Categories with a meaningful order Education level, income bracket, disease stage, Likert scale

A further distinction is:

  • Dichotomous (k = 2): Only two categories (e.g., sex: Male/Female)
  • Multichotomous (k > 2): Three or more categories (e.g., education: 4 levels)

Note that categorical variables can also be created by grouping continuous variables (e.g., age groups from continuous age), though this generally results in a loss of information.

1.2 The Naive Approach: Why Numeric Codes Fail

Suppose education has been coded as: 1 = Less than HS, 2 = HS graduate, 3 = Some college, 4 = College graduate.

If we include this numeric code directly in a regression model, we are assuming:

\[Y = \beta_0 + \beta_1 X_1 + \beta_2 \cdot \text{educ_numeric} + \varepsilon\]

This forces the model to assume that the difference in expected \(Y\) between “Less than HS” and “HS graduate” is the same as the difference between “HS graduate” and “Some college,” and the same again between “Some college” and “College graduate.” In other words, we are assuming equally spaced, linear increments.

# The WRONG way: treating education as a continuous numeric variable
naive_mod <- lm(menthlth_days ~ age + educ_numeric, data = brfss_dv)

tidy(naive_mod, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Naive Model: Education Treated as Continuous",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Naive Model: Education Treated as Continuous
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 9.5601 0.5039 18.9727 0 8.5723 10.5479
age -0.0661 0.0063 -10.5135 0 -0.0784 -0.0538
educ_numeric -0.7168 0.1158 -6.1917 0 -0.9437 -0.4898

This model estimates a single coefficient for education, meaning each step up the education ladder is associated with the same change in mentally unhealthy days. This constraint is problematic for two reasons:

  1. For nominal variables (like race or marital status), numeric codes are entirely arbitrary. The “distance” between code 1 and code 2 has no meaning.
  2. Even for ordinal variables (like education), the assumption of equal spacing is rarely justified. The jump from “Less than HS” to “HS graduate” is substantively different from “Some college” to “College graduate.”

Let’s visualize why this matters:

# Compute observed group means
group_means <- brfss_dv |>
  summarise(mean_days = mean(menthlth_days), .by = c(education, educ_numeric))

# Generate predictions from the naive model
pred_naive <- tibble(
  educ_numeric = 1:4,
  predicted = predict(naive_mod, newdata = tibble(age = mean(brfss_dv$age), educ_numeric = 1:4))
)

ggplot() +
  geom_point(data = group_means,
             aes(x = educ_numeric, y = mean_days),
             size = 4, color = "steelblue") +
  geom_line(data = pred_naive,
            aes(x = educ_numeric, y = predicted),
            color = "tomato", linewidth = 1.2, linetype = "dashed") +
  geom_point(data = pred_naive,
             aes(x = educ_numeric, y = predicted),
             size = 3, color = "tomato", shape = 17) +
  scale_x_continuous(
    breaks = 1:4,
    labels = c("Less than HS", "HS graduate", "Some college", "College graduate")
  ) +
  labs(
    title = "Observed Group Means (blue) vs. Naive Linear Fit (red)",
    subtitle = "The naive model forces equal spacing between education levels",
    x = "Education Level",
    y = "Mean Mentally Unhealthy Days"
  ) +
  theme_minimal(base_size = 13)
Naive Linear Fit vs. Actual Group Means by Education

Naive Linear Fit vs. Actual Group Means by Education

Key takeaway: The blue dots (observed means) do not fall along a straight line. The naive linear model (red) misrepresents the actual pattern. We need a more flexible approach.

2. Dummy (Indicator) Variables

2.1 Definition and the k - 1 Rule

A dummy variable (also called an indicator variable) is a variable that takes on only two possible values:

  • 1 to indicate the presence of a condition
  • 0 to indicate its absence

If a categorical predictor has \(k\) categories, we need exactly \(k - 1\) dummy variables when the model includes an intercept. The omitted category becomes the reference group (also called the control group or baseline group).

Why \(k - 1\) and not \(k\)? Because the intercept already captures the mean for the reference group. Including all \(k\) dummies plus the intercept would create perfect multicollinearity (the dummy variables would sum to equal the intercept column), and the model could not be estimated.

2.2 The Dichotomous Case (k = 2)

The simplest example is a variable with two categories, such as sex.

With \(k = 2\), we need \(2 - 1 = 1\) dummy variable. If we choose Female as the reference group:

\[\text{male} = \begin{cases} 1 & \text{if male} \\ 0 & \text{if female} \end{cases}\]

The regression model becomes:

\[Y = \beta_0 + \beta_1 \cdot \text{age} + \beta_2 \cdot \text{male} + \varepsilon\]

For males (\(\text{male} = 1\)): \[E(Y | \text{age}, \text{male}) = (\beta_0 + \beta_2) + \beta_1 \cdot \text{age}\]

For females (\(\text{male} = 0\)): \[E(Y | \text{age}, \text{female}) = \beta_0 + \beta_1 \cdot \text{age}\]

Both groups share the same slope for age but have different intercepts. The coefficient \(\beta_2\) is the expected difference in \(Y\) between males and females, holding age constant.

# Fit model with sex as a dummy variable
mod_sex <- lm(menthlth_days ~ age + sex, data = brfss_dv)

tidy(mod_sex, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Model with Dichotomous Dummy Variable: Sex",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Model with Dichotomous Dummy Variable: Sex
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 6.6262 0.3730 17.7666 0 5.8951 7.3574
age -0.0698 0.0063 -11.1011 0 -0.0821 -0.0575
sexFemale 1.8031 0.2210 8.1585 0 1.3698 2.2364 
b_sex <- round(coef(mod_sex), 3)

Interpretation:

  • Intercept (6.626): The estimated mean number of mentally unhealthy days for a male of age 0. (This is a mathematical artifact, not a substantive value.)
  • Age (-0.07): Each additional year of age is associated with a 0.07 day change in mentally unhealthy days, holding sex constant.
  • Sex: Female (1.803): Compared to males (the reference group), females report an estimated 1.803 more mentally unhealthy days on average, holding age constant.

Note that R automatically creates dummy variables when a factor is included in lm(). It uses alphabetical or level order to set the reference group, which is why Male (the first level) is the reference here.

pred_sex <- ggpredict(mod_sex, terms = c("age [20:80]", "sex"))

ggplot(pred_sex, aes(x = x, y = predicted, color = group, fill = group)) +
  geom_line(linewidth = 1.2) +
  geom_ribbon(aes(ymin = conf.low, ymax = conf.high), alpha = 0.15, color = NA) +
  labs(
    title    = "Predicted Mental Health Days by Age and Sex",
    subtitle = "Parallel lines: same slope, different intercepts",
    x        = "Age (years)",
    y        = "Predicted Mentally Unhealthy Days",
    color    = "Sex",
    fill     = "Sex"
  ) +
  theme_minimal(base_size = 13) +
  scale_color_brewer(palette = "Set1")
Parallel Regression Lines: Males vs. Females

Parallel Regression Lines: Males vs. Females

Geometrically: Dummy variables produce parallel regression lines. The intercept shifts by \(\beta_2\) for the non-reference group, but the slope remains the same.

