Time: ~30 minutes
Goal: Practice one-way ANOVA analysis from start to finish using real public health data
Learning Objectives:
Structure:
Submission: Upload your completed .Rmd file and published to Brightspace by the end of class.
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.
# 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)")| 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 |
##
## Normal Overweight Obese
## 1939 1937 2150
Interpretation: We have 6026 adults with complete BP and BMI data across three BMI categories.
# 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")| 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).
# 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:
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
# 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:
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.
## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = BPSysAve ~ bmi_category, data = bp_bmi_data)
##
## $bmi_category
## diff lwr upr p adj
## Overweight-Normal 4.507724 3.397134 5.618314 0
## Obese-Normal 7.391744 6.309024 8.474464 0
## Obese-Overweight 2.884019 1.801006 3.967033 0
Interpretation:
| Comparison | Mean Diff | 95% CI | p-value | Significant? |
|---|---|---|---|---|
| Overweight - Normal | 4.51 | [3.4, 5.62] | 3.82e-12 | Yes |
| Obese - Normal | 7.39 | [6.31, 8.47] | < 0.001 | Yes |
| Obese - Overweight | 2.88 | [1.8, 3.97] | 1.38e-09 | Yes |
Conclusion: All three pairwise comparisons are statistically significant. Obese adults have higher SBP than overweight adults, who in turn have higher SBP than normal-weight adults.
# Extract sum of squares from ANOVA table
anova_summary <- summary(anova_model)[[1]]
ss_treatment <- anova_summary$`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
## Percentage of variance explained: 4.11 %
Interpretation: BMI category explains 4.11% of the variance in systolic BP.
While statistically significant, the practical effect is modest—BMI category alone doesn’t explain most of the variation in blood pressure.
ANOVA Assumptions:
Diagnostic Plot Interpretation:
# 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:
Overall Assessment: With n > 2000, ANOVA is robust to minor violations. Our assumptions are reasonably satisfied.
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.
Your Task: Complete the same 9-step analysis workflow you just practiced, but now on a different outcome and predictor.
# 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 the first 6 rows
head(mental_health_data) %>%
kable(caption = "Meantal Health Dataset (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 |
##
## None Moderate Vigorous
## 3139 768 1850
YOUR TURN - Answer these questions:
# 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
summary_stats <- 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 %>%
kable(digits = 2,
caption = "Descriptive Statistics: Bad 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:
Participants in the non activiy level group has the highest mean number of bad mental health days (5.08)
# 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
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 = "Set2") +
labs(
title = "Bad Mental Health Days by Activity Level",
subtitle = "NHANES Data, Adults aged 18+",
x = "Activity Level",
y = "Bad Mental Health Days",
fill = "Activity Level"
) +
theme_minimal(base_size = 12) +
theme(legend.position = "none")YOUR TURN - Describe what you see:
There appears to be a trend: Participants in the higher level physical activity level categories (vigorous) have less bad mental health days.
The variability (box heights) are similar across the none and moderate activity levels. However, the vigorous group has less variability in comparison.
YOUR TURN - Write the hypotheses:
Null Hypothesis (H₀): (H₀):μ_None = μ_Moderate = μ_Vigorous
Alternative Hypothesis (H₁): (H₁):At least one population mean differs from the others.
Significance level: α = 0.05
# YOUR TURN: Fit the ANOVA model
# Outcome: DaysMentHlthBad
# Predictor: activity_level
# Fit the one-way ANOVA model
anova_model <- aov(DaysMentHlthBad ~ activity_level, data = mental_health_data)
# Display the ANOVA table
summary(anova_model)## 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:
# YOUR TURN: Conduct Tukey HSD test
# Only if your ANOVA p-value < 0.05
# Conduct Turkey HSD Test as Anova p-value <0.5
tukey_results <- TukeyHSD(anova_model)
print(tukey_results)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = DaysMentHlthBad ~ activity_level, data = mental_health_data)
##
## $activity_level
## diff lwr upr p adj
## Moderate-None -1.2725867 -2.045657 -0.4995169 0.0003386
## Vigorous-None -1.5464873 -2.109345 -0.9836298 0.0000000
## Vigorous-Moderate -0.2739006 -1.098213 0.5504114 0.7159887
YOUR TURN - Complete the table:
| Comparison | Mean Difference | 95% CI Lower | 95% CI Upper | p-value | Significant? |
|---|---|---|---|---|---|
| Moderate - None | -1.2725867 | -2.045657 | -0.4995169 | 0.0003386 | Yes |
| Vigorous - None | -1.5464873 | -2.109345 | -0.9836298 | 0.0000000 | Yes |
| Vigorous - Moderate | -0.2739006 | -1.098213 | 0.5504114 | 0.7159887 | No |
Interpretation:
Which specific groups differ significantly?
