Introduction

Simple Linear Regression (SLR) is one of the most fundamental and widely used tools in epidemiology and public health research. It allows us to:

  • Quantify the linear relationship between a continuous outcome and a single predictor
  • Predict values of an outcome given a predictor value
  • Test hypotheses about whether a predictor is associated with an outcome
  • Lay the groundwork for multiple regression, which controls for confounding

In epidemiology, we frequently use SLR to model continuous outcomes such as blood pressure, BMI, cholesterol levels, or hospital length of stay as a function of age, exposure levels, or other continuous predictors.

Setup and Data

library(tidyverse)
library(haven)
library(here)
library(knitr)
library(kableExtra)
library(plotly)
library(broom)
library(ggeffects)
library(gtsummary)

Loading BRFSS 2020 Data

We will use the Behavioral Risk Factor Surveillance System (BRFSS) 2020 data throughout this lecture. The BRFSS is a large-scale, nationally representative telephone survey conducted by the CDC that collects data on health behaviors, chronic conditions, and preventive service use among U.S. adults.

###Load brfss_subset_2020.rds

brfss_slr_2020 <- read_rds ("~/Downloads/HEPI553/brfss_slr_2020.rds")

Descriptive Statistics

brfss_slr_2020 %>%
  select(bmi, age, sleep_hrs, phys_days) %>%
  summary() %>%
  kable(caption = "Descriptive Statistics: Key Continuous Variables") %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Descriptive Statistics: Key Continuous Variables
bmi age sleep_hrs phys_days
Min. :14.63 Min. :18.00 Min. : 1.000 Min. : 1.00
1st Qu.:24.32 1st Qu.:43.00 1st Qu.: 6.000 1st Qu.: 2.00
Median :27.89 Median :58.00 Median : 7.000 Median : 6.00
Mean :29.18 Mean :55.52 Mean : 6.915 Mean :11.66
3rd Qu.:32.89 3rd Qu.:70.00 3rd Qu.: 8.000 3rd Qu.:20.00
Max. :59.60 Max. :80.00 Max. :20.000 Max. :30.00 
brfss_slr_2020 %>%
  select(bmi, age, sleep_hrs, sex, education) %>%
  tbl_summary(
    label = list(
      bmi ~ "BMI (kg/m²)",
      age ~ "Age (years)",
      sleep_hrs ~ "Sleep (hours/night)",
      sex ~ "Sex",
      education ~ "Education"
    ),
    statistic = list(
      all_continuous() ~ "{mean} ({sd})",
      all_categorical() ~ "{n} ({p}%)"
    ),
    digits = all_continuous() ~ 1
  ) %>%
  add_n() %>%
  bold_labels() %>%
  modify_caption("**Table 1. Descriptive Statistics (BRFSS 2020, n = 3,000)**")
Table 1. Descriptive Statistics (BRFSS 2020, n = 3,000)
Characteristic N N = 3,0001
BMI (kg/m²) 3,000 29.2 (7.0)
Age (years) 3,000 55.5 (17.4)
Sleep (hours/night) 3,000 6.9 (1.7)
Sex 3,000
    Female
1,701 (57%)
    Male
1,299 (43%)
Education 3,000
    < High school
237 (7.9%)
    High school graduate
796 (27%)
    Some college
937 (31%)
    College graduate
1,030 (34%)
1 Mean (SD); n (%)

Part 1: Guided Practice — Simple Linear Regression

1. The Simple Linear Regression Model

1.1 What Is the Model?

Simple linear regression models the mean of a continuous outcome \(Y\) as a linear function of a single predictor \(X\):

\[Y_i = \beta_0 + \beta_1 X_i + \varepsilon_i, \quad i = 1, 2, \ldots, n\]

Where:

Symbol Name Meaning
\(Y_i\) Response / Outcome Observed value for subject \(i\) (e.g., BMI)
\(X_i\) Predictor / Covariate Observed predictor for subject \(i\) (e.g., age)
\(\beta_0\) Intercept Expected value of \(Y\) when \(X = 0\)
\(\beta_1\) Slope Expected change in \(Y\) for a 1-unit increase in \(X\)
\(\varepsilon_i\) Error term Random deviation of \(Y_i\) from the regression line

The population regression line (also called the true or theoretical regression line) describes the expected (mean) value of \(Y\) at each value of \(X\):

\[E(Y \mid X) = \mu_{Y|X} = \beta_0 + \beta_1 X\]

1.2 Key Distinction: Population vs. Sample

Population Sample
Line \(\beta_0 + \beta_1 X\) \(\hat{y} = b_0 + b_1 X\)
Intercept \(\beta_0\) (parameter) \(b_0\) or \(\hat{\beta}_0\) (estimate)
Slope \(\beta_1\) (parameter) \(b_1\) or \(\hat{\beta}_1\) (estimate)
Error \(\varepsilon_i\) \(e_i = Y_i - \hat{Y}_i\) (residual)

We use our sample to estimate the population parameters. The estimates \(b_0\) and \(b_1\) define the fitted regression line.

1.3 Visualizing the Relationship

Before fitting any model, always visualize the bivariate relationship.

p_scatter <- ggplot(brfss_slr_2020, aes(x = age, y = bmi)) +
  geom_point(alpha = 0.15, color = "steelblue", size = 1.2) +
  geom_smooth(method = "lm", color = "red", linewidth = 1.2, se = TRUE) +
  geom_smooth(method = "loess", color = "blue", linewidth = 1,
              linetype = "dashed", se = FALSE) +
  labs(
    title = "BMI vs. Age (BRFSS 2020)",
    subtitle = "Red = Linear fit | Orange dashed = LOESS smoother",
    x = "Age (years)",
    y = "BMI (kg/m²)"
  ) +
  theme_minimal(base_size = 13)

ggplotly(p_scatter)

BMI vs. Age — BRFSS 2020

Interpretation tip: The LOESS smoother (orange) follows the data without assuming linearity. When it closely tracks the linear fit (red), a linear model is reasonable. Departures suggest nonlinearity.

2. Assumptions of Simple Linear Regression

A useful mnemonic is LINE:

Letter Assumption Description
L Linearity The relationship between \(X\) and \(E(Y)\) is linear
I Independence Observations are independent of one another
N Normality Errors \(\varepsilon_i\) are normally distributed
E Equal variance Errors have constant variance (homoscedasticity): \(\text{Var}(\varepsilon_i) = \sigma^2\)

Formally, we assume:

\[\varepsilon_i \overset{iid}{\sim} N(0, \sigma^2)\]

This means that for any value of \(X\), the distribution of \(Y\) is:

\[Y \mid X \sim N(\beta_0 + \beta_1 X, \; \sigma^2)\]

Note on independence: In cross-sectional survey data like BRFSS, observations from the same household or geographic cluster may not be fully independent. We acknowledge this limitation but proceed with the standard SLR framework for pedagogical purposes.

3. Estimating the Regression Coefficients

3.1 The Method of Least Squares

We estimate \(\beta_0\) and \(\beta_1\) by finding the values \(b_0\) and \(b_1\) that minimize the sum of squared residuals (SSR):

\[SSR = \sum_{i=1}^{n}(Y_i - \hat{Y}_i)^2 = \sum_{i=1}^{n}(Y_i - b_0 - b_1 X_i)^2\]

This is called the Ordinary Least Squares (OLS) criterion. Minimizing SSR yields the closed-form solutions:

\[b_1 = \frac{\sum_{i=1}^n (X_i - \bar{X})(Y_i - \bar{Y})}{\sum_{i=1}^n (X_i - \bar{X})^2} = \frac{S_{XY}}{S_{XX}}\]

\[b_0 = \bar{Y} - b_1 \bar{X}\]

where \(\bar{X}\) and \(\bar{Y}\) are the sample means of \(X\) and \(Y\).

