hu— title: “Simple Linear Regression” subtitle: “EPI 553 - Advanced Epidemiologic Methods” author: “Muntasir Masum” date: “February 24 & 26, 2026” output: html_document: toc: true toc_float: true toc_depth: 3 theme: flatly highlight: tango code_folding: show —

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:

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 raw BRFSS 2020 data
brfss_full <- read_xpt("C:/Users/tahia/OneDrive/Desktop/UAlbany PhD/Epi 553/LLCP2020XPT/LLCP2020.XPT") %>%
  janitor::clean_names()

# Select variables of interest
brfss_slr <- brfss_full %>%
  select(bmi5, age80, sex, educag, genhlth, physhlth, sleptim1)

# Recode variables
brfss_slr <- brfss_slr %>%
  mutate(
    bmi       = bmi5 / 100,
    age       = age80,
    sex       = factor(ifelse(sex == 1, "Male", "Female")),
    education = factor(case_when(
      educag == 1 ~ "< High school",
      educag == 2 ~ "High school graduate",
      educag == 3 ~ "Some college",
      educag == 4 ~ "College graduate"
    ), levels = c("< High school", "High school graduate",
                  "Some college", "College graduate")),
    gen_health_num = ifelse(genhlth  %in% 1:5,  genhlth,  NA_real_),
    sleep_hrs      = ifelse(sleptim1 %in% 1:24, sleptim1, NA_real_),
    phys_days      = ifelse(physhlth %in% 0:30, physhlth, NA_real_)
  )

# Select recoded variables, apply filters, drop missing, take sample
set.seed(553)
brfss_slr <- brfss_slr %>%
  select(bmi, age, sex, education, gen_health_num, sleep_hrs, phys_days) %>%
  filter(bmi > 14.5, bmi < 60, age >= 18, age <= 80) %>%
  drop_na() %>%
  slice_sample(n = 3000)

# Save analytic dataset
saveRDS(brfss_slr, ("C:/Users/tahia/OneDrive/Desktop/UAlbany PhD/Epi 553/Lab 6/new_data.rds"))

tibble(
  Metric = c("Observations", "Variables"),
  Value  = c(nrow(brfss_slr), ncol(brfss_slr))
) %>%
  kable(caption = "Analytic Dataset Dimensions") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Analytic Dataset Dimensions
Metric Value
Observations 3000
Variables 7

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

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

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

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

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

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, 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)

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 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, 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.

brfss_lab <- read_rds("C:/Users/tahia/OneDrive/Desktop/UAlbany PhD/Epi 553/Lab 6/new_data.rds")

Task 1: Explore the Variables (15 points)

# (a) Create a summary table of phys_days and bmi

brfss_lab %>%
  select(bmi, phys_days) %>%
  summary() %>%
  kable(caption = "Descriptive Statistics: BMi and Physical activity") %>%
  kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)
Descriptive Statistics: BMi and Physical activity
bmi phys_days
Min. :14.63 Min. : 1.00
1st Qu.:24.32 1st Qu.: 2.00
Median :27.89 Median : 6.00
Mean :29.18 Mean :11.66
3rd Qu.:32.89 3rd Qu.:20.00
Max. :59.60 Max. :30.00 
brfss_lab %>%
  select(bmi, phys_days) %>%
  tbl_summary(
    label = list(
      bmi ~ "BMI (kg/m²)",
    phys_days ~ "Physical activity in days"
    ),
    statistic = list(
      all_continuous() ~ "{mean} ({sd})",
      all_categorical() ~ "{n} ({p}%)"
    ),
    digits = all_continuous() ~ 1
  ) %>%
  add_n() %>%
  bold_labels() %>%
  modify_caption("**Table 1. Descriptive Statistics of BMI and Physical Activity (BRFSS 2020, n = 3,000)**")
Table 1. Descriptive Statistics of BMI and Physical Activity (BRFSS 2020, n = 3,000)
Characteristic N N = 3,0001
BMI (kg/m²) 3,000 29.2 (7.0)
Physical activity in days 3,000 11.7 (11.2)
1 Mean (SD) 
# (b) Create a histogram of phys_days — describe the distribution

ggplot(brfss_lab, aes(x = phys_days)) +
  geom_histogram(bins = 30, fill = "steelblue", color = "black") +
  labs(
    title = "Histogram of Days Physically Active (Last 30 Days)",
    x = "Number of Days Physically Active",
    y = "Frequency"
  ) +
  theme_minimal()

# (c) Create a scatter plot of phys_days (Y) vs bmi (X)

ggplot(brfss_lab, aes(x = bmi, y = phys_days)) +
  geom_point(color = "steelblue", alpha = 0.5) +
  geom_smooth(method = "lm", se = FALSE, color = "darkred") +
  labs(
    title = "Physically Active Days vs BMI",
    x = "BMI (kg/m²)",
    y = "Number of Physically Active Days"
  ) +
  theme_minimal()

Questions:

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

Mean & SD

BMI: Mean = 29.2, SD = 7.0

Physical activity days (phys_days): Mean = 11.7 days, SD = 11.2 days

The mean (11.7) is much higher than the median (6), and the 3rd quartile is 20 with a max of 30. So the distribution looks right-skewed, with many people reporting fewer active days and some reporting very high values (close to 30).

