STAT 204 Exam Study Guide & Practice Problems

Author

Fall 2024

Published

23 November 2025

Overview

This study guide will help you prepare for the STAT 204 exam covering:

  1. Exploratory Data Analysis (EDA) - Interpreting visualizations
  2. Linear Regression - Model fitting, interpretation, and diagnostics
  3. Analysis of Variance (ANOVA) - Two-way ANOVA with interactions
  4. Logistic Regression - Model fitting, interpretation, and diagnostics

Exam Format:

  • 90 minutes
  • Closed book, one page of handwritten notes allowed (front side only)
  • Calculator permitted
Code 
library(tidyverse)
library(knitr)
library(broom)

# Set theme
theme_set(theme_minimal(base_size = 11))

What to Study

Key Topics

Linear Regression

  • Writing regression equations
  • Interpreting coefficients (slope and intercept)
  • Hypothesis testing for regression coefficients
  • Making predictions from fitted models
  • Checking model assumptions (linearity, constant variance, normality, independence)
  • Interpreting diagnostic plots (Residuals vs. Fitted, Q-Q plots)
  • Understanding R² and adjusted R²

ANOVA

  • One-way and two-way ANOVA models
  • Main effects and interaction effects
  • Writing ANOVA model equations
  • Interpreting ANOVA tables (F-statistics, p-values)
  • Calculating fitted/predicted values from model output
  • Testing hypotheses about main effects and interactions
  • Understanding interaction plots

Logistic Regression

  • Understanding binary outcomes
  • Writing logistic regression equations (log-odds and probability forms)
  • Interpreting coefficients as log-odds ratios
  • Converting between log-odds, odds, and probabilities
  • Making predictions with logistic regression
  • Testing significance of predictors

EDA

  • Interpreting boxplots (median, quartiles, outliers, variability)
  • Interpreting scatterplots (direction, strength, form of relationships)
  • Describing distributions and relationships in data
  • Identifying patterns and unusual observations

Important Formulas

Simple Linear Regression

Model: \(Y_i = \beta_0 + \beta_1 X_i + \varepsilon_i\) where \(\varepsilon_i \sim N(0, \sigma^2)\)

Fitted equation: \(\hat{Y} = b_0 + b_1 X\)

Two-Way ANOVA with Interaction

Model: \(Y_{ijk} = \mu + \alpha_i + \beta_j + (\alpha\beta)_{ij} + \varepsilon_{ijk}\)

where:

  • \(\mu\) = overall mean
  • \(\alpha_i\) = effect of factor A at level i
  • \(\beta_j\) = effect of factor B at level j
  • \((\alpha\beta)_{ij}\) = interaction effect
  • \(\varepsilon_{ijk}\) = random error

Logistic Regression

Model (log-odds): \(\log\left(\frac{p}{1-p}\right) = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \ldots\)

Model (probability): \(p = \frac{e^{\beta_0 + \beta_1 X_1 + \beta_2 X_2 + \ldots}}{1 + e^{\beta_0 + \beta_1 X_1 + \beta_2 X_2 + \ldots}}\)

where: - \(p\) = probability that \(Y = 1\) - \(\beta_1\) = change in log-odds for one-unit increase in \(X_1\)

Practice Problems

Section 1: Exploratory Data Analysis

Practice Problem 1.1: Study Methods

Consider a boxplot showing the distribution of exam scores across four different study methods (A, B, C, D).

