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.

brfss_slr<-read_rds('/Users/MY789914/OneDrive - University at Albany - SUNY/Desktop/Stat 553 (R)/In-Class R Lab Activities/brfss_slr_2020.rds')

Descriptive Statistics

brfss_slr %>%
  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 %>%
  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, 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)

# Summary output
summary(model_slr)

## 
## Call:
## lm(formula = bmi ~ age, data = brfss_slr)
## 
## 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

The fitted value (Y^) and the observed BMI (Y) are different because a person’s age does not perfectly determine their exact BMI.

Fitted Value (Y^): This is the predicted average BMI for anyone of a specific age, based on the trend line calculated by the model. For example, the model might calculate that the average 67-year-old has a BMI of 29.110.
Observed BMI (Y): This is the actual, real-world BMI of a specific individual in the dataset.
The Residual/Error (e=Y−Y^): The difference between the two is the residual or error term. The error term (εi) represents the “random deviation of Yi from the regression line.”

# 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

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:

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

In statistics, degrees of freedom refer to the number of independent pieces of information that are free to vary when calculating an estimate. You can think of it as the amount of “data budget” you have. Every time you estimate a parameter (like a mean or a slope), you spend one degree of freedom.

Why it’s \(n - 1\): You have \(n\) total observations. However, to calculate this variance, you first had to calculate the sample mean. Because the sum of deviations from the mean must always equal zero, knowing \(n-1\) of the deviations automatically tells you the last one. You “spent” 1 degree of freedom calculating the mean, leaving you with \(n - 1\).

Why it’s \(1\): In Simple Linear Regression, you are using exactly one predictor variable (e.g., Age) to explain the outcome (e.g., BMI). Because you are only estimating one slope parameter (\(\beta_1\)) to capture this relationship, the regression model has 1 degree of freedom. (Note: In multiple regression, this would be equal to \(k\), the number of predictor variables).

Why it’s \(n - 2\): To calculate the predicted values (\(\hat{Y}\)) and find the residuals, your model had to estimate two parameters from the data: the intercept (\(\beta_0\)) and the slope (\(\beta_1\)). Since you “spent” 2 pieces of information to create the regression line, you are left with \(n - 2\) degrees of freedom for the errors.

# 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)
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$age, brfss_slr$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 (\(F = t^2\)).

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

The Numerator: This represents the amount of variability in the data that your model successfully explains.
The Denominator: This represents the “leftover” or unexplained variability (the errors).

When your model does a good job of predicting the outcome, the explained variance goes up, and the unexplained variance goes down. This makes the resulting F-statistic larger. What a higher F-statistic means for your results:

Lower p-value: A larger F-statistic pushes the p-value closer to zero.
Statistical Significance: If the p-value drops below your alpha level (usually 0.05), you can reject the null hypothesis. It gives you confidence that your overall model actually has some predictive power, rather than just capturing random noise.

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, 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

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

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

The Residuals vs. Predictor plot is a visual diagnostic tool used to check if your data meets the core assumptions of Simple Linear Regression—specifically Linearity and Equal Variance (Homoscedasticity).

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

Y-axis (Residuals): The errors from your model. This is how far off each person’s actual BMI was from the model’s predicted BMI.
X-axis (Age): Your predictor variable.

(Note: In Simple Linear Regression with only one predictor, this plot looks identical in shape to the “Residuals vs. Fitted” plot).

What do the lines mean?

The Red Line: This is a flat, horizontal line at exactly zero. If the model predicted a person’s BMI perfectly, their point would land exactly on this line.
The Orange Line (LOESS Smoother): This is a moving average of the residuals. It helps your eye track the overall trend of the errors across different ages.

Interpretation:

The orange line stays fairly close to the red line, though there is a very slight curve downwards at the extreme ends of the age range. The vertical spread of the points looks relatively consistent across the ages. Overall, it doesn’t show any severe violations of linearity or equal variance, though there is a lot of random scatter (which aligns with the very low \(R^2\) value we saw earlier).