  1. Based on the scatter plot, does the relationship between BMI and poor physical health days appear to be linear? Are there any obvious outliers?

Scatter plot interpretation: The relationship between BMI and physical activity appears weak and not strongly linear. If anything, it may show a slight negative trend (higher BMI → fewer active days), but it doesn’t look very strong. There may be a few extreme BMI values (like near 60) that could act as possible outliers.

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

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

model_lab <- lm(phys_days ~ bmi, data = brfss_lab)

# Summary output
summary(model_lab)
## 
## Call:
## lm(formula = phys_days ~ bmi, data = brfss_lab)
## 
## 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_lab, conf.int = TRUE) %>%
  mutate(across(where(is.numeric), ~ round(., 4))) %>%
  kable(
    caption = "Simple Linear Regression: Physically Sctive 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: Physically Sctive 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

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


slope_test <- tidy(model_lab, conf.int = TRUE) %>% filter(term == "bmi")

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) 0.1451
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:

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

Fitted equation:

  1. Interpret the slope (\(b_1\)) in context — what does it mean in plain English?

For each 1 unit increase in BMI (kg/m²), the model predicts about 0.145 more physically active days (on average), holding nothing else constant (since it’s a simple model).

  1. Is the intercept (\(b_0\)) interpretable in this context? Why or why not?

Not really, b0=7.4228 is the predicted active days when BMI = 0, which isn’t realistic for humans, so it doesn’t have a practical meaning here.

  1. Is the association statistically significant at \(\alpha = 0.05\)? State the null hypothesis, test statistic, and p-value.

To test whether BMI is associated with physically active days, we tested the null hypothesis H0:β1=0, meaning there is no linear relationship between BMI and physically active days. The t-statistic for the slope was 5.013 with 2998 degrees of freedom, and the p-value was 5.66 × 10⁻⁷. Since this p-value is much smaller than α = 0.05, we reject the null hypothesis and conclude that there is a statistically significant association between BMI and physically active days in this sample.

Task 3: ANOVA Decomposition and R² (15 points)

# (a) Display the ANOVA table

anova_lab <- anova(model_lab)

anova_lab %>%
  kable(
    caption = "ANOVA Table: Physycally ACtive 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: Physycally ACtive Days ~ BMI
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

SSRegression <- anova_lab$`Sum Sq`[1]
SSResidual   <- anova_lab$`Sum Sq`[2]

# Total Sum of Squares
SSTotal <- SSRegression + SSResidual

# Print results
SSRegression
## [1] 3105.365
SSResidual
## [1] 370411.7
SSTotal
## [1] 373517.1
# (c) Compute R² two ways: from the model object and from the SS decomposition

#From Model object

R2_model <- summary(model_lab)$r.squared

R2_model
## [1] 0.008313849
#From SS decomposition

R2_ss <- SSRegression / SSTotal

R2_ss
## [1] 0.008313849

Questions:

  1. Fill in the ANOVA table components: \(SS_{Total}\), \(SS_{Regression}\), \(SS_{Residual}\), \(df\), and \(F\)-statistic.

From the ANOVA table, the SSRegression (SSR) is 3105.365, the SSResidual (SSE) is 370,411.743, and the SSTotal (SST) is 373,517.1. The regression has 1 degree of freedom, the residual has 2998 df, and the total df is 2999. The F-statistic is 25.134, which reflects the ratio of the regression mean square (3105.365) to the residual mean square (123.553).

  1. What is the \(R^2\) value? Interpret it in plain English.

The R² value is 0.0083, meaning BMI explains only about 0.83% of the variation in physically active days. In plain English, BMI barely explains any of the differences in how many days people are physically active. Even though the relationship is statistically significant, it’s very weak in terms of practical explanation.

  1. 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 tells us that BMI alone does a poor job explaining variation in physically active days. About 99% of the variation is due to other factors not included in the model. Things like age, sex, overall health status, disability, motivation, work schedule, environment, or even reporting error could explain most of the remaining variation.

