🎲 Module 2: Chi-Square Goodness-of-Fit Test

Testing if Data Matches Expected Distributions

📚 Learning Objectives

By the end of this module, you will be able to:

Understand what the goodness-of-fit test tells you
Set up hypotheses for goodness-of-fit tests
Run chisq.test() in R with specified expected proportions
Interpret Chi-square output (χ², df, p-value)
Create frequency tables and visualizations
Make informed conclusions from goodness-of-fit results

🎯 What is a Goodness-of-Fit Test?

A goodness-of-fit test answers the question:

"Does my observed data fit an expected pattern?"

Key characteristics:

One categorical variable with 2+ categories
You have an expected distribution to test against
Compares observed frequencies to expected frequencies

🐀 Example: Do Rats Prefer Certain Maze Arms?

Research Question: Do rats have equal preference for three maze arms?

Setup: Test 60 rats, each chooses one of three arms (Left, Center, Right)

Null Hypothesis (H₀): Rats choose arms equally (33.3% each)

Expected frequencies: 20 rats per arm (60 ÷ 3 = 20)

Maze Arm	Observed	Expected
Left	15	20
Center	18	20
Right	27	20
Total	60	60

Chi-square asks: Is the difference between observed (15, 18, 27) and expected (20, 20, 20) statistically significant, or just random variation?

📝 Setting Up Hypotheses

Null Hypothesis (H₀): The observed distribution matches the expected distribution

Alternative Hypothesis (H₁): The observed distribution differs from the expected distribution

⚠️ Important Note About Hypotheses

Unlike t-tests or ANOVA, Chi-square goodness-of-fit is non-directional. We don't predict which category will be higher - just that the pattern differs from expectation.

Common Expected Distributions:

Expected Pattern	When to Use	Example
Equal proportions	Testing if all categories are equally likely	Coin flips (50/50), die rolls (1/6 each)
Specified ratios	Testing Mendelian ratios, known distributions	Genetics (3:1 ratio), survey norms
Population proportions	Comparing sample to known population	Census data, disease prevalence

The Chi-Square Formula

χ² = Σ [(O - E)² / E]

Where:

χ² = Chi-square statistic (pronounced "kai-square")
O = Observed frequency in each category
E = Expected frequency in each category
Σ = Sum across all categories

Interpretation: Larger χ² values = bigger difference between observed and expected = more evidence against H₀

🧮 Interactive Chi-Square Calculator

Calculate χ² by hand to understand how it works!

Category	Observed (O)	Expected (E)	(O-E)²/E
Category 1			-
Category 2			-
Category 3			-

💻 Running Chi-Square Goodness-of-Fit in R

Now let's run this analysis in R! The function is chisq.test().

1Create Your Observed Data

# Vector of observed frequencies
observed <- c(15, 18, 27)

# Give them meaningful names
names(observed) <- c("Left", "Center", "Right")

# Check your data
observed

2Run the Chi-Square Test

# Test against equal proportions (default)
chisq.test(observed)

# OR specify custom expected proportions
# For equal: p = c(1/3, 1/3, 1/3)
chisq.test(observed