EPI 553 — Homework 2: Simple Linear Regression
Tahia Sufyani
Submission: Knit this file to HTML, publish to RPubs with the title
epi553_hw02_lastname_firstname, and paste the RPubs URL in the Brightspace submission comments. Also upload this.Rmdfile to Brightspace.AI Policy: AI tools are NOT permitted on this assignment. See the assignment description for full details.
Part 0: Data Preparation (10 points)
# Import the dataset — update the path if needed
bmd <- read.csv("C:/Users/tahia/OneDrive/Desktop/UAlbany PhD/Epi 553/HW/HW1/bmd.csv")
# Quick check
glimpse(bmd)## Rows: 2,898
## Columns: 14
## $ SEQN <int> 93705, 93708, 93709, 93711, 93713, 93714, 93715, 93716, 93721…
## $ RIAGENDR <int> 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2…
## $ RIDAGEYR <int> 66, 66, 75, 56, 67, 54, 71, 61, 60, 60, 64, 67, 70, 53, 57, 7…
## $ RIDRETH1 <int> 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 <int> 2, 3, 1, 3, 1, 2, 1, 2, 3, 3, 3, 2, 2, 3, 3, 2, 3, 1, 2, 1, 1…
## $ totmet <int> 240, 120, 720, 840, 360, NA, 6320, 2400, NA, NA, 1680, 240, 4…
## $ metcat <int> 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 <int> 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, …# Recode RIDRETH1 as a labeled factor
bmd <- bmd %>%
mutate(
RIDRETH1 = factor(RIDRETH1,
levels = 1:5,
labels = c("Mexican American", "Other Hispanic",
"Non-Hispanic White", "Non-Hispanic Black", "Other")),
# Recode RIAGENDR as a labeled factor
RIAGENDR = factor(RIAGENDR,
levels = c(1, 2),
labels = c("Male", "Female")),
# Recode smoker as a labeled factor
smoker = factor(smoker,
levels = c(1, 2, 3),
labels = c("Current", "Past", "Never"))
)## Total N: 2898## Missing DXXOFBMD: 612## Missing calcium: 293Answer:
- Total N: 2898
- Missing DXXOFBMD: 612
- Missing calcium: 293
# Create the analytic dataset (exclude missing DXXOFBMD or calcium)
bmd_analytic <- bmd %>%
filter(!is.na(DXXOFBMD), !is.na(calcium))
cat("Final analytic N:", nrow(bmd_analytic), "\n")## Final analytic N: 2129Answer:
| Quantity | Value |
|---|---|
| Missing DXXOFBMD only | 476 |
| Missing calcium only | 157 |
| Missing BOTH (overlap) | 136 |
| Rows dropped (missing either) | 769 |
| Final analytic N | 2,129 |
769 rows were dropped in total. The overlap of 136 rows (missing on both variables) is why the dropped rows (769) is less than the sum of individual missingness counts (612 + 293 = 905). The correct Final analytic N is exactly 2,129.
Part 1: Exploratory Visualization (15 points)
Research Question: Is there a linear association between dietary calcium intake (calcium, mg/day) and total femur bone mineral density (DXXOFBMD, g/cm²)?
# Create a scatterplot with a fitted regression line
ggplot(bmd_analytic, aes(x = calcium, y = DXXOFBMD)) +
geom_point(alpha = 0.3) +
geom_smooth(method = "lm", se = FALSE, color = "steelblue") +
labs(
title = "Association between dietary calcium intake and total femur bone mineral density",
x = "Dietary Calcium Intake (calcium, mg/day)",
y = "Femur Bone Mineral Density (DXXOFBMD, g/cm²)"
) +
theme_minimal()
Written interpretation (3–5 sentences):
[Describe what the scatterplot reveals. Is there a visible linear trend? In which direction? Does the relationship appear strong or weak? Are there any notable outliers or non-linearities?]
Answer
The intercept (0.8992 g/cm²) indicates the predicted total femur BMD when dietary calcium intake is zero. However, because no participants in the dataset report zero calcium intake, this value falls outside the observed data range and is not substantively meaningful. Instead, it functions mainly as a mathematical baseline that anchors the regression equation.
Part 2: Simple Linear Regression (40 points)
Step 1 — Fit the Model (5 points)
# Fit the simple linear regression model
model <- lm(DXXOFBMD ~ calcium, data = bmd_analytic)
# Display the full model summary
summary(model)##
## Call:
## lm(formula = DXXOFBMD ~ calcium, data = bmd_analytic)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.55653 -0.10570 -0.00561 0.10719 0.62624
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.992e-01 7.192e-03 125.037 < 2e-16 ***
## calcium 3.079e-05 7.453e-06 4.131 3.75e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.1582 on 2127 degrees of freedom
## Multiple R-squared: 0.007959, Adjusted R-squared: 0.007493
## F-statistic: 17.07 on 1 and 2127 DF, p-value: 3.751e-05Step 2 — Interpret the Coefficients (10 points)
A. Intercept (β₀):
[Interpret the intercept in 2–4 sentences. What does it represent numerically? Is it a meaningful quantity in this context?]