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_lab, newdata = new_bmi, interval = "confidence", level=0.95) %>%
  as.data.frame() %>%
  rename(Fitted = fit, CI_Lower = lwr, CI_Upper = upr)
ci_pred
##     Fitted CI_Lower CI_Upper
## 1 11.05108 10.58774 11.51441
# (b) Calculate the 95% prediction interval for a person with BMI = 25


pi_pred <- predict(model_lab, newdata = new_bmi, interval = "prediction", level = 0.95) %>%
  as.data.frame() %>%
  rename(PI_Lower = lwr, PI_Upper = upr) %>%
  select(-fit)
pi_pred
##   PI_Lower PI_Upper
## 1 -10.7485 32.85066
results_table <- bind_cols(new_bmi, ci_pred, pi_pred) %>%
  mutate(across(where(is.numeric), ~ round(., 2)))

results_table %>%
  kable(
    caption = "Fitted Values, 95% Confidence Intervals, and Prediction Intervals by BMI",
    col.names = c("BMI= 25", "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 BMI
95% CI for Mean
95% PI for Individual
BMI= 25 Fitted BMI 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

# Create a grid of BMI values across the observed range
bmi_grid <- data.frame(bmi = seq(min(brfss_lab$bmi), max(brfss_lab$bmi), length.out = 200))

ci_band <- predict(model_lab, newdata = bmi_grid, interval = "confidence", level=0.95) %>%
  as.data.frame() %>%
  bind_cols(bmi_grid)

pi_band <- predict(model_lab, newdata = bmi_grid, interval = "prediction", level=0.95) %>%
  as.data.frame() %>%
  bind_cols(bmi_grid)

p_ci_pi <- ggplot() +
  geom_point(data = brfss_lab, 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: Physical Activite Days ~ BMI",
    subtitle = "Dark band = 95% CI for mean response | Light band = 95% PI for individual observation",
    x = "BMI",
    y = "Physically Active in Days",
    caption = "BRFSS 2020, n = 3,000"
  ) +
  theme_minimal(base_size = 13)

p_ci_pi

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 physically active days is 11.05 days. The 95% confidence interval (CI) for this mean is 10.59 to 11.51 days, meaning we are 95% confident that the true average number of active days for people with BMI = 25 falls within this range.

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

For a specific new individual with a BMI of 25, the 95% prediction interval (PI) is much wider: –10.75 to 32.85 days. This interval reflects where an individual person’s number of active days could realistically fall.

  1. 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 confidence interval because it accounts for both the uncertainty in estimating the mean and the natural person-to-person variability in physically active days. The confidence interval is used when we care about the average response for a group, while the prediction interval is used when we want to predict the outcome for one specific individual.

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_lab, which = 1:4,
     col = adjustcolor("steelblue", 0.4),
     pch = 19, cex = 0.6)

# (b) Create a residuals vs. fitted plot using ggplot

# Augment model to get fitted values and residuals
aug_lab <- augment(model_lab)

ggplot(aug_lab, aes(x = .fitted, y = .resid)) +
  geom_point(alpha = 0.2, color = "steelblue") +
  geom_hline(yintercept = 0, color = "red", linewidth = 1) +
  geom_smooth(method = "loess", se = FALSE, color = "orange", linewidth = 1) +
  labs(
    title = "Residuals vs Fitted Values",
    subtitle = "Checking linearity and equal variance (homoscedasticity)",
    x = "Fitted values",
    y = "Residuals"
  ) +
  theme_minimal()

# (c) Create a normal Q-Q plot of residuals using ggplot


p_qq <- ggplot(aug_lab, 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

aug_lab <- aug_lab %>%
  mutate(
    obs_num = row_number(),
    cooks_d = cooks.distance(model_lab),
    influential = ifelse(cooks_d > 4 / n, "Potentially influential", "Not influential")
  )

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

p_cooks <- ggplot(aug_lab, 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.

From the Residuals vs. Fitted plot, there is some evidence of nonlinearity. The orange LOESS curve is slightly curved rather than flat, especially at lower fitted values, suggesting the relationship may not be perfectly linear. The spread of residuals also changes a bit across fitted values, which hints at mild heteroscedasticity (non-constant variance). The striped pattern you see is likely because phys_days is a count variable (0–30), so residuals fall in horizontal bands.

  1. 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?

Looking at the Q-Q plot, the residuals clearly deviate from the red line, especially in the tails. This suggests the residuals are not normally distributed. In this context, that makes sense because phys_days is a bounded count variable (0 to 30 days), not a continuous normal variable. The departures from normality suggest the outcome is skewed and discrete rather than bell-shaped.

