Homework 1 overview

Due: Thursday, February 20, 2026 (11:59 pm ET)
Topics: One-way ANOVA + Correlation (based on Lectures/Labs 02–03)
Dataset: bmd.csv (NHANES 2017–2018 DEXA subsample)
What to submit (2 items):

  1. Your .Rmd file
  2. Your RPubs link (published HTML)

AI policy (Homework 1): AI tools (ChatGPT, Gemini, Claude, Copilot, etc.) are not permitted for coding assistance on HW1. You must write/debug your code independently.

Required filename for your submission:

EPI553_HW01_LastName_FirstName.Rmd

Required RPubs title (use this exact pattern):

epi553_hw01_lastname_firstname

This will create an RPubs URL that ends in something like:
https://rpubs.com/your_username/epi553_hw01_lastname_firstname

Data description

This homework uses bmd.csv, a subset of NHANES 2017–2018 respondents who completed a DEXA scan.

Key variables:

Variable Description Type
DXXOFBMD Total femur bone mineral density Continuous (g/cm²)
RIDRETH1 Race/ethnicity Categorical (1–5)
RIAGENDR Sex Categorical (1=Male, 2=Female)
RIDAGEYR Age in years Continuous
BMXBMI Body mass index Continuous (kg/m²)
smoker Smoking status Categorical (1=Current, 2=Past, 3=Never)
calcium Dietary calcium intake Continuous (mg/day)
DSQTCALC Supplemental calcium Continuous (mg/day)
vitd Dietary vitamin D intake Continuous (mcg/day)
DSQTVD Supplemental vitamin D Continuous (mcg/day)
totmet Total physical activity (MET-min/week) Continuous

Part 0 — Load and prepare the data (10 points)

0.1 Import

Load required packages and import the dataset.

# Load required packages
library(tidyverse)
library(car)
library(kableExtra)
library(broom)
library(dplyr)
library(patchwork)
# Import data (adjust path as needed)
bmd <- readr::read_csv("/Users/jingjunyang/Desktop/EPI553 Project/bmd.csv", show_col_types = FALSE)

# Inspect the data
glimpse(bmd)
## Rows: 2,898
## Columns: 14
## $ SEQN     <dbl> 93705, 93708, 93709, 93711, 93713, 93714, 93715, 93716, 93721…
## $ RIAGENDR <dbl> 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2…
## $ RIDAGEYR <dbl> 66, 66, 75, 56, 67, 54, 71, 61, 60, 60, 64, 67, 70, 53, 57, 7…
## $ RIDRETH1 <dbl> 4, 5, 4, 5, 3, 4, 5, 5, 1, 3, 3, 1, 5, 4, 2, 3, 2, 4, 4, 3, 3…
## $ BMXBMI   <dbl> 31.7, 23.7, 38.9, 21.3, 23.5, 39.9, 22.5, 30.7, 35.9, 23.8, 2…
## $ smoker   <dbl> 2, 3, 1, 3, 1, 2, 1, 2, 3, 3, 3, 2, 2, 3, 3, 2, 3, 1, 2, 1, 1…
## $ totmet   <dbl> 240, 120, 720, 840, 360, NA, 6320, 2400, NA, NA, 1680, 240, 4…
## $ metcat   <dbl> 0, 0, 1, 1, 0, NA, 2, 2, NA, NA, 2, 0, 0, 0, 1, NA, 0, NA, 1,…
## $ DXXOFBMD <dbl> 1.058, 0.801, 0.880, 0.851, 0.778, 0.994, 0.952, 1.121, NA, 0…
## $ tbmdcat  <dbl> 0, 1, 0, 1, 1, 0, 0, 0, NA, 1, 0, 0, 1, 0, 0, 1, NA, NA, 0, N…
## $ calcium  <dbl> 503.5, 473.5, NA, 1248.5, 660.5, 776.0, 452.0, 853.5, 929.0, …
## $ vitd     <dbl> 1.85, 5.85, NA, 3.85, 2.35, 5.65, 3.75, 4.45, 6.05, 6.45, 3.3…
## $ DSQTVD   <dbl> 20.557, 25.000, NA, 25.000, NA, NA, NA, 100.000, 50.000, 46.6…
## $ DSQTCALC <dbl> 211.67, 820.00, NA, 35.00, 13.33, NA, 26.67, 1066.67, 35.00, …