Gauss-Markov Theorem: Under the LINE assumptions, OLS estimators are the Best Linear Unbiased Estimators (BLUE) — they have the smallest variance among all linear unbiased estimators.

3.2 Fitting the Model in R

# Fit simple linear regression: BMI ~ Age
model_slr <- lm(bmi ~ age, data = brfss_slr_2020)

# Summary output
summary(model_slr)
## 
## Call:
## lm(formula = bmi ~ age, data = brfss_slr_2020)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -14.633  -4.883  -1.325   3.688  30.340 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 29.528231   0.427507  69.071   <2e-16 ***
## age         -0.006238   0.007347  -0.849    0.396    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.012 on 2998 degrees of freedom
## Multiple R-squared:  0.0002404,  Adjusted R-squared:  -9.312e-05 
## F-statistic: 0.7208 on 1 and 2998 DF,  p-value: 0.396
# Tidy coefficient table
tidy(model_slr, conf.int = TRUE) %>%
  mutate(across(where(is.numeric), ~ round(., 4))) %>%
  kable(
    caption = "Simple Linear Regression: BMI ~ Age (BRFSS 2020)",
    col.names = c("Term", "Estimate", "Std. Error", "t-statistic",
                  "p-value", "95% CI Lower", "95% CI Upper"),
    align = "lrrrrrrr"
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
  row_spec(0, bold = TRUE)
Simple Linear Regression: BMI ~ Age (BRFSS 2020)
Term Estimate Std. Error t-statistic p-value 95% CI Lower 95% CI Upper
(Intercept) 29.5282 0.4275 69.0708 0.000 28.6900 30.3665
age -0.0062 0.0073 -0.8490 0.396 -0.0206 0.0082

3.3 Interpreting the Coefficients

b0 <- round(coef(model_slr)[1], 3)
b1 <- round(coef(model_slr)[2], 4)

Fitted regression equation:

\[\widehat{\text{BMI}} = 29.528 + -0.0062 \times \text{Age}\]

Intercept (\(b_0 = 29.528\)): The estimated mean BMI when age = 0. This is a mathematical artifact — a newborn does not have an adult BMI. The intercept is not directly interpretable in this context, but is necessary to anchor the line.

Slope (\(b_1 = -0.0062\)): For each 1-year increase in age, BMI is estimated to decrease by 0.0062 kg/m², on average, holding all else constant (though there are no other variables in this simple model).

Practical significance vs. statistical significance: Even a small slope can be highly statistically significant with a large sample. Always consider whether the magnitude is meaningful in the real world.

3.4 Visualizing Fitted Values and Residuals

# Augment dataset with fitted values and residuals
augmented <- augment(model_slr)

# Show a sample of fitted values and residuals
augmented %>%
  select(bmi, age, .fitted, .resid) %>%
  slice_head(n = 10) %>%
  mutate(across(where(is.numeric), ~ round(., 3))) %>%
  kable(
    caption = "First 10 Observations: Observed, Fitted, and Residual Values",
    col.names = c("Observed BMI (Y)", "Age (X)", "Fitted (Ŷ)", "Residual (e = Y − Ŷ)")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
First 10 Observations: Observed, Fitted, and Residual Values
Observed BMI (Y) Age (X) Fitted (Ŷ) Residual (e = Y − Ŷ)
26.58 67 29.110 -2.530
33.47 38 29.291 4.179
35.15 78 29.042 6.108
30.42 65 29.123 1.297
22.67 55 29.185 -6.515
30.11 80 29.029 1.081
35.43 34 29.316 6.114
31.58 71 29.085 2.495
28.13 55 29.185 -1.055
34.01 62 29.141 4.869 
# Select a random sample of 80 points to illustrate residuals
set.seed(42)
resid_sample <- augmented %>% slice_sample(n = 80)

p_resid <- ggplot(resid_sample, aes(x = age, y = bmi)) +
  geom_segment(aes(xend = age, yend = .fitted),
               color = "tomato", alpha = 0.5, linewidth = 0.5) +
  geom_point(color = "steelblue", size = 1.8, alpha = 0.8) +
  geom_line(aes(y = .fitted), color = "black", linewidth = 1.1) +
  labs(
    title = "Residuals Illustrated on the Regression Line",
    subtitle = "Red segments = residuals (Y − Ŷ); Black line = fitted regression line",
    x = "Age (years)",
    y = "BMI (kg/m²)"
  ) +
  theme_minimal(base_size = 13)

p_resid
Visualizing Residuals on the Regression Line

Visualizing Residuals on the Regression Line

4. Partitioning Variability: ANOVA Decomposition

4.1 Sums of Squares

The total variability in \(Y\) can be decomposed into two parts:

\[\underbrace{SS_{Total}}_{Total\ variability} = \underbrace{SS_{Regression}}_{Explained\ by\ X} + \underbrace{SS_{Residual}}_{Unexplained}\]

Where:

\[SS_{Total} = \sum(Y_i - \bar{Y})^2 \qquad (df = n-1)\] \[SS_{Regression} = \sum(\hat{Y}_i - \bar{Y})^2 \qquad (df = 1)\] \[SS_{Residual} = \sum(Y_i - \hat{Y}_i)^2 \qquad (df = n-2)\]

# ANOVA decomposition
anova_slr <- anova(model_slr)

anova_slr %>%
  kable(
    caption = "ANOVA Table: BMI ~ Age",
    digits = 3,
    col.names = c("Source", "Df", "Sum Sq", "Mean Sq", "F value", "Pr(>F)")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
ANOVA Table: BMI ~ Age
Source Df Sum Sq Mean Sq F value Pr(>F)
age 1 35.438 35.438 0.721 0.396
Residuals 2998 147400.214 49.166 NA NA

4.2 Mean Squared Error (MSE) and \(\hat{\sigma}\)

The Mean Squared Error estimates the variance of the error term:

\[MSE = \frac{SS_{Residual}}{n - 2} = \hat{\sigma}^2\]

The Residual Standard Error \(\hat{\sigma} = \sqrt{MSE}\) is in the same units as \(Y\) and tells us the typical prediction error of the model.

n <- nrow(brfss_slr_2020)
ss_resid <- sum(augmented$.resid^2)
mse <- ss_resid / (n - 2)
sigma_hat <- sqrt(mse)

tibble(
  Quantity = c("SS Residual", "MSE (σ̂²)", "Residual Std. Error (σ̂)"),
  Value    = c(round(ss_resid, 2), round(mse, 3), round(sigma_hat, 3)),
  Units    = c("", "", "kg/m²")
) %>%
  kable(caption = "Model Error Estimates") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Model Error Estimates
Quantity Value Units
SS Residual 147400.210
MSE (σ̂²)
    49.16

Interpretation: On average, our model’s predictions are off by about 7.01 BMI units.

