This lecture reviews the foundational biostatistical concepts covered in HSTA/HEPI 552. These concepts are essential for understanding the advanced statistical modeling techniques we’ll explore in EPI 553.
By the end of this review, you should be able to:
Most research aims to assess relationships among a set of variables. The choice of analysis depends on several factors:
A random variable is a variable whose observed values are possible outcomes of a random experiment.
Variables can serve different roles depending on your research question:
Important note: A variable can be a predictor in one study and a response in another. For example: - Stroke predicted by systolic blood pressure (SBP) - SBP predicted by age
Nominal variables have categories without a natural hierarchy: - Gender - Ethnic identity - Cancer type
Ordinal variables have categories with a natural ordering: - Level of pain (mild, moderate, severe) - Response frequency (never, sometimes, always)
Example in R:
# Creating a nominal variable
gender <- factor(c("Male", "Female", "Female", "Male", "Male"))
print(gender)[1] Male Female Female Male Male
Levels: Female Male# Creating an ordinal variable
pain_level <- factor(c("Mild", "Severe", "Moderate", "Mild", "Severe"),
levels = c("Mild", "Moderate", "Severe"),
ordered = TRUE)
print(pain_level)[1] Mild Severe Moderate Mild Severe
Levels: Mild < Moderate < SevereDiscrete variables take on only countable distinct values: - Number of children in a family - Number of cancers in a population
Continuous variables can take on infinite possible values: - Weight, height, time - Can be interval (no true zero) or ratio (has true zero)
Example in R:
# Discrete variable
num_children <- c(0, 2, 1, 3, 2, 1, 0, 4)
hist(num_children,
main = "Distribution of Number of Children",
xlab = "Number of Children",
col = "lightblue",
breaks = seq(-0.5, 4.5, 1))
# Continuous variable
weights <- rnorm(100, mean = 70, sd = 10)
hist(weights,
main = "Distribution of Weights (kg)",
xlab = "Weight (kg)",
col = "lightgreen",
breaks = 20)
The probability distribution of a random variable gives the relative frequencies associated with all possible values in a population.
Probability distributions are mathematical functions that provide the probability corresponding to each value or range of values taken by a random variable.
A binomial random variable describes the number of occurrences of an event in a series of n trials.
Parameters: - n: fixed number of trials - p: probability of success on each trial
Properties: - Each trial outcome is independent - Dichotomous outcome (success/failure) - Probability of success remains constant across trials
Notation: If X follows a binomial distribution with parameters n and p, the probability mass function is:
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k \in \{0, 1, \ldots, n\}\]
R Example:
# Binomial distribution with n=10, p=0.3
n <- 10
p <- 0.3
x <- 0:10
# Probability mass function
prob <- dbinom(x, size = n, prob = p)
# Visualize
barplot(prob,
names.arg = x,
main = "Binomial Distribution (n=10, p=0.3)",
xlab = "Number of Successes",
ylab = "Probability",
col = "steelblue")
# Example: Probability of exactly 3 successes
dbinom(3, size = 10, prob = 0.3)[1] 0.2668279# Cumulative probability: P(X <= 3)
pbinom(3, size = 10, prob = 0.3)[1] 0.6496107# Generate random sample
set.seed(123)
random_sample <- rbinom(1000, size = 10, prob = 0.3)
hist(random_sample,
main = "Random Sample from Binomial(10, 0.3)",
xlab = "Number of Successes",
col = "coral",
breaks = seq(-0.5, 10.5, 1))
A Poisson random variable describes the number of events during a fixed interval of time.