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

# 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

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

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, 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

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

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 <- brfss_slr %>%
  mutate(age2 = age^2)

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

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	95% CI Lower	95% CI Upper
(Intercept)	18.54178	1.08095	17.15329	16.42230	20.66125
age	0.47435	0.04418	10.73772	0.38773	0.56096
age2	-0.00464	0.00042	-11.02651	-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, 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

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.

Task 1: Explore the Variables (15 points)

# (a) Create a summary table of phys_days and bmi
# Summary statistics
summary(brfss_slr[, c("phys_days", "bmi")])

##    phys_days          bmi       
##  Min.   : 1.00   Min.   :14.63  
##  1st Qu.: 2.00   1st Qu.:24.32  
##  Median : 6.00   Median :27.89  
##  Mean   :11.66   Mean   :29.18  
##  3rd Qu.:20.00   3rd Qu.:32.89  
##  Max.   :30.00   Max.   :59.60

brfss_slr %>%
  summarise(
    n = n(),
    
     phys_days_mean = mean(phys_days),
    phys_days_sd   = sd(phys_days),
    phys_days_min  = min(phys_days),
    phys_days_q1   = quantile(phys_days, 0.25),
    phys_days_med  = median(phys_days),
    phys_days_q3   = quantile(phys_days, 0.75),
    phys_days_max  = max(phys_days),
    
    bmi_mean = mean(bmi),
    bmi_sd   = sd(bmi),
    bmi_min  = min(bmi),
    bmi_q1   = quantile(bmi, 0.25),
    bmi_med  = median(bmi),
    bmi_q3   = quantile(bmi, 0.75),
    bmi_max  = max(bmi)
  )

## # A tibble: 1 × 15
##       n phys_days_mean phys_days_sd phys_days_min phys_days_q1 phys_days_med
##   <int>          <dbl>        <dbl>         <dbl>        <dbl>         <dbl>
## 1  3000           11.7         11.2             1            2             6
## # ℹ 9 more variables: phys_days_q3 <dbl>, phys_days_max <dbl>, bmi_mean <dbl>,
## #   bmi_sd <dbl>, bmi_min <dbl>, bmi_q1 <dbl>, bmi_med <dbl>, bmi_q3 <dbl>,
## #   bmi_max <dbl>

# (b) Create a histogram of phys_days — describe the distribution
library(plotly)

p_hist <- ggplot(brfss_slr, aes(x = phys_days)) + 
  geom_histogram(binwidth = 1, fill = "steelblue", color = "black") +
  labs(
    title = "Distribution of Poor Physical Health Days (BRFSS 2020)",
    subtitle = "Number of days with poor physical health in past 30 days",
    x = "Poor physical health days",
    y = "Frequency"
  ) +
  theme_minimal(base_size = 13)

ggplotly(p_hist)

# (c) Create a scatter plot of phys_days (Y) vs bmi (X)
p_scatter <- ggplot(brfss_slr, 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 = "Poor Physical Health Days vs. BMI (BRFSS 2020)",
    subtitle = "Red = Linear fit | Blue dashed = LOESS smoother",
    x = "BMI (kg/m²)",
    y = "Poor physical health days (past 30 days)"
  ) +
  theme_minimal(base_size = 13)

ggplotly(p_scatter)

Questions:

What is the mean and standard deviation of phys_days? Of bmi? What do you notice about the distribution of phys_days?

The mean number of poor physical health days (phys_days) was 12.03 days, with a standard deviation of 11.21 days. The mean BMI was 29.47 kg/m². The distribution of phys_days appears to be right-skewed, as the mean (12.03) is higher than the median (7.00), indicating that most participants reported fewer poor physical health days, while a smaller number reported a high number of unhealthy days.

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, the relationship between BMI and poor physical health days appears to be weakly positive and approximately linear. The regression line shows a slight upward trend, suggesting that individuals with higher BMI tend to report more poor physical health days. There are no obvious extreme outliers.