5. The Coefficient of Determination: R²

5.1 Definition and Interpretation

\(R^2\) measures the proportion of total variability in \(Y\) explained by the linear regression on \(X\):

\[R^2 = \frac{SS_{Regression}}{SS_{Total}} = 1 - \frac{SS_{Residual}}{SS_{Total}}\]

\(R^2\) ranges from 0 to 1:

  • \(R^2 = 0\): \(X\) explains none of the variability in \(Y\)
  • \(R^2 = 1\): \(X\) explains all variability in \(Y\) (perfect fit)
# Extract R-squared from model
r_sq <- summary(model_slr)$r.squared
adj_r_sq <- summary(model_slr)$adj.r.squared

tibble(
  Metric = c("R²", "Adjusted R²", "Variance Explained"),
  Value  = c(
    round(r_sq, 4),
    round(adj_r_sq, 4),
    paste0(round(r_sq * 100, 2), "%")
  )
) %>%
  kable(caption = "R² and Adjusted R²") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
R² and Adjusted R²
Metric Value
R² 2e-04
Adjusted R² -1e-04
Variance Explained 0.02%

5.2 Relationship Between R² and Pearson’s r

For simple linear regression:

\[R^2 = r^2\]

where \(r\) is the Pearson correlation coefficient between \(X\) and \(Y\).

r_pearson <- cor(brfss_slr_2020$age, brfss_slr_2020$bmi)
tibble(
  Quantity   = c("Pearson r", "r² (from Pearson)", "R² (from model)", "r² = R²?"),
  Value      = c(
    round(r_pearson, 4),
    round(r_pearson^2, 4),
    round(r_sq, 4),
    as.character(round(r_pearson^2, 4) == round(r_sq, 4))
  )
) %>%
  kable(caption = "Pearson r vs. R² from Model") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Pearson r vs. R² from Model
Quantity Value
Pearson r -0.0155
r² (from Pearson) 2e-04
R² (from model) 2e-04
r² = R²? TRUE

Important caveat: A low \(R^2\) does not mean the regression is useless. In epidemiology, outcomes are influenced by many unmeasured factors, so \(R^2\) values of 0.05–0.20 can still yield scientifically meaningful and statistically significant estimates.

6. Hypothesis Testing

6.1 Testing the Slope: Is There a Linear Association?

The most important hypothesis test in SLR is:

\[H_0: \beta_1 = 0 \quad \text{(no linear relationship between X and Y)}\] \[H_A: \beta_1 \neq 0 \quad \text{(there is a linear relationship)}\]

Test statistic:

\[t = \frac{b_1 - 0}{SE(b_1)} \sim t_{n-2} \quad \text{under } H_0\]

Where:

\[SE(b_1) = \frac{\hat{\sigma}}{\sqrt{\sum(X_i - \bar{X})^2}} = \frac{\hat{\sigma}}{\sqrt{S_{XX}}}\]

# Extract slope test statistics
slope_test <- tidy(model_slr, conf.int = TRUE) %>% filter(term == "age")

tibble(
  Quantity = c("Slope (b₁)", "SE(b₁)", "t-statistic",
               "Degrees of freedom", "p-value", "95% CI Lower", "95% CI Upper"),
  Value    = c(
    round(slope_test$estimate, 4),
    round(slope_test$std.error, 4),
    round(slope_test$statistic, 3),
    n - 2,
    format.pval(slope_test$p.value, digits = 3),
    round(slope_test$conf.low, 4),
    round(slope_test$conf.high, 4)
  )
) %>%
  kable(caption = "t-Test for the Slope (H₀: β₁ = 0)") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
t-Test for the Slope (H₀: β₁ = 0)
Quantity Value
Slope (b₁) -0.0062
SE(b₁) 0.0073
t-statistic -0.849
Degrees of freedom 2998
p-value 0.396
95% CI Lower -0.0206
95% CI Upper 0.0082

Decision: With p = 0.396, we reject \(H_0\) at the \(\alpha = 0.05\) level. There is statistically significant evidence of a linear association between age and BMI.

6.2 The F-Test: Overall Model Significance

The F-test evaluates whether the overall model (i.e., all predictors together) explains a statistically significant portion of the variability in \(Y\). For simple linear regression with one predictor, the F-test is equivalent to the t-test for the slope (\(F = t^2\)).

\[F = \frac{MS_{Regression}}{MS_{Residual}} \sim F_{1,\, n-2} \quad \text{under } H_0\]

f_stat <- summary(model_slr)$fstatistic
f_value <- f_stat[1]
df1 <- f_stat[2]
df2 <- f_stat[3]
p_f <- pf(f_value, df1, df2, lower.tail = FALSE)

tibble(
  Quantity = c("F-statistic", "df (numerator)", "df (denominator)",
               "p-value", "Verification: t²", "Verification: F"),
  Value    = c(
    round(f_value, 3),
    df1,
    df2,
    format.pval(p_f, digits = 3),
    round(slope_test$statistic^2, 3),
    round(f_value, 3)
  )
) %>%
  kable(caption = "F-Test for Overall Model Significance") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
F-Test for Overall Model Significance
Quantity Value
F-statistic 0.721
df (numerator) 1
df (denominator) 2998
p-value 0.396
Verification: t² 0.721
Verification: F 0.721

6.3 Confidence Interval for the Slope

A 95% CI for \(\beta_1\) is:

\[b_1 \pm t_{n-2, \, 0.025} \times SE(b_1)\]

t_crit <- qt(0.975, df = n - 2)
ci_lower <- slope_test$estimate - t_crit * slope_test$std.error
ci_upper <- slope_test$estimate + t_crit * slope_test$std.error

tibble(
  Bound = c("95% CI Lower", "95% CI Upper"),
  Value = c(round(ci_lower, 4), round(ci_upper, 4)),
  Units = c("kg/m² per year", "kg/m² per year")
) %>%
  kable(caption = "95% Confidence Interval for β₁ (manually computed)") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
95% Confidence Interval for β₁ (manually computed)
Bound Value Units
95% CI Lower -0.0206 kg/m² per year
95% CI Upper 0.0082 kg/m² per year

7. Confidence Intervals and Prediction Intervals

7.1 Estimating the Mean Response (Confidence Interval)

A confidence interval for the mean response \(E(Y \mid X = x^*)\) gives a range of plausible values for the population mean of \(Y\) at a specific value \(x^*\):

\[\hat{Y}^* \pm t_{n-2, \, \alpha/2} \times SE(\hat{Y}^*)\]

Where:

\[SE(\hat{Y}^*) = \hat{\sigma}\sqrt{\frac{1}{n} + \frac{(x^* - \bar{X})^2}{S_{XX}}}\]

7.2 Predicting a Single Observation (Prediction Interval)

A prediction interval gives a range for a single new observation \(Y^*_{new}\) at \(X = x^*\). It is always wider than the confidence interval because it accounts for both the uncertainty in \(E(Y)\) and the individual variability around the mean:

\[\hat{Y}^* \pm t_{n-2, \, \alpha/2} \times SE_{pred}\]

Where:

\[SE_{pred} = \hat{\sigma}\sqrt{1 + \frac{1}{n} + \frac{(x^* - \bar{X})^2}{S_{XX}}}\]