0.2 Recode to readable factors

Create readable labels for the main categorical variables:

bmd <- bmd %>%
  mutate(
    sex = factor(RIAGENDR, 
                 levels = c(1, 2), 
                 labels = c("Male", "Female")),
    ethnicity = factor(RIDRETH1, 
                       levels = c(1, 2, 3, 4, 5),
                       labels = c("Mexican American", 
                                  "Other Hispanic",
                                  "Non-Hispanic White", 
                                  "Non-Hispanic Black", 
                                  "Other")),
    smoker_f = factor(smoker, 
                      levels = c(1, 2, 3),
                      labels = c("Current", "Past", "Never"))
  )

0.3 Missingness + analytic sample

Report the total sample size and create the analytic dataset for ANOVA by removing missing values:

# Total observations
n_total <- nrow(bmd)

# Create analytic dataset for ANOVA (complete cases for BMD and ethnicity)
anova_df <- bmd %>%
  filter(!is.na(DXXOFBMD), !is.na(ethnicity))

# Sample size for ANOVA analysis
n_anova <- nrow(anova_df)

# Display sample sizes
n_total
## [1] 2898
n_anova
## [1] 2286

Write 2–4 sentences summarizing your analytic sample here.

[Total sample size: 2898; 612 observations were removed due to missing data; the final analytic sample size is 2286 for the ANOVA analysis.]

Part 1 — One-way ANOVA: BMD by ethnicity (50 points)

Research question: Do mean BMD values differ across ethnicity groups?

1.1 Hypotheses (required)

State your null and alternative hypotheses in both statistical notation and plain language.

Statistical notation:

  • H₀: [μ_1 = μ_2 = μ_3 = μ_4 = μ_5 ]
  • Hₐ: [At least one μ is different]

Plain language:

  • H₀: [The mean BMD is the same across all ethnicity groups]
  • Hₐ: [At least one group mean is different from others]

1.2 Exploratory plot and group summaries

Create a table showing sample size, mean, and standard deviation of BMD for each ethnicity group:

anova_df %>%
  group_by(ethnicity) %>%
  summarise(
    n = n(),
    mean_bmd = mean(DXXOFBMD, na.rm = TRUE),
    sd_bmd = sd(DXXOFBMD, na.rm = TRUE)
  ) %>%
  arrange(desc(n)) %>%
  kable(digits = 3)
ethnicity n mean_bmd sd_bmd
Non-Hispanic White 846 0.888 0.160
Non-Hispanic Black 532 0.973 0.161
Other 409 0.897 0.148
Mexican American 255 0.950 0.149
Other Hispanic 244 0.946 0.156

Create a visualization comparing BMD distributions across ethnicity groups:

ggplot(anova_df, aes(x = ethnicity, y = DXXOFBMD)) +
  geom_boxplot(outlier.shape = NA) +
  geom_jitter(width = 0.15, alpha = 0.25) +
  labs(
    x = "Ethnicity group",
    y = "Bone mineral density (g/cm²)"
  ) +
  theme(axis.text.x = element_text(angle = 25, hjust = 1))

Interpret the plot: What patterns do you observe in BMD across ethnicity groups?