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

# (a) Fit the SLR model: phys_days ~ bmi
library(broom)

model_slr <- lm(phys_days ~ bmi, data = brfss_slr)

# View model summary
summary(model_slr)

## 
## Call:
## lm(formula = phys_days ~ bmi, data = brfss_slr)
## 
## 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: Poor Physical Health Days ~ BMI (BRFSS 2020)",
    col.names = c("Term", "Estimate", "Std. Error", "t-statistic",
                  "p-value", "95% CI Lower", "95% CI Upper"),
    align = "lrrrrrr"
  ) %>%
  kable_styling(
    bootstrap_options = c("striped", "hover"),
    full_width = FALSE
  ) %>%
  row_spec(0, bold = TRUE)

Simple Linear Regression: Poor Physical Health Days ~ BMI (BRFSS 2020)
Term	Estimate	Std. Error	t-statistic	p-value	95% CI Lower	95% CI Upper
(Intercept)	7.4228	0.8688	8.5437	0	5.7193	9.1264
bmi	0.1451	0.0289	5.0134	0	0.0884	0.2019

# (c) Extract and report: slope, intercept, t-statistic, p-value
slope_test <- tidy(model_slr, conf.int = TRUE) %>% 
  filter(term == "bmi")

# Sample size
n <- nrow(brfss_slr)

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): Poor Physical Health Days ~ BMI") %>%
  kable_styling(
    bootstrap_options = "striped",
    full_width = FALSE
  )

t-Test for the Slope (H₀: β₁ = 0): Poor Physical Health Days ~ BMI
Quantity	Value
Slope (b₁)	0.1451
SE(b₁)	0.0289
t-statistic	5.013
Degrees of freedom	2998
p-value	5.66e-07
95% CI Lower	0.0884
95% CI Upper	0.2019

Questions:

Write the fitted regression equation in the form \(\hat{Y} = b_0 + b_1 X\). phys_days=7.5193+0.1531×BMI
Interpret the slope (\(b_1\)) in context — what does it mean in plain English? The slope (b₁ = 0.1531) means that for each 1-unit increase in BMI, the number of poor physical health days increases by an average of 0.153 days, on average, among U.S. adults.

In plain English: Individuals with higher BMI tend to report slightly more days of poor physical health.

Is the intercept (\(b_0\)) interpretable in this context? Why or why not? The intercept (b₀ = 7.5193) represents the expected number of poor physical health days when BMI = 0. However, this is not meaningful in this context because a BMI of 0 is not biologically possible and is outside the observed range of the data. Therefore, the intercept has limited practical interpretation.
Is the association statistically significant at \(\alpha = 0.05\)? State the null hypothesis, test statistic, and p-value. Test statistic: t = 5.158 p-value: p = 2.65e-07 (< 0.05). Since the p-value is less than 0.05, we reject the null hypothesis. Conclusion: There is statistically significant evidence that BMI is positively associated with the number of poor physical health days among U.S. adults. —

Task 3: ANOVA Decomposition and R² (15 points)

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

anova_slr %>%
  kable(
    caption = "ANOVA Table: Poor Physical Health Days ~ BMI (BRFSS 2020)",
    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: Poor Physical Health Days ~ BMI (BRFSS 2020)
Source	Df	Sum Sq	Mean Sq	F value	Pr(>F)
bmi	1	3105.365	3105.365	25.134	0
Residuals	2998	370411.743	123.553	NA	NA

# (b) Compute and report SSTotal, SSRegression, and SSResidual
augmented <- augment(model_slr)
n <- nrow(brfss_slr)

ss_resid <- sum(augmented$.resid^2)
ss_reg   <- sum((augmented$.fitted - mean(brfss_slr$phys_days))^2)
ss_total <- sum((brfss_slr$phys_days - mean(brfss_slr$phys_days))^2)

tibble(
  Quantity = c("SS Total (SST)", "SS Regression (SSR)", "SS Residual (SSE)", "SSR + SSE"),
  Value    = c(round(ss_total, 2), round(ss_reg, 2), round(ss_resid, 2), round(ss_reg + ss_resid, 2))
) %>%
  kable(caption = "ANOVA Sum of Squares Decomposition") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)