3. Multichotomous Dummy Variables (k > 2)

3.1 Constructing the Dummies

Education has \(k = 4\) categories, so we need \(4 - 1 = 3\) dummy variables. If we choose “Less than HS” as the reference group:

\[\text{HS_graduate} = \begin{cases} 1 & \text{if HS graduate} \\ 0 & \text{otherwise} \end{cases}\]

\[\text{Some_college} = \begin{cases} 1 & \text{if Some college} \\ 0 & \text{otherwise} \end{cases}\]

\[\text{College_graduate} = \begin{cases} 1 & \text{if College graduate} \\ 0 & \text{otherwise} \end{cases}\]

The data matrix looks like this:

Dummy Variable Encoding for Education (Reference: Less than HS)
Observation Education HS_graduate Some_college College_graduate
1 Less than HS 0 0 0
2 HS graduate 1 0 0
3 Some college 0 1 0
4 College graduate 0 0 1
5 Less than HS 0 0 0

Notice that the reference group is identified by having all dummy variables equal to zero.

3.2 The Reference Group

The reference group is the category against which all others are compared. Key points:

  • The choice of reference group affects the interpretation of coefficients but not the model’s predictive ability
  • The predicted values are identical regardless of which reference group is chosen
  • Choose the reference group based on your research question (e.g., the most common category, the “control” condition, or the group of primary interest)
  • You can always test differences between any pair of groups, not just comparisons to the reference

3.3 Fitting the Model in R

When we include a factor variable in lm(), R automatically creates the dummy variables. The first level of the factor is used as the reference group by default.

# Fit model with education as a factor (R creates dummies automatically)
mod_educ <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs + education,
               data = brfss_dv)

tidy(mod_educ, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Model with Education Dummy Variables (Reference: Less than HS)",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Model with Education Dummy Variables (Reference: Less than HS)
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 11.1377 0.7390 15.0709 0.0000 9.6889 12.5865
age -0.0772 0.0060 -12.9522 0.0000 -0.0888 -0.0655
sexFemale 1.6813 0.2075 8.1038 0.0000 1.2745 2.0880
physhlth_days 0.3112 0.0133 23.3334 0.0000 0.2850 0.3373
sleep_hrs -0.6281 0.0771 -8.1463 0.0000 -0.7793 -0.4770
educationHS graduate -0.5873 0.4719 -1.2445 0.2134 -1.5125 0.3379
educationSome college -0.1289 0.4735 -0.2723 0.7854 -1.0572 0.7993
educationCollege graduate -1.1429 0.4607 -2.4805 0.0132 -2.0461 -0.2396 
b_educ <- round(coef(mod_educ), 3)
ci_educ <- round(confint(mod_educ), 3)

3.4 Interpreting Each Dummy Coefficient

The model is:

\[\widehat{\text{Mental Health Days}} = 11.138 + -0.077(\text{Age}) + 1.681(\text{Female}) + 0.311(\text{Phys Days}) + -0.628(\text{Sleep}) + -0.587(\text{HS grad}) + -0.129(\text{Some college}) + -1.143(\text{College grad})\]

Each education coefficient represents the estimated difference in mentally unhealthy days between that group and the reference group (Less than HS), holding all other variables constant:

  • HS graduate (\(\hat{\beta}\) = -0.587): Compared to those with less than a high school education, HS graduates report an estimated 0.587 fewer mentally unhealthy days, holding age, sex, physical health days, and sleep constant.

  • Some college (\(\hat{\beta}\) = -0.129): Compared to those with less than a high school education, those with some college report an estimated 0.129 fewer mentally unhealthy days, holding all else constant.

  • College graduate (\(\hat{\beta}\) = -1.143): Compared to those with less than a high school education, college graduates report an estimated 1.143 fewer mentally unhealthy days, holding all else constant.

Key pattern: All comparisons are made relative to the reference group. The coefficients do NOT directly tell us the difference between, say, HS graduates and college graduates. We would need to compute \(\hat{\beta}_{\text{HS grad}} - \hat{\beta}_{\text{College grad}}\) for that comparison (or change the reference group).

3.5 Visualizing the Parallel Lines

pred_educ <- ggpredict(mod_educ, terms = c("age [20:80]", "education"))

ggplot(pred_educ, aes(x = x, y = predicted, color = group, fill = group)) +
  geom_line(linewidth = 1.1) +
  geom_ribbon(aes(ymin = conf.low, ymax = conf.high), alpha = 0.1, color = NA) +
  labs(
    title    = "Predicted Mental Health Days by Age and Education",
    subtitle = "Parallel lines: same slopes for age, different intercepts by education",
    x        = "Age (years)",
    y        = "Predicted Mentally Unhealthy Days",
    color    = "Education",
    fill     = "Education"
  ) +
  theme_minimal(base_size = 13) +
  scale_color_brewer(palette = "Set2")
Predicted Mental Health Days by Age and Education Level

Predicted Mental Health Days by Age and Education Level

These are a series of parallel lines, one for each education level. The slope for age is the same across all groups; only the intercept differs. Each education dummy shifts the intercept up or down relative to the reference group.

4. Changing the Reference Group

4.1 Using relevel() in R

We may want to change the reference group to a category that is more epidemiologically meaningful. For instance, “College graduate” is the largest group and could serve as a natural comparison.

# Change reference group to College graduate
brfss_dv$education_reref <- relevel(brfss_dv$education, ref = "College graduate")

mod_educ_reref <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs + education_reref,
                     data = brfss_dv)

tidy(mod_educ_reref, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Same Model, Different Reference Group (Reference: College graduate)",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Same Model, Different Reference Group (Reference: College graduate)
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 9.9948 0.6272 15.9349 0.0000 8.7652 11.2245
age -0.0772 0.0060 -12.9522 0.0000 -0.0888 -0.0655
sexFemale 1.6813 0.2075 8.1038 0.0000 1.2745 2.0880
physhlth_days 0.3112 0.0133 23.3334 0.0000 0.2850 0.3373
sleep_hrs -0.6281 0.0771 -8.1463 0.0000 -0.7793 -0.4770
education_rerefLess than HS 1.1429 0.4607 2.4805 0.0132 0.2396 2.0461
education_rerefHS graduate 0.5556 0.2574 2.1586 0.0309 0.0510 1.0601
education_rerefSome college 1.0139 0.2566 3.9507 0.0001 0.5108 1.5171