Information from the boxplot:

  • Method A: median = 75, Q1 = 68, Q3 = 82, min = 55, max = 95
  • Method B: median = 82, Q1 = 77, Q3 = 88, min = 70, max = 98
  • Method C: median = 70, Q1 = 65, Q3 = 76, min = 60, max = 85
  • Method D: median = 78, Q1 = 70, Q3 = 85, min = 50, max = 100 (with outliers at 50 and 100)

Questions:

  1. Which study method has the highest median exam score?
  2. Which method shows the most variability in scores?
  3. Are there any outliers? If so, for which method(s)?
  4. Based on this visualization, which study method would you recommend and why?
Code 
# Create the data for visualization
study_data <- data.frame(
  Method = rep(c("A", "B", "C", "D"), each = 30),
  Score = c(
    rnorm(30, mean = 75, sd = 7),
    rnorm(30, mean = 82, sd = 5),
    rnorm(30, mean = 70, sd = 5),
    c(50, 100, rnorm(28, mean = 78, sd = 7))
  )
)

# Adjust to match the five-number summaries given
study_data <- study_data %>%
  group_by(Method) %>%
  mutate(Score = case_when(
    Method == "A" ~ pmin(pmax(Score, 55), 95),
    Method == "B" ~ pmin(pmax(Score, 70), 98),
    Method == "C" ~ pmin(pmax(Score, 60), 85),
    TRUE ~ Score
  )) %>%
  ungroup()
Code 
# Create boxplot
ggplot(study_data, aes(x = Method, y = Score, fill = Method)) +
  geom_boxplot(alpha = 0.7) +
  labs(
    title = "Distribution of Exam Scores by Study Method",
    x = "Study Method",
    y = "Exam Score"
  ) +
  theme(legend.position = "none") +
  scale_fill_brewer(palette = "Set2")

Practice Problem 1.2: Office Hours and Study Time

A scatterplot shows the relationship between hours studied (x-axis) and exam score (y-axis) for 200 students, with points colored by whether the student attended office hours.

Code 
# Create simulated data
set.seed(123)
n <- 200
office_hours_data <- data.frame(
  Hours_Studied = runif(n, 0, 20),
  Office_Hours = sample(c("Yes", "No"), n, replace = TRUE, prob = c(0.4, 0.6))
) %>%
  mutate(
    # Students who attend office hours get a boost
    Exam_Score = 50 + 2.5 * Hours_Studied + 
                 ifelse(Office_Hours == "Yes", 10, 0) + 
                 rnorm(n, 0, 8)
  ) %>%
  mutate(Exam_Score = pmin(pmax(Exam_Score, 30), 100))
Code 
# Create scatterplot
ggplot(office_hours_data, aes(x = Hours_Studied, y = Exam_Score, color = Office_Hours)) +
  geom_point(alpha = 0.6, size = 2) +
  geom_smooth(method = "lm", se = FALSE, linetype = "dashed") +
  labs(
    title = "Relationship between Study Time and Exam Score",
    subtitle = "Colored by Office Hours Attendance",
    x = "Hours Studied",
    y = "Exam Score",
    color = "Attended\nOffice Hours"
  ) +
  scale_color_manual(values = c("No" = "#e74c3c", "Yes" = "#3498db")) +
  theme(legend.position = "top")

Questions:

  1. Describe the relationship between hours studied and exam score.
  2. Does attendance at office hours appear to be associated with exam scores? Explain.
  3. What variables would you include in a statistical model to predict exam scores based on this visualization?

Section 2: Linear Regression

Practice Problem 2.1: Exercise and Heart Rate

A researcher studies the relationship between daily exercise time (in minutes) and resting heart rate (in beats per minute) for 150 adults.

Code 
# Create exercise data
set.seed(456)
n <- 150
exercise_data <- data.frame(
  ExerciseTime = runif(n, 0, 90)
) %>%
  mutate(
    HeartRate = 78.5423 - 0.2134 * ExerciseTime + rnorm(n, 0, 8.92)
  )
Code 
# Fit the model
M1 <- lm(HeartRate ~ ExerciseTime, data = exercise_data)
summary(M1)

Call:
lm(formula = HeartRate ~ ExerciseTime, data = exercise_data)