Parameter: - λ (lambda): expected number of events (λ > 0)
Properties: - Occurrences of events are independent - Probability of a single event is proportional to interval length - Two events cannot occur at exactly the same time
Notation: If X follows a Poisson distribution with parameter λ, the probability mass function is:
\[P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k \in \{0, 1, 2, \ldots\}\]
R Example:
# Poisson distribution with lambda = 3
lambda <- 3
x <- 0:10
# Probability mass function
prob <- dpois(x, lambda = lambda)
# Visualize
barplot(prob,
names.arg = x,
main = "Poisson Distribution (λ=3)",
xlab = "Number of Events",
ylab = "Probability",
col = "darkgreen")
# Example: Number of hospital admissions per hour
# If λ = 5 admissions per hour
lambda_admissions <- 5
# Probability of exactly 3 admissions in an hour
dpois(3, lambda = lambda_admissions)[1] 0.1403739# Probability of 5 or fewer admissions
ppois(5, lambda = lambda_admissions)[1] 0.6159607# Generate random sample
set.seed(456)
random_admissions <- rpois(1000, lambda = lambda_admissions)
hist(random_admissions,
main = "Random Sample: Hospital Admissions per Hour",
xlab = "Number of Admissions",
col = "purple",
breaks = seq(-0.5, max(random_admissions) + 0.5, 1))
The normal distribution is the most common continuous probability distribution.
Parameters: - μ (mu): mean - σ² (sigma squared): variance
Properties: - Bell-shaped - Symmetric around the mean - Completely determined by mean and variance
Notation: If X follows a normal distribution with mean μ and variance σ², the probability density function is:
\[f(x; \mu, \sigma^2) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right)\]
R Functions for Normal Distribution:
# Standard normal distribution (μ=0, σ=1)
x <- seq(-4, 4, length = 100)
y <- dnorm(x, mean = 0, sd = 1)
plot(x, y,
type = "l",
lwd = 2,
col = "blue",
main = "Standard Normal Distribution",
xlab = "x",
ylab = "Density")
abline(v = 0, lty = 2, col = "red")
# Density at specific point
dnorm(0, mean = 0, sd = 1)[1] 0.3989423# Cumulative probability: P(Z < 1.96)
pnorm(1.96, mean = 0, sd = 1)[1] 0.9750021# Find the value where P(Z < z) = 0.975 (95th percentile)
qnorm(0.975, mean = 0, sd = 1)[1] 1.959964# Generate random sample
set.seed(789)
random_normal <- rnorm(1000, mean = 100, sd = 15)
hist(random_normal,
probability = TRUE,
main = "Random Sample from N(100, 15²)",
xlab = "Value",
col = "skyblue",
breaks = 30)
curve(dnorm(x, mean = 100, sd = 15),
add = TRUE,
col = "red",
lwd = 2)
The normal distribution is fundamental because of the Central Limit Theorem:
The sample mean of observations from a random variable with finite mean and variance converges to a normal distribution as the sample size increases.
This means that for large sample sizes, many distributions become approximately normal, which is why the normal distribution is so important for statistical inference.
All hypothesis tests follow this same framework:
State hypotheses:
Choose significance level (α): Usually 0.05
Calculate test statistic: Depends on the type of test
Find the critical value or p-value: Using the appropriate distribution
Make a decision: Reject or fail to reject H₀
Tests whether a sample mean differs from a hypothesized population mean.
Hypotheses: - H₀: μ = μ₀ - H₁: μ ≠ μ₀ (two-tailed test)
Assumptions: - Sample is random - Population is approximately normally distributed (or large sample size)
Test statistic:
\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]
where: - \(\bar{x}\) is the sample mean - s is the sample standard deviation - n is the sample size - This follows a t-distribution with n - 1 degrees of freedom
R Example:
# Example: Is average systolic BP different from 130?
set.seed(101)
sbp <- rnorm(50, mean = 125, sd = 15)
# Conduct one-sample t-test
t_test_result <- t.test(sbp, mu = 130)
print(t_test_result)
One Sample t-test
data: sbp
t = -3.469, df = 49, p-value = 0.001098
alternative hypothesis: true mean is not equal to 130
95 percent confidence interval:
119.1667 127.1141
sample estimates:
mean of x
123.1404 # Visualize
hist(sbp,
main = "Systolic Blood Pressure",
xlab = "SBP (mmHg)",
col = "lightblue",
breaks = 15)
abline(v = mean(sbp), col = "red", lwd = 2, lty = 2, label = "Sample mean")
abline(v = 130, col = "green", lwd = 2, lty = 3, label = "Hypothesized mean")
Compares means between two independent groups.