[Different ethnicity groups have different mean BMD; The plot shows that Non-Hispanic Black have the highest median BMD compare to any other groups, while non-hispanic white participants have the lowest. Variability (box heights) looks similar across groups except for Non-Hispanic White]

1.3 Fit ANOVA model + report the F-test

Fit a one-way ANOVA with:

  • Outcome: BMD (DXXOFBMD)
  • Predictor: Ethnicity (5 groups)
# Fit ANOVA model
fit_aov <- aov(DXXOFBMD ~ ethnicity, data = anova_df)

# Display ANOVA table
summary(fit_aov)
##               Df Sum Sq Mean Sq F value Pr(>F)    
## ethnicity      4   2.95  0.7371   30.18 <2e-16 ***
## Residuals   2281  55.70  0.0244                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Report the F-statistic, degrees of freedom, and p-value in a formatted table:

# Create tidy ANOVA table
broom::tidy(fit_aov) %>%
  kable(digits = 4)
term df sumsq meansq statistic p.value
ethnicity 4 2.9482 0.7371 30.185 0
Residuals 2281 55.6973 0.0244 NA NA

Write your interpretation here:

[Since p<0.05, we reject Ho; there is statistically significant evidence that mean BMD differs across at least two ethnicity groups.]

1.4 Assumption checks (normality + equal variances)

ANOVA requires three assumptions: independence, normality of residuals, and equal variances. Check the latter two graphically and with formal tests.

If assumptions are not satisfied, say so clearly, but still proceed with the ANOVA and post-hoc tests.

Normality of residuals

Create diagnostic plots:

par(mfrow = c(1, 2))
plot(fit_aov, which = 1)  # Residuals vs fitted
plot(fit_aov, which = 2)  # QQ plot

Equal variances (Levene’s test)

car::leveneTest(DXXOFBMD ~ ethnicity, data = anova_df)
## Levene's Test for Homogeneity of Variance (center = median)
##         Df F value Pr(>F)
## group    4  1.5728 0.1788
##       2281

Summarize what you see in 2–4 sentences:

[The Q-Q plot of residuals shows that the points fall along the diagonal reference line, which suggests that the residuals are approximately normally distributed. Levene’s test for homogeneity of variaty was not statistically significant. Since p value 0.1788 >0.05, we fail to reject the null hypothesis. F (4,2281)=1.57. This indicates that there is no evidence of unequal variances across ethnicity groups. Therefore, the assumption of homogeneity of variance is satisfied. ]

1.5 Post-hoc comparisons (pairwise tests)

Because ethnicity has 5 groups, you must do a multiple-comparisons procedure.

Use Tukey HSD and report:

  • Pairwise comparisons
  • Mean differences
  • 95% confidence intervals
  • Adjusted p-values
tukey_results <- TukeyHSD(fit_aov)
print(tukey_results)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = DXXOFBMD ~ ethnicity, data = anova_df)
## 
## $ethnicity
##                                               diff          lwr         upr
## Other Hispanic-Mexican American       -0.004441273 -0.042644130  0.03376158
## Non-Hispanic White-Mexican American   -0.062377249 -0.092852618 -0.03190188
## Non-Hispanic Black-Mexican American    0.022391619 -0.010100052  0.05488329
## Other-Mexican American                -0.053319344 -0.087357253 -0.01928144
## Non-Hispanic White-Other Hispanic     -0.057935976 -0.088934694 -0.02693726
## Non-Hispanic Black-Other Hispanic      0.026832892 -0.006150151  0.05981593
## Other-Other Hispanic                  -0.048878071 -0.083385341 -0.01437080
## Non-Hispanic Black-Non-Hispanic White  0.084768868  0.061164402  0.10837333
## Other-Non-Hispanic White               0.009057905 -0.016633367  0.03474918
## Other-Non-Hispanic Black              -0.075710963 -0.103764519 -0.04765741
##                                           p adj
## Other Hispanic-Mexican American       0.9978049
## Non-Hispanic White-Mexican American   0.0000003
## Non-Hispanic Black-Mexican American   0.3276613
## Other-Mexican American                0.0001919
## Non-Hispanic White-Other Hispanic     0.0000036
## Non-Hispanic Black-Other Hispanic     0.1722466
## Other-Other Hispanic                  0.0010720
## Non-Hispanic Black-Non-Hispanic White 0.0000000
## Other-Non-Hispanic White              0.8719109
## Other-Non-Hispanic Black              0.0000000

