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).
It is difficult to estimate the statistical significance, as we do not have the p value or the confidence interval to conclude the significance level.
# 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)## # A tibble: 6 × 6
## ID Age Gender DaysMentHlthBad PhysActive activity_level
## <int> <int> <fct> <int> <fct> <fct>
## 1 51624 34 male 15 No None
## 2 51624 34 male 15 No None
## 3 51624 34 male 15 No None
## 4 51630 49 female 10 No None
## 5 51647 45 female 3 Yes Vigorous
## 6 51647 45 female 3 Yes Vigorous
## [1] 5757
##
## 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
# Calculate summary statistics by BMI category
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:
# 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 = "Set2") +
labs(
title = "Days of Bad Mental Health by Activity Levels",
subtitle = "NHANES Data, Adults aged 18+",
x = "Activity Levels",
y = "Days of Bad Mental Health",
fill = "Activity Category"
) +
theme_minimal(base_size = 12) +
theme(legend.position = "none")YOUR TURN - Describe what you see:
YOUR TURN - Write the hypotheses:
Null Hypothesis (H₀): Null Hypothesis (H₀): μ_None = μ_Moderate = μ_Vigorous (All three population means are equal) There is no difference in the mean number of days of bad mental health across the three groups of physical activity ” none”, “moderate”, “vigorous”. Alternative Hypothesis (H₁): Alternative Hypothesis (H₁): At least one population mean differs from the others There is significant difference in the mean number of days of bad mental health across the three groups of physical activity ” none”, “moderate”, “vigorous”. Significance level: α = Significance level: α = 0.05
# YOUR TURN: Fit the ANOVA model
# Outcome: DaysMentHlthBad
# Predictor: activity_level
anova_model <- aov(DaysMentHlthBad ~ activity_level, data = mental_health_data)
#Display 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
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)
##
## $activity_level
## diff lwr upr p adj
## Moderate-None -1.2725867 -2.045657 -0.4995169 0.0003386
## Vigorous-None -1.5464873 -2.109345 -0.9836298 0.0000000
## Vigorous-Moderate -0.2739006 -1.098213 0.5504114 0.7159887
YOUR TURN - Complete the table:
| Comparison | Mean Difference | 95% CI Lower | 95% CI Upper | p-value | Significant? |
|---|---|---|---|---|---|
| Moderate - None -1.2725867. -2.045657. -0.4995169. 0.0003386. yes | |||||
| Vigorous - None -1.5464873. -2.109345. -0.9836298. 0.0000000. yes | |||||
| Vigorous - Moderate -0.2739006. -1.098213. 0.5504114. 0.7159887. no |
Interpretation:
Which specific groups differ significantly?
The activity group with the moderate-none group and vigorous-none group differs statistically significant with a p-value of -0.0003386. (CI:-2.045657 to-0.4995169) and p value 0.0000000 (CI: -2.109345 to -0.9836298) respectively.
# YOUR TURN: Calculate eta-squared
# Hint: Extract Sum Sq from the ANOVA summary
# Extract sum of squares from ANOVA table
anova_summary <- summary(anova_model)[[1]]
ss_treatment <- anova_summary$`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:
YOUR TURN - Evaluate each plot:
Residuals vs Fitted: The points across the three groups has the same height. All the clusters are mostly above 0. This shows the effect size is small, therefore there is less of group differences.
Q-Q Plot: The points follow the diagonal line to some extent and then there is a sudden shift on the upper end. This does not reflects normality.
Scale-Location: Scale plot reflect the homogeneity of variance assumption. In this case the red line is horizontally tilted. Moreover, the data points shows a upward trend from the red line. This means that equal variance assumption is not reasonable.
Residuals vs Leverage: Identifies outliers and influential observations. No data points are beyond the cook’s distance lines, thus no data points are influencing the estimates.
# YOUR TURN: Conduct Levene's test
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
YOUR TURN - Overall assessment:
Are assumptions reasonably met? The levene value is 0.063, which is more than 0.05 therefore, we will fail to reject null and will conclude that assumption of equal variance across the three groups of physical activity was met.
Do any violations threaten your conclusions?
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:
This study examines the effect of physical activity on the number of days with bad mental health among the adults from NHANES data adults aged adults aged 18-65. The overall sample size was 5757. The physical activity done by the individuals were divided into three groups with “none” (n=3139), “moderate” (n=768) and “vigorous” (n=1850). The mean physical activity performed by the three groups were reported as 5.08 (SD=9.01) in the none group. In the moderate group of physical activity mean was reported as 3.81 (SD=6.87). In the vigorous activity group the mean was reported as 3.54 (SD=7.17). The median was reported as 0 for all the three groups. A one way ANOVA was conducted to measure the difference among the three groups on the effect of physical activity on the number of days with bad mental health. It was found there is a significant difference among the three groups of physical activity (p value= 9.52 x 10^(-11). Indicating the number of days of physical activity is related with bad mental health. Although the effect size was small (η² = 0.008), explaining only 0.8% of the variance. Levene’s test confirmed that the assumption of equal variances was reasonably met (F = 2.76, p = 0.063). Through this study we can conclude that engaging in physical activity is associated with a modest reduction in days with bad mental health. Public health interventions that promote physical activity may therefore help improve mental health outcomes at the population level.
1. How does the effect size help you understand the practical vs. statistical significance? Effect size quantifies the magnitude of the relationship or difference, independent of sample size. In this study, η² = 0.008 represents the proportion of total variance in days with bad mental health explained by physical activity. While the ANOVA shows that differences among groups are statistically significant, the small effect size indicates that physical activity accounts for only a modest portion (0.8%) of the variability in mental health outcomes. Reporting effect size allows researchers to assess practical significance, showing whether differences are meaningful in real-world terms, not just statistically detectable.
2. Why is it important to check ANOVA assumptions? What might happen if they’re violated? Checking ANOVA assumptions is crucial because ANOVA relies on certain conditions to ensure that the results are valid, interpretable, and reliable. If assumptions are violated, the statistical conclusions (p-values, F-statistics, confidence intervals) can be misleading.
3. In public health practice, when might you choose to use ANOVA?
ANOVA would be applicable when there are more than one treatment groups, exposure levels. It is used when the predictor is categorical in nature and outcome is continuous in nature.
4. What was the most challenging part of this lab activity?
I really like the flow of questions. It covers all the aspects which we need to think in order to conclude the results. The assignment was well designed and systematic. However, Its a lengthy assignment. I wish few more days could be given for the submission.
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):