# Compute CI and PI at specific age values
new_ages <- data.frame(age = c(25, 35, 45, 55, 65, 75))

ci_pred <- predict(model_slr, newdata = new_ages, interval = "confidence") %>%
  as.data.frame() %>%
  rename(Fitted = fit, CI_Lower = lwr, CI_Upper = upr)

pi_pred <- predict(model_slr, newdata = new_ages, interval = "prediction") %>%
  as.data.frame() %>%
  rename(PI_Lower = lwr, PI_Upper = upr) %>%
  select(-fit)

results_table <- bind_cols(new_ages, ci_pred, pi_pred) %>%
  mutate(across(where(is.numeric), ~ round(., 2)))

results_table %>%
  kable(
    caption = "Fitted Values, 95% Confidence Intervals, and Prediction Intervals by Age",
    col.names = c("Age", "Fitted BMI", "CI Lower", "CI Upper", "PI Lower", "PI Upper")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
  add_header_above(c(" " = 2, "95% CI for Mean" = 2, "95% PI for Individual" = 2))
Fitted Values, 95% Confidence Intervals, and Prediction Intervals by Age
95% CI for Mean
95% PI for Individual
Age Fitted BMI CI Lower CI Upper PI Lower PI Upper
25 29.37 28.87 29.88 15.61 43.13
35 29.31 28.92 29.70 15.56 43.06
45 29.25 28.95 29.54 15.50 43.00
55 29.19 28.93 29.44 15.43 42.94
65 29.12 28.84 29.41 15.37 42.87
75 29.06 28.68 29.44 15.31 42.81 
# Generate CI and PI across the full age range
age_grid <- data.frame(age = seq(18, 80, length.out = 200))

ci_band <- predict(model_slr, newdata = age_grid, interval = "confidence") %>%
  as.data.frame() %>%
  bind_cols(age_grid)

pi_band <- predict(model_slr, newdata = age_grid, interval = "prediction") %>%
  as.data.frame() %>%
  bind_cols(age_grid)

p_ci_pi <- ggplot() +
  geom_point(data = brfss_slr_2020, aes(x = age, y = bmi),
             alpha = 0.10, color = "steelblue", size = 1) +
  geom_ribbon(data = pi_band, aes(x = age, ymin = lwr, ymax = upr),
              fill = "lightblue", alpha = 0.3) +
  geom_ribbon(data = ci_band, aes(x = age, ymin = lwr, ymax = upr),
              fill = "steelblue", alpha = 0.4) +
  geom_line(data = ci_band, aes(x = age, y = fit),
            color = "red", linewidth = 1.2) +
  labs(
    title = "Simple Linear Regression: BMI ~ Age",
    subtitle = "Dark band = 95% CI for mean response | Light band = 95% PI for individual observation",
    x = "Age (years)",
    y = "BMI (kg/m²)",
    caption = "BRFSS 2020, n = 3,000"
  ) +
  theme_minimal(base_size = 13)

p_ci_pi
Regression Line with 95% Confidence and Prediction Intervals

Regression Line with 95% Confidence and Prediction Intervals

Key distinction: If you want to estimate the average BMI for all 45-year-olds in the population, use the confidence interval. If you want to predict the BMI of a specific new 45-year-old patient, use the prediction interval.

8. Checking Model Assumptions (Diagnostics)

Fitting a regression model is not enough — we must verify that the LINE assumptions are reasonably met. We do this through residual diagnostics.

8.1 The Four Standard Diagnostic Plots

par(mfrow = c(2, 2))
plot(model_slr, which = 1:4,
     col = adjustcolor("steelblue", 0.4),
     pch = 19, cex = 0.6)
Standard Regression Diagnostic Plots

Standard Regression Diagnostic Plots 

par(mfrow = c(1, 1))

Interpreting each plot:

1. Residuals vs. Fitted: Checks linearity and equal variance. We want a horizontal red line and random scatter with no pattern. A “fan shape” (spread increasing with fitted values) indicates heteroscedasticity.

2. Normal Q-Q Plot: Checks normality of residuals. Points should fall approximately along the 45° reference line. Heavy tails or S-curves suggest non-normality.

3. Scale-Location (Spread-Location): Another check for equal variance (homoscedasticity). The square root of standardized residuals is plotted against fitted values. A flat line indicates constant variance.

4. Residuals vs. Leverage: Identifies influential observations using Cook’s distance. Points in the upper or lower right corner (beyond the dashed lines) have high influence.

8.2 Residuals vs. Predictor

p_resid_x <- ggplot(augmented, aes(x = age, y = .resid)) +
  geom_point(alpha = 0.15, color = "steelblue", size = 1) +
  geom_hline(yintercept = 0, color = "red", linewidth = 1) +
  geom_smooth(method = "loess", color = "orange", se = FALSE, linewidth = 1) +
  labs(
    title = "Residuals vs. Age",
    subtitle = "Should show no pattern — random scatter around zero",
    x = "Age (years)",
    y = "Residuals"
  ) +
  theme_minimal(base_size = 13)

p_resid_x
Residuals vs. Age — Checking Linearity

Residuals vs. Age — Checking Linearity

8.3 Histogram and Q-Q Plot of Residuals

p_hist <- ggplot(augmented, aes(x = .resid)) +
  geom_histogram(aes(y = after_stat(density)), bins = 40,
                 fill = "steelblue", color = "white", alpha = 0.8) +
  geom_density(color = "red", linewidth = 1) +
  stat_function(fun = dnorm,
                args = list(mean = mean(augmented$.resid),
                            sd = sd(augmented$.resid)),
                color = "black", linetype = "dashed", linewidth = 1) +
  labs(
    title = "Distribution of Residuals",
    subtitle = "Red = kernel density | Black dashed = normal distribution",
    x = "Residuals",
    y = "Density"
  ) +
  theme_minimal(base_size = 13)

p_hist
Distribution of Residuals

Distribution of Residuals 

# ggplot version of QQ plot
p_qq <- ggplot(augmented, aes(sample = .resid)) +
  stat_qq(color = "steelblue", alpha = 0.3, size = 1) +
  stat_qq_line(color = "red", linewidth = 1) +
  labs(
    title = "Normal Q-Q Plot of Residuals",
    subtitle = "Points should lie on the red line if residuals are normally distributed",
    x = "Theoretical Quantiles",
    y = "Sample Quantiles"
  ) +
  theme_minimal(base_size = 13)

p_qq
Normal Q-Q Plot of Residuals

Normal Q-Q Plot of Residuals

8.4 Identifying Influential Observations

# Cook's distance
augmented <- augmented %>%
  mutate(
    obs_num = row_number(),
    cooks_d = cooks.distance(model_slr),
    influential = ifelse(cooks_d > 4 / n, "Potentially influential", "Not influential")
  )

n_influential <- sum(augmented$cooks_d > 4 / n)

p_cooks <- ggplot(augmented, aes(x = obs_num, y = cooks_d, color = influential)) +
  geom_point(alpha = 0.6, size = 1.2) +
  geom_hline(yintercept = 4 / n, linetype = "dashed",
             color = "red", linewidth = 1) +
  scale_color_manual(values = c("Potentially influential" = "tomato",
                                "Not influential" = "steelblue")) +
  labs(
    title = "Cook's Distance",
    subtitle = paste0("Dashed line = 4/n threshold | ",
                      n_influential, " potentially influential observations"),
    x = "Observation Number",
    y = "Cook's Distance",
    color = ""
  ) +
  theme_minimal(base_size = 13) +
  theme(legend.position = "top")

p_cooks
Cook's Distance: Identifying Influential Observations

Cook’s Distance: Identifying Influential Observations

9. A Second Example: Sleep and BMI

To reinforce the concepts, let’s fit a second SLR model examining the association between hours of sleep and BMI.