Hypotheses (two-tailed): - H₀: μ₁ = μ₂ - H₁: μ₁ ≠ μ₂
Assumptions: - Both samples are random - Both populations are approximately normally distributed - Equal variances (or can be adjusted)
Test statistic (assuming equal variances):
\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{1/n_1 + 1/n_2}}\]
where \(s_p\) is the pooled standard deviation.
R Example:
# Generate sample data
set.seed(202)
control_bp <- rnorm(40, mean = 140, sd = 12)
treatment_bp <- rnorm(40, mean = 130, sd = 10)
# Create data frame
bp_comparison <- data.frame(
bp = c(control_bp, treatment_bp),
group = factor(rep(c("Control", "Treatment"), c(40, 40)))
)
# Two-sample t-test
t_test_comparison <- t.test(bp ~ group, data = bp_comparison, var.equal = TRUE)
print(t_test_comparison)
Two Sample t-test
data: bp by group
t = 4.1685, df = 78, p-value = 7.87e-05
alternative hypothesis: true difference in means between group Control and group Treatment is not equal to 0
95 percent confidence interval:
5.515281 15.599483
sample estimates:
mean in group Control mean in group Treatment
141.0522 130.4948 # Boxplot with points
boxplot(bp ~ group,
data = bp_comparison,
main = "Blood Pressure by Group",
ylab = "Systolic BP (mmHg)",
col = c("lightcoral", "lightblue"))
# Add individual data points
stripchart(bp ~ group,
data = bp_comparison,
vertical = TRUE,
method = "jitter",
add = TRUE,
pch = 20,
col = rgb(0, 0, 0, 0.3))
Tests whether a population variance equals a hypothesized value.
Hypotheses (two-tailed): - H₀: σ² = σ₀² - H₁: σ² ≠ σ₀²
Test statistic:
\[\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\]
This follows a chi-square distribution with n - 1 degrees of freedom.
R Example:
# Test if population variance differs from 100
set.seed(303)
weight_data <- rnorm(30, mean = 75, sd = 12)
n <- length(weight_data)
s2 <- var(weight_data)
sigma2_0 <- 100
chi_sq_stat <- (n - 1) * s2 / sigma2_0
# Critical values for two-tailed test at α = 0.05
alpha <- 0.05
lower_critical <- qchisq(alpha/2, df = n - 1)
upper_critical <- qchisq(1 - alpha/2, df = n - 1)
cat("Test statistic:", chi_sq_stat, "\n")Test statistic: 29.6684 cat("Critical values: [", lower_critical, ",", upper_critical, "]\n")Critical values: [ 16.04707 , 45.72229 ]cat("Sample variance:", s2, "\n")Sample variance: 102.3048 cat("P-value:", 2 * min(pchisq(chi_sq_stat, df = n - 1),
1 - pchisq(chi_sq_stat, df = n - 1)), "\n")P-value: 0.8613845 Tests whether there is a relationship between two categorical variables.
