Simple Linear Regression

EPI 553 - Advanced Epidemiologic Methods

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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.

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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_slr <- read_rds('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.

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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.

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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

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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.

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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.

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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	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

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

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

peak_age <- -b1_q / (2 * b2_q)

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

Quadratic Model Coefficients and Implied Turning Point

Quantity	Value
b₁ (Age)	0.474350
b₂ (Age²)	-0.004635
Peak / Trough Age (-b₁ / 2b₂)	51.200000

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

---

11. Summary of Key Formulas

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

---

Part 2: Lab Activity

Overview

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

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

Setup Instructions

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

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

# Load raw BRFSS 2020 data
brfss_slr <- read_rds('brfss_slr_2020.rds')

---

Task 1: Explore the Variables (15 points)

# Load analytic dataset
brfss_slr <- readRDS(here::here("brfss_slr_2020.rds"))

# -------------------------
# (a) Summary table
# -------------------------

summary_table <- brfss_slr %>%
  summarise(
    mean_phys_days = mean(phys_days),
    sd_phys_days   = sd(phys_days),
    min_phys_days  = min(phys_days),
    max_phys_days  = max(phys_days),

    mean_bmi = mean(bmi),
    sd_bmi   = sd(bmi),
    min_bmi  = min(bmi),
    max_bmi  = max(bmi)
  )

summary_table

## # A tibble: 1 × 8
##   mean_phys_days sd_phys_days min_phys_days max_phys_days mean_bmi sd_bmi
##            <dbl>        <dbl>         <dbl>         <dbl>    <dbl>  <dbl>
## 1           11.7         11.2             1            30     29.2   7.01
## # ℹ 2 more variables: min_bmi <dbl>, max_bmi <dbl>

# -------------------------
# (b) Histogram of phys_days
# -------------------------

ggplot(brfss_slr, aes(x = phys_days)) +
  geom_histogram(binwidth = 1, fill = "steelblue", color = "black") +
  labs(
    title = "Distribution of Physically Unhealthy Days",
    x = "Physically Unhealthy Days (past 30)",
    y = "Count"
  ) +
  theme_minimal()

# -------------------------
# (c) Scatter plot: phys_days vs bmi
# -------------------------

ggplot(brfss_slr, aes(x = bmi, y = phys_days)) +
  geom_point(alpha = 0.4) +
  geom_smooth(method = "lm", se = FALSE, color = "red") +
  labs(
    title = "Physically Unhealthy Days vs BMI",
    x = "BMI",
    y = "Physically Unhealthy 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?

   Physically Unhealthy Days (phys_days)

   Mean: 11.66 days

   Standard deviation: 11.16 days

   BMI Mean: 29.18

   Standard deviation: 7.01

   The distribution of physically unhealthy days is right-skewed, with most participants reporting relatively few unhealthy days and a smaller group reporting many unhealthy days.

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

   The distribution of physically unhealthy days is right-skewed, with most individuals reporting relatively few unhealthy days and a smaller number reporting many unhealthy days. The majority of observations cluster at lower values, while a tail extends toward higher values (up to 30 days). This suggests that frequent poor physical health days are less common but present in a subset of the sample.

   The scatterplot shows a weak positive relationship between BMI and physically unhealthy days, with considerable variability and no strong linear pattern.

---

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

# Load dataset
brfss_slr <- readRDS(here::here("brfss_slr_2020.rds"))

# -------------------------
# (a) Fit SLR model
# -------------------------
model <- lm(phys_days ~ bmi, data = brfss_slr)


# -------------------------
# (b) Tidy coefficient table with 95% CI
# -------------------------
library(broom)

tidy_table <- tidy(model, conf.int = TRUE)
tidy_table

## # A tibble: 2 × 7
##   term        estimate std.error statistic  p.value conf.low conf.high
##   <chr>          <dbl>     <dbl>     <dbl>    <dbl>    <dbl>     <dbl>
## 1 (Intercept)    7.42     0.869       8.54 2.04e-17   5.72       9.13 
## 2 bmi            0.145    0.0289      5.01 5.66e- 7   0.0884     0.202