Residuals:
     Min       1Q   Median       3Q      Max 
-21.1794  -6.2669  -0.2039   7.3226  18.7637 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)  77.31281    1.45890  52.994  < 2e-16 ***
ExerciseTime -0.19434    0.02637  -7.369 1.12e-11 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 8.331 on 148 degrees of freedom
Multiple R-squared:  0.2684,    Adjusted R-squared:  0.2635 
F-statistic: 54.31 on 1 and 148 DF,  p-value: 1.119e-11

Questions:

  1. Write the fitted regression equation.
  2. Interpret the slope coefficient in context.
  3. What is the predicted resting heart rate for someone who exercises 30 minutes per day?
  4. The t-value corresponds to a hypothesis test. Write the null and alternative hypotheses, and state your conclusion using α = 0.05.
  5. What does the R-squared value tell us about this model?

Practice Problem 2.2: Model Diagnostics

Below are residual diagnostic plots for the model in Problem 2.1.

Code 
# Create diagnostic plots
par(mfrow = c(1, 2))
plot(M1, which = 1, main = "Residuals vs Fitted")
plot(M1, which = 2, main = "Q-Q Residuals")

Code 
par(mfrow = c(1, 1))

Questions:

  1. Based on the Residuals vs Fitted plot, does the linearity assumption appear satisfied? Explain.
  2. Based on the Residuals vs Fitted plot, does the constant variance assumption appear satisfied? Explain.
  3. Based on the Q-Q plot, does the normality assumption appear satisfied? Explain.
  4. Overall, do you have concerns about using this model? Why or why not?

Section 3: Analysis of Variance (ANOVA)

Practice Problem 3.1: Plant Growth Study

A study examines plant growth (in cm) under three different fertilizer types (A, B, C) and two watering schedules (Daily, Weekly).

Code 
# Create plant growth data
set.seed(789)
plants <- expand.grid(
  Fertilizer = c("A", "B", "C"),
  WateringSchedule = c("Daily", "Weekly"),
  Rep = 1:10
) %>%
  mutate(
    Fertilizer = factor(Fertilizer, levels = c("A", "B", "C")),
    WateringSchedule = factor(WateringSchedule, levels = c("Daily", "Weekly"))
  ) %>%
  mutate(
    # Create growth based on model with interaction
    Growth = 15.234 + 
      ifelse(Fertilizer == "B", 3.456, 0) +
      ifelse(Fertilizer == "C", 5.678, 0) +
      ifelse(WateringSchedule == "Weekly", -2.345, 0) +
      ifelse(Fertilizer == "B" & WateringSchedule == "Weekly", 1.234, 0) +
      ifelse(Fertilizer == "C" & WateringSchedule == "Weekly", 4.567, 0) +
      rnorm(n(), 0, 4.12)
  ) %>%
  select(-Rep)
Code 
# Fit the model
M2 <- lm(Growth ~ Fertilizer * WateringSchedule, data = plants)
summary(M2)

Call:
lm(formula = Growth ~ Fertilizer * WateringSchedule, data = plants)

Residuals:
     Min       1Q   Median       3Q      Max 
-11.7408  -2.2954  -0.5498   2.7841   7.3430 

Coefficients:
                                   Estimate Std. Error t value Pr(>|t|)    
(Intercept)                         14.5326     1.2410  11.710  < 2e-16 ***
FertilizerB                          2.8567     1.7551   1.628 0.109417    
FertilizerC                          6.8893     1.7551   3.925 0.000247 ***
WateringScheduleWeekly              -1.5863     1.7551  -0.904 0.370088    
FertilizerB:WateringScheduleWeekly   0.7959     2.4820   0.321 0.749709    
FertilizerC:WateringScheduleWeekly   3.4522     2.4820   1.391 0.169966    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 3.924 on 54 degrees of freedom
Multiple R-squared:  0.4872,    Adjusted R-squared:  0.4397 
F-statistic: 10.26 on 5 and 54 DF,  p-value: 5.987e-07

Questions:

  1. Write the two-way ANOVA model equation. Define all terms.
  2. What is the predicted growth for a plant receiving Fertilizer A with daily watering?
  3. What is the predicted growth for a plant receiving Fertilizer C with weekly watering?