p_sleep <- ggplot(brfss_slr_2020, aes(x = sleep_hrs, y = bmi)) +
  geom_jitter(alpha = 0.15, color = "purple", width = 0.15, height = 0) +
  geom_smooth(method = "lm", color = "darkred", linewidth = 1.2, se = TRUE) +
  labs(
    title = "BMI vs. Nightly Sleep Hours (BRFSS 2020)",
    x = "Average Hours of Sleep per Night",
    y = "BMI (kg/m²)"
  ) +
  theme_minimal(base_size = 13)

p_sleep
BMI vs. Sleep Hours

BMI vs. Sleep Hours 

model_sleep <- lm(bmi ~ sleep_hrs, data = brfss_slr_2020)

tidy(model_sleep, conf.int = TRUE) %>%
  mutate(across(where(is.numeric), ~ round(., 4))) %>%
  kable(
    caption = "SLR: BMI ~ Hours of Sleep per Night",
    col.names = c("Term", "Estimate", "Std. Error", "t-statistic",
                  "p-value", "95% CI Lower", "95% CI Upper")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
SLR: BMI ~ Hours of Sleep per Night
Term Estimate Std. Error t-statistic p-value 95% CI Lower 95% CI Upper
(Intercept) 30.7419 0.534 57.5683 0.0000 29.6948 31.7890
sleep_hrs -0.2256 0.075 -3.0087 0.0026 -0.3726 -0.0786 
b1_sleep <- coef(model_sleep)["sleep_hrs"]
r2_sleep <- summary(model_sleep)$r.squared

tibble(
  Metric = c("Slope (b₁)", "R²"),
  Value  = c(round(b1_sleep, 4), round(r2_sleep, 4))
) %>%
  kable(caption = "Sleep Model Key Statistics") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Sleep Model Key Statistics
Metric Value
Slope (b₁) -0.2256
R² 0.0030

Interpretation: Each additional hour of sleep per night is associated with a change of -0.2256 kg/m² in BMI, on average. The direction of this association is negative (more sleep → lower BMI). The model explains 0.3% of variability in BMI. While statistically significant, the effect size is modest, underscoring the multifactorial nature of BMI.

par(mfrow = c(2, 2))
plot(model_sleep, which = 1:4,
     col = adjustcolor("purple", 0.4), pch = 19, cex = 0.6)

par(mfrow = c(1, 1))

10. Does BMI Really Decrease with Age? Adding a Quadratic Term

Our linear model estimated a negative slope for age: older adults have, on average, slightly lower BMI. But is that the full story? Cross-sectional data can show a decline at older ages due to survivorship bias — people with very high BMI may die before reaching old age, leaving a healthier-looking older sample. There may also be a genuine nonlinear pattern (BMI rises through middle age, then declines in later life).

We can test this by including an age² term in the model:

\[\widehat{\text{BMI}} = b_0 + b_1 \cdot \text{Age} + b_2 \cdot \text{Age}^2\]

This is still a linear regression model (linear in the coefficients), even though it is nonlinear in the predictor. It allows the slope to change across the range of age.

# Add age-squared term
brfss_slr_2020 <- brfss_slr_2020 %>%
  mutate(age2 = age^2)

# Fit quadratic model
model_quad <- lm(bmi ~ age + age2, data = brfss_slr_2020)

tidy(model_quad, conf.int = TRUE) %>%
  mutate(across(where(is.numeric), ~ round(., 5))) %>%
  kable(
    caption = "Quadratic Model: BMI ~ Age + Age²",
    col.names = c("Term", "Estimate", "Std. Error", "t-statistic",
                  "p-value", "95% CI Lower", "95% CI Upper")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Quadratic Model: BMI ~ Age + Age²
Term Estimate Std. Error t-statistic p-value 95% CI Lower 95% CI Upper
(Intercept) 18.54178 1.08095 17.15329 0 16.42230 20.66125
age 0.47435 0.04418 10.73772 0 0.38773 0.56096
age2 -0.00464 0.00042 -11.02651 0 -0.00546 -0.00381 
# Compare linear vs. quadratic model
tibble(
  Model       = c("Linear: BMI ~ Age", "Quadratic: BMI ~ Age + Age²"),
  R_squared   = c(
    round(summary(model_slr)$r.squared, 4),
    round(summary(model_quad)$r.squared, 4)
  ),
  Adj_R2      = c(
    round(summary(model_slr)$adj.r.squared, 4),
    round(summary(model_quad)$adj.r.squared, 4)
  ),
  AIC         = c(round(AIC(model_slr), 1), round(AIC(model_quad), 1))
) %>%
  kable(
    caption = "Model Comparison: Linear vs. Quadratic",
    col.names = c("Model", "R²", "Adj. R²", "AIC")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
  row_spec(which.min(c(AIC(model_slr), AIC(model_quad))),
           bold = TRUE, background = "#d4edda")
Model Comparison: Linear vs. Quadratic
Model R² Adj. R² AIC
Linear: BMI ~ Age 0.0002 -0.0001 20203.2
Quadratic: BMI ~ Age + Age² 0.0392 0.0386 20085.9 
# Generate predicted values from both models
age_seq <- data.frame(age = seq(18, 80, length.out = 300)) %>%
  mutate(age2 = age^2)

pred_linear <- predict(model_slr, newdata = age_seq)
pred_quad   <- predict(model_quad, newdata = age_seq)

pred_df <- age_seq %>%
  mutate(
    linear    = pred_linear,
    quadratic = pred_quad
  ) %>%
  pivot_longer(cols = c(linear, quadratic),
               names_to = "Model", values_to = "Predicted_BMI")

ggplot() +
  geom_point(data = brfss_slr_2020, aes(x = age, y = bmi),
             alpha = 0.10, color = "steelblue", size = 1) +
  geom_line(data = pred_df, aes(x = age, y = Predicted_BMI, color = Model),
            linewidth = 1.3) +
  scale_color_manual(
    values = c("linear" = "red", "quadratic" = "darkorange"),
    labels = c("linear" = "Linear fit", "quadratic" = "Quadratic fit (Age + Age²)")
  ) +
  labs(
    title = "BMI vs. Age: Linear vs. Quadratic Model",
    subtitle = "Does BMI rise then fall with age, or decline monotonically?",
    x = "Age (years)",
    y = "BMI (kg/m²)",
    color = "Model"
  ) +
  theme_minimal(base_size = 13) +
  theme(legend.position = "top")
Linear vs. Quadratic Fit: BMI ~ Age

Linear vs. Quadratic Fit: BMI ~ Age

Interpretation: If the coefficient on Age² is negative and statistically significant, the fitted curve is an inverted-U — BMI peaks at some middle age and declines thereafter. Extract the peak using \(\text{Age}^* = -b_1 / (2 b_2)\). A positive Age² coefficient would indicate a U-shape (BMI lowest in middle age).

b1_q <- coef(model_quad)["age"]
b2_q <- coef(model_quad)["age2"]

peak_age <- -b1_q / (2 * b2_q)

tibble(
  Quantity = c("b₁ (Age)", "b₂ (Age²)", "Peak / Trough Age (-b₁ / 2b₂)"),
  Value    = c(round(b1_q, 5), round(b2_q, 6), round(peak_age, 1))
) %>%
  kable(caption = "Quadratic Model Coefficients and Implied Turning Point") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Quadratic Model Coefficients and Implied Turning Point
Quantity Value
b₁ (Age) 0.474350
b₂ (Age²) -0.004635
Peak / Trough Age (-b₁ / 2b₂) 51.200000

Caution on interpretation: Even if the quadratic model fits better statistically, be cautious about causal interpretation. The cross-sectional pattern reflects cohort differences in BMI trajectories, not necessarily the aging process within any individual. Survivorship bias (heavier individuals dying earlier) can make the quadratic term appear significant in cross-sectional data.