4.2 What Changes and What Stays the Same?

tribble(
  ~Quantity, ~`Ref: Less than HS`, ~`Ref: College graduate`,
  "Intercept", round(coef(mod_educ)[1], 3), round(coef(mod_educ_reref)[1], 3),
  "Age coefficient", round(coef(mod_educ)[2], 3), round(coef(mod_educ_reref)[2], 3),
  "Sex coefficient", round(coef(mod_educ)[3], 3), round(coef(mod_educ_reref)[3], 3),
  "Physical health days", round(coef(mod_educ)[4], 3), round(coef(mod_educ_reref)[4], 3),
  "Sleep hours", round(coef(mod_educ)[5], 3), round(coef(mod_educ_reref)[5], 3),
  "R-squared", round(summary(mod_educ)$r.squared, 4), round(summary(mod_educ_reref)$r.squared, 4),
  "Residual SE", round(summary(mod_educ)$sigma, 3), round(summary(mod_educ_reref)$sigma, 3)
) |>
  kable(caption = "Comparing Models with Different Reference Groups") |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Comparing Models with Different Reference Groups
Quantity Ref: Less than HS Ref: College graduate
Intercept 11.1380 9.9950
Age coefficient -0.0770 -0.0770
Sex coefficient 1.6810 1.6810
Physical health days 0.3110 0.3110
Sleep hours -0.6280 -0.6280
R-squared 0.1553 0.1553
Residual SE 7.2690 7.2690

What changes:

  • The intercept
  • The education dummy variable coefficients (they now represent comparisons to the new reference group)

What stays the same:

  • Coefficients for age, sex, physical health days, and sleep
  • \(R^2\), Adjusted \(R^2\), and Residual Standard Error
  • All predicted values
  • The overall F-test

This is a critical point: Changing the reference group does not change the model’s fit or predictions. It only changes the interpretation of the dummy variable coefficients.

# Verify that predicted values are identical
pred_orig <- predict(mod_educ)
pred_reref <- predict(mod_educ_reref)

tibble(
  Check = c("Maximum absolute difference in predictions",
            "Correlation between predictions"),
  Value = c(max(abs(pred_orig - pred_reref)),
            cor(pred_orig, pred_reref))
) |>
  kable(caption = "Verification: Predicted Values Are Identical") |>
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Verification: Predicted Values Are Identical
Check Value
Maximum absolute difference in predictions 0
Correlation between predictions 1

5. The Dummy Variable Trap (Perfect Multicollinearity)

5.1 Why We Cannot Include All k Dummies

If we include \(k\) dummy variables and an intercept for a variable with \(k\) categories, the columns of the design matrix \(X\) are linearly dependent. Specifically:

\[\text{Intercept} = D_1 + D_2 + \cdots + D_k\]

where \(D_1, \ldots, D_k\) are the \(k\) dummy variables (one for each category). This means the matrix \(X^TX\) is singular and cannot be inverted, so the OLS estimator \(\hat{\beta} = (X^TX)^{-1}X^TY\) does not exist.

This is called the dummy variable trap.

The Dummy Variable Trap: Intercept = A + B + C (perfect linear dependence)
Obs Intercept A B C A + B + C
1 1 1 0 0 1
2 1 0 1 0 1
3 1 0 0 1 1
4 1 1 0 0 1

Solutions:

  1. Reference cell coding (default in R): Include \(k - 1\) dummies; the omitted category is absorbed into the intercept
  2. Remove the intercept: Fit the model with - 1 in the formula and include all \(k\) dummies. Then each coefficient is the group mean (adjusted for other predictors) rather than a difference from a reference.
# Model without intercept: all k dummies included
mod_no_int <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs + education - 1,
                 data = brfss_dv)

tidy(mod_no_int, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Model Without Intercept: All k Education Dummies Included",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Model Without Intercept: All k Education Dummies Included
Term Estimate SE t p-value CI Lower CI Upper
age -0.0772 0.0060 -12.9522 0.0000 -0.0888 -0.0655
sexMale 11.1377 0.7390 15.0709 0.0000 9.6889 12.5865
sexFemale 12.8190 0.7524 17.0365 0.0000 11.3439 14.2941
physhlth_days 0.3112 0.0133 23.3334 0.0000 0.2850 0.3373
sleep_hrs -0.6281 0.0771 -8.1463 0.0000 -0.7793 -0.4770
educationHS graduate -0.5873 0.4719 -1.2445 0.2134 -1.5125 0.3379
educationSome college -0.1289 0.4735 -0.2723 0.7854 -1.0572 0.7993
educationCollege graduate -1.1429 0.4607 -2.4805 0.0132 -2.0461 -0.2396

Caution: Removing the intercept changes the interpretation of \(R^2\) and should only be done when there is a substantive reason. In most epidemiological applications, reference cell coding (the default) is preferred.

6. Testing Whether a Categorical Variable is Significant

6.1 The Partial F-Test (Type I and Type III)

When a categorical variable with \(k\) levels enters the model as \(k - 1\) dummies, we cannot assess its overall significance by looking at individual t-tests for each dummy. A single dummy might not be statistically significant on its own, yet the variable as a whole might be.

To test whether education as a whole is associated with the outcome, we use a partial F-test (also called an extra sum of squares F-test):

\[H_0: \beta_{\text{HS grad}} = \beta_{\text{Some college}} = \beta_{\text{College grad}} = 0\] \[H_A: \text{At least one } \beta_j \neq 0\]

This compares the full model (with education) to a reduced model (without education):

# Reduced model (no education)
mod_reduced <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs, data = brfss_dv)

# Partial F-test
f_test <- anova(mod_reduced, mod_educ)

f_test |>
  tidy() |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(caption = "Partial F-test: Does Education Improve the Model?") |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Partial F-test: Does Education Improve the Model?
term df.residual rss df sumsq statistic p.value
menthlth_days ~ age + sex + physhlth_days + sleep_hrs 4995 264715.2 NA NA NA NA
menthlth_days ~ age + sex + physhlth_days + sleep_hrs + education 4992 263744.4 3 970.7509 6.1246 4e-04

6.2 Using car::Anova() for Type III Tests

The car::Anova() function with type = "III" provides a convenient way to test the overall significance of each predictor, including categorical variables:

Anova(mod_educ, type = "III") |>
  tidy() |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(caption = "Type III ANOVA: Testing Each Predictor's Contribution") |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Type III ANOVA: Testing Each Predictor’s Contribution
term sumsq df statistic p.value
(Intercept) 12000.1867 1 227.1325 0e+00
age 8863.3522 1 167.7603 0e+00
sex 3469.6448 1 65.6714 0e+00
physhlth_days 28765.1139 1 544.4492 0e+00
sleep_hrs 3506.1243 1 66.3619 0e+00
education 970.7509 3 6.1246 4e-04
Residuals 263744.4348 4992 NA NA

Type I vs. Type III: Type I (sequential) sums of squares depend on the order variables enter the model. Type III (partial) sums of squares test each variable after all others, regardless of order. For unbalanced observational data (the norm in epidemiology), Type III is preferred.

