Lab Overview

Time: ~30 minutes

Goal: Practice one-way ANOVA analysis from start to finish using real public health data

Learning Objectives:

  • Understand when and why to use ANOVA instead of multiple t-tests
  • Set up hypotheses for ANOVA
  • Conduct and interpret the F-test
  • Perform post-hoc tests when appropriate
  • Check ANOVA assumptions
  • Calculate and interpret effect size (η²)

Structure:

  • Part A: Guided Example (follow along)
  • Part B: Your Turn (independent practice)

Submission: Upload your completed .Rmd file and published to Brightspace by the end of class.


PART A: GUIDED EXAMPLE

Example: Blood Pressure and BMI Categories

Research Question: Is there a difference in mean systolic blood pressure (SBP) across three BMI categories (Normal weight, Overweight, Obese)?

Why ANOVA? We have one continuous outcome (SBP) and one categorical predictor with THREE groups (BMI category). Using multiple t-tests would inflate our Type I error rate.


Step 1: Setup and Data Preparation

# Load necessary libraries
library(tidyverse)   # For data manipulation and visualization
library(knitr)       # For nice tables
library(car)         # For Levene's test
library(NHANES)      # NHANES dataset

# Load the NHANES data
data(NHANES)

Create analysis dataset:

# Set seed for reproducibility
set.seed(553)

# Create BMI categories and prepare data
bp_bmi_data <- NHANES %>%
  filter(Age >= 18 & Age <= 65) %>%  # Adults 18-65
  filter(!is.na(BPSysAve) & !is.na(BMI)) %>%
  mutate(
    bmi_category = case_when(
      BMI < 25 ~ "Normal",
      BMI >= 25 & BMI < 30 ~ "Overweight",
      BMI >= 30 ~ "Obese",
      TRUE ~ NA_character_
    ),
    bmi_category = factor(bmi_category, 
                         levels = c("Normal", "Overweight", "Obese"))
  ) %>%
  filter(!is.na(bmi_category)) %>%
  select(ID, Age, Gender, BPSysAve, BMI, bmi_category)

# Display first few rows
head(bp_bmi_data) %>% 
  kable(caption = "Blood Pressure and BMI Dataset (first 6 rows)")
Blood Pressure and BMI Dataset (first 6 rows)
ID Age Gender BPSysAve BMI bmi_category
51624 34 male 113 32.22 Obese
51624 34 male 113 32.22 Obese
51624 34 male 113 32.22 Obese
51630 49 female 112 30.57 Obese
51647 45 female 118 27.24 Overweight
51647 45 female 118 27.24 Overweight
# Check sample sizes
table(bp_bmi_data$bmi_category)
## 
##     Normal Overweight      Obese 
##       1939       1937       2150

Interpretation: We have 6026 adults with complete BP and BMI data across three BMI categories.


Step 2: Descriptive Statistics

# Calculate summary statistics by BMI category
summary_stats <- bp_bmi_data %>%
  group_by(bmi_category) %>%
  summarise(
    n = n(),
    Mean = mean(BPSysAve),
    SD = sd(BPSysAve),
    Median = median(BPSysAve),
    Min = min(BPSysAve),
    Max = max(BPSysAve)
  )

summary_stats %>% 
  kable(digits = 2, 
        caption = "Descriptive Statistics: Systolic BP by BMI Category")
Descriptive Statistics: Systolic BP by BMI Category
bmi_category n Mean SD Median Min Max
Normal 1939 114.23 15.01 113 78 221
Overweight 1937 118.74 13.86 117 83 186
Obese 2150 121.62 15.27 120 82 226

Observation: The mean SBP appears to increase from Normal (114.2) to Overweight (118.7) to Obese (121.6).

But is this difference statistically significant?