Hypotheses: - H₀: No relationship between variables - H₁: There is a relationship
Example: 2×2 Contingency Table
| A = 1 | A = 0 | Total | |
|---|---|---|---|
| B = 1 | O₁₁ | O₁₂ | R₁ |
| B = 0 | O₂₁ | O₂₂ | R₂ |
| Total | C₁ | C₂ | N |
Expected count under H₀:
\[E_{ij} = \frac{R_i \times C_j}{N}\]
Test statistic:
\[\chi^2 = \sum_{i=1}^k \sum_{j=1}^m \frac{(O_{ij} - E_{ij})^2}{E_{ij}}\]
Rejection region: χ² > χ²₍ₖ₋₁₎₍ₘ₋₁₎,₁₋α
Assumption: Expected counts ≥ 5 in at least 80 percent of cells
R Example:
# Example: Relationship between smoking and lung disease
smoking <- c(rep("Smoker", 150), rep("Non-smoker", 150))
disease <- c(rep("Disease", 80), rep("No Disease", 70),
rep("Disease", 30), rep("No Disease", 120))
# Create contingency table
contingency_table <- table(smoking, disease)
print(contingency_table) disease
smoking Disease No Disease
Non-smoker 30 120
Smoker 80 70# Chi-square test
chi_test <- chisq.test(contingency_table)
print(chi_test)
Pearson's Chi-squared test with Yates' continuity correction
data: contingency_table
X-squared = 34.464, df = 1, p-value = 4.342e-09# Show expected counts
print(chi_test$expected) disease
smoking Disease No Disease
Non-smoker 55 95
Smoker 55 95# Visualize
mosaicplot(contingency_table,
main = "Smoking Status vs Disease",
color = c("lightcoral", "lightblue"),
xlab = "Smoking Status",
ylab = "Disease Status")
The F-distribution is right-skewed and depends on two degrees of freedom parameters: ν₁ and ν₂
Key property: If X₁ follows a chi-square distribution with ν₁ degrees of freedom and X₂ follows a chi-square distribution with ν₂ degrees of freedom and both are independent, then:
\[F = \frac{X_1/\nu_1}{X_2/\nu_2} \sim F_{\nu_1, \nu_2}\]
Symmetric property:
\[F_{\nu_1, \nu_2, \alpha} = \frac{1}{F_{\nu_2, \nu_1, 1-\alpha}}\]
R Example:
# F-distributions with different degrees of freedom
x <- seq(0, 5, length = 200)
plot(x, df(x, df1 = 5, df2 = 10),
type = "l",
lwd = 2,
col = "red",
main = "F-Distributions",
xlab = "x",
ylab = "Density",
ylim = c(0, 1))
lines(x, df(x, df1 = 10, df2 = 10), col = "blue", lwd = 2)
lines(x, df(x, df1 = 20, df2 = 30), col = "green", lwd = 2)
legend("topright",
legend = c("F(5,10)", "F(10,10)", "F(20,30)"),
col = c("red", "blue", "green"),
lwd = 2)
# Computing lower percentile using symmetry property
# F(5,10,0.05) = 1/F(10,5,0.95)
qf(0.05, df1 = 5, df2 = 10)[1] 0.21119041 / qf(0.95, df1 = 10, df2 = 5)[1] 0.2111904Tests whether population variances from two samples are equal.
Hypotheses (two-tailed): - H₀: σ₁² = σ₂² - H₁: σ₁² ≠ σ₂²
Test statistic:
\[F = \frac{s_1^2}{s_2^2}\]
Rejection region: F < F_{ν₁,ν₂,α/2} or F > F_{ν₁,ν₂,1-α/2}
Assumptions: Random samples from two normally distributed populations
R Example:
# Compare variances of two groups
set.seed(404)
group1 <- rnorm(30, mean = 50, sd = 10)
group2 <- rnorm(25, mean = 50, sd = 15)
# F-test for equal variances
var_test <- var.test(group1, group2)
print(var_test)
F test to compare two variances
data: group1 and group2
F = 0.36542, num df = 29, denom df = 24, p-value = 0.01061
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.1647913 0.7871065
sample estimates:
ratio of variances
0.3654152 # Manual calculation
f_stat <- var(group1) / var(group2)
cat("F-statistic:", f_stat, "\n")F-statistic: 0.3654152 cat("Group 1 variance:", var(group1), "\n")Group 1 variance: 74.15993 cat("Group 2 variance:", var(group2), "\n")Group 2 variance: 202.947 # Visualize
boxplot(group1, group2,
names = c("Group 1", "Group 2"),
main = "Comparison of Two Groups",
ylab = "Value",
col = c("lightblue", "lightgreen"))
It is crucial to plan your analysis before you collect data:
# Complete example: Analyzing a dataset
# Generate sample data: Effect of treatment on blood pressure
set.seed(505)