Practice Problem 3.2: ANOVA Table Interpretation

The ANOVA table for the model in Problem 3.1:

Code 
# Create ANOVA table
anova(M2) %>%
  tidy() %>%
  kable(digits = 4, caption = "ANOVA Table for Plant Growth Model")
ANOVA Table for Plant Growth Model
term df sumsq meansq statistic p.value
Fertilizer 2 757.0331 378.5165 24.5771 0.0000
WateringSchedule 1 0.4350 0.4350 0.0282 0.8672
Fertilizer:WateringSchedule 2 32.6788 16.3394 1.0609 0.3532
Residuals 54 831.6641 15.4012 NA NA

Questions:

  1. Write the null and alternative hypotheses for testing the interaction effect. What is your conclusion using α = 0.05?
  2. Does Fertilizer type have a statistically significant main effect on Growth? Support your answer with the appropriate statistics.
  3. Interpret what a significant interaction means in the context of this problem.
Code 
# Create interaction plot
plants %>%
  group_by(Fertilizer, WateringSchedule) %>%
  summarise(Mean_Growth = mean(Growth), .groups = "drop") %>%
  ggplot(aes(x = WateringSchedule, y = Mean_Growth, 
             color = Fertilizer, group = Fertilizer)) +
  geom_line(size = 1.2) +
  geom_point(size = 3) +
  labs(
    title = "Interaction Plot: Plant Growth by Fertilizer and Watering Schedule",
    x = "Watering Schedule",
    y = "Mean Growth (cm)",
    color = "Fertilizer"
  ) +
  scale_color_brewer(palette = "Set1") +
  theme_minimal(base_size = 12)

Section 4: Comprehensive Practice Problems

Practice Problem 4.1: Ice Cream Sales

An ice cream shop owner wants to predict daily sales based on temperature. They collect data for 60 days.

Code 
# Create ice cream sales data
set.seed(321)
ice_cream <- data.frame(
  Temperature = runif(60, 60, 95)
) %>%
  mutate(
    Sales = 45.67 + 2.34 * Temperature + rnorm(60, 0, 15.4)
  )
Code 
# Fit the model
M3 <- lm(Sales ~ Temperature, data = ice_cream)
summary(M3)

Call:
lm(formula = Sales ~ Temperature, data = ice_cream)

Residuals:
    Min      1Q  Median      3Q     Max 
-38.674  -9.787   0.416   9.137  25.275 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  63.6480    15.1981   4.188 9.70e-05 ***
Temperature   2.1044     0.1939  10.855 1.36e-15 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 14.42 on 58 degrees of freedom
Multiple R-squared:  0.6701,    Adjusted R-squared:  0.6644 
F-statistic: 117.8 on 1 and 58 DF,  p-value: 1.361e-15
Code 
# Visualize
ggplot(ice_cream, aes(x = Temperature, y = Sales)) +
  geom_point(alpha = 0.6, size = 2) +
  geom_smooth(method = "lm", se = TRUE, color = "#e74c3c") +
  labs(
    title = "Ice Cream Sales vs. Temperature",
    x = "Temperature (°F)",
    y = "Sales ($)"
  )

Questions:

  1. Write the fitted equation and interpret both coefficients.
  2. Predict sales when the temperature is 85°F.
  3. Test whether temperature is a significant predictor of sales at α = 0.05.
  4. What percentage of variation in sales is explained by temperature?

Practice Problem 4.2: Coffee Shop Satisfaction

A coffee shop studies customer satisfaction (scale 1-10) based on wait time (Short, Medium, Long) and drink type (Hot, Cold).