Step 3: Visualize the Data

# Create boxplots with individual points
ggplot(bp_bmi_data, 
  aes(x = bmi_category, y = BPSysAve, fill = bmi_category)) +
  geom_boxplot(alpha = 0.7, outlier.shape = NA) +
  geom_jitter(width = 0.2, alpha = 0.1, size = 0.5) +
  scale_fill_brewer(palette = "Set2") +
  labs(
    title = "Systolic Blood Pressure by BMI Category",
    subtitle = "NHANES Data, Adults aged 18-65",
    x = "BMI Category",
    y = "Systolic Blood Pressure (mmHg)",
    fill = "BMI Category"
  ) +
  theme_minimal(base_size = 12) +
  theme(legend.position = "none")

What the plot tells us:

  • There appears to be a trend: higher BMI categories have higher median SBP
  • The boxes overlap, but the obese group appears shifted upward
  • Variability (box heights) looks similar across groups

Step 4: Set Up Hypotheses

Null Hypothesis (H₀): μ_Normal = μ_Overweight = μ_Obese
(All three population means are equal)

Alternative Hypothesis (H₁): At least one population mean differs from the others

Significance level: α = 0.05


Step 5: Fit the ANOVA Model

# Fit the one-way ANOVA model
anova_model <- aov(BPSysAve ~ bmi_category, data = bp_bmi_data)

# Display the ANOVA table
summary(anova_model)
##                Df  Sum Sq Mean Sq F value Pr(>F)    
## bmi_category    2   56212   28106   129.2 <2e-16 ***
## Residuals    6023 1309859     217                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Interpretation:

  • F-statistic: 129.24
  • Degrees of freedom: df₁ = 2 (k-1 groups), df₂ = 6023 (n-k)
  • p-value: < 2e-16 (very small)
  • Decision: Since p < 0.05, we reject H₀
  • Conclusion: There is statistically significant evidence that mean systolic BP differs across at least two BMI categories.

Step 6: Post-Hoc Tests (Tukey HSD)

Why do we need this? The F-test tells us that groups differ, but not which groups differ. Tukey’s Honest Significant Difference controls the family-wise error rate for multiple pairwise comparisons.

# Conduct Tukey HSD test
tukey_results <- TukeyHSD(anova_model)
print(tukey_results)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = BPSysAve ~ bmi_category, data = bp_bmi_data)
## 
## $bmi_category
##                       diff      lwr      upr p adj
## Overweight-Normal 4.507724 3.397134 5.618314     0
## Obese-Normal      7.391744 6.309024 8.474464     0
## Obese-Overweight  2.884019 1.801006 3.967033     0
# Visualize the confidence intervals
plot(tukey_results, las = 0)

Interpretation:

Comparison Mean Diff 95% CI p-value Significant?
Overweight - Normal 4.51 [3.4, 5.62] 1.98e-13 Yes
Obese - Normal 7.39 [6.31, 8.47] < 0.001 Yes
Obese - Overweight 2.88 [1.8, 3.97] 1.38e-09 Yes

Conclusion: All three pairwise comparisons are statistically significant. Obese adults have higher SBP than overweight adults, who in turn have higher SBP than normal-weight adults.


Step 7: Calculate Effect Size

# Extract sum of squares from ANOVA table
anova_summary <- summary(anova_model)[[1]]

ss_treatment <- anova_summary$`Sum Sq`[1]
ss_total <- sum(anova_summary$`Sum Sq`)

# Calculate eta-squared
eta_squared <- ss_treatment / ss_total

cat("Eta-squared (η²):", round(eta_squared, 4), "\n")
## Eta-squared (η²): 0.0411
cat("Percentage of variance explained:", round(eta_squared * 100, 2), "%")
## Percentage of variance explained: 4.11 %

Interpretation: BMI category explains 4.11% of the variance in systolic BP.

  • Effect size guidelines: Small (0.01), Medium (0.06), Large (0.14)
  • Our effect: Small

While statistically significant, the practical effect is modest—BMI category alone doesn’t explain most of the variation in blood pressure.