Create and present a readable table (you can tidy the Tukey output):

# YOUR CODE HERE to create a tidy table from Tukey results
# Hint: Convert to data frame, add rownames as column, filter for significance
tukey_df <- as.data.frame(tukey_results$ethnicity)

tukey_table <- tukey_df %>%
  rownames_to_column(var = "Comparison") %>%

  rename(
    'Difference in Means' =diff,
    'Lower CI' = lwr,
    'Upper CI' = upr,
    'Adjusted P-value' = 'p adj'
  ) 
  
tukey_table %>% kable()
Comparison Difference in Means Lower CI Upper CI Adjusted P-value
Other Hispanic-Mexican American -0.0044413 -0.0426441 0.0337616 0.9978049
Non-Hispanic White-Mexican American -0.0623772 -0.0928526 -0.0319019 0.0000003
Non-Hispanic Black-Mexican American 0.0223916 -0.0101001 0.0548833 0.3276613
Other-Mexican American -0.0533193 -0.0873573 -0.0192814 0.0001919
Non-Hispanic White-Other Hispanic -0.0579360 -0.0889347 -0.0269373 0.0000036
Non-Hispanic Black-Other Hispanic 0.0268329 -0.0061502 0.0598159 0.1722466
Other-Other Hispanic -0.0488781 -0.0833853 -0.0143708 0.0010720
Non-Hispanic Black-Non-Hispanic White 0.0847689 0.0611644 0.1083733 0.0000000
Other-Non-Hispanic White 0.0090579 -0.0166334 0.0347492 0.8719109
Other-Non-Hispanic Black -0.0757110 -0.1037645 -0.0476574 0.0000000

Write a short paragraph: “The groups that differed were …”

[ The groups that differed were Non-Hispanic Black and Non-Hispanic White, with Non-Hispanic Black adults showing higher mean BMD. Non-Hispanic White Adults also had lower BMD than Mexican American and other Hispanic adults. Additionally, other adults had lower BMD than Mexican American, Other Hispanic, and Non-Hispanic Black Adults. All other pairwise comparisons did not show statistically significant differences in mean BMD. ]

1.6 Effect size (eta-squared)

Compute η² and interpret it (small/moderate/large is okay as a qualitative statement).

Formula: η² = SS_between / SS_total

# Get ANOVA table with sum of squares
anova_tbl <- broom::tidy(fit_aov)
anova_tbl
## # A tibble: 2 × 6
##   term         df sumsq meansq statistic   p.value
##   <chr>     <dbl> <dbl>  <dbl>     <dbl>     <dbl>
## 1 ethnicity     4  2.95 0.737       30.2  1.65e-24
## 2 Residuals  2281 55.7  0.0244      NA   NA
anova_summary <- summary(fit_aov) [[1]]
ss_between <- anova_summary$`Sum Sq`[1]
ss_total <- sum(anova_summary$`Sum Sq`)

eta_squared<- ss_between / ss_total

cat("Eta-squared (η²):", round(eta_squared, 4), "\n")
## Eta-squared (η²): 0.0503
cat("Percentage of variance explained:", round (eta_squared* 100, 2), "%")
## Percentage of variance explained: 5.03 %

Interpret in 1–2 sentences:

[The effect size for ethnicity was 5.03%, indicating that approximately 5% of the total variance in BMD is explained by ethnicity. This represents a small-to-moderate effect, suggesting that while differences between groups are statistically significant, other factors contribute substantially to BMD variation]

Part 2 — Correlation analysis (40 points)

In this section you will:

  1. Create total intake variables (dietary + supplemental)
  2. Assess whether transformations are needed
  3. Produce three correlation tables examining relationships between:
    • Table A: Predictors vs. outcome (BMD)
    • Table B: Predictors vs. exposure (physical activity)
    • Table C: Predictors vs. each other