7. Contrasts and Alternative Coding Schemes

7.1 Reference (Treatment) Coding — The Default

This is what R uses by default (contr.treatment). Each coefficient represents the difference between a group and the reference group.

contrasts(brfss_dv$education)
##                  HS graduate Some college College graduate
## Less than HS               0            0                0
## HS graduate                1            0                0
## Some college               0            1                0
## College graduate           0            0                1

7.2 Effect (Deviation) Coding

In effect coding (contr.sum), each coefficient represents the difference between a group’s mean and the grand mean (the unweighted average of all group means). This is common in ANOVA contexts.

# Set effect coding
brfss_dv$education_effect <- brfss_dv$education
contrasts(brfss_dv$education_effect) <- contr.sum(4)

mod_effect <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs + education_effect,
                 data = brfss_dv)

tidy(mod_effect, conf.int = TRUE) |>
  mutate(
    term = case_when(
      str_detect(term, "education_effect1") ~ "Education: Less than HS vs. Grand Mean",
      str_detect(term, "education_effect2") ~ "Education: HS graduate vs. Grand Mean",
      str_detect(term, "education_effect3") ~ "Education: Some college vs. Grand Mean",
      TRUE ~ term
    ),
    across(where(is.numeric), \(x) round(x, 4))
  ) |>
  kable(
    caption = "Effect Coding: Each Education Coefficient vs. Grand Mean",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Effect Coding: Each Education Coefficient vs. Grand Mean
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 10.6729 0.6172 17.2911 0.0000 9.4628 11.8830
age -0.0772 0.0060 -12.9522 0.0000 -0.0888 -0.0655
sexFemale 1.6813 0.2075 8.1038 0.0000 1.2745 2.0880
physhlth_days 0.3112 0.0133 23.3334 0.0000 0.2850 0.3373
sleep_hrs -0.6281 0.0771 -8.1463 0.0000 -0.7793 -0.4770
Education: Less than HS vs. Grand Mean 0.4648 0.3323 1.3988 0.1619 -0.1866 1.1162
Education: HS graduate vs. Grand Mean -0.1225 0.1939 -0.6319 0.5275 -0.5026 0.2576
Education: Some college vs. Grand Mean 0.3358 0.1946 1.7257 0.0845 -0.0457 0.7174

With effect coding, the intercept is the grand mean (adjusted for covariates), and each education coefficient shows how far that group deviates from the grand mean. The omitted group’s deviation is the negative sum of the others.

7.3 Polynomial (Orthogonal) Contrasts for Ordinal Variables

When a categorical variable is truly ordinal (like education), we can test for specific patterns using orthogonal polynomial contrasts (contr.poly). These decompose the group differences into linear, quadratic, and cubic trends.

# Ordinal polynomial contrasts
brfss_dv$education_ord <- brfss_dv$education
contrasts(brfss_dv$education_ord) <- contr.poly(4)

mod_ord <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs + education_ord,
              data = brfss_dv)

tidy(mod_ord, conf.int = TRUE) |>
  mutate(
    term = case_when(
      str_detect(term, "\\.L$") ~ "Education: Linear trend",
      str_detect(term, "\\.Q$") ~ "Education: Quadratic trend",
      str_detect(term, "\\.C$") ~ "Education: Cubic trend",
      TRUE ~ term
    ),
    across(where(is.numeric), \(x) round(x, 4))
  ) |>
  kable(
    caption = "Polynomial Contrasts: Testing Linear, Quadratic, and Cubic Trends",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Polynomial Contrasts: Testing Linear, Quadratic, and Cubic Trends
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 10.6729 0.6172 17.2911 0.0000 9.4628 11.8830
age -0.0772 0.0060 -12.9522 0.0000 -0.0888 -0.0655
sexFemale 1.6813 0.2075 8.1038 0.0000 1.2745 2.0880
physhlth_days 0.3112 0.0133 23.3334 0.0000 0.2850 0.3373
sleep_hrs -0.6281 0.0771 -8.1463 0.0000 -0.7793 -0.4770
Education: Linear trend -0.6642 0.3158 -2.1028 0.0355 -1.2833 -0.0450
Education: Quadratic trend -0.2133 0.2682 -0.7954 0.4264 -0.7391 0.3125
Education: Cubic trend -0.5630 0.2142 -2.6282 0.0086 -0.9830 -0.1431

Interpretation:

  • The linear contrast tests whether there is a consistent trend (higher education = fewer/more days)
  • The quadratic contrast tests whether the relationship curves (U-shaped or inverted-U)
  • The cubic contrast tests for an S-shaped pattern

Polynomial contrasts are most useful when the categories have a clear, meaningful order and you want to characterize the shape of the trend rather than compare individual groups to a reference.

7.4 Coding Scheme Comparison Summary

Summary of Dummy Variable Coding Schemes
Coding Scheme R Function Intercept Each β represents Best for
Treatment (Reference) contr.treatment (default) Reference group mean Difference from reference group Group comparisons to baseline
Effect (Deviation) contr.sum Grand mean Deviation from grand mean ANOVA-style analyses
Polynomial (Ordinal) contr.poly Grand mean Linear/quadratic/cubic trend Ordinal variables with ordered levels

8. Practical Considerations

8.1 Choosing the Reference Group

Guidelines for choosing the reference group:

  1. Most common category — maximizes the precision of comparisons (largest reference sample)
  2. Control or baseline condition — natural in experimental or clinical settings
  3. Epidemiologically meaningful comparator — the group to which you want to compare all others
  4. Consistency with published literature — facilitates comparison across studies

8.2 When as.factor() Is Required

If a categorical variable is stored as numeric in your data (e.g., coded 0, 1, 2, 3), R will treat it as continuous by default. You must use as.factor() or factor() to tell R it is categorical:

# WRONG: R treats educ_numeric as continuous
mod_wrong <- lm(menthlth_days ~ educ_numeric, data = brfss_dv)

# RIGHT: Convert to factor first
mod_right <- lm(menthlth_days ~ factor(educ_numeric), data = brfss_dv)

# Compare: 1 coefficient (wrong) vs. 3 coefficients (right)
tribble(
  ~Model, ~`Number of education coefficients`, ~`Degrees of freedom used`,
  "Numeric (wrong)", 1, 1,
  "Factor (correct)", 3, 3
) |>
  kable(caption = "Numeric vs. Factor Treatment of Categorical Variables") |>
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Numeric vs. Factor Treatment of Categorical Variables
Model Number of education coefficients Degrees of freedom used
Numeric (wrong) 1 1
Factor (correct) 3 3

8.3 Comparing Non-Reference Groups

What if we want to compare HS graduates to college graduates, but neither is the reference group? We have two options:

Option 1: Change the reference group with relevel().