Only the moderate - none and vigorous none comparisons differ significantly. The vigorous - moderate comparison does not differ significantly.
# YOUR TURN: Calculate eta-squared
# Hint: Extract Sum Sq from the ANOVA summary
# Extract sum of squares from ANOVA table
anova_summary <- summary(anova_model)[[1]]
ss_treatment <- anova_summary$`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): 0.008
## Percentage of variance explained: 0.8 %
YOUR TURN - Interpret:
η² = 0.008
Percentage of variance explained: Activity level category explains 0.8% of the variance in bad mental health days.
Effect size classification (small/medium/large): Small (less than 0.01)
What does this mean practically?
While statistically significant, the practical effect is modest-Activity level category alone doesn’t explain most of the variation in bad mental health days.
# YOUR TURN: Create diagnostic plots
# Create diagnostic plots
par(mfrow = c(2, 2))
plot(anova_model)YOUR TURN - Evaluate each plot:
Residuals vs Fitted: Points begin to show random scatter around zero but then slowly move away from zero. There appears to be no clear pattern.
Q-Q Plot: Points do not follow the diagonal line well -> Normality assumption is not reasonable (Skewed Right)
Scale-Location: Red line is relatively flat but shows a gradual increase -> Equal variance assumption is somewhat reasonable.
Residuals vs Leverage: No points beyond Cook’s distance lines -> No highly influential outliers.
# YOUR TURN: Conduct Levene's test
levene_test <- leveneTest(DaysMentHlthBad ~ activity_level, data = mental_health_data)
print(levene_test)## 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? p-value:9.517e-11 If p < 0.05, we reject equal variances Here: Equal variances assumption is not met. = Our assumptions are not reasonably satisfied.
Do any violations threaten your conclusions? = No, with n > 2000, ANOVA is robust to minor violations.
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 the mean number of bad mental health days differs across activity level categories (None, Moderate, Vigorous) among 5757 adults aged 18+ from NHANES.Descriptive statistics showed mean bad mental health days of 5.08 days (SD = 9.01) for the none physical activity level, 3.81 days (SD = 6.87) for moderate physical activity level, and 3.54 days (SD = 7.17) for vigorous activity level.
The ANOVA revealed a statistically significant difference in mean bad mental health days across at least two activity level categories, F(2,5754) = 23.17, p < 0.001. Turkey’s HSD post-hoc test indicated that only two of the three pairwise comparisons were significant (Moderate - None) and (Vigorous - None) (p < 0.001). The last comparison (Vigorous - Moderate) did not differ significantly (p > 0.05): adults participating in moderate physical activity experienced on average 1.2 less bad mental health days than adults participating in no physical activity, and adults participating in vigorous activity level experienced on average 1.5 less bad mental health days than adults participating in no physical activity.
The effect size (η² = 0.008) indicates that activity level category explains 0.8% of the variance in bad mental health days, representing a small practical effect. These findings support the well-established relationship between engaging in some form of physical activity and experiencing lower bad mental health days, though other factors account for most of the variation in bad mental health days
1. How does the effect size help you understand the practical vs. statistical significance?
Effect size can be used to understand practical vs. statistical significance because it measures the magnitude of the relationship difference between variables, rather than just indicating whether a difference exists. Therefore, effect size helps determine whether a statistically significant result is large enough to have a meaningful implication in the real-world.
2. Why is it important to check ANOVA assumptions? What might happen if they’re violated?
It is important to check ANOVA assumptions to ensure the validity of your p-values and inferences. When these assumptions are violated, ANOVA may produce misleading results, such as an increased risk of Type I errors.
3. In public health practice, when might you choose to use ANOVA?
You would choose to use ANOVA when you want to compare 3 or more group means. In public health practice, you may use ANOVA to compare average HGA1C levels across 3 or more different activity level groups.
4. What was the most challenging part of this lab activity?
The most challenging part of this lab activity was step 8: Check Assumptions as I have never interpreted these types of plots before.
Before submitting, verify you have:
To submit: Upload both your .Rmd file and the HTML output to Brightspace.
Lab completed on: February 05, 2026
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) |
Code Execution (4 points):
Completion (4 points):
Interpretation (4 points):
Results Section (3 points):