2.1 Create total intake variables

Calculate total nutrient intake by adding dietary and supplemental sources:

# Create total calcium and vitamin D variables
# Note: Replace NA in supplement variables with 0 (no supplementation)
bmd2 <- bmd %>%
  mutate(
    DSQTCALC_0 = replace_na(DSQTCALC, 0),
    DSQTVD_0 = replace_na(DSQTVD, 0),
    calcium_total = calcium + DSQTCALC_0,
    vitd_total = vitd + DSQTVD_0
  )

2.2 Decide on transformations (show evidence)

Before calculating correlations, examine the distributions of continuous variables. Based on skewness and the presence of outliers, you may need to apply transformations.

Create histograms to assess distributions:

# Recommended: BMXBMI, calcium_total, vitd_total, totmet
# Example structure:


p1 <- ggplot(bmd2, aes(x = BMXBMI)) + 
  geom_histogram(bins = 30, fill= "steelblue", color = "black") +
  labs (title = "BMI Distribution", x="BMI (kg/m^2)", y = "Count")

p2 <- ggplot(bmd2, aes(x = calcium_total)) + 
  geom_histogram(bins = 30, fill ="salmon", color = "black")+
  labs (title = "Total Calcium Intake", x="mg/day", y = "Count")

p3 <- ggplot(bmd2, aes( x=vitd_total)) +
  geom_histogram(bins = 30, fill = "green", color = "black") +
  labs (title = "Total Vitamin D Intake", x= "mcg/day", y= "Count")

p4 <-ggplot(bmd2, aes(x= totmet)) +
  geom_histogram(bins = 30, fill = "purple", color = "black") +
  labs (title = "Total Physical Activity", x="MET-min/week", y="Count")

(p1 | p2) / (p3 | p4)

Based on your assessment, apply transformations as needed:

bmd2 <- bmd2 %>%
  mutate(
    log_bmi= log(BMXBMI),
    log_calcium_total = log1p(calcium_total),
    sqrt_vitd_total = sqrt(vitd_total),
    sqrt_totmet = sqrt(totmet),
  )

Document your reasoning: “I applied a log transformations to BMI and total calcium because their distributions were right-skewed with some extreme values in the histograms, I applied a square root transforamtion to total Vitamin D and Total MET to reduce skewness while handling zeros..”

2.3 Create three correlation tables (Pearson)

For each correlation, report:

  • Correlation coefficient (r)
  • p-value
  • Sample size (n)

Helper function for correlation pairs

# Function to calculate Pearson correlation for a pair of variables
corr_pair <- function(data, x, y) {
  # Select variables and remove missing values
  d <- data %>%
    select(all_of(c(x, y))) %>%
    drop_na()
  
  # Need at least 3 observations for correlation
  if(nrow(d) < 3) {
    return(tibble(
      x = x,
      y = y,
      n = nrow(d),
      estimate = NA_real_,
      p_value = NA_real_
    ))
  }
  
  # Calculate Pearson correlation
  ct <- cor.test(d[[x]], d[[y]], method = "pearson")
  
  # Return tidy results
  tibble(
    x = x,
    y = y,
    n = nrow(d),
    estimate = unname(ct$estimate),
    p_value = ct$p.value
  )
}

Table A: Variables associated with the outcome (BMD)

Examine correlations between potential predictors and BMD:

  • Age vs. BMD
  • BMI vs. BMD
  • Total calcium vs. BMD
  • Total vitamin D vs. BMD

Use transformed versions where appropriate.