Option 2: Compute the difference manually from the model output.

# Difference between HS graduate and College graduate
# = β_HS_grad - β_College_grad
diff_est <- coef(mod_educ)["educationHS graduate"] - coef(mod_educ)["educationCollege graduate"]

# Use linearHypothesis() for a formal test with SE and p-value
lin_test <- linearHypothesis(mod_educ, "educationHS graduate - educationCollege graduate = 0")

cat("Estimated difference (HS grad - College grad):", round(diff_est, 3), "days\n")
## Estimated difference (HS grad - College grad): 0.556 days
cat("F-statistic:", round(lin_test$F[2], 3), "\n")
## F-statistic: 4.66
cat("p-value:", round(lin_test$`Pr(>F)`[2], 4), "\n")
## p-value: 0.0309

car::linearHypothesis() is a powerful function for testing any linear combination of coefficients, not just comparisons to the reference group.

Summary of Key Concepts

Concept Key Point
Categorical predictors Cannot be included as raw numeric codes in regression
Dummy variables Binary (0/1) indicators; need \(k - 1\) for \(k\) categories
Reference group The omitted category; all comparisons are relative to it
Changing reference Use relevel(); predictions unchanged, interpretation changes
Partial F-test Tests whether the categorical variable as a whole is significant
Dummy variable trap Including \(k\) dummies + intercept = perfect multicollinearity
as.factor() Required when categorical variable is stored as numeric
Coding schemes Treatment (default), effect, polynomial — each answers a different question
Type III ANOVA Preferred for unbalanced observational data
Linear hypothesis linearHypothesis() tests comparisons between non-reference groups

Part 2: In-Class Lab Activity

EPI 553 — Dummy Variables Lab Due: End of class, March 23, 2026

Instructions

In this lab, you will practice constructing, fitting, and interpreting regression models with dummy variables using the BRFSS 2020 analytic dataset. Work through each task systematically. You may discuss concepts with classmates, but your written answers and R code must be your own.

Submission: Knit your .Rmd to HTML and upload to Brightspace by end of class.

Data for the Lab

Use the saved analytic dataset from today’s lecture. It contains 5,000 randomly sampled BRFSS 2020 respondents with the following variables:

Variable Description Type
menthlth_days Mentally unhealthy days in past 30 Continuous (0–30)
physhlth_days Physically unhealthy days in past 30 Continuous (0–30)
sleep_hrs Sleep hours per night Continuous (1–14)
age Age in years (capped at 80) Continuous
sex Sex (Male/Female) Factor
education Education level (4 categories) Factor
gen_health General health status (5 categories) Factor
marital_status Marital status (6 categories) Factor
educ_numeric Education as numeric code (1–4) Numeric 
# Load the dataset
library(tidyverse)
library(haven)
library(janitor)
library(knitr)
library(kableExtra)
library(broom)
library(gtsummary)
library(GGally)
library(car)
library(ggeffects)
library(plotly)

brfss_dv <- readRDS(
  "/Users/zoya_hayes/Desktop/EPI553/EPI553_Coding/data/brfss_dv_2020.rds"
)

Task 1: Exploratory Data Analysis (15 points)

1a. (5 pts) Create a descriptive statistics table using tbl_summary() that includes menthlth_days, age, sex, gen_health, and marital_status. Show means (SD) for continuous variables and n (%) for categorical variables.

# Create a descriptive statistics table with means and n 
brfss_dv |>
  select(menthlth_days, age, sex, gen_health, marital_status) |>
  tbl_summary(
    label = list(
      menthlth_days  ~ "Mentally unhealthy days (past 30)",
      age            ~ "Age (years)",
      sex            ~ "Sex",
      gen_health     ~ "General health status",
      marital_status ~ "Marital Status"
    ),
    statistic = list(
      all_continuous()  ~ "{mean} ({sd})",
      all_categorical() ~ "{n} ({p}%)"
    ),
    digits = all_continuous() ~ 1,
    missing = "no"
  ) |>
  add_n() |>
  bold_labels() |>
  italicize_levels() |>
  modify_caption("**Table 1. Descriptive Statistics — BRFSS 2020 Analytic Sample (n = 5,000)**") |>
  as_flex_table()
**Table 1. Descriptive Statistics — BRFSS 2020 Analytic Sample (n = 5,000)**

Characteristic

N

N = 5,0001

Mentally unhealthy days (past 30)

5,000

3.8 (7.9)

Age (years)

5,000

54.9 (17.5)

Sex

5,000

Male

2,303 (46%)

Female

2,697 (54%)

General health status

5,000

Excellent

1,065 (21%)

Very good

1,803 (36%)

Good

1,426 (29%)

Fair

523 (10%)

Poor

183 (3.7%)

Marital Status

5,000

Married

2,708 (54%)

Divorced

622 (12%)

Widowed

534 (11%)

Separated

109 (2.2%)

Never married

848 (17%)

Unmarried couple

179 (3.6%)

1Mean (SD); n (%)

1b. (5 pts) Create a boxplot of menthlth_days by gen_health. Which group reports the most mentally unhealthy days? Does the pattern appear consistent with what you would expect?

# Create boxplot for mentlth_days by gen_health
ggplot(brfss_dv, aes(x = gen_health, y = menthlth_days, fill = gen_health)) +
  geom_boxplot(alpha = 0.7, outlier.alpha = 0.2) +
  scale_fill_brewer(palette = "Blues") +
  labs(
    title = "Mentally Unhealthy Days by General Health Status",
    subtitle = "BRFSS 2020 (n = 5,000)",
    x = "General Health Status",
    y = "Mentally Unhealthy Days (Past 30)"
  ) +
  theme_minimal(base_size = 13) +
  theme(legend.position = "none")

The group that reports the most mentally unhealthy days are those who have a poor general health status. The group that reports the least mentally unhealthy days are those who have an excellent general health status. This pattern is consistent with what I would expect because poor health for me includes mental, physical, and social health meaning that they would likely have more mentally unhealthy days since they have a lower overall health status.

1c. (5 pts) Create a grouped bar chart or table showing the mean number of mentally unhealthy days by marital_status. Which marital status group has the highest mean? The lowest?