n_control <- 50
n_treatment <- 50
control_bp <- rnorm(n_control, mean = 140, sd = 15)
treatment_bp <- rnorm(n_treatment, mean = 130, sd = 12)
# Combine into data frame
bp_data <- data.frame(
bp = c(control_bp, treatment_bp),
group = factor(rep(c("Control", "Treatment"), c(n_control, n_treatment)))
)
# 1. Descriptive statistics
library(dplyr)
bp_data %>%
group_by(group) %>%
summarise(
n = n(),
mean_bp = mean(bp),
sd_bp = sd(bp),
se_bp = sd_bp / sqrt(n)
)# A tibble: 2 × 5
group n mean_bp sd_bp se_bp
<fct> <int> <dbl> <dbl> <dbl>
1 Control 50 138. 14.2 2.01
2 Treatment 50 130. 13.5 1.90# 2. Visualize
boxplot(bp ~ group,
data = bp_data,
main = "Blood Pressure by Treatment Group",
ylab = "Systolic BP (mmHg)",
col = c("lightcoral", "lightblue"))
# Add points
stripchart(bp ~ group,
data = bp_data,
vertical = TRUE,
method = "jitter",
add = TRUE,
pch = 20,
col = "darkgray")
# 3. Test for equal variances
var.test(bp ~ group, data = bp_data)
F test to compare two variances
data: bp by group
F = 1.1151, num df = 49, denom df = 49, p-value = 0.7045
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.6327925 1.9650160
sample estimates:
ratio of variances
1.1151 # 4. Two-sample t-test
t.test(bp ~ group, data = bp_data, var.equal = TRUE)
Two Sample t-test
data: bp by group
t = 2.8379, df = 98, p-value = 0.00552
alternative hypothesis: true difference in means between group Control and group Treatment is not equal to 0
95 percent confidence interval:
2.362059 13.346979
sample estimates:
mean in group Control mean in group Treatment
137.8893 130.0348 # 5. Effect size (Cohen's d)
mean_diff <- mean(treatment_bp) - mean(control_bp)
pooled_sd <- sqrt(((n_control - 1) * var(control_bp) +
(n_treatment - 1) * var(treatment_bp)) /
(n_control + n_treatment - 2))
cohens_d <- mean_diff / pooled_sd
cat("Cohen's d =", cohens_d, "\n")Cohen's d = -0.5675794 Probability distributions: - dnorm(), pnorm(), qnorm(), rnorm() - Normal distribution - dbinom(), pbinom(), rbinom() - Binomial distribution - dpois(), ppois(), rpois() - Poisson distribution - dt(), pt(), qt(), rt() - t-distribution - dchisq(), pchisq(), qchisq() - Chi-square distribution - df(), pf(), qf() - F-distribution
Statistical tests: - t.test() - One and two-sample t-tests - var.test() - F-test for equal variances - chisq.test() - Chi-square test of independence
In upcoming lectures, we’ll build on these foundations to explore: - Linear regression models - Logistic regression for binary outcomes - Generalized linear models - Mixed effects models
Kleinbaum, D. G., Kupper, L. L., Nizam, A., and Rosenberg, E. S. (2013). Applied Regression Analysis and Other Multivariable Methods. Cengage Learning.
sessionInfo()R version 4.5.2 (2025-10-31)
Platform: aarch64-apple-darwin20
Running under: macOS Tahoe 26.2
Matrix products: default
BLAS: /System/Library/Frameworks/Accelerate.framework/Versions/A/Frameworks/vecLib.framework/Versions/A/libBLAS.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/4.5-arm64/Resources/lib/libRlapack.dylib; LAPACK version 3.12.1
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
time zone: America/New_York
tzcode source: internal
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] dplyr_1.1.4
loaded via a namespace (and not attached):
[1] digest_0.6.39 utf8_1.2.6 R6_2.6.1 fastmap_1.2.0
[5] tidyselect_1.2.1 xfun_0.55 magrittr_2.0.4 glue_1.8.0
[9] tibble_3.3.0 knitr_1.51 pkgconfig_2.0.3 htmltools_0.5.9
[13] generics_0.1.4 rmarkdown_2.30 lifecycle_1.0.4 cli_3.6.5
[17] vctrs_0.6.5 compiler_4.5.2 rstudioapi_0.17.1 tools_4.5.2
[21] pillar_1.11.1 evaluate_1.0.5 yaml_2.3.12 otel_0.2.0
[25] rlang_1.1.6 jsonlite_2.0.0 htmlwidgets_1.6.4