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

# Sample sizes by physical activity group
mental_health_data %>%
  count(activity_level)
## # A tibble: 3 × 2
##   activity_level     n
##   <fct>          <int>
## 1 None            3139
## 2 Moderate         768
## 3 Vigorous        1850

YOUR TURN - Answer these questions:

  • How many people are in each physical activity group?
    • None: 3139
    • Moderate: 768
    • Vigorous: 1850

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
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)
  )
## # A tibble: 3 × 7
##   activity_level     n  Mean    SD Median   Min   Max
##   <fct>          <int> <dbl> <dbl>  <dbl> <int> <int>
## 1 None            3139  5.08  9.01      0     0    30
## 2 Moderate         768  3.81  6.87      0     0    30
## 3 Vigorous        1850  3.54  7.17      0     0    30

YOUR TURN - Interpret:

  • Which group has the highest mean number of bad mental health days? None
  • Which group has the lowest? Vigorous

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
library(ggplot2)

ggplot(mental_health_data, aes(x = activity_level, y = DaysMentHlthBad)) +
  geom_boxplot(fill = "lightblue") +
  labs(
    title = "Days of Poor Mental Health by Physical Activity Level",
    x = "Physical Activity Level",
    y = "Days of Poor Mental Health (Past 30 Days)"
  ) +
  theme_minimal()

YOUR TURN - Describe what you see:

  • Do the groups appear to differ? Yes. The None physical activity group shows a higher median and higher overall distribution of bad mental health days compared to the Moderate and Vigorous groups. The Moderate and Vigorous groups appear more similar to each other and both have lower central values than the None group.
  • Are the variances similar across groups? The variances appear somewhat similar, though the None group shows slightly greater spread, with more extreme values and a wider interquartile range. However, the differences in variability do not appear extreme, suggesting the homogeneity of variance assumption is reasonably met.

Step 4: Set Up Hypotheses

YOUR TURN - Write the hypotheses:

Null Hypothesis (H₀): The mean number of days of poor mental health is equal across all physical activity levels (None, Moderate, and Vigorous).

H0: μNone=μModerate=μVigorous

Alternative Hypothesis (H₁): At least one physical activity group has a mean number of days of poor mental health that is different from the others.

H1: At least one μ differs

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 <- 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:

  • F-statistic: 23.17
  • Degrees of freedom: F(2, 5754) Between groups: 2 Within groups: 5754
  • p-value: p = 9.52 × 10⁻¹¹ or p < 0.001
  • Decision (reject or fail to reject H₀): Reject H₀
  • Statistical conclusion in words: There is a statistically significant difference in the mean number of poor mental health days (DaysMentHlthBad) across physical activity levels, F(2, 5754) = 23.17, p < 0.001. We reject the null hypothesis that the mean number of poor mental health days is equal across activity level groups.

Step 6: Post-Hoc Tests

# YOUR TURN: Conduct Tukey HSD test
# Only if your ANOVA p-value < 0.05
anova_model <- aov(DaysMentHlthBad ~ activity_level, data = mental_health_data)
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
TukeyHSD(anova_model)
##   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 −1.27 −2.05 −0.50 0.00034 Yes
Vigorous - None −1.55 −2.11 −0.98 < 0.001 Yes
Vigorous - Moderate −0.27 −1.10 0.55 0.716 No

Interpretation:

Which specific groups differ significantly? Tukey post-hoc comparisons showed that participants in the Moderate and Vigorous physical activity groups reported significantly fewer days of poor mental health than those in the None group. There was no statistically significant difference in the mean number of bad mental health days between the Moderate and Vigorous activity groups.


Step 7: Calculate Effect Size

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

anova_summary <- summary(anova_model)[[1]]

eta_sq <- anova_summary["activity_level", "Sum Sq"] /
          sum(anova_summary[, "Sum Sq"])

eta_sq
## [1] 0.007988564

YOUR TURN - Interpret:

  • η² = 0.008
  • Percentage of variance explained: 0.008×100=0.8%
  • Effect size classification (small/medium/large): According to common guidelines for η²: Small ≈ 0.01 Medium ≈ 0.06 Large ≈ 0.14. Small effect
  • What does this mean practically? Although physical activity level is statistically significantly associated with the number of days of poor mental health, it explains less than 1% of the total variance. This suggests that while physical activity matters, many other factors also contribute to mental health outcomes.