# Calculate mean mental health days 
mean_mtl_hlt <- brfss_dv |> 
  group_by(marital_status) |> 
  summarize(mean_mtl_hlt = mean(menthlth_days, na.rm = TRUE))

# Create grouped bar chart
ggplot(mean_mtl_hlt, aes(x = marital_status, y = mean_mtl_hlt, fill = marital_status)) +
  geom_col(alpha = 0.85) +
  geom_bar(stat = "identity", position = "dodge") +
  geom_text(aes(label = round(mean_mtl_hlt, 2)), vjust = -0.3) +
  scale_fill_brewer(palette = "Greens") +
  labs(
    title = "Mean Mentally Unhealthy Days by Marital Status",
    x = "Marital Status", 
    y = "Mean Mentally Unhealthy Days Over Past 30 Days"
  ) + 
  theme_minimal(base_size = 13) + 
  theme(legend.position = "none")

The marital status group with the highest mean were separated couples with 6.22 mean mentally unhealthy days. The marital status group with the lowest mean were widowed individuals with 2.67 mean mentally unhealthy days. —

Task 2: The Naive Approach (10 points)

2a. (5 pts) Using the gen_health variable, create a numeric version coded as: Excellent = 1, Very good = 2, Good = 3, Fair = 4, Poor = 5. Fit a simple regression model: menthlth_days ~ gen_health_numeric. Report the coefficient and interpret it.

# Create a new dataset with gen_health as a numeric code
brfss_dv_naive <- brfss_dv |>  
  mutate(
  gen_health_num = case_when(
  gen_health == "Excellent" ~ 1,
  gen_health == "Very good" ~ 2, 
  gen_health == "Good" ~ 3,
  gen_health == "Fair" ~ 4,
  gen_health == "Poor" ~ 5, 
))

# Run a model with this new numeric version
naive_mod <- lm(menthlth_days ~ gen_health_num, data = brfss_dv_naive)

# Extract and create a table with values from the model
tidy(naive_mod, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Naive Model: Gen Health Treated as Continuous",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Naive Model: Gen Health Treated as Continuous
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) -0.6718 0.2705 -2.4840 0.013 -1.2021 -0.1416
gen_health_num 1.8578 0.1036 17.9259 0.000 1.6547 2.0610

According to the naive version, the coefficient using gen_health_num is 1.8578. Each additional unit of general health is associated with an increase of 1.8578 mentally unhealthy days.

2b. (5 pts) Now fit the same model but treating gen_health as a factor: menthlth_days ~ gen_health. Compare the two models. Why does the factor version use 4 coefficients instead of 1? Explain why the naive numeric approach may be misleading.

# Fit linear model
factor_mod <- lm(menthlth_days ~ gen_health, data = brfss_dv_naive)

# Create tidy table 
tidy(factor_mod, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Gen Health Treated as Factor",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Gen Health Treated as Factor
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 2.1174 0.2332 9.0790 0.0000 1.6602 2.5746
gen_healthVery good 0.5903 0.2941 2.0070 0.0448 0.0137 1.1670
gen_healthGood 1.9535 0.3082 6.3375 0.0000 1.3492 2.5577
gen_healthFair 5.0624 0.4064 12.4572 0.0000 4.2657 5.8590
gen_healthPoor 9.6640 0.6090 15.8678 0.0000 8.4701 10.8580

For the naive model, there is only one coefficient because the model is assessing general health as a numeric variable. This makes the model believe that the difference in expected Y is the same from one category to another (fair to poor or excellent to very good). This is an issue because the distance between code 1 and code 2 has no true meaning and is unjustified. The factor model instead provides four coefficients (each one compared to a reference group of excellent). The factor model creates dummy categories with one reference group. Which in this case, as we look at the coefficient, we see that very good has the lowest increase in mentally unhealthy days and poor has the highest increase. The numeric or naive way is misleading because it lumps all of these levels together because it assumes the distance between them is the same causing a confusing and invalid explanation of the variable. —

Task 3: Dummy Variable Regression with General Health (25 points)

3a. (5 pts) Fit the following model with gen_health as a factor:

menthlth_days ~ age + sex + physhlth_days + sleep_hrs + gen_health

Write out the fitted regression equation.

# Fit full linear model
full_mod <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs + gen_health, data = brfss_dv_naive)

# Create tidy table
tidy(full_mod, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Regression with Age, Sex, Physical Health, Sleep Hours, and General Health (Reference: Excellent)",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Regression with Age, Sex, Physical Health, Sleep Hours, and General Health (Reference: Excellent)
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 9.5930 0.6304 15.2163 0.0000 8.3570 10.8289
age -0.0867 0.0060 -14.4888 0.0000 -0.0984 -0.0749
sexFemale 1.7254 0.2055 8.3971 0.0000 1.3226 2.1282
physhlth_days 0.2314 0.0162 14.3057 0.0000 0.1997 0.2631
sleep_hrs -0.5866 0.0766 -7.6607 0.0000 -0.7367 -0.4365
gen_healthVery good 0.7899 0.2797 2.8247 0.0048 0.2417 1.3382
gen_healthGood 1.8436 0.2973 6.2020 0.0000 1.2608 2.4264
gen_healthFair 3.3953 0.4180 8.1234 0.0000 2.5759 4.2147
gen_healthPoor 5.3353 0.6829 7.8122 0.0000 3.9965 6.6742

The model is:

\[\widehat{\text{Mental Health Days}} = 9.5930 + -0.0867(\text{Age}) + 1.7254 (\text{Female}) + 0.2314 (\text{Phys Days}) + -0.5866 (\text{Sleep}) + 0.7899 (\text{Very good}) + 1.8436(\text{Good}) + 3.3953(\text{Fair}) + 5.3353(\text{Poor})\]

3b. (10 pts) Interpret every dummy variable coefficient for gen_health in plain language. Be specific about the reference group, the direction and magnitude of each comparison, and include the phrase “holding all other variables constant.”

  • Very Good (\(\hat{\beta}\) = 0.7899): Compared to those with excellent general health, those with very good general health status report an estimated 0.7899 more mentally unhealthy days, holding age, sex, physical health days, and sleep constant.
  • Good (\(\hat{\beta}\) = 1.8436): Compared to those with excellent general heath, those with good general health status report an estimated 1.8436 more mentally unhealthy days, holding age, sex, physical health days, and sleep constant.
  • Fair (\(\hat{\beta}\) = 3.3953): Compared to those with excellent general health, those with fair general health status report an estimated 3.3953 more mentally unhealthy days, holding age, sex, physical health days, and sleep constant.
  • Poor (\(\hat{\beta}\) = 5.3353): Compared to those with excellent general health, those with poor general health status report an estimated 5.3353 more mentally unhealthy days, holding age, sex, physical health days, and sleep constant.

3c. (10 pts) Create a coefficient plot (forest plot) showing the estimated coefficients and 95% confidence intervals for the gen_health dummy variables only. Which group differs most from the reference group?