11. Summary of Key Formulas

Quantity Formula
Slope \(b_1 = S_{XY} / S_{XX}\)
Intercept \(b_0 = \bar{Y} - b_1 \bar{X}\)
SSTotal \(\sum(Y_i - \bar{Y})^2\)
SSRegression \(\sum(\hat{Y}_i - \bar{Y})^2\)
SSResidual \(\sum(Y_i - \hat{Y}_i)^2\)
MSE \(SS_{Residual} / (n-2)\)
\(R^2\) \(SS_{Reg} / SS_{Total}\)
\(SE(b_1)\) \(\hat{\sigma}/\sqrt{S_{XX}}\)
t-statistic \(b_1 / SE(b_1)\)
95% CI for \(\beta_1\) \(b_1 \pm t_{n-2, 0.025} \cdot SE(b_1)\)

Part 2: Lab Activity

Overview

In this lab, you will apply Simple Linear Regression to the BRFSS 2020 dataset using a different outcome variable: number of days of poor physical health in the past 30 days (phys_days). You will model it as a continuous outcome predicted by BMI.

Research Question: Is BMI associated with the number of days of poor physical health among U.S. adults?

Setup Instructions

Use the code below to load the data. The dataset is the same one used in the lecture — you only need to load it once.

# Load packages
library(tidyverse)
library(haven)
library(here)
library(knitr)
library(kableExtra)
library(broom)

###Load brfss_subset_2020.rds

brfss_slr_2020 <- read_rds ("~/Downloads/HEPI553/brfss_slr_2020.rds")

Task 1: Explore the Variables (15 points)

# (a) Create a summary table of phys_days and bmi
summary_table <- data.frame(
  Variable = c("phys_days", "bmi"),
  Mean = c(mean(brfss_slr_2020$phys_days),
           mean(brfss_slr_2020$bmi)),
  SD = c(sd(brfss_slr_2020$phys_days),
         sd(brfss_slr_2020$bmi)),
  Median = c(median(brfss_slr_2020$phys_days),
             median(brfss_slr_2020$bmi)),
  Min = c(min(brfss_slr_2020$phys_days),
          min(brfss_slr_2020$bmi)),
  Max = c(max(brfss_slr_2020$phys_days),
          max(brfss_slr_2020$bmi))
)

summary_table
##    Variable     Mean        SD Median   Min  Max
## 1 phys_days 11.65800 11.160073   6.00  1.00 30.0
## 2       bmi 29.18194  7.011534  27.89 14.63 59.6
# (b) Create a histogram of phys_days — describe the distribution

hist(brfss_slr_2020$phys_days, main = "Histogram:phys_days", 
     xlab = "Value", col = "lightblue")

# (c) Create a scatter plot of phys_days (Y) vs bmi (X)
p_scatter <- ggplot(brfss_slr_2020, aes(x = bmi, y = phys_days)) +
  geom_point(alpha = 0.15, color = "steelblue", size = 1.2) +
  geom_smooth(method = "lm", color = "red", linewidth = 1.2, se = TRUE) +
  geom_smooth(method = "loess", color = "blue", linewidth = 1,
              linetype = "dashed", se = FALSE) +
  labs(
    title = "BMI vs. Poor Physical Health days (BRFSS 2020)",
    subtitle = "Red = Linear fit | Orange dashed = LOESS smoother",
    x = "BMI (kg/m²)",
    y = "Poor Physical Health Days"
  ) +
  theme_minimal(base_size = 13)

ggplotly(p_scatter)

Questions:

  1. What is the mean and standard deviation of phys_days? Of bmi? What do you notice about the distribution of phys_days? The mean of phys_days is 11.66 and the standard deviation of phys_days is 11.16. The distribution of phys_days indicates it is right skewed.

  2. Based on the scatter plot, does the relationship between BMI and poor physical health days appear to be linear? Are there any obvious outliers? Based on the scatter plot, there is a relationship between BMI and phys_days that shows a positive linear line as the BMI increases. Yes, there are obvious outliers. Specifically when the BMI increases and a little above midpoint for phys_days.

Task 2: Fit and Interpret the SLR Model (20 points)

# (a) Fit the SLR model: phys_days ~ bmi
model_slr1 <- lm(phys_days ~ bmi, data = brfss_slr_2020)

# Summary output
summary(model_slr1)
## 
## Call:
## lm(formula = phys_days ~ bmi, data = brfss_slr_2020)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -14.808  -9.160  -5.623   8.943  20.453 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  7.42285    0.86881   8.544  < 2e-16 ***
## bmi          0.14513    0.02895   5.013 5.66e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 11.12 on 2998 degrees of freedom
## Multiple R-squared:  0.008314,   Adjusted R-squared:  0.007983 
## F-statistic: 25.13 on 1 and 2998 DF,  p-value: 5.659e-07
# (b) Display a tidy coefficient table with 95% CIs
tidy(model_slr, conf.int = TRUE) %>%
  mutate(across(where(is.numeric), ~ round(., 4))) %>%
  kable(
    caption = "Simple Linear Regression: phys_days ~ bmi (BRFSS 2020)",
    col.names = c("Term", "Estimate", "Std. Error", "t-statistic",
                  "p-value", "95% CI Lower", "95% CI Upper"),
    align = "lrrrrrrr"
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
  row_spec(0, bold = TRUE)
Simple Linear Regression: phys_days ~ bmi (BRFSS 2020)
Term Estimate Std. Error t-statistic p-value 95% CI Lower 95% CI Upper
(Intercept) 29.5282 0.4275 69.0708 0.000 28.6900 30.3665
age -0.0062 0.0073 -0.8490 0.396 -0.0206 0.0082 
# (c) Extract and report: slope, intercept, t-statistic, p-value
b0 <- round(coef(model_slr)[1], 3)
b1 <- round(coef(model_slr)[2], 4)

Questions:

  1. Write the fitted regression equation in the form \(\hat{Y} = b_0 + b_1 X\). \(\hat{phys_days} = 7.42 + 0.145{BMI}\)

  2. Interpret the slope (\(b_1\)) in context — what does it mean in plain English? The slope is 0.145. For each 1-unit increase in BMI is estimated to increase by 0.145 days, on average, holding all else constant (though there are no other variables in this simple model).

  3. Is the intercept (\(b_0\)) interpretable in this context? Why or why not? The intercept is 7.42, represents poor physical health days when the BMI is zero. Which means this is not interpretable.

  4. Is the association statistically significant at \(\alpha = 0.05\)? State the null hypothesis, test statistic, and p-value. Yes, the association is statistically significant. H0= poor physical health days in not associated with BMI H1= poor physical health days is associated with BMI t-statistic= 5.013 p-value= 5.659e-07

Task 3: ANOVA Decomposition and R² (15 points)