Step 8: Check Assumptions

# YOUR TURN: Create diagnostic plots
par(mfrow = c(2, 2))
plot(anova_model)

par(mfrow = c(1, 1))

YOUR TURN - Evaluate each plot:

  1. Residuals vs Fitted: The residuals appear to be randomly scattered around zero with no strong systematic pattern. This suggests that the linearity and independence assumptions are reasonably met, although slight clustering is expected with a discrete outcome variable.

  2. Q-Q Plot: The points follow the reference line fairly closely, with some deviation in the tails. Given the large sample size, these minor departures from normality are not concerning, and the normality assumption is reasonably satisfied.

  3. Scale-Location: The spread of residuals is relatively consistent across fitted values, with no strong funnel shape. This suggests that the homogeneity of variance assumption is adequately met.

  4. Residuals vs Leverage: No observations appear to have both high leverage and large residuals. There are no influential points that would unduly affect the ANOVA results.

# YOUR TURN: Conduct Levene's test
library(car)

leveneTest(DaysMentHlthBad ~ activity_level, data = mental_health_data)
## 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. Levene’s test is not statistically significant (p > 0.05), indicating that the variances of days of poor mental health are not significantly different across physical activity groups. This supports the homogeneity of variance assumption required for ANOVA.
  • Do any violations threaten your conclusions? No. Even if minor deviations from normality or equal variances were present, the very large sample size makes ANOVA robust to these violations. Therefore, none of the assumptions violations meaningfully threaten the validity of the ANOVA results.

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:

The analytic sample included 5,757 adults aged 18 years and older with complete data on physical activity and mental health. Participants were categorized into three physical activity groups: None (n = 3,139), Moderate (n = 768), and Vigorous (n = 1,850). Descriptive statistics indicated that the mean number of days of poor mental health in the past 30 days was highest among individuals reporting no physical activity (M = 5.08, SD = 9.01), followed by those engaging in moderate physical activity (M = 3.81, SD = 6.87), and lowest among those engaging in vigorous physical activity (M = 3.54, SD = 7.17). Median values were zero across all groups, reflecting a right-skewed distribution with many participants reporting no days of poor mental health.

A one-way analysis of variance (ANOVA) was conducted to examine differences in mean days of poor mental health across physical activity levels. The ANOVA revealed a statistically significant effect of physical activity level on days of poor mental health, F(2, 5754) = 23.17, p < .001. Tukey post-hoc comparisons showed that individuals in both the Moderate (mean difference = −1.27, p < .001) and Vigorous (mean difference = −1.55, p < .001) activity groups reported significantly fewer days of poor mental health compared to those in the None group. However, there was no statistically significant difference between the Moderate and Vigorous activity groups (p = .716).

The effect size for the ANOVA was small (η² = 0.008), indicating that physical activity level explained approximately 0.8% of the variance in days of poor mental health. While the magnitude of the effect was modest, the findings have important public health implications. Even small reductions in poor mental health days at the population level may translate into meaningful improvements in overall well-being. These results suggest that engaging in regular physical activity—whether moderate or vigorous—is associated with fewer days of poor mental health compared to no physical activity, highlighting physical activity as a potentially important and accessible target for mental health promotion.


Reflection Questions

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

The effect size provides information about the magnitude of the relationship, not just whether it is statistically significant. In this analysis, the ANOVA was statistically significant, but the effect size (η² ≈ 0.008) was small, indicating that physical activity level explains less than 1% of the variation in days of poor mental health. This helps distinguish statistical significance, which can be influenced by large sample sizes, from practical significance, which reflects how meaningful the difference is in real-world terms.

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

Checking ANOVA assumptions is important because violations can lead to invalid results, such as inflated Type I error rates or reduced statistical power. If assumptions like normality or homogeneity of variance are severely violated, the F-test may not accurately reflect true group differences. Although ANOVA is fairly robust to minor violations, especially with large samples, serious violations could result in incorrect conclusions about statistical significance.

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

ANOVA is useful in public health when comparing the mean value of a continuous outcome across three or more groups, such as comparing average blood pressure across different treatment groups or mental health outcomes across levels of health behaviors. It allows researchers to test for overall group differences before conducting more detailed comparisons, making it an efficient tool for evaluating group-based interventions or population subgroups.

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

The most challenging part of this lab was interpreting the results beyond statistical significance, particularly understanding how a small effect size can still be meaningful in a large public health context. Additionally, correctly setting up and interpreting post-hoc comparisons required careful attention to reference groups and confidence intervals.


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)