Step 8: Check Assumptions

ANOVA Assumptions:

  1. Independence: Observations are independent (assumed based on study design)
  2. Normality: Residuals are approximately normally distributed
  3. Homogeneity of variance: Equal variances across groups
# Create diagnostic plots
par(mfrow = c(2, 2))
plot(anova_model)

par(mfrow = c(1, 1))

Diagnostic Plot Interpretation:

  1. Residuals vs Fitted: Points show random scatter around zero with no clear pattern → Good!
  2. Q-Q Plot: Points follow the diagonal line reasonably well → Normality assumption is reasonable
  3. Scale-Location: Red line is relatively flat → Equal variance assumption is reasonable
  4. Residuals vs Leverage: No points beyond Cook’s distance lines → No highly influential outliers
# Levene's test for homogeneity of variance
levene_test <- leveneTest(BPSysAve ~ bmi_category, data = bp_bmi_data)
print(levene_test)
## Levene's Test for Homogeneity of Variance (center = median)
##         Df F value  Pr(>F)  
## group    2  2.7615 0.06328 .
##       6023                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Levene’s Test Interpretation:

  • p-value: 0.0633
  • If p < 0.05, we would reject equal variances
  • Here: Equal variance assumption is met

Overall Assessment: With n > 2000, ANOVA is robust to minor violations. Our assumptions are reasonably satisfied.


Step 9: Report Results

Example Results Section:

We conducted a one-way ANOVA to examine whether mean systolic blood pressure (SBP) differs across BMI categories (Normal, Overweight, Obese) among 6,026 adults aged 18-65 from NHANES. Descriptive statistics showed mean SBP of 114.2 mmHg (SD = 15) for normal weight, 118.7 mmHg (SD = 13.9) for overweight, and 121.6 mmHg (SD = 15.3) for obese individuals.

The ANOVA revealed a statistically significant difference in mean SBP across BMI categories, F(2, 6023) = 129.24, p < 0.001. Tukey’s HSD post-hoc tests indicated that all pairwise comparisons were significant (p < 0.05): obese adults had on average 7.4 mmHg higher SBP than normal-weight adults, and 2.9 mmHg higher than overweight adults.

The effect size (η² = 0.041) indicates that BMI category explains 4.1% of the variance in systolic blood pressure, representing a small practical effect. These findings support the well-established relationship between higher BMI and elevated blood pressure, though other factors account for most of the variation in SBP.


PART B: YOUR TURN - INDEPENDENT PRACTICE

Practice Problem: Physical Activity and Depression

Research Question: Is there a difference in the number of days with poor mental health across three physical activity levels (None, Moderate, Vigorous)?

Your Task: Complete the same 9-step analysis workflow you just practiced, but now on a different outcome and predictor.


Step 1: Data Preparation

# Prepare the dataset
set.seed(553)

mental_health_data <- NHANES %>%
  filter(Age >= 18) %>%
  filter(!is.na(DaysMentHlthBad) & !is.na(PhysActive)) %>%
  mutate(
    activity_level = case_when(
      PhysActive == "No" ~ "None",
      PhysActive == "Yes" & !is.na(PhysActiveDays) & PhysActiveDays < 3 ~ "Moderate",
      PhysActive == "Yes" & !is.na(PhysActiveDays) & PhysActiveDays >= 3 ~ "Vigorous",
      TRUE ~ NA_character_
    ),
    activity_level = factor(activity_level, 
                           levels = c("None", "Moderate", "Vigorous"))
  ) %>%
  filter(!is.na(activity_level)) %>%
  select(ID, Age, Gender, DaysMentHlthBad, PhysActive, activity_level)

# YOUR TURN: Display the first 6 rows and check sample sizes

# Display first few rows
head(mental_health_data) %>% 
  kable(caption = "Mental Health Days and Physical Activity (first 6 rows)")
Mental Health Days and Physical Activity (first 6 rows)
ID Age Gender DaysMentHlthBad PhysActive activity_level
51624 34 male 15 No None
51624 34 male 15 No None
51624 34 male 15 No None
51630 49 female 10 No None
51647 45 female 3 Yes Vigorous
51647 45 female 3 Yes Vigorous
# Check sample sizes
table(mental_health_data$activity_level)
## 
##     None Moderate Vigorous 
##     3139      768     1850

YOUR TURN - Answer these questions:

  • How many people are in each physical activity group?
    • None: 3,139
    • Moderate: 768
    • Vigorous: 1,850

Step 2: Descriptive Statistics

# YOUR TURN: Calculate summary statistics by activity level
# Hint: Follow the same structure as the guided example
# Variables to summarize: n, Mean, SD, Median, Min, Max

