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?

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.

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] 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.


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 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
# Overall sample size
nrow(mental_health_data)
## [1] 5757
# Sample size by activity level
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: 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

# 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")
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:

  • Which group has the highest mean number of bad mental health days? Participants reporting no physical activity (“None”) has the highest mean number of days of bad mental health (Mean = 5.08).
  • Which group has the lowest? Participants reporting vigorous physical activity (Vigorous) has the lowest mean (Mean = 3.54).

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 = "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:

  • Do the groups appear to differ? There is difference across each of the group. The distribution of Days of Poor Mental Health for the “None” and “Moderate” activity groups is quite similar, with both showing a median of 0 and similar spread. The “Vigorous” activity group, however, shows slightly lower mean value (3.54 days), indicating fewer days of poor mental health compared with the other two groups.Suggesting higher levels of physical activity may be associated with better mental health, even though the difference between “None” and “Moderate” is minimal.
  • Are the variances similar across groups? The variance across the three groups “None”, “Moderate” and “Vigorous” activity groups is not same. The “None” and “Moderate” physical activity groups show greater variability, with wider boxes (IQRs). There is a less of variability reported for the “vigorous” activity group. —

Step 4: Set Up Hypotheses

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


Step 5: Fit the ANOVA Model

# 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:

  • F-statistic: 23.17
  • Degrees of freedom: 2
  • p-value: 9.52 x 10^(-11)
  • Decision (reject or fail to reject H₀): We reject the null hypothesis
  • Statistical conclusion in words: The p value is 9.52 x 10^(-11), which suggest, there is statistically significant evidence that the mean number of bad mental health days differs across at least two physical activity groups. Indicating the low physical activity is related bad mental health. —

Step 6: Post-Hoc Tests

# 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
# Visualize the confidence intervals
plot(tukey_results, las = 0)

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.

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 <- 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
cat("Percentage of variance explained:", round(eta_squared * 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): small
  • What does this mean practically? There is a difference in the groups. The outcome variability η² = 0.008 which is relatively small and signifies that only 0.8% of variability in the outcome (number of bad mental health days) is explained by the exposure (physical activity). This implies there is a statistically difference between the three groups of physical activity. However, the magnitude of effect is limited.

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 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.

  2. 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.

  3. 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.

  4. 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?

The ANOVA results are statistically significant, since the sample size is large, it does not violates any conclusions.

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:

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.


Reflection Questions

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.


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)