# -------------------------
# (c) Extract key statistics
# -------------------------

intercept <- tidy_table$estimate[tidy_table$term == "(Intercept)"]
slope     <- tidy_table$estimate[tidy_table$term == "bmi"]
t_stat    <- tidy_table$statistic[tidy_table$term == "bmi"]
p_value   <- tidy_table$p.value[tidy_table$term == "bmi"]

list(
  Intercept = intercept,
  Slope = slope,
  t_statistic = t_stat,
  p_value = p_value
)

## $Intercept
## [1] 7.422846
## 
## $Slope
## [1] 0.1451293
## 
## $t_statistic
## [1] 5.01337
## 
## $p_value
## [1] 5.659364e-07

Questions:

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

   \(\hat{Y}\)= 7.48+0.14X

   Where:

   • \(\hat{Y}\) = predicted physically unhealthy days

   • \(X = \text{BMI}\)

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

   For every one-unit increase in BMI, the expected number of physically unhealthy days increases by approximately 0.14 days, on average.

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

   The intercept (7.48) represents the predicted number of unhealthy days when BMI = 0. This is not interpretable because: A BMI of 0 is biologically impossible. It falls far outside the observed range of BMI in the data. So while mathematically necessary, it has no real-world meaning

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

   t = 4.95, p < 0.001

   At α = 0.05, there is a statistically significant association between BMI and physically unhealthy days. However, the effect size is small, indicating a weak positive relationship.

---

Task 3: ANOVA Decomposition and R² (15 points)

# Load dataset
brfss_slr <- readRDS(here::here("brfss_slr_2020.rds"))

# Fit model
model <- lm(phys_days ~ bmi, data = brfss_slr)

# -------------------------
# (a) ANOVA table
# -------------------------
anova_table <- anova(model)
anova_table

## Analysis of Variance Table
## 
## Response: phys_days
##             Df Sum Sq Mean Sq F value    Pr(>F)    
## bmi          1   3105 3105.36  25.134 5.659e-07 ***
## Residuals 2998 370412  123.55                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

# -------------------------
# (b) Compute SS components
# -------------------------

SS_regression <- anova_table$`Sum Sq`[1]
SS_residual   <- anova_table$`Sum Sq`[2]
SS_total      <- SS_regression + SS_residual

list(
  SSTotal = SS_total,
  SSRegression = SS_regression,
  SSResidual = SS_residual
)

## $SSTotal
## [1] 373517.1
## 
## $SSRegression
## [1] 3105.365
## 
## $SSResidual
## [1] 370411.7

# -------------------------
# (c) Compute R² two ways
# -------------------------

# Method 1: From model summary
R2_model <- summary(model)$r.squared

# Method 2: From sum of squares
R2_manual <- SS_regression / SS_total

list(
  R2_from_model = R2_model,
  R2_from_SS = R2_manual
)

## $R2_from_model
## [1] 0.008313849
## 
## $R2_from_SS
## [1] 0.008313849

Questions:

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

   df Regression ​ =1

   dfResidual=n−2=2998

   dfTotal=n−1=2999

   F≈t2≈4.952≈24.50

   SS Regression ​ ≈3,027.95, df=1 df=1, F≈24.50 F≈24.50

   SS Residual ​ ≈370,484.31, df=2998 df=2998

   SS Total ​ ≈373,512.25, df=2999 df=2999

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

   \(R^ 2\)≈0.008 (about 0.8%)

   BMI explains less than 1% of the variation in physically unhealthy days in this sample, so BMI alone is a very weak predictor of poor physical health days.

3. What does this \(R^2\) tell you about how well BMI alone explains variation in poor physical health days? What might explain the remaining variation?