Code 
# Create coffee shop data
set.seed(654)
coffee_data <- expand.grid(
  WaitTime = c("Short", "Medium", "Long"),
  DrinkType = c("Hot", "Cold"),
  Rep = 1:20
) %>%
  mutate(
    WaitTime = factor(WaitTime, levels = c("Short", "Medium", "Long")),
    DrinkType = factor(DrinkType, levels = c("Hot", "Cold"))
  ) %>%
  mutate(
    Satisfaction = 8.234 +
      ifelse(WaitTime == "Medium", -0.987, 0) +
      ifelse(WaitTime == "Long", -2.456, 0) +
      ifelse(DrinkType == "Cold", 0.543, 0) +
      ifelse(WaitTime == "Medium" & DrinkType == "Cold", -0.234, 0) +
      ifelse(WaitTime == "Long" & DrinkType == "Cold", -1.123, 0) +
      rnorm(n(), 0, 0.8)
  ) %>%
  mutate(Satisfaction = pmin(pmax(Satisfaction, 1), 10)) %>%
  select(-Rep)
Code 
# Fit the model
M4 <- lm(Satisfaction ~ WaitTime * DrinkType, data = coffee_data)
summary(M4)

Call:
lm(formula = Satisfaction ~ WaitTime * DrinkType, data = coffee_data)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.23266 -0.35797 -0.04237  0.41916  1.90237 

Coefficients:
                             Estimate Std. Error t value Pr(>|t|)    
(Intercept)                    8.0115     0.1634  49.018  < 2e-16 ***
WaitTimeMedium                -1.1453     0.2311  -4.955 2.53e-06 ***
WaitTimeLong                  -2.5314     0.2311 -10.952  < 2e-16 ***
DrinkTypeCold                  0.9064     0.2311   3.922 0.000151 ***
WaitTimeMedium:DrinkTypeCold   0.1432     0.3269   0.438 0.662155    
WaitTimeLong:DrinkTypeCold    -1.2227     0.3269  -3.741 0.000289 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.7309 on 114 degrees of freedom
Multiple R-squared:  0.7865,    Adjusted R-squared:  0.7771 
F-statistic: 83.99 on 5 and 114 DF,  p-value: < 2.2e-16
Code 
# Interaction plot
coffee_data %>%
  group_by(WaitTime, DrinkType) %>%
  summarise(Mean_Satisfaction = mean(Satisfaction), .groups = "drop") %>%
  ggplot(aes(x = WaitTime, y = Mean_Satisfaction, 
             color = DrinkType, group = DrinkType)) +
  geom_line(size = 1.2) +
  geom_point(size = 3) +
  labs(
    title = "Customer Satisfaction by Wait Time and Drink Type",
    x = "Wait Time",
    y = "Mean Satisfaction (1-10)",
    color = "Drink Type"
  ) +
  scale_color_manual(values = c("Hot" = "#e74c3c", "Cold" = "#3498db"))

Questions:

  1. What is the predicted satisfaction for a customer who waits a short time for a hot drink?
  2. What is the predicted satisfaction for a customer who waits a long time for a cold drink?
  3. Based on the coefficients, how does wait time affect satisfaction? Does this effect differ by drink type?

Section 5: Logistic Regression

Practice Problem 5.1: Legendary Pokémon Classification

A researcher wants to predict whether a Pokémon is Legendary based on its total stats. They create a variable called TotalStats (sum of HP, Attack, Defense, Speed, Sp. Atk, and Sp. Def) and fit a logistic regression model.

Code 
# Create total stats variable
set.seed(999)
# Load Pokemon data
pokemon <- read_csv("https://gist.githubusercontent.com/armgilles/194bcff35001e7eb53a2a8b441e8b2c6/raw/92200bc0a673d5ce2110aaad4544ed6c4010f687/pokemon.csv") %>%
  rename(
    Type1 = `Type 1`,
    Type2 = `Type 2`
  ) %>%
  mutate(
    Generation = as.factor(Generation),
    Type1 = as.factor(Type1),
    Legendary = as.logical(Legendary)
  )

pokemon_log <- pokemon %>%
  mutate(
    TotalStats = HP + Attack + Defense + Speed + `Sp. Atk` + `Sp. Def`
  )