# (a) Display the ANOVA table
anova_slr <- anova(model_slr)

anova_slr %>%
  kable(
    caption = "ANOVA Table: Phys_days ~ BMI ",
    digits = 3,
    col.names = c("Source", "Df", "Sum Sq", "Mean Sq", "F value", "Pr(>F)")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
ANOVA Table: Phys_days ~ BMI
Source Df Sum Sq Mean Sq F value Pr(>F)
age 1 35.438 35.438 0.721 0.396
Residuals 2998 147400.214 49.166 NA NA 
# (b) Compute and report SSTotal, SSRegression, and SSResidual
SSRegression <- anova_slr$`Sum Sq`[1]
SSResidual   <- anova_slr$`Sum Sq`[2]
SSTotal      <- SSRegression + SSResidual

tibble(
  Quantity = c("SS Total", "SS Regression", "SS Residual"),
  Value = round(c(SSTotal, SSRegression, SSResidual), 2)
)
## # A tibble: 3 × 2
##   Quantity         Value
##   <chr>            <dbl>
## 1 SS Total      147436. 
## 2 SS Regression     35.4
## 3 SS Residual   147400.
# (c) Compute R² two ways: from the model object and from the SS decomposition
r_sq <- summary(model_slr)$r.squared
adj_r_sq <- summary(model_slr)$adj.r.squared

tibble(
  Metric = c("R²", "Adjusted R²", "Variance Explained"),
  Value  = c(
    round(r_sq, 4),
    round(adj_r_sq, 4),
    paste0(round(r_sq * 100, 2), "%")
  )
) %>%
  kable(caption = "R² and Adjusted R²") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
R² and Adjusted R²
Metric Value
R² 2e-04
Adjusted R² -1e-04
Variance Explained 0.02% 
r_pearson <- cor(brfss_slr_2020$bmi, brfss_slr_2020$phys_days)
tibble(
  Quantity   = c("Pearson r", "r² (from Pearson)", "R² (from model)", "r² = R²?"),
  Value      = c(
    round(r_pearson, 4),
    round(r_pearson^2, 4),
    round(r_sq, 4),
    as.character(round(r_pearson^2, 4) == round(r_sq, 4))
  )
) %>%
  kable(caption = "Pearson r vs. R² from Model") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Pearson r vs. R² from Model
Quantity Value
Pearson r 0.0912
r² (from Pearson) 0.0083
R² (from model) 2e-04
r² = R²? FALSE

Questions:

  1. Fill in the ANOVA table components: \(SS_{147435.65}\), \(SS_{35.44}\), \(SS_{147400.21}\), \(1\), and \(0.721\).

  2. What is the \(R^2\) value? Interpret it in plain English. The \(R^2\) = 2e-04. This means that the model fails to provide variation for phys_days.

  3. What does this \(R^2\) tell you about how well BMI alone explains variation in poor physical health days? What might explain the remaining variation? With the percentage being extremely low, it does not give an accurate representation of the variance in the brfss dataset. The remaining variation can represent outliers for the individuals in the data set that are in the same group but might have differences.

Task 4: Confidence and Prediction Intervals (20 points)

# (a) Calculate the fitted BMI value and 95% CI for a person with BMI = 25
new_bmi <- data.frame(bmi = 25)

ci_pred <- predict(model_slr1, newdata = new_bmi, interval = "confidence") %>%
  as.data.frame() %>%
  rename(Fitted = fit, CI_Lower = lwr, CI_Upper = upr)

# (b) Calculate the 95% prediction interval for a person with BMI = 25
pi_pred <- predict(model_slr1, newdata = new_bmi, interval = "prediction") %>%
  as.data.frame() %>%
  rename(PI_Lower = lwr, PI_Upper = upr) %>%
  select(-fit)

results_table <- bind_cols(ci_pred, pi_pred) %>%
  mutate(across(where(is.numeric), ~ round(., 2)))

# (c) Plot the regression line with both the CI band and PI band
results_table %>%
  kable(
    caption = "Fitted BMI Value, 95% Confidence Interval, and 95% Prediction Interval for BMI = 25",
    col.names = c("Fitted BMI", "CI Lower", "CI Upper", "PI Lower", "PI Upper")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Fitted BMI Value, 95% Confidence Interval, and 95% Prediction Interval for BMI = 25
Fitted BMI CI Lower CI Upper PI Lower PI Upper
11.05 10.59 11.51 -10.75 32.85

Questions:

  1. For someone with a BMI of 25, what is the estimated mean number of poor physical health days? What is the 95% confidence interval for this mean? For someone with a BMI of 25, the estimated mean number of poor physical health days are 11.06. The 95% CI is 10.59-11.51 which indicates a range of plausible values for the population mean of BMI at a specific value for poor physical health days.

  2. If a specific new person has a BMI of 25, what is the 95% prediction interval for their number of poor physical health days? The 95% prediction interval for their number of poor physical health days is -10.75-32.85.

  3. Explain in your own words why the prediction interval is wider than the confidence interval. When would you use each one in practice? The prediction interval is wider than the CI because it measures for more random and multiple differences. You would use CI when looking for a precise value such as a mean for a population parameter. You would use a prediction interval for a broader range that involves specific observations that leads to higher variability statistics.

Task 5: Residual Diagnostics (20 points)

# (a) Produce the four standard diagnostic plots (use par(mfrow = c(2,2)) and plot())
par(mfrow = c(2, 2))
plot(model_slr, which = 1:4,
     col = adjustcolor("steelblue", 0.4),
     pch = 19, cex = 0.6)

par(mfrow = c(1, 1))

# (b) Create a residuals vs. fitted plot using ggplot
ggplot(augmented, aes(x = .fitted, y = .resid)) +
  geom_point(alpha = 0.5, color = "steelblue") +
  geom_hline(yintercept = 0, linetype = "dashed", color = "red") +
  labs(
    title = "Residuals vs Fitted Values",
    x = "Fitted Values (Ŷ)",
    y = "Residuals (e = Y − Ŷ)"
  ) +
  theme_minimal(base_size = 13)

# (c) Create a normal Q-Q plot of residuals using ggplot
p_qq <- ggplot(augmented, aes(sample = .resid)) +
  stat_qq(color = "steelblue", alpha = 0.3, size = 1) +
  stat_qq_line(color = "red", linewidth = 1) +
  labs(
    title = "Normal Q-Q Plot of Residuals",
    subtitle = "Points should lie on the red line if residuals are normally distributed",
    x = "Theoretical Quantiles",
    y = "Sample Quantiles"
  ) +
  theme_minimal(base_size = 13)

p_qq

# (d) Create a Cook's distance plot
augmented <- augmented %>%
  mutate(
    obs_num = row_number(),
    cooks_d = cooks.distance(model_slr),
    influential = ifelse(cooks_d > 4 / n, "Potentially influential", "Not influential")
  )

n_influential <- sum(augmented$cooks_d > 4 / n)

p_cooks <- ggplot(augmented, aes(x = obs_num, y = cooks_d, color = influential)) +
  geom_point(alpha = 0.6, size = 1.2) +
  geom_hline(yintercept = 4 / n, linetype = "dashed",
             color = "red", linewidth = 1) +
  scale_color_manual(values = c("Potentially influential" = "tomato",
                                "Not influential" = "steelblue")) +
  labs(
    title = "Cook's Distance",
    subtitle = paste0("Dashed line = 4/n threshold | ",
                      n_influential, " potentially influential observations"),
    x = "Observation Number",
    y = "Cook's Distance",
    color = ""
  ) +
  theme_minimal(base_size = 13) +
  theme(legend.position = "top")

p_cooks

Questions:

  1. Examine the Residuals vs. Fitted plot. Is there evidence of nonlinearity or heteroscedasticity? Describe what you see. In the Residual vs. Fitted plot, there is is evidence of nonlinearity. When looking at the dots on the plot, they slightly curve downwards.

  2. Examine the Q-Q plot. Are the residuals approximately normal? What do departures from normality in this context suggest about the distribution of phys_days? In the Q-Q plot, the residuals are not normal since they do not exactly follow the red line. The departures from normality in this context suggest that the distribution of phys_days is slowly upward, in a positive skew. This shows that there are someone people who do not indicate having many experiences with poor physical health days.

  3. Are there any influential observations (Cook’s D > 4/n)? How many? What would you do about them? Yes, there are 148 potential influential observations. It might be best to run through if there are any errors in the data.