Answer:
The intercept (0.8992 g/cm²) represents the model’s predicted total femur BMD when dietary calcium intake is zero. Although it provides the baseline for the regression line, this value has limited clinical or biological relevance because no individuals in the dataset report consuming zero dietary calcium. Since it falls outside the observed range of the data, interpreting it as a real-world prediction would involve unreliable extrapolation. Therefore, the intercept should be viewed primarily as a mathematical constant that anchors the regression model rather than as a substantively meaningful estimate.
B. Slope (β₁):
[Interpret the slope in 2–4 sentences. For every 1-unit increase in calcium (mg/day), what is the estimated change in BMD (g/cm²)? State the direction. Is the effect large or small given typical calcium intake ranges?]
Answer:
The slope (3.079 × 10⁻⁵ g/cm² per mg/day, or 0.00003079 g/cm² per mg/day) indicates that for each additional 1 mg/day increase in dietary calcium intake, total femur BMD is expected to increase by about 0.00003079 g/cm² on average. This reflects a positive association between calcium intake and BMD. However, the effect per unit is extremely small. When considering a more meaningful difference in intake—such as 500 mg/day—the estimated increase in BMD is approximately 0.0154 g/cm². Even at this level, the change remains modest compared with the typical variability in BMD within the population (IQR ≈ 0.21 g/cm² based on the residual distribution).
Step 3 — Statistical Inference (15 points)
## 2.5 % 97.5 %
## (Intercept) 8.851006e-01 9.133069e-01
## calcium 1.617334e-05 4.540649e-05State your hypotheses:
H₀: β₁ = 0 — There is no linear association between dietary calcium intake and total femur BMD in the population; the slope equals zero.
H₁: β₁ ≠ 0 — There is a linear association between dietary calcium intake and total femur BMD; the slope is different from zero (two-sided test).
Report the test results:
[Report the t-statistic, degrees of freedom, and p-value from the summary() output above. State your conclusion: do you reject H₀? What does this mean for the association between calcium and BMD?]
MODEL TEST RESULTS:
The t-statistic for the slope is t = 4.131 on 2,127 degrees of freedom, with a p-value of 3.75 × 10⁻⁵ (p < 0.001). Because p < 0.05, we reject H₀. This provides statistically significant evidence that dietary calcium intake is positively associated with total femur BMD in this sample.
Interpret the 95% confidence interval for β₁:
[Interpret the CI in plain language — what range of values is plausible for the true slope?]
Answer:
The 95% confidence interval for the slope is (1.617 × 10⁻⁵, 4.541 × 10⁻⁵) g/cm² per mg/day. This means we are 95% confident that the true population slope falls within this range. In practical terms, for each additional 1 mg/day increase in dietary calcium intake, the true average increase in total femur BMD is likely between 0.00001617 and 0.00004541 g/cm². Because the entire interval lies above zero, the result is consistent with rejecting the null hypothesis (H₀) of no association.
Step 4 — Model Fit: R² and Residual Standard Error (10 points)
R² (coefficient of determination):
[What proportion of the variance in BMD is explained by dietary calcium intake? Based on this R², how well does the model fit the data? What does this suggest about the importance of other predictors?]
Answer: R² INTERPRETATION:
The R² is 0.007959 (approximately 0.80%), meaning that dietary calcium intake explains less than 1% of the variance in total femur BMD in this sample. While the association is statistically significant (p < 0.001), the model fit is very poor by conventional standards. This low R² strongly suggests that BMD is influenced by many other factors beyond dietary calcium including age, sex, body mass index, physical activity, smoking status, race/ethnicity, and supplemental calcium none of which are accounted for in this simple model.
Residual Standard Error (RSE):
[Report the RSE. Express it in the units of the outcome and explain what it tells you about the average prediction error of the model.]
Answer RSE INTERPRETATION:
The residual standard error is 0.1582 g/cm². This represents the average distance between observed BMD values and the model’s predicted values the typical prediction error of the model is about 0.1582 g/cm². Given that BMD values in this population range from roughly 0.4 to 1.6 g/cm² (as visible in the scatterplot), an RSE of 0.1582 g/cm² represents substantial unexplained variability, consistent with the near-zero R². —
Part 3: Prediction (20 points)
# Create a new data frame with the target predictor value
new_data <- data.frame(calcium = 1000)
# 95% confidence interval for the mean response at calcium = 1000
predict(model, newdata = new_data, interval = "confidence")## fit lwr upr
## 1 0.9299936 0.9229112 0.937076# 95% prediction interval for a new individual at calcium = 1000
predict(model, newdata = new_data, interval = "prediction")## fit lwr upr
## 1 0.9299936 0.6195964 1.240391Written interpretation (3–6 sentences):
[Answer all four questions from the assignment description:
- What is the predicted BMD at calcium = 1,000 mg/day? (Report with units.)