  1. Are there any influential observations (Cook’s D > 4/n)? How many? What would you do about them?

For Cook’s distance, there are influential observations above the 4/n threshold. Based on your plot, there are 136 potentially influential observations. However, their Cook’s D values are still relatively small overall. I would not automatically remove them. Instead, I would check whether they are data errors or legitimate extreme values and possibly run a sensitivity analysis to see if results change when excluding them.

  1. 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.)

Ihe LINE assumptions are only partially met. Linearity is questionable, normality is clearly violated, and there is some mild heteroscedasticity. The most problematic assumption is normality, mainly because the outcome variable (phys_days) is a discrete count variable bounded between 0 and 30. A linear regression may not be the best model here is a Poisson regression might be more appropriate given the nature of the outcome.

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

p_phys_days <- ggplot(brfss_lab, aes(x = phys_days, y = age)) +
  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 = "Age vs. Physical Activity in Day (BRFSS 2020)",
    x = "Physical Activity in Days",
    y = "Age (years)"
  ) +
  theme_minimal(base_size = 13)

p_phys_days

# (b) Display results and compare to the BMI model

model_Phys_days <- lm(bmi ~ phys_days, data = brfss_lab)

tidy(model_Phys_days, conf.int = TRUE) %>%
  mutate(across(where(is.numeric), ~ round(., 4))) %>%
  kable(
    caption = "SLR: Age ~ Physical Activity in Days",
    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: Age ~ Physical Activity in Days
Term Estimate Std. Error t-statistic p-value 95% CI Lower 95% CI Upper
(Intercept) 28.5141 0.1844 154.6357 0 28.1525 28.8757
phys_days 0.0573 0.0114 5.0134 0 0.0349 0.0797 
# (c) Which predictor has the stronger association? Compare R² values.

b1_phys_days <- coef(model_Phys_days)["sleep_hrs"]
r2_phys_days <- summary(model_Phys_days)$r.squared

tibble(
  Metric = c("Slope (b₁)", "R²"),
  Value  = c(round(b1_phys_days, 4), round(r2_phys_days, 4))
) %>%
  kable(caption = "Physical Activity Model Key Statistics") %>%
  kable_styling(bootstrap_options = "striped", full_width = FALSE)
Physical Activity Model Key Statistics
Metric Value
Slope (b₁) NA
R² 0.0083

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?

The association between age and physically active days appears to be positive, meaning as age increases, physically active days slightly increase (based on the positive slope ≈ 0.057). The magnitude is small, similar to what we saw with BMI (BMI slope ≈ 0.145). Both associations are statistically significant (p < 0.001), but the effect sizes are very small in practical terms. So direction-wise both are positive, magnitude is weak, and both are statistically significant.

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

The R² for the BMI model was 0.0083, and the R² for this model is also about 0.0083. That means both predictors explain only about 0.83% of the variability in physically active days. Neither predictor meaningfully explains the outcome. They perform almost identically and very poorly in terms of explained variance.

  1. 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 these two simple linear regression models, neither BMI nor age alone is a strong predictor of physically active days. Even though both associations are statistically significant, they explain less than 1% of the variability, which is practically negligible. A key limitation here is that phys_days is a bounded count variable (0–30 days), which violates normality assumptions and makes linear regression less appropriate. A count-based model like Poisson regression would likely be more suitable. Also, using simple models ignores other important factors (health status, disability, socioeconomic status, etc.) that probably explain much more of the variation.

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 ucrt)
## Platform: x86_64-w64-mingw32/x64
## Running under: Windows 11 x64 (build 26200)
## 
## 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_1.2.0     
## [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_1.4.4      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  janitor_2.2.1      snakecase_0.11.1   litedown_0.9      
## [25] htmltools_0.5.9    sass_0.4.10        yaml_2.3.12        lazyeval_0.2.2    
## [29] pillar_1.11.1      jquerylib_0.1.4    cachem_1.1.0       nlme_3.1-168      
## [33] commonmark_2.0.0   tidyselect_1.2.1   digest_0.6.39      stringi_1.8.7     
## [37] labeling_0.4.3     splines_4.5.2      rprojroot_2.1.1    fastmap_1.2.0     
## [41] grid_4.5.2         cli_3.6.5          magrittr_2.0.4     cards_0.7.1       
## [45] withr_3.0.2        scales_1.4.0       backports_1.5.0    timechange_0.3.0  
## [49] rmarkdown_2.30     httr_1.4.7         otel_0.2.0         hms_1.1.4         
## [53] evaluate_1.0.5     viridisLite_0.4.2  mgcv_1.9-3         markdown_2.0      
## [57] rlang_1.1.7        glue_1.8.0         xml2_1.5.2         svglite_2.2.2     
## [61] rstudioapi_0.18.0  jsonlite_2.0.0     R6_2.6.1           systemfonts_1.3.1 
## [65] fs_1.6.6