   The \(R^ 2\) value of approximately 0.008 indicates that BMI explains less than 1% of the variation in physically unhealthy days, meaning BMI alone is a very poor predictor of poor physical health days. Most of the variability is due to other factors not included in the model, such as age, chronic health conditions, mental health, lifestyle behaviors (e.g., physical activity, smoking), socioeconomic status, and natural individual differences. This suggests that poor physical health is influenced by many variables, not just BMI.

---

Task 4: Confidence and Prediction Intervals (20 points)

# Load data + fit model
brfss_slr <- readRDS(here::here("brfss_slr_2020.rds"))
model <- lm(phys_days ~ bmi, data = brfss_slr)

# -------------------------
# (a) Fitted value + 95% CI at BMI = 25
# -------------------------
new25 <- tibble(bmi = 25)

ci_25 <- predict(model, newdata = new25, interval = "confidence", level = 0.95)
ci_25 <- as.data.frame(ci_25) %>% mutate(bmi = 25)
ci_25   # fit, lwr, upr

##        fit      lwr      upr bmi
## 1 11.05108 10.58774 11.51441  25

# -------------------------
# (b) 95% Prediction interval at BMI = 25
# -------------------------
pi_25 <- predict(model, newdata = new25, interval = "prediction", level = 0.95)
pi_25 <- as.data.frame(pi_25) %>% mutate(bmi = 25)
pi_25   # fit, lwr, upr

##        fit      lwr      upr bmi
## 1 11.05108 -10.7485 32.85066  25

# Optional: print as a clean little table
bind_rows(
  ci_25 %>% mutate(type = "95% Confidence Interval (mean)"),
  pi_25 %>% mutate(type = "95% Prediction Interval (individual)")
) %>%
  select(type, bmi, fit, lwr, upr)

##                                       type bmi      fit       lwr      upr
## 1...1       95% Confidence Interval (mean)  25 11.05108  10.58774 11.51441
## 1...2 95% Prediction Interval (individual)  25 11.05108 -10.74850 32.85066

# -------------------------
# (c) Plot regression line + CI band + PI band
# -------------------------
# Create a grid of BMI values across the observed range
bmi_grid <- tibble(
  bmi = seq(min(brfss_slr$bmi), max(brfss_slr$bmi), length.out = 200)
)

# Get confidence and prediction intervals across grid
ci_grid <- as.data.frame(predict(model, newdata = bmi_grid, interval = "confidence")) %>%
  bind_cols(bmi_grid) %>%
  rename(ci_fit = fit, ci_lwr = lwr, ci_upr = upr)

pi_grid <- as.data.frame(predict(model, newdata = bmi_grid, interval = "prediction")) %>%
  bind_cols(bmi_grid) %>%
  rename(pi_fit = fit, pi_lwr = lwr, pi_upr = upr)

plot_df <- left_join(ci_grid, pi_grid, by = "bmi")

library(ggplot2)

ggplot(brfss_slr, aes(x = bmi, y = phys_days)) +
  geom_point(alpha = 0.25) +

  # Prediction interval band
  geom_ribbon(
    data = plot_df,
    aes(x = bmi, ymin = pi_lwr, ymax = pi_upr),
    inherit.aes = FALSE,
    alpha = 0.15
  ) +

  # Confidence interval band
  geom_ribbon(
    data = plot_df,
    aes(x = bmi, ymin = ci_lwr, ymax = ci_upr),
    inherit.aes = FALSE,
    alpha = 0.25
  ) +

  # Regression line
  geom_line(
    data = plot_df,
    aes(x = bmi, y = ci_fit),
    inherit.aes = FALSE,
    linewidth = 1
  ) +

  labs(
    title = "SLR with Confidence and Prediction Bands",
    x = "BMI",
    y = "Physically Unhealthy Days"
  ) +
  theme_minimal()

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?

   Y ^ =7.48+0.14(25)

   Y^=7.48+3.50=10.98

   Y ^ =7.48+3.50=10.98

   Estimated mean:
   

   ≈ 11 poor physical health days

   We are 95% confident that the average number of physically unhealthy days among individuals with BMI = 25 is between about 10.5 and 11.5 days.

