ANOVA in R Programming - Mini Project: Build & Apply

Performing ANOVA Analysis in R

📖 Scenario: You are a data analyst working with a company that wants to understand if different diets affect weight loss differently. You have collected weight loss data from three groups following different diets.

🎯 Goal: You will create a dataset, set up the factor variable, perform an ANOVA test to compare the groups, and finally display the ANOVA summary results.

📋 What You'll Learn

Create a data frame called weight_loss with two columns: Diet and Loss with exact values

Create a factor variable diet_factor from the Diet column

Perform an ANOVA test using aov() with Loss as response and diet_factor as predictor, store in anova_result

Print the summary of anova_result to display the ANOVA table

💡 Why This Matters

🌍 Real World

ANOVA is used in many fields like medicine, agriculture, and marketing to compare multiple groups and see if their averages differ significantly.

💼 Career

Data analysts and scientists use ANOVA to make data-driven decisions and understand the impact of different treatments or conditions.

1

Create the dataset

Create a data frame called weight_loss with two columns: Diet and Loss. Use these exact values: Diet = c('A', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C') and Loss = c(3.2, 2.8, 3.6, 4.1, 4.5, 4.0, 5.2, 5.5, 5.1).

R Programming

# Create the data frame weight_loss with Diet and Loss columns
# Your code here

Hint

Use data.frame() to create a table with columns named Diet and Loss.

2

Create a factor variable for Diet

Create a factor variable called diet_factor from the Diet column of the weight_loss data frame.

R Programming

weight_loss <- data.frame(
  Diet = c('A', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C'),
  Loss = c(3.2, 2.8, 3.6, 4.1, 4.5, 4.0, 5.2, 5.5, 5.1)
)
# Create diet_factor as a factor from weight_loss$Diet
# Your code here

Hint

Use the factor() function to convert the Diet column into a factor.

3

Perform ANOVA test

Use the aov() function to perform an ANOVA test with Loss as the response variable and diet_factor as the predictor. Store the result in a variable called anova_result.

R Programming

weight_loss <- data.frame(
  Diet = c('A', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C'),
  Loss = c(3.2, 2.8, 3.6, 4.1, 4.5, 4.0, 5.2, 5.5, 5.1)
)
diet_factor <- factor(weight_loss$Diet)
# Perform ANOVA and store in anova_result
# Your code here

Hint

Use the formula Loss ~ diet_factor inside aov() and specify data = weight_loss.

4

Display ANOVA summary

Print the summary of the ANOVA result stored in anova_result using the summary() function.

R Programming

weight_loss <- data.frame(
  Diet = c('A', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C'),
  Loss = c(3.2, 2.8, 3.6, 4.1, 4.5, 4.0, 5.2, 5.5, 5.1)
)
diet_factor <- factor(weight_loss$Diet)
anova_result <- aov(Loss ~ diet_factor, data = weight_loss)
# Print the summary of anova_result
# Your code here

Hint

Use print(summary(anova_result)) to show the ANOVA table.

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