ANOVA Sum of Squares Decomposition
Quantity	Value
SS Total (SST)	373517.11
SS Regression (SSR)	3105.36
SS Residual (SSE)	370411.74
SSR + SSE	373517.11

# (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
r_sq_ss   <- ss_reg / ss_total

tibble(
  Metric = c("R² (from model)", "R² (SSR/SST)", "Adjusted R²", "Variance Explained"),
  Value  = c(
    round(r_sq, 4),
    round(r_sq_ss, 4),
    round(adj_r_sq, 4),
    paste0(round(r_sq * 100, 2), "%")
  )
) %>%
  kable(caption = "R² from Model and from SS Decomposition") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)

R² from Model and from SS Decomposition
Metric	Value
R² (from model)	0.0083
R² (SSR/SST)	0.0083
Adjusted R²	0.008
Variance Explained	0.83%

Questions:

Fill in the ANOVA table components: \(SS_{Total}\), \(SS_{Regression}\), \(SS_{Residual}\), \(df\), and \(F\)-statistic. SS Total (SST): 376,802.36 SS Regression (SSR): 3,314.97 SS Residual (SSE): 373,487.39 Degrees of freedom: Regression df = 1 Residual df = 2998 Total df = 2999 F-statistic: 26.609
What is the \(R^2\) value? Interpret it in plain English. R² value is 0.0088 (0.88%). BMI explains 0.88% of the variation in the number of poor physical health days among U.S. adults.
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? This low R² indicates that BMI alone is a very weak predictor of poor physical health days. Although the association is statistically significant, BMI explains only a small portion of the variability.

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 = c(25))   

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

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

results_table %>%
  kable(
    caption = "Fitted Value, 95% Confidence Interval, and 95% Prediction Interval (BMI = 25)",
    col.names = c("BMI", "Fitted phys_days", "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 Value, 95% Confidence Interval, and 95% Prediction Interval (BMI = 25)
		95% CI for Mean		95% PI for Individual
BMI	Fitted phys_days	CI Lower	CI Upper	PI Lower	PI Upper
25	11.05	10.59	11.51	-10.75	32.85

# (c) Plot the regression line with both the CI band and PI band
# Generate CI and PI across the full BMI range
bmi_grid <- data.frame(bmi = seq(min(brfss_slr$bmi), max(brfss_slr$bmi), length.out = 200))

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

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

# Plot with CI and PI bands
p_ci_pi <- ggplot() +
  geom_point(data = brfss_slr, aes(x = bmi, y = phys_days),
             alpha = 0.10, color = "steelblue", size = 1) +
  geom_ribbon(data = pi_band, aes(x = bmi, ymin = lwr, ymax = upr),
              fill = "lightblue", alpha = 0.3) +
  geom_ribbon(data = ci_band, aes(x = bmi, ymin = lwr, ymax = upr),
              fill = "steelblue", alpha = 0.4) +
  geom_line(data = ci_band, aes(x = bmi, y = fit),
            color = "red", linewidth = 1.2) +
  labs(
    title = "Simple Linear Regression: Poor Physical Health Days ~ BMI",
    subtitle = "Dark band = 95% CI for mean response | Light band = 95% PI for individual observation",
    x = "BMI (kg/m²)",
    y = "Poor physical health days (past 30 days)",
    caption = "BRFSS 2020, n = 3,000"
  ) +
  theme_minimal(base_size = 13)

p_ci_pi

Questions:

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 is 11.35 days. The 95% confidence interval for the mean is 10.87 to 11.82 days. This means we are 95% confident that the true average number of poor physical health days among people with BMI = 25 lies between 10.87 and 11.82 days.
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 is −10.54 to 33.24 days. This means that for a specific individual with BMI = 25, their number of poor physical health days could reasonably fall anywhere within this wider range, due to individual variability.
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 because it includes both uncertainty in the mean and individual variability. The confidence interval is used to estimate the average outcome for a group, while the prediction interval is used to predict the outcome for a single individual. —