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

   95% PI ≈ (−10.8 , 32.8)

   For a specific individual with BMI = 25, we predict their number of physically unhealthy days will fall between approximately −11 and 33 days with 95% confidence.

3. Explain in your own words why the prediction interval is wider than the confidence interval. When would you use each one in practice?

   The prediction interval is wider than the confidence interval because it accounts for both the uncertainty in estimating the regression line and the natural variability among individual people. A confidence interval estimates the average number of poor health days for all individuals with BMI = 25, while a prediction interval estimates the likely range for one specific person. Because individuals vary more than group averages, prediction intervals are always wider.

---

Task 5: Residual Diagnostics (20 points)

# Load data and fit model
brfss_slr <- readRDS(here::here("brfss_slr_2020.rds"))
model <- lm(phys_days ~ bmi, data = brfss_slr)

# --------------------------------------------------
# (a) Four standard diagnostic plots (base R)
# --------------------------------------------------
par(mfrow = c(2, 2))
plot(model)

par(mfrow = c(1,1))  # reset layout


# --------------------------------------------------
# Prepare residual data for ggplot plots
# --------------------------------------------------
library(ggplot2)

diag_df <- data.frame(
  fitted = fitted(model),
  residuals = resid(model),
  std_resid = rstandard(model),
  cooks = cooks.distance(model)
)


# --------------------------------------------------
# (b) Residuals vs Fitted (ggplot)
# --------------------------------------------------
ggplot(diag_df, aes(x = fitted, y = residuals)) +
  geom_point(alpha = 0.4) +
  geom_hline(yintercept = 0, linetype = "dashed") +
  labs(
    title = "Residuals vs Fitted",
    x = "Fitted Values",
    y = "Residuals"
  ) +
  theme_minimal()

# --------------------------------------------------
# (c) Normal Q-Q plot (ggplot)
# --------------------------------------------------
ggplot(diag_df, aes(sample = residuals)) +
  stat_qq() +
  stat_qq_line() +
  labs(title = "Normal Q-Q Plot of Residuals") +
  theme_minimal()

# --------------------------------------------------
# (d) Cook's Distance Plot
# --------------------------------------------------
ggplot(diag_df, aes(x = seq_along(cooks), y = cooks)) +
  geom_col() +
  geom_hline(yintercept = 4/nrow(diag_df), linetype = "dashed") +
  labs(
    title = "Cook's Distance",
    x = "Observation",
    y = "Cook's Distance"
  ) +
  theme_minimal()

Questions:

1. Examine the Residuals vs. Fitted plot. Is there evidence of nonlinearity or heteroscedasticity? Describe what you see.

   The residuals vs fitted plot shows a clear pattern rather than random scatter. The residuals form diagonal banding and a slight funnel shape, indicating nonlinearity and heteroscedasticity. The banding occurs because the outcome variable (phys_days) is discrete and bounded (0–30), which violates the constant variance assumption.

2. Examine the Q-Q plot. Are the residuals approximately normal? What do departures from normality in this context suggest about the distribution of phys_days?

   The residuals are not normally distributed. The Q-Q plot shows strong curvature and clear departures from the diagonal line, especially in the tails. This suggests the residual distribution is skewed and bounded, which is expected because physically unhealthy days is a count-like variable rather than a continuous normally distributed variable.

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

   The dashed reference line represents 4/n≈0.0013 4/n≈0.0013. Several observations exceed this threshold, indicating the presence of multiple potentially influential points. However, none appear extremely large, so they are likely not overly influential individually. In practice, these observations should be examined for data errors or unusual values before deciding whether to remove them.

4. Overall, do the LINE assumptions appear to be met? Which assumption(s) may be most problematic for this model, and why? (Hint: think about the nature of the outcome variable.)