# Visualize relationship
ggplot(pokemon_log, aes(x = TotalStats, y = as.numeric(Legendary), color = Legendary)) +
  geom_point(alpha = 0.3, position = position_jitter(height = 0.02)) +
  geom_smooth(method = "glm", method.args = list(family = "binomial"), 
              se = TRUE, color = "black") +
  labs(
    title = "Probability of Being Legendary vs Total Stats",
    x = "Total Stats",
    y = "Legendary (0 = No, 1 = Yes)"
  ) +
  scale_color_manual(values = c("FALSE" = "#3498db", "TRUE" = "#e74c3c")) +
  theme(legend.position = "top")

Code 
# Fit logistic regression model
M5 <- glm(Legendary ~ TotalStats, data = pokemon_log, family = binomial)
summary(M5)

Call:
glm(formula = Legendary ~ TotalStats, family = binomial, data = pokemon_log)

Coefficients:
              Estimate Std. Error z value Pr(>|z|)    
(Intercept) -19.738819   2.065737  -9.555   <2e-16 ***
TotalStats    0.031795   0.003513   9.051   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 450.90  on 799  degrees of freedom
Residual deviance: 189.59  on 798  degrees of freedom
AIC: 193.59

Number of Fisher Scoring iterations: 8

Questions:

  1. Write the logistic regression model equation in terms of log-odds (logit form).
  2. Interpret the coefficient for TotalStats. What does it tell us about the relationship between total stats and the odds of being Legendary?
  3. Use the model to predict the probability that a Pokémon with TotalStats = 600 is Legendary.
  4. Based on the output, is TotalStats a significant predictor of Legendary status at α = 0.05? Support your answer.

Practice Problem 5.2: Type and Legendary Status

Now consider a model that includes both TotalStats and Type1 (simplified to Water vs. Non-Water for this example).

Code 
# Create simplified type variable
pokemon_log2 <- pokemon_log %>%
  mutate(
    IsWater = ifelse(Type1 == "Water", "Water", "Non-Water"),
    IsWater = factor(IsWater, levels = c("Non-Water", "Water"))
  )
Code 
# Fit model with Type
M6 <- glm(Legendary ~ TotalStats + IsWater, data = pokemon_log2, family = binomial)
summary(M6)

Call:
glm(formula = Legendary ~ TotalStats + IsWater, family = binomial, 
    data = pokemon_log2)

Coefficients:
               Estimate Std. Error z value Pr(>|z|)    
(Intercept)  -19.841208   2.117647  -9.369   <2e-16 ***
TotalStats     0.032182   0.003612   8.910   <2e-16 ***
IsWaterWater  -1.369323   0.756276  -1.811   0.0702 .  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 450.90  on 799  degrees of freedom
Residual deviance: 185.68  on 797  degrees of freedom
AIC: 191.68

Number of Fisher Scoring iterations: 8
Code 
# Visualize
ggplot(pokemon_log2, aes(x = TotalStats, y = as.numeric(Legendary), color = IsWater)) +
  geom_point(alpha = 0.3, position = position_jitter(height = 0.02)) +
  geom_smooth(method = "glm", method.args = list(family = "binomial"), 
              se = FALSE) +
  labs(
    title = "Probability of Being Legendary by Total Stats and Type",
    x = "Total Stats",
    y = "Legendary (0 = No, 1 = Yes)",
    color = "Type"
  ) +
  scale_color_manual(values = c("Non-Water" = "#e74c3c", "Water" = "#3498db"))

Questions:

  1. What is the predicted log-odds of being Legendary for a Non-Water type Pokémon with TotalStats = 500?
  2. What is the predicted probability of being Legendary for a Water type Pokémon with TotalStats = 500?
  3. Does Type (Water vs. Non-Water) have a significant effect on Legendary status after controlling for TotalStats? Use α = 0.05.