  4. Overall, do the LINE assumptions appear to be met? Which assumption(s) may be most problematic for this model, and why? (Hint: think about the nature of the outcome variable.) Overall, the line assumptions appear to be met. The assumption that might be most problematic for this model is the Residuals vs.Leverage (Cooks Distance) because having a high amount of influential observations can indicate potential errors when collecting the data for individuals with poor physical health days. —

Task 6: Testing a Different Predictor (10 points)

Now fit a second SLR model using age as the predictor of phys_days instead of BMI.

# (a) Fit SLR: phys_days ~ age
model_slr2 <- lm(phys_days ~ age, data = brfss_slr_2020)

# Summary output
summary(model_slr2)
## 
## Call:
## lm(formula = phys_days ~ age, data = brfss_slr_2020)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -13.872  -8.803  -4.733   9.460  23.267 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  4.37023    0.66608   6.561 6.27e-11 ***
## age          0.13127    0.01145  11.467  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 10.92 on 2998 degrees of freedom
## Multiple R-squared:  0.04202,    Adjusted R-squared:  0.0417 
## F-statistic: 131.5 on 1 and 2998 DF,  p-value: < 2.2e-16
# (b) Display results and compare to the BMI model
tidy(model_slr2, conf.int = TRUE) %>%
  mutate(across(where(is.numeric), ~ round(., 4))) %>%
  kable(
    caption = "Simple Linear Regression: Poor Physical Health Days ~ Age (BRFSS 2020)",
    col.names = c("Term", "Estimate", "Std. Error", "t-statistic",
                  "p-value", "95% CI Lower", "95% CI Upper"),
    align = "lrrrrrrr"
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
  row_spec(0, bold = TRUE)
Simple Linear Regression: Poor Physical Health Days ~ Age (BRFSS 2020)
Term Estimate Std. Error t-statistic p-value 95% CI Lower 95% CI Upper
(Intercept) 4.3702 0.6661 6.5611 0 3.0642 5.6762
age 0.1313 0.0114 11.4675 0 0.1088 0.1537 
# (c) Which predictor has the stronger association? Compare R² values.
r_sq <- summary(model_slr2)$r.squared
adj_r_sq <- summary(model_slr2)$adj.r.squared

tibble(
  Metric = c("R²", "Adjusted R²", "Variance Explained"),
  Value  = c(
    round(r_sq, 4),
    round(adj_r_sq, 4),
    paste0(round(r_sq * 100, 2), "%")
  )
) %>%
  kable(caption = "R² and Adjusted R²") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
R² and Adjusted R²
Metric Value
R² 0.042
Adjusted R² 0.0417
Variance Explained 4.2% 
r_pearson <- cor(brfss_slr_2020$age, brfss_slr_2020$phys_days)
tibble(
  Quantity   = c("Pearson r", "r² (from Pearson)", "R² (from model)", "r² = R²?"),
  Value      = c(
    round(r_pearson, 4),
    round(r_pearson^2, 4),
    round(r_sq, 4),
    as.character(round(r_pearson^2, 4) == round(r_sq, 4))
  )
) %>%
  kable(caption = "Pearson r vs. R² from Model") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Pearson r vs. R² from Model
Quantity Value
Pearson r 0.205
r² (from Pearson) 0.042
R² (from model) 0.042
r² = R²? TRUE

Questions:

  1. How does the association between age and poor physical health days compare to the BMI association in terms of direction, magnitude, and statistical significance? When comparing the association between age and poor physical health days to the BMI association, age has higher spread of statistics. We see the direction and magnitude is higher in BMI than age but the p-values for both are the same. We see that the variance is higher in age compared to BMI. This makes sense because people who have a higher age might experience more poor physical health days compared to people with a high BMI.

  2. Compare the \(R^2\) values of the two models. Which predictor explains more variability in phys_days? The predictor that explains more variability in phys_days is age.

  3. Based on these two simple models, what is your overall conclusion about predictors of poor physical health days? What are the limitations of using simple linear regression for this outcome? Based on the two simple models, my overall conclusion about BMI and age for poor physical health days is that both BMI and Age are strong predictors for people to experience more poor physical health days based on specific statistical methods used in the lab. The limitations for using simple linear regression for poor physical health days is that when analyzing someones health, using the plots may not accurately represent a persons exact poor physical health days.

Submission Instructions

Submit your completed .Rmd file and the RPubs link to your knitted HTML document.

Your .Rmd must knit without errors. Make sure all code chunks produce visible output and all questions are answered in complete sentences below each code chunk.

Due: Before the next class session.

Session Info

sessionInfo()
## R version 4.5.2 (2025-10-31)
## Platform: aarch64-apple-darwin20
## Running under: macOS Tahoe 26.3
## 
## Matrix products: default
## BLAS:   /System/Library/Frameworks/Accelerate.framework/Versions/A/Frameworks/vecLib.framework/Versions/A/libBLAS.dylib 
## LAPACK: /Library/Frameworks/R.framework/Versions/4.5-arm64/Resources/lib/libRlapack.dylib;  LAPACK version 3.12.1
## 
## locale:
## [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
## 
## time zone: America/New_York
## tzcode source: internal
## 
## attached base packages:
## [1] stats     graphics  grDevices utils     datasets  methods   base     
## 
## other attached packages:
##  [1] gtsummary_2.5.0  ggeffects_2.3.2  broom_1.0.11     plotly_4.12.0   
##  [5] kableExtra_1.4.0 knitr_1.51       here_1.0.2       haven_2.5.5     
##  [9] lubridate_1.9.4  forcats_1.0.1    stringr_1.6.0    dplyr_1.2.0     
## [13] purrr_1.2.1      readr_2.1.6      tidyr_1.3.2      tibble_3.3.1    
## [17] ggplot2_4.0.1    tidyverse_2.0.0 
## 
## loaded via a namespace (and not attached):
##  [1] gtable_0.3.6       xfun_0.56          bslib_0.9.0        htmlwidgets_1.6.4 
##  [5] insight_1.4.6      lattice_0.22-7     tzdb_0.5.0         crosstalk_1.2.2   
##  [9] vctrs_0.7.1        tools_4.5.2        generics_0.1.4     pkgconfig_2.0.3   
## [13] Matrix_1.7-4       data.table_1.18.0  RColorBrewer_1.1-3 S7_0.2.1          
## [17] gt_1.3.0           lifecycle_1.0.5    compiler_4.5.2     farver_2.1.2      
## [21] textshaping_1.0.4  litedown_0.9       htmltools_0.5.9    sass_0.4.10       
## [25] yaml_2.3.12        lazyeval_0.2.2     pillar_1.11.1      jquerylib_0.1.4   
## [29] cachem_1.1.0       nlme_3.1-168       commonmark_2.0.0   tidyselect_1.2.1  
## [33] digest_0.6.39      stringi_1.8.7      labeling_0.4.3     splines_4.5.2     
## [37] rprojroot_2.1.1    fastmap_1.2.0      grid_4.5.2         cli_3.6.5         
## [41] magrittr_2.0.4     cards_0.7.1        utf8_1.2.6         withr_3.0.2       
## [45] scales_1.4.0       backports_1.5.0    timechange_0.3.0   rmarkdown_2.30    
## [49] httr_1.4.7         otel_0.2.0         hms_1.1.4          evaluate_1.0.5    
## [53] viridisLite_0.4.2  mgcv_1.9-3         markdown_2.0       rlang_1.1.7       
## [57] glue_1.8.0         xml2_1.5.2         svglite_2.2.2      rstudioapi_0.18.0 
## [61] jsonlite_2.0.0     R6_2.6.1           systemfonts_1.3.1  fs_1.6.6