# Create data group from tidy
model_data <- tidy(factor_mod, conf.int = TRUE)
print(model_data)
## # A tibble: 5 × 7
##   term                estimate std.error statistic  p.value conf.low conf.high
##   <chr>                  <dbl>     <dbl>     <dbl>    <dbl>    <dbl>     <dbl>
## 1 (Intercept)            2.12      0.233      9.08 1.55e-19   1.66        2.57
## 2 gen_healthVery good    0.590     0.294      2.01 4.48e- 2   0.0137      1.17
## 3 gen_healthGood         1.95      0.308      6.34 2.54e-10   1.35        2.56
## 4 gen_healthFair         5.06      0.406     12.5  4.23e-35   4.27        5.86
## 5 gen_healthPoor         9.66      0.609     15.9  2.34e-55   8.47       10.9
# Remove the intercept so only dummy variables are displayed
model_data_no_intercept <- model_data |> 
  filter(term != "(Intercept)") 

# Create a coefficient plot with confidence intervals 
ggplot(model_data_no_intercept, aes(x = estimate, y = term)) +
  geom_point(size = 3) +
  geom_errorbar(
    aes(xmin = conf.low, xmax = conf.high),
    width = 0.2,
    orientation = "y"
  ) +
  geom_vline(xintercept = 0, linetype = "dashed", color = "grey40") +
  labs(
    title = "Estimated Coefficients and 95% Confidence Intervals",
    x = "Estimate",
    y = "Dummy Variable"
  ) +
  theme_classic(base_size = 14)

The group that differs most from the reference group is those who are in the gen_health poor group. —

Task 4: Changing the Reference Group (15 points)

4a. (5 pts) Use relevel() to change the reference group for gen_health to “Good.” Refit the model from Task 3a.

# Change reference group to Good
brfss_dv_naive$gen_health_reref <- relevel(brfss_dv_naive$gen_health, ref = "Good")

# Refit the model with the relevel
full_mod_relevel <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs + gen_health_reref, data = brfss_dv_naive)

# Create tidy table
tidy(full_mod_relevel, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Same Model, Different Reference Group (Reference: Good)",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Same Model, Different Reference Group (Reference: Good)
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 11.4366 0.6298 18.1584 0e+00 10.2019 12.6713
age -0.0867 0.0060 -14.4888 0e+00 -0.0984 -0.0749
sexFemale 1.7254 0.2055 8.3971 0e+00 1.3226 2.1282
physhlth_days 0.2314 0.0162 14.3057 0e+00 0.1997 0.2631
sleep_hrs -0.5866 0.0766 -7.6607 0e+00 -0.7367 -0.4365
gen_health_rerefExcellent -1.8436 0.2973 -6.2020 0e+00 -2.4264 -1.2608
gen_health_rerefVery good -1.0537 0.2581 -4.0819 0e+00 -1.5597 -0.5476
gen_health_rerefFair 1.5517 0.3861 4.0186 1e-04 0.7947 2.3087
gen_health_rerefPoor 3.4917 0.6506 5.3673 0e+00 2.2164 4.7671

4b. (5 pts) Compare the education and other continuous variable coefficients between the two models (original reference vs. new reference). Are they the same? Why or why not?

# Make table to show education and other continuous variables coefficients
tribble(
  ~Quantity, ~`Ref: Excellent`, ~`Ref: Good`,
  "Intercept", round(coef(full_mod)[1], 3), round(coef(full_mod_relevel)[1],3),
  "Age coefficient", round(coef(full_mod)[2], 3), round(coef(full_mod_relevel)[2],3),
  "Sex coefficient", round(coef(full_mod)[3], 3), round(coef(full_mod_relevel)[3],3),
  "Physical health days",round(coef(full_mod)[4], 3), round(coef(full_mod_relevel)[4],3),
  "Sleep hours", round(coef(full_mod)[5], 3), round(coef(full_mod_relevel)[5],3),
  "R-squared", round(summary(full_mod)$r.squared, 4), round(summary(full_mod_relevel)$r.squared, 4),
  "Residual SE", round(summary(full_mod)$sigma, 3), round(summary(full_mod_relevel)$sigma, 3), 
) |> 
  kable(caption = "Comparing Models with Different Reference Groups") |> 
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Comparing Models with Different Reference Groups
Quantity Ref: Excellent Ref: Good
Intercept 9.5930 11.4370
Age coefficient -0.0870 -0.0870
Sex coefficient 1.7250 1.7250
Physical health days 0.2310 0.2310
Sleep hours -0.5870 -0.5870
R-squared 0.1694 0.1694
Residual SE 7.2080 7.2080

The other variable coefficients between the two models are the same. The only differences are within the general health and intercept because we are using a different reference group. By using a different reference group, these coefficients now represent comparisons to the new reference group. The other coefficients for age, sex, physical health, and sleep hours will stay the same.

4c. (5 pts) Verify that the predicted values from both models are identical by computing the correlation between the two sets of predictions. Explain in your own words why changing the reference group does not change predictions.

# Verify that predicted values are identical
pred_orig <- predict(full_mod)
pred_reref <- predict(full_mod_relevel)

# Create tibble table
tibble(
  Check = c("Maximum absolute difference in predictions",
            "Correlation between predictions"),
  Value = c(max(abs(pred_orig - pred_reref)),
            cor(pred_orig, pred_reref))
) |>
  kable(caption = "Verification: Predicted Values Are Identical") |>
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Verification: Predicted Values Are Identical
Check Value
Maximum absolute difference in predictions 0
Correlation between predictions 1

The predicted values from both models are identical as demonstrated by a 0 maximum difference and a correlation of 1 between the predictions. This is because changing the reference group will not change the model’s fit or predictions. It does change the interpretation of the dummy variable coefficients. The model predictions are not any different because the releveling only changes the baseline instead of altering the relationship that was already defined. —

Task 5: Partial F-Test for General Health (20 points)

5a. (5 pts) Fit a reduced model without gen_health:

menthlth_days ~ age + sex + physhlth_days + sleep_hrs

Report \(R^2\) and Adjusted \(R^2\) for both the reduced model and the full model (from Task 3a).

# Create reduced model
reduced_mod <- lm(menthlth_days ~ age + sex + physhlth_days + sleep_hrs, data = brfss_dv_naive)

# Create tidy table for reduced model
tidy(reduced_mod, conf.int = TRUE) |>
  mutate(across(where(is.numeric), \(x) round(x, 4))) |>
  kable(
    caption = "Reduced Regression with Age, Sex, Physical Health, Sleep Hours",
    col.names = c("Term", "Estimate", "SE", "t", "p-value", "CI Lower", "CI Upper")
  ) |>
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Reduced Regression with Age, Sex, Physical Health, Sleep Hours
Term Estimate SE t p-value CI Lower CI Upper
(Intercept) 10.5041 0.6124 17.1531 0 9.3036 11.7046
age -0.0773 0.0060 -12.9721 0 -0.0890 -0.0656
sexFemale 1.6864 0.2074 8.1296 0 1.2797 2.0931
physhlth_days 0.3172 0.0132 24.0152 0 0.2913 0.3431
sleep_hrs -0.6330 0.0772 -8.2029 0 -0.7843 -0.4817 
# Create table with R² and adjusted R² for full model
glance(full_mod) |> 
  select(r.squared, adj.r.squared) |> 
  pivot_longer(everything(), names_to = "Metric", values_to = "Value") |> 
  mutate(
    Value = round(Value, 4),
    Metric = dplyr::recode(Metric,
      "r.squared"     = "R²",
      "adj.r.squared" = "Adjusted R²"
    )
  ) |> 
  kable(caption = "Full Model Summary") |> 
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Full Model Summary
Metric Value
R² 0.1694
Adjusted R² 0.1681 
# Create table with R² and adjusted R² for reduced model
glance(reduced_mod) |> 
  select(r.squared, adj.r.squared) |> 
  pivot_longer(everything(), names_to = "Metric", values_to = "Value") %>%
  mutate(
    Value = round(Value, 4),
    Metric = dplyr::recode(Metric,
      "r.squared"     = "R²",
      "adj.r.squared" = "Adjusted R²"
    )
  ) |> 
  kable(caption = "Reduced Model Summary") |> 
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Reduced Model Summary
Metric Value
R² 0.1522
Adjusted R² 0.1515