# YOUR CODE HERE
# Use the corr_pair function to create correlations
# Use bind_rows() to combine multiple correlation results
# Format with kable()

tableA <- bind_rows(
  corr_pair(bmd2, "RIDAGEYR", "DXXOFBMD"),
  corr_pair(bmd2, "log_bmi", "DXXOFBMD"),
  corr_pair(bmd2, "log_calcium_total", "DXXOFBMD"),
  corr_pair(bmd2, "sqrt_vitd_total", "DXXOFBMD")
 ) %>%
   mutate(
     estimate = round(estimate, 3),
     p_value = signif(p_value, 3)
  )
 
 tableA %>% kable()
x y n estimate p_value
RIDAGEYR DXXOFBMD 2286 -0.232 0.00000
log_bmi DXXOFBMD 2275 0.425 0.00000
log_calcium_total DXXOFBMD 2129 0.012 0.59200
sqrt_vitd_total DXXOFBMD 2129 -0.062 0.00426

Interpret the results:

[Age: Negatively correlated with BMD (r = -0.232, p < 0.001). This indicates that as age increases, bone mineral density tends to decrease. The strength of this relationship is small to moderate. BMI (log-transformed): Positively correlated with BMD (r = 0.425, p < 0.001). Individuals with higher BMI tend to have higher BMD. This is a moderate positive association and is the strongest correlation observed in the table. Total calcium intake (log-transformed): Not significantly correlated with BMD (r = 0.012, p = 0.592). This suggests no meaningful relationship between calcium intake and BMD in this dataset. Total vitamin D intake (square-root transformed): Negatively correlated with BMD (r = -0.062, p = 0.004). Although statistically significant, the relationship is very weak and likely not practically meaningful. Summary: Age and BMI are the most relevant predictors of BMD, with BMI showing a moderate positive association. Calcium intake does not show a significant association, and vitamin D intake shows a very weak negative correlation.]

Table B: Variables associated with the exposure (MET)

Examine correlations between potential confounders and physical activity:

  • Age vs. MET
  • BMI vs. MET
  • Total calcium vs. MET
  • Total vitamin D vs. MET
# YOUR CODE HERE
# Follow the same structure as Table A
# Use sqrt_totmet (or transformed version) as the outcome variable
bmd2 <- bmd2 %>% 
  mutate(sqrt_totmet = sqrt(totmet))

tableB <- bind_rows(
  corr_pair(bmd2, "RIDAGEYR", "sqrt_totmet"),
  corr_pair(bmd2, "log_bmi", "sqrt_totmet"),
  corr_pair(bmd2, "log_calcium_total", "sqrt_totmet"),
  corr_pair(bmd2, "sqrt_vitd_total", "sqrt_totmet")
)%>%
  mutate(
    estimate= round(estimate, 3),
    p_value =signif(p_value, 3)
  )
tableB %>% knitr::kable()
x y n estimate p_value
RIDAGEYR sqrt_totmet 1934 -0.097 1.96e-05
log_bmi sqrt_totmet 1912 0.001 9.51e-01
log_calcium_total sqrt_totmet 1777 0.086 2.82e-04
sqrt_vitd_total sqrt_totmet 1777 -0.002 9.32e-01

Interpret the results:

[Age and total calcium intake are associated with the physical activity levels. Age is negatively correlated with physical activity (r=-0.097, p<0.001), meaning older participants tend to be slightly less active. Total calcium intake is positively correlated with physical activity (r=0.086, p<0.001), meaning that participants with higher calcium intake tend to be slightly more active. BMI and total vitamin intake are not significantly associated with physical activity in this sample]

Table C: Predictors associated with each other

Examine correlations among all predictor variables (all pairwise combinations):

