ANOVA in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of one-way ANOVA summary

What is the output of the following R code snippet that performs a one-way ANOVA on the given data?

R Programming

group <- factor(c('A','A','B','B','C','C'))
values <- c(5,7,8,9,6,7)
model <- aov(values ~ group)
summary(model)

A

Df Sum Sq Mean Sq F value Pr(>F)
group 2 8.67 4.33 1.3 0.35
Residuals 3 10 3.33

B

Df Sum Sq Mean Sq F value Pr(>F)
group 2 10 5 2 0.2
Residuals 3 7.5 2.5

C

Df Sum Sq Mean Sq F value Pr(>F)
group 2 12 6 3 0.1
Residuals 3 6 2

D Error in aov(values ~ group): object 'values' not found

💡 Hint

Check the degrees of freedom and sums of squares calculated by the aov function.

The one-way ANOVA partitions variance into group and residual components. The summary output shows degrees of freedom (Df), sum of squares (Sum Sq), mean squares (Mean Sq), F value, and p-value (Pr(>F)). The correct output matches the calculated values for this data.

🧠 Conceptual

intermediate

Understanding assumptions of ANOVA

Which of the following is NOT an assumption required for a valid one-way ANOVA test?

A The residuals are normally distributed within each group.

B The observations are independent of each other.

C The dependent variable is measured on a nominal scale.

D The groups have equal variances (homogeneity of variance).

💡 Hint

Think about the type of data ANOVA analyzes.

ANOVA requires the dependent variable to be continuous (interval or ratio scale), not nominal. Nominal data cannot be analyzed with ANOVA because it does not have meaningful numeric values.

🔧 Debug

advanced

Identify the error in two-way ANOVA code

What error will the following R code produce when run, and why?

R Programming

factor1 <- factor(c('Low','Low','High','High'))
factor2 <- factor(c('A','B','A','B'))
response <- c(10,15,20,25)
model <- aov(response ~ factor1 + factor2 + factor1:factor3)
summary(model)

A Error: object 'factor3' not found

B Error: response variable must be numeric

C Error: factors must have the same length

D No error, model runs successfully

💡 Hint

Check the variables used in the formula.

The formula references 'factor3' which is not defined in the code, causing an error.

❓ Predict Output

advanced

Output of Tukey HSD post-hoc test

What is the output of the following R code performing Tukey's Honest Significant Difference test after ANOVA?

R Programming

group <- factor(c('X','X','Y','Y','Z','Z'))
values <- c(4,5,7,8,6,7)
model <- aov(values ~ group)
tukey <- TukeyHSD(model)
print(tukey)

A Error in TukeyHSD(model): model must be a fitted aov object

B

  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = values ~ group)

$group
       diff        lwr       upr     p adj
Y-X 2.5 0.1 4.9 0.04
Z-X 1.0 -1.4 3.4 0.5
Z-Y -1.5 -3.9 0.9 0.2

C

  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = values ~ group)

$group
       diff        lwr       upr     p adj
Y-X 3.0 1.0 5.0 0.01
Z-X 1.5 -0.5 3.5 0.3
Z-Y -1.5 -3.5 0.5 0.2

D No significant differences found between groups

💡 Hint

Check the differences and confidence intervals in the TukeyHSD output.

The TukeyHSD output shows pairwise group differences with confidence intervals and adjusted p-values. Option B matches the expected output for this data.

🧠 Conceptual

expert

Interpreting interaction effects in two-way ANOVA

In a two-way ANOVA, what does a significant interaction effect between two factors imply?

A Both factors independently affect the response variable with no influence on each other.

B Neither factor has any effect on the response variable.

C The response variable is not normally distributed.

D The effect of one factor on the response variable depends on the level of the other factor.

💡 Hint

Think about how factors influence each other in interaction.

A significant interaction means the effect of one factor changes depending on the level of the other factor, showing they do not act independently.

Practice

1. What is the main purpose of using aov() function in R?

easy

A. To create a scatter plot of data

B. To calculate correlation between two variables

C. To perform ANOVA and test if group means differ

D. To compute the median of a numeric vector

Solution

1. Step 1: Understand the function purpose

   The aov() function in R is designed to perform Analysis of Variance (ANOVA).

2. Step 2: Identify what ANOVA tests

   ANOVA tests if the means of different groups are significantly different from each other.