# Calculate summary statistics by Activity Level
summary_stats_pb <- mental_health_data %>%
  group_by(activity_level) %>%
  summarise(
    n = n(),
    Mean = mean(DaysMentHlthBad),
    SD = sd(DaysMentHlthBad),
    Median = median(DaysMentHlthBad),
    Min = min(DaysMentHlthBad),
    Max = max(DaysMentHlthBad)
  )

summary_stats_pb %>% 
  kable(digits = 2, 
        caption = "Descriptive Statistics: Poor Mental Health Days by Activity Level")
Descriptive Statistics: Poor Mental Health Days by Activity Level
activity_level n Mean SD Median Min Max
None 3139 5.08 9.01 0 0 30
Moderate 768 3.81 6.87 0 0 30
Vigorous 1850 3.54 7.17 0 0 30

YOUR TURN - Interpret: - Which group has the highest mean number of bad mental health days?

–The group that had no or none physical activity had the highest mean number of bad mental health days with a mean of 5.08 days.

  • Which group has the lowest?

–The group that had vigorous phyisical activity had the lowest mean number of bad mental health days, with a mean of 3.54 days.


Step 3: Visualization

# YOUR TURN: Create boxplots comparing DaysMentHlthBad across activity levels
# Hint: Use the same ggplot code structure as the example
# Change variable names and labels appropriately

# Create boxplots with individual points
ggplot(mental_health_data, 
  aes(x = activity_level, y = DaysMentHlthBad, fill = activity_level)) +
  geom_boxplot(alpha = 0.7, outlier.shape = NA) +
  geom_jitter(width = 0.2, alpha = 0.1, size = 0.5) +
  scale_fill_brewer(palette = "Set1") +
  labs(
    title = "Poor Mental Health Days by Activity Level",
    subtitle = "NHANES Data, Adults aged 18-65",
    x = "Activity Level",
    y = "Poor Mental Health Days (Days)",
    fill = "Activity Level"
  ) +
  theme_minimal(base_size = 12) +
  theme(legend.position = "none")

YOUR TURN - Describe what you see:

  • Do the groups appear to differ? – Yes, the groups appear to be different. The vigorous group is smaller compared to the None and Moderate group.

  • Are the variances similar across groups? –The varainces don’t seem to be similar across groups.The None group seem to have a noticeable grouping at around 15 days and 30 days unlike Moderate and Vigorous. —

Step 4: Set Up Hypotheses

YOUR TURN - Write the hypotheses:

Null Hypothesis (H₀): μ_None = μ_Moderate = μ_Vigorous (All three population means are equal)

Alternative Hypothesis (H₁): At least one population mean differs from the others

Significance level: α = 0.05


Step 5: Fit the ANOVA Model

# YOUR TURN: Fit the ANOVA model
# Outcome: DaysMentHlthBad
# Predictor: activity_level

# Fit the one-way ANOVA model
anova_model_pb <- aov(DaysMentHlthBad ~ activity_level, data = mental_health_data)

# Display the ANOVA table
summary(anova_model_pb)
##                  Df Sum Sq Mean Sq F value   Pr(>F)    
## activity_level    2   3109  1554.6   23.17 9.52e-11 ***
## Residuals      5754 386089    67.1                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

YOUR TURN - Extract and interpret the results:

  • F-statistic: 23.17
  • Degrees of freedom: 2
  • p-value: 9.52e-11
  • Decision (reject or fail to reject H₀): Reject the null hypothesis
  • Statistical conclusion in words: There is statistically significant evidence which means poor mental health days differs across at least two physical activity categories.