   Overall, while the model detects a statistically significant association, the diagnostic plots indicate that a simple linear regression is not ideal for this outcome. A model designed for bounded or count data (e.g., Poisson or negative binomial regression) would likely be more appropriate.

---

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.

# Load dataset
brfss_slr <- readRDS(here::here("brfss_slr_2020.rds"))

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

# -------------------------
# (b) Display results
# -------------------------
summary(model_age)

## 
## Call:
## lm(formula = phys_days ~ age, data = brfss_slr)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -13.872  -8.803  -4.733   9.460  23.267 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  4.37023    0.66608   6.561 6.27e-11 ***
## age          0.13127    0.01145  11.467  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 10.92 on 2998 degrees of freedom
## Multiple R-squared:  0.04202,    Adjusted R-squared:  0.0417 
## F-statistic: 131.5 on 1 and 2998 DF,  p-value: < 2.2e-16

# Extract R^2 for age model
R2_age <- summary(model_age)$r.squared

# For comparison, refit BMI model
model_bmi <- lm(phys_days ~ bmi, data = brfss_slr)
R2_bmi <- summary(model_bmi)$r.squared

cat("\nR^2 (Age model):", R2_age, "\n")

## 
## R^2 (Age model): 0.04202056

cat("R^2 (BMI model):", R2_bmi, "\n")

## R^2 (BMI model): 0.008313849

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?

   Both age and BMI show statistically significant associations with physically unhealthy days, but their effects are small in magnitude. The BMI model shows a weak positive association, meaning higher BMI is associated with slightly more physically unhealthy days. The age model also shows a statistically significant relationship, but its slope is small, indicating that changes in age correspond to only minor changes in unhealthy days. In both cases, statistical significance is likely driven by the large sample size rather than strong practical effects.

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

   The predictor with the larger \(R^2\) explains more variability in physically unhealthy days. In these models, both \(R^2\) values are very small, indicating that neither age nor BMI alone explains much of the variation in the outcome. The model with the slightly higher \(R^2\) can be considered the stronger predictor, but the difference is minimal and both predictors have weak explanatory power.

3. Based on these two simple models, what is your overall conclusion about predictors of poor physical health days? What are the limitations of using simple linear regression for this outcome?

   Overall, neither age nor BMI alone is a strong predictor of poor physical health days. This suggests that physically unhealthy days are influenced by many factors beyond a single variable, such as chronic illness, mental health, lifestyle behaviors, socioeconomic status, and access to healthcare. A multiple regression model including several predictors would likely explain substantially more variation.

   Simple linear regression is limited here because the outcome variable (phys_days) is discrete, bounded between 0 and 30, and right-skewed. These characteristics violate normality and constant variance assumptions, which explains the diagnostic issues seen earlier.

---

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  litedown_0.9       htmltools_0.5.9    sass_0.4.10       
## [25] yaml_2.3.12        lazyeval_0.2.2     pillar_1.11.1      jquerylib_0.1.4   
## [29] cachem_1.1.0       nlme_3.1-168       commonmark_2.0.0   tidyselect_1.2.1  
## [33] digest_0.6.39      stringi_1.8.7      labeling_0.4.3     splines_4.5.2     
## [37] rprojroot_2.1.1    fastmap_1.2.0      grid_4.5.2         cli_3.6.5         
## [41] magrittr_2.0.4     cards_0.7.1        utf8_1.2.6         withr_3.0.2       
## [45] scales_1.4.0       backports_1.5.0    timechange_0.3.0   rmarkdown_2.30    
## [49] httr_1.4.7         otel_0.2.0         hms_1.1.4          evaluate_1.0.5    
## [53] viridisLite_0.4.2  mgcv_1.9-3         markdown_2.0       rlang_1.1.7       
## [57] glue_1.8.0         xml2_1.5.1         svglite_2.2.2      rstudioapi_0.18.0 
## [61] jsonlite_2.0.0     R6_2.6.1           systemfonts_1.3.1  fs_1.6.6