  • Age, BMI, Total calcium, Total vitamin D
# YOUR CODE HERE
# Create all pairwise combinations of predictors
# Hint: Use combn() to generate pairs, then map_dfr() with corr_pair()


preds <- c("RIDAGEYR", "log_bmi", "log_calcium_total", "sqrt_vitd_total")
pairs <- combn(preds, 2, simplify = FALSE)
tableC <- map_dfr(pairs, ~corr_pair(bmd2, .x[1], .x[2])) %>%
   mutate(
     estimate = round(estimate, 3),
     p_value = signif(p_value, 3)
   )
 
 tableC %>% kable()
x y n estimate p_value
RIDAGEYR log_bmi 2834 -0.098 2.00e-07
RIDAGEYR log_calcium_total 2605 0.048 1.35e-02
RIDAGEYR sqrt_vitd_total 2605 0.153 0.00e+00
log_bmi log_calcium_total 2569 0.000 9.83e-01
log_bmi sqrt_vitd_total 2569 0.007 7.31e-01
log_calcium_total sqrt_vitd_total 2605 0.314 0.00e+00

Interpret the results: [Most predictor pairs show weak correlations |r|<0.2, suggests they are independent and unlikely cause multicollinearity problems in regression models. The only moderate correlation is between log_calcium_total and log_vita_total r=0.314, which indicates some shared variation but generally would not be a serious concern.]

Optional: Visualization examples

Create scatterplots to visualize key relationships:

# Example: BMD vs BMI
ggplot(bmd2, aes(x = log_bmi, y = DXXOFBMD)) +
  geom_point(alpha = 0.25) +
  geom_smooth(method = "lm", se = FALSE) +
  labs(
    x = "log(BMI)",
    y = "BMD (g/cm²)",
    title = "BMD vs BMI (transformed)"
  )

# Example: BMD vs Physical Activity
ggplot(bmd2, aes(x = sqrt_totmet, y = DXXOFBMD)) +
  geom_point(alpha = 0.25) +
  geom_smooth(method = "lm", se = FALSE) +
  labs(
    x = "sqrt(totmet)",
    y = "BMD (g/cm²)",
    title = "BMD vs MET (transformed)"
  )

Part 3 — Reflection (10 points)

Write a structured reflection (200–400 words) addressing the following:

A. Comparing Methods (ANOVA vs Correlation)

  • When would you use ANOVA versus correlation in epidemiological research?
  • What are the key differences in what these methods tell us?
  • Give one example of a research question better suited to ANOVA and one better suited to correlation.

B. Assumptions and Limitations

  • Which assumptions were most challenging to meet in this assignment?
  • How might violations of assumptions affect your conclusions?
  • What are the limitations of cross-sectional correlation analysis for understanding relationships between nutrition/activity and bone health?

C. R Programming Challenge

  • What was the most difficult part of the R coding for this assignment?
  • How did you solve it? (Describe your problem-solving process)
  • What R skill did you strengthen through completing this assignment?

YOUR REFLECTION HERE (200–400 words)

[A. Comparing Methods (ANOVA vs Correlation) ANOVA and correlation serve different purposes in epidemiological research. ANOVA is used to compare the means of a continuous outcome across two or more categorical groups. Correlation assesses the strength and direction of a linear relationship between two continuous variables. For example, a research question like “Does mean bone mineral density differ by ethnicity?” is best addressed with ANOVA. On the other hand, a question such as “Is higher BMI associated with higher bone mineral density?” is better suited for correlation analysis.

B. The most challenging assumptions in this assignment were normality and linearity for the correlation analyses. Many variables were skewed, which required transformations to better meet these assumptions. Violating assumptions, such as non-normal distributions or strong outliers, could bias the correlation coefficients or p-values, leading to misleading conclusions. A major limitation of cross-sectional correlation analysis is that it cannot establish causality. While I can observe associations between nutrition, physical activity, and BMD, I cannot determine whether changes in calcium intake or exercise cause changes in bone health..

C. The hardest part of the R coding was applying variable transformations consistently across tables and plots. I addressed this by checking each variable’s distribution with histograms, choosing the right transformation, and using mutate() systematically. This helped me work better with dplyr, use custom functions like corr_pair(), and create reproducible tables and plots. I also became more confident fixing errors from missing or misnamed columns.]

Submission checklist

Before submitting, verify:

Good luck! Remember to start early and use office hours if you need help.