Step 6: Post-Hoc Tests

# YOUR TURN: Conduct Tukey HSD test
# Only if your ANOVA p-value < 0.05

# Conduct Tukey HSD test
tukey_results_pb <- TukeyHSD(anova_model_pb)
print(tukey_results_pb)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = DaysMentHlthBad ~ activity_level, data = mental_health_data)
## 
## $activity_level
##                         diff       lwr        upr     p adj
## Moderate-None     -1.2725867 -2.045657 -0.4995169 0.0003386
## Vigorous-None     -1.5464873 -2.109345 -0.9836298 0.0000000
## Vigorous-Moderate -0.2739006 -1.098213  0.5504114 0.7159887
# Visualize the confidence intervals
plot(tukey_results_pb, las = 0)

YOUR TURN - Complete the table:

Interpretation:

Comparison Mean Diff 95% CI p-value Significant?
Moderate - None -1.27 [3.4, -0.5] 0.000339 Yes
Vigorous - None -1.55 [-2.11, -0.98] < 0.001 Yes
Vigorous - Moderate -0.27 [-1.1, 0.55] 0.716 No

Conclusion: Two of the three pairwise comparisons are statistically significant. Individuals with no or none physical activity have more poor mental health days compared to individuals with moderate or vigorous physical activity. There are no statistical difference between those who with vigorous or moderate physical activity when it comes to poor mental health days.


Step 7: Calculate Effect Size

# YOUR TURN: Calculate eta-squared
# Hint: Extract Sum Sq from the ANOVA summary

# Extract sum of squares from ANOVA table
anova_summary_pb <- summary(anova_model_pb)[[1]]

ss_treatment_pb <- anova_summary_pb$`Sum Sq`[1]
ss_total_pb <- sum(anova_summary_pb$`Sum Sq`)

# Calculate eta-squared
eta_squared_pb <- ss_treatment_pb / ss_total_pb

cat("Eta-squared (η²):", round(eta_squared_pb, 4), "\n")
## Eta-squared (η²): 0.008
cat("Percentage of variance explained:", round(eta_squared_pb * 100, 2), "%")
## Percentage of variance explained: 0.8 %

YOUR TURN - Interpret:

  • η² = 0.008

  • Percentage of variance explained: –0.8%

  • Effect size classification (small/medium/large): –Large effect size

  • What does this mean practically? – While statistically significant, the practical effect is none-physical activity category alone can explain most of the variation in poor mentak health days.


Step 8: Check Assumptions

  1. Independence: Observations are independent (assumed based on study design)
  2. Normality: Residuals are approximately normally distributed
  3. Homogeneity of variance: Equal variances across groups
# YOUR TURN: Create diagnostic plots
# Create diagnostic plots
par(mfrow = c(2, 2))
plot(anova_model_pb)

par(mfrow = c(1, 1))

YOUR TURN - Evaluate each plot:

Diagnostic Plot Interpretation:

  1. Residuals vs Fitted: Points show random scatter around zero with no clear pattern → Good!
  2. Q-Q Plot: Points dont exactly follow the diagonal line reasonably well → Normality assumption is not reasonable
  3. Scale-Location: Red line is relatively flat → Equal variance assumption is reasonable
  4. Residuals vs Leverage: No points beyond Cook’s distance lines → No highly influential outliers
# YOUR TURN: Conduct Levene's test

levene_test_pb <- leveneTest(DaysMentHlthBad ~ activity_level, data = mental_health_data)
print(levene_test_pb)
## Levene's Test for Homogeneity of Variance (center = median)
##         Df F value    Pr(>F)    
## group    2  23.168 9.517e-11 ***
##       5754                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

YOUR TURN - Overall assessment:

  • Are assumptions reasonably met? – yes, I believe that the assumptions are reasonably met.

  • Do any violations threaten your conclusions?

  • p-value: 9.52e-11

  • If p < 0.05, we would reject equal variances

  • Here: Variances differ significantly, but with large n, ANOVA is robust

Overall Assessment: –With n > 2000, ANOVA is robust to minor violations. Our assumptions are reasonably satisfied. —

Step 9: Write Up Results

YOUR TURN - Write a complete 2-3 paragraph results section:

Include: 1. Sample description and descriptive statistics 2. F-test results 3. Post-hoc comparisons (if applicable) 4. Effect size interpretation 5. Public health significance

Your Results Section:

We conducted a one-way ANOVA to examine whether poor mental health days differs across physical activity levels (None, Moderate, Vigorous) among 5,757 adults aged 18-65 from NHANES. Descriptive statistics showed poor mental health days of 5.1 Days (SD = 9) for None, 3.8 Days (SD = 6.9) for Moderate, and 3.5 Days (SD = 7.2) for Vigorous individuals.