Answer:
The model predicts a total femur BMD of 0.9229112 g/cm² for an individual with a dietary calcium intake of 1,000 mg/day.
- What does the 95% confidence interval represent?
Answer:
The 95% confidence interval for the mean BMD at calcium = 1,000 mg/day is (0.9229, 0.9371) g/cm². This interval estimates the average BMD across all individuals in the population who consume 1,000 mg/day of dietary calcium. It is narrow (width ≈ 0.014 g/cm²) because it reflects only the uncertainty in estimating the mean regression line at this point, not variability between individuals.
- What is the 95% prediction interval, and why is it wider than the CI?
Answer:
The 95% prediction interval is (0.6196, 1.2404) g/cm². This is substantially wider than the confidence interval (width ≈ 0.621 g/cm²) because it must account for two sources of uncertainty simultaneously: the uncertainty in estimating the mean response, and the natural individual-to-individual variability in BMD. While the CI tells us where the average BMD falls for people consuming 1,000 mg/day, the PI tells us where a single new individual’s BMD would likely fall and people vary considerably even at the same calcium intake level.
- Is 1,000 mg/day a meaningful value to predict at, given the data?]
Answer:
A calcium intake of 1,000 mg/day is a clinically meaningful and appropriate value to predict at. It corresponds to the Recommended Dietary Allowance (RDA) for adults aged 19–50, and it falls well within the observed range of calcium values in the dataset making this an interpolation rather than an extrapolation, so the prediction is on solid footing.
Part 4: Reflection (15 points)
Write 200–400 words in continuous prose (not bullet points) addressing all three areas below.
A. Statistical Insight (6 points)
[What does the regression model tell you about the calcium–BMD relationship? Were the results surprising? What are the key limitations of interpreting SLR from a cross-sectional survey as causal evidence? What confounders might explain the observed association?]
Answer
The regression model shows a statistically significant but very small positive relationship between dietary calcium intake and total femur BMD. Although the association is significant, calcium explains less than 1% of the variation in BMD, which means its practical impact is very limited. This is not surprising because bone health is influenced by many other factors such as age, sex, hormones, physical activity, and overall health. A key limitation of this analysis is that it is based on cross-sectional data, so we cannot determine cause and effect. Since calcium intake and BMD were measured at the same time, we cannot say that higher calcium intake leads to higher BMD. There may also be confounding factors. For example, older individuals tend to have lower BMD and may also consume less calcium. Similarly, sex and physical activity can influence both calcium intake and bone density, which may partly explain the observed relationship.
B. From ANOVA to Regression (5 points)
[Homework 1 used one-way ANOVA to compare mean BMD across ethnic groups. Now you have used SLR to model BMD as a function of a continuous predictor. Compare these two approaches: what kinds of questions does each method answer? What does regression give you that ANOVA does not? When would you prefer one over the other?]
Answer One-way ANOVA and simple linear regression both analyze a continuous outcome, but they answer different types of questions. ANOVA is used when the predictor is categorical, such as race or ethnicity, and it tests whether the average BMD differs across groups. However, it does not show the direction or size of the relationship. In contrast, simple linear regression is used with a continuous predictor, like calcium intake, and it estimates how much BMD changes with each unit increase in calcium. Regression provides more detailed information, including the slope, predicted values, and R², which tells us how well the model explains the outcome. ANOVA is best when comparing group means, while regression is more useful when studying relationships or trends between variables and making predictions.
C. R Programming Growth (4 points)
[What was the most challenging part of this assignment from a programming perspective? How did you work through it? What R skill do you feel more confident about after completing this homework?]
Answer The most challenging part of this assignment was understanding the difference between the confidence interval and the prediction interval when using the predict() function. Since both are created with similar code and look very similar, it was easy to confuse them at first. I worked through this by focusing on their meaning: the confidence interval estimates the average outcome, while the prediction interval estimates the value for an individual, which is why it is wider. Another part that required careful thinking was interpreting the residual standard error. At first, the value did not seem very meaningful, but comparing it to the range of BMD values helped me understand that the model has a large amount of unexplained variation. After completing this assignment, I feel more confident using lm(), interpreting regression output, and using functions like confint() and predict() to better understand my results.
End of Homework 2