3. Final Answer:

   To perform ANOVA and test if group means differ -> Option C

4. Quick Check:

   aov() = test group means difference [OK]

Hint: Remember: aov() is for comparing group averages [OK]

Common Mistakes:

• Confusing aov() with correlation functions
• Thinking aov() creates plots
• Using aov() to calculate medians

2. Which of the following is the correct syntax to run a one-way ANOVA in R on variable score grouped by group in data frame df?

easy

A. anova(score, group, data = df)

B. aov(score ~ group, data = df)

C. aov(data = df, score ~ group)

D. anova(df$score ~ df$group)

Solution

1. Step 1: Recall correct aov() syntax

   The correct syntax is aov(response ~ factor, data = dataframe).

2. Step 2: Match syntax to options

   aov(score ~ group, data = df) matches this exactly: aov(score ~ group, data = df).

3. Final Answer:

   aov(score ~ group, data = df) -> Option B

4. Quick Check:

   Correct aov() syntax = aov(score ~ group, data = df) [OK]

Hint: Use formula: response ~ group, with data= argument [OK]

Common Mistakes:

• Swapping formula and data arguments
• Using anova() instead of aov() for model fitting
• Incorrect formula order or missing data argument

3. What will be the output of the following R code?

df <- data.frame(group = rep(c('A', 'B'), each=3), score = c(5,6,7,8,9,10))
model <- aov(score ~ group, data = df)
summary(model)

medium

A. Shows significant difference between groups with low p-value

B. Shows no significant difference with high p-value

C. Returns an error due to incorrect data frame

D. Prints the raw data frame without analysis

Solution

1. Step 1: Understand the data and groups

   Group A has scores 5,6,7; Group B has 8,9,10. Group B scores are clearly higher.

2. Step 2: Interpret ANOVA summary output

   Since group means differ noticeably, summary(model) will show a low p-value indicating significant difference.

3. Final Answer:

   Shows significant difference between groups with low p-value -> Option A

4. Quick Check:

   Different group means = significant ANOVA result [OK]

Hint: If group means differ clearly, expect low p-value [OK]

Common Mistakes:

• Assuming no difference despite clear mean gap
• Expecting error from correct code
• Confusing summary output with raw data print

4. Identify the error in this R code snippet for ANOVA:

data <- data.frame(treatment = c('X','Y','X','Y'), value = c(10, 15, 12, 14))
result <- aov(value, treatment, data=data)
summary(result)

medium

A. Variable names are not quoted properly

B. Data frame is missing required columns

C. summary() cannot be used on aov objects

D. Incorrect argument order in aov() function

Solution

1. Step 1: Check aov() function usage

   aov() requires a formula as first argument, not separate variables.

2. Step 2: Identify correct syntax

   The correct call is aov(value ~ treatment, data=data), not aov(value, treatment, data=data).

3. Final Answer:

   Incorrect argument order in aov() function -> Option D

4. Quick Check:

   aov() needs formula first [OK]

Hint: Always use formula: response ~ group in aov() [OK]

Common Mistakes:

• Passing variables separately instead of formula
• Misusing summary() on wrong object types
• Incorrect data frame structure assumptions

5. You have a data frame df with three groups in column treatment and numeric values in response. You want to check if the group means differ significantly and then perform a post-hoc test to find which groups differ. Which sequence of R commands is correct?

hard

A. model <- aov(response ~ treatment, data=df)
summary(model)
TukeyHSD(model)

B. summary(aov(response, treatment, data=df))
posthoc <- TukeyHSD(response)

C. anova(response ~ treatment, data=df)
posthoc <- aov(response ~ treatment, data=df)

D. model <- aov(treatment ~ response, data=df)
summary(model)
posthoc <- TukeyHSD(model)

Solution

1. Step 1: Fit ANOVA model correctly

   Use aov(response ~ treatment, data=df) to fit the model.

2. Step 2: Summarize and perform post-hoc test

   Use summary(model) to check significance, then TukeyHSD(model) for pairwise group comparisons.

3. Final Answer:

   model <- aov(response ~ treatment, data=df) summary(model) TukeyHSD(model) -> Option A

4. Quick Check:

   ANOVA then TukeyHSD = model <- aov(response ~ treatment, data=df) summary(model) TukeyHSD(model) [OK]

Hint: Run aov(), then summary(), then TukeyHSD() for post-hoc [OK]

Common Mistakes:

• Using wrong formula order in aov()
• Trying TukeyHSD on raw data instead of model
• Confusing anova() with aov() usage