From the full model, R² is 0.1694 and adjusted R² is 0.1681. From the reduced model, R² is 0.1522 and adjusted R² is 0.1515.

5b. (10 pts) Conduct a partial F-test using anova() to test whether gen_health as a whole significantly improves the model. State the null and alternative hypotheses. Report the F-statistic, degrees of freedom, and p-value. State your conclusion.

# Compare using anova() and perform the partial F-test
anova(reduced_mod, full_mod) |> 
  as.data.frame() |> 
  rownames_to_column("Model") |> 
  mutate(
    Model = c("Reduced (no general health)", "Full (with general health)"),
    across(where(is.numeric), ~ round(., 4))
  ) |> 
  kable(
    caption = "Table 6. Partial F-Test: Does gen_health add to the model?",
    col.names = c("Model", "Res. df", "RSS", "df", "Sum of Sq", "F", "p-value")
  ) |> 
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Table 6. Partial F-Test: Does gen_health add to the model?
Model Res. df RSS df Sum of Sq F p-value
Reduced (no general health) 4995 264715.2 NA NA NA NA
Full (with general health) 4991 259335.4 4 5379.751 25.8838 0

\[H_0: \beta_{\text{General Health}} = 0\] \[H_A: \beta_{\text{General Health}}\neq 0\] The f-statistic value is 25.8838, with 4991 degrees of freedom, and the p-value is less than .001. Based on this f-statistic and p-value, we would reject the null hypothesis that general health is not associated with mentally unhealthy days. We would turn to the alternative hypothesis that there is an association between general health and mentally unhealthy days and determine what evidence there is.

5c. (5 pts) Use car::Anova() with type = "III" on the full model. Compare the result for gen_health to your partial F-test. Are they consistent?

# Type III using car::Anova()
anova_type3 <- Anova(full_mod, type = "III")

# Partial sum of squares Type III 
anova_type3 |> 
  as.data.frame() |> 
  rownames_to_column("Source") |> 
  mutate(across(where(is.numeric), ~ round(., 4))) |> 
  kable(
    caption = "Type III (Partial) Sums of Squares — car::Anova()",
    col.names = c("Source", "Sum of Sq", "df", "F value", "p-value")
  ) |> 
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Type III (Partial) Sums of Squares — car::Anova()
Source Sum of Sq df F value p-value
(Intercept) 12030.737 1 231.5357 0
age 10907.874 1 209.9258 0
sex 3663.847 1 70.5120 0
physhlth_days 10633.920 1 204.6535 0
sleep_hrs 3049.400 1 58.6868 0
gen_health 5379.751 4 25.8838 0
Residuals 259335.435 4991 NA NA 
# Partial F-Tests using Type III 
Anova(full_mod, type = "III") |> 
  as.data.frame() |> 
  rownames_to_column("Source") |> 
  mutate(
    Significant = ifelse(`Pr(>F)` < 0.05, "Yes", "No"),
    across(where(is.numeric), ~ round(., 4))
  ) |> 
  kable(
    caption = "Type III Partial F-Tests for Each Predictor",
    col.names = c("Source", "Sum of Sq", "df", "F value", "p-value", "Significant (α = 0.05)")
  ) |> 
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) |> 
  column_spec(6, bold = TRUE)
Type III Partial F-Tests for Each Predictor
Source Sum of Sq df F value p-value Significant (α = 0.05)
(Intercept) 12030.737 1 231.5357 0 Yes
age 10907.874 1 209.9258 0 Yes
sex 3663.847 1 70.5120 0 Yes
physhlth_days 10633.920 1 204.6535 0 Yes
sleep_hrs 3049.400 1 58.6868 0 Yes
gen_health 5379.751 4 25.8838 0 Yes
Residuals 259335.435 4991 NA NA NA

Yes, the result from type III partial F-test is consistent with my partial F-test. This is because they are both determining if the predictor (gen_health) is significant with the same nested models. Both the f-statistic (25.8839) and p-value (<.001) are the same in both tables. —

Task 6: Public Health Interpretation (15 points)

6a. (5 pts) Using the full model from Task 3a, write a 3–4 sentence paragraph summarizing the association between general health status and mental health days for a non-statistical audience. Your paragraph should:

  • Identify which general health groups differ most from the reference
  • State the direction and approximate magnitude of the association
  • Appropriately acknowledge the cross-sectional nature of the data
  • Not use any statistical jargon (no “significant,” “coefficient,” “p-value,” “confidence interval”)

The general health groups that differ most form the reference are those who identify with poor and fair general health status. Compared to excellent health, poor health status individuals have an estimated 5.3353 more mentally unhealthy days, holding age, sex, physical health days, and sleep constant while compared to excellent health, fair health status individuals have an estimated 3.3953 more mentally unhealthy days, holding age, sex, physical health days, and sleep constant. It is important to note that the data employed in this study is cross-sectional so the data may not be fully independent and does not include weights meaning it may be unrepresentative.

6b. (10 pts) Now consider both the education model (from the guided practice) and the general health model (from your lab). Discuss: Which categorical predictor appears to be more strongly associated with mental health days? How would you decide which to include if you were building a final model? Write 3–4 sentences addressing this comparison. —

I would say say that after completing this lab, it is suggested that the general health model appears to be more strongly associated with mental health days than the education model. College graduates with a reference of less than HS only suggested a negative association with mentally unhealthy days of 1.14 when holding age, sex, physical health days, and sleep constant whereas poor general health with a reference of excellent health suggested a positive association of 5.34 more mentally unhealthy days when holding age, sex, physical health days, and sleep constant. I would decide which one to include in my model by employing the calculation of R² and adjusted R² because the adjusted R² will inform whether or not a predictor adds to the model since R² simply increase as more predictors are added. I would also consult p-values to determine statistical significant of certain predictors.

End of Lab Activity