The ANOVA revealed a statistically significant difference in poor mental health between no physical activity and both moderate and vigorous physical activity, F(2, 5754) = 23.17, p < 0.001. Tukey’s HSD post-hoc tests indicated that all pairwise comparisons were significant (p < 0.05): No physical activity individuals had on average -1.5 more poor mental health days than moderate physical activity individuals, and -0.3 more poor mental health than physical activity individuals.

The effect size (η² = 0.008) indicates that physical activity level explains 0.8% of the variance in poor mental health days, representing a small practical effect. These findings support the well-established relationship between physical activity and poor mental health days, though other factors account for some of the variation in poor mental health days.


Reflection Questions

1. How does the effect size help you understand the practical vs. statistical significance?

The effect size helps in understand whether or not the difference that are seen can truly be attributed to the relationship between the variables or if there is another factor that accounts for some or most of the difference that are present. There could be significant statistical difference between groups but the effect size could be small which would indicate that something else is occurring.

2. Why is it important to check ANOVA assumptions? What might happen if they’re violated?

It is important to check ANOVA assumptions because if they are not met they are violated and you have a small sample size then any difference between groups can not be attributed to the predictor variable.

3. In public health practice, when might you choose to use ANOVA?

ANOVA is an acceptable test in public health practice when you have 3 or more group means that you want to test if there are any difference between them. The health outcome must be a continuous variable while the predictor variable must be categorical.

4. What was the most challenging part of this lab activity?

The most challenging part of this lab was working through the an error in the data visualization chuck. For whatever reason the chuck was giving an error when nothing was wrong so troubleshooting that was challenging to deal with while working through the activity.


Submission Checklist

Before submitting, verify you have:

To submit: Upload both your .Rmd file and the HTML output to Brightspace.


Lab completed on: February 05, 2026


GRADING RUBRIC (For TA Use)

Total Points: 15

Category Criteria Points Notes
Code Execution All code chunks run without errors 4 - Deduct 1 pt per major error
- Deduct 0.5 pt per minor warning
Completion All “YOUR TURN” sections attempted 4 - Part B Steps 1-9 completed
- All fill-in-the-blank answered
- Tukey table filled in
Interpretation Correct statistical interpretation 4 - Hypotheses correctly stated (1 pt)
- ANOVA results interpreted (1 pt)
- Post-hoc results interpreted (1 pt)
- Assumptions evaluated (1 pt)
Results Section Professional, complete write-up 3 - Includes descriptive stats (1 pt)
- Reports F-test & post-hoc (1 pt)
- Effect size & significance (1 pt)

Detailed Grading Guidelines

Code Execution (4 points):

  • 4 pts: All code runs perfectly, produces correct output
  • 3 pts: Minor issues (1-2 small errors or warnings)
  • 2 pts: Several errors but demonstrates understanding
  • 1 pt: Major errors, incomplete code
  • 0 pts: Code does not run at all

Completion (4 points):

  • 4 pts: All sections attempted thoughtfully
  • 3 pts: 1-2 sections incomplete or minimal effort
  • 2 pts: Several sections missing
  • 1 pt: Only partial completion
  • 0 pts: Little to no work completed

Interpretation (4 points):

  • 4 pts: All interpretations correct and well-explained
  • 3 pts: Minor errors in interpretation
  • 2 pts: Several interpretation errors
  • 1 pt: Significant misunderstanding of concepts
  • 0 pts: No interpretation provided

Results Section (3 points):

  • 3 pts: Publication-quality, complete results section
  • 2 pts: Good but missing some elements
  • 1 pt: Incomplete or poorly written
  • 0 pts: No results section written

Common Deductions

  • -0.5 pts: Missing sample sizes in write-up
  • -0.5 pts: Not reporting confidence intervals
  • -1 pt: Incorrect hypothesis statements
  • -1 pt: Misinterpreting p-values
  • -1 pt: Not checking assumptions
  • -0.5 pts: Poor formatting (no tables, unclear output)