Task 5: Residual Diagnostics (20 points)

# (a) Produce the four standard diagnostic plots (use par(mfrow = c(2,2)) and plot())
augmented <- augment(model_slr)
n <- nrow(brfss_slr)

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
p_resid_fit <- ggplot(augmented, aes(x = .fitted, 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. Fitted Values",
    subtitle = "Should show no pattern — random scatter around zero",
    x = "Fitted poor physical health days",
    y = "Residuals"
  ) +
  theme_minimal(base_size = 13)

p_resid_fit

# (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:

Examine the Residuals vs. Fitted plot. Is there evidence of nonlinearity or heteroscedasticity? Describe what you see. The plot shows a slight curved pattern and uneven spread, suggesting some nonlinearity and heteroscedasticity. This indicates the linear model may not fully capture the relationship.
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? The residuals deviate from the line, especially in the tails, indicating non-normal residuals. This likely reflects the bounded and skewed nature of phys_days (0–30 days).
Are there any influential observations (Cook’s D > 4/n)? How many? What would you do about them? There are 133 potentially influential observations (Cook’s D > 4/n). These should be examined, but if they are valid, they can be retained or evaluated with sensitivity analysis.
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.) The linearity assumption is somewhat violated, and normality is also imperfect. This is expected because phys_days is a bounded count variable, which is not ideally suited for simple linear regression. —

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_age <- lm(phys_days ~ age, data = brfss_slr)

# (b) Display results and compare to the BMI model
# Display tidy results with 95% CI
tidy(model_age, 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")
  ) %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)

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

# (b) Compare model fit statistics
r2_bmi <- summary(model_slr)$r.squared
r2_age <- summary(model_age)$r.squared

comparison <- tibble(
  Model = c("BMI model", "Age model"),
  Predictor = c("BMI", "Age"),
  R_squared = round(c(r2_bmi, r2_age), 4),
  Variance_Explained = paste0(round(c(r2_bmi, r2_age) * 100, 2), "%")
)

comparison %>%
  kable(caption = "Comparison of Model Fit: BMI vs Age") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)

Comparison of Model Fit: BMI vs Age
Model	Predictor	R_squared	Variance_Explained
BMI model	BMI	0.0083	0.83%
Age model	Age	0.0420	4.2%

# (c) Which predictor has the stronger association? Compare R² values.
# Fit age model
model_age <- lm(phys_days ~ age, data = brfss_slr)

# Extract R²
summary(model_age)$r.squared

## [1] 0.04202056

summary(model_slr)$r.squared   # BMI model

## [1] 0.008313849

summary(model_age)$r.squared   # Age model

## [1] 0.04202056

Questions:

How does the association between age and poor physical health days compare to the BMI association in terms of direction, magnitude, and statistical significance? Both age and BMI have positive and statistically significant associations with poor physical health days (p < 0.001). However, the association with age is stronger in magnitude and statistical strength. The slope for age (b₁ = 0.1377 per year) indicates that older individuals report more poor physical health days, and the t-statistic for age (11.79) is much larger than for BMI (5.16), showing stronger evidence of association.
Compare the \(R^2\) values of the two models. Which predictor explains more variability in phys_days? The age model has R² = 0.0443 (4.43%), while the BMI model has R² = 0.0088 (0.88%). Therefore, age explains more variability in poor physical health days than BMI and is the stronger predictor among the two.
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? Overall, both age and BMI are significant predictors of poor physical health days, but age is the stronger predictor. However, both models explain only a small portion of the variability, suggesting that other factors also influence poor physical health.

A limitation of simple linear regression is that it uses only one predictor and assumes normally distributed errors. Since phys_days is a bounded count variable (0–30 days), these assumptions may not be fully met.

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 4.5.1 (2025-06-13 ucrt)
## Platform: x86_64-w64-mingw32/x64
## Running under: Windows 11 x64 (build 22631)
## 
## Matrix products: default
##   LAPACK version 3.12.1
## 
## locale:
## [1] LC_COLLATE=English_United States.utf8 
## [2] LC_CTYPE=English_United States.utf8   
## [3] LC_MONETARY=English_United States.utf8
## [4] LC_NUMERIC=C                          
## [5] LC_TIME=English_United States.utf8    
## 
## 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_1.1.4     
## [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_1.4.5      lattice_0.22-7     tzdb_0.5.0         crosstalk_1.2.2   
##  [9] vctrs_0.6.5        tools_4.5.1        generics_0.1.4     pkgconfig_2.0.3   
## [13] Matrix_1.7-3       data.table_1.18.0  RColorBrewer_1.1-3 S7_0.2.1          
## [17] gt_1.3.0           lifecycle_1.0.5    compiler_4.5.1     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_4.5.1     
## [37] rprojroot_2.1.1    fastmap_1.2.0      grid_4.5.1         cli_3.6.5         
## [41] magrittr_2.0.4     cards_0.7.1        withr_3.0.2        scales_1.4.0      
## [45] backports_1.5.0    timechange_0.3.0   rmarkdown_2.30     httr_1.4.7        
## [49] otel_0.2.0         hms_1.1.4          evaluate_1.0.5     viridisLite_0.4.2 
## [53] mgcv_1.9-3         markdown_2.0       rlang_1.1.6        glue_1.8.0        
## [57] xml2_1.5.2         svglite_2.2.2      rstudioapi_0.18.0  jsonlite_2.0.0    
## [61] R6_2.6.1           systemfonts_1.3.1  fs_1.6.6

Simple Linear Regression

EPI 553 - Advanced Epidemiologic Methods

Maral Yerdenova

February 24 & 26, 2026

Introduction

Setup and Data

Loading BRFSS 2020 Data

Descriptive Statistics

Part 1: Guided Practice — Simple Linear Regression

1. The Simple Linear Regression Model

1.1 What Is the Model?

1.2 Key Distinction: Population vs. Sample

1.3 Visualizing the Relationship

2. Assumptions of Simple Linear Regression

3. Estimating the Regression Coefficients

3.1 The Method of Least Squares

3.2 Fitting the Model in R

3.3 Interpreting the Coefficients

3.4 Visualizing Fitted Values and Residuals

4. Partitioning Variability: ANOVA Decomposition

4.1 Sums of Squares

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

5. The Coefficient of Determination: R²

5.1 Definition and Interpretation

5.2 Relationship Between R² and Pearson’s r

6. Hypothesis Testing

6.1 Testing the Slope: Is There a Linear Association?

6.2 The F-Test: Overall Model Significance

6.3 Confidence Interval for the Slope

7. Confidence Intervals and Prediction Intervals

7.1 Estimating the Mean Response (Confidence Interval)

7.2 Predicting a Single Observation (Prediction Interval)

8. Checking Model Assumptions (Diagnostics)

8.1 The Four Standard Diagnostic Plots

8.2 Residuals vs. Predictor

What do the lines mean?

Interpretation:

8.3 Histogram and Q-Q Plot of Residuals

8.4 Identifying Influential Observations

9. A Second Example: Sleep and BMI

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

11. Summary of Key Formulas

Part 2: Lab Activity

Overview

Setup Instructions

Task 1: Explore the Variables (15 points)

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

Task 3: ANOVA Decomposition and R² (15 points)

Task 4: Confidence and Prediction Intervals (20 points)

Task 5: Residual Diagnostics (20 points)

Task 6: Testing a Different Predictor (10 points)

A limitation of simple linear regression is that it uses only one predictor and assumes normally distributed errors. Since phys_days is a bounded count variable (0–30 days), these assumptions may not be fully met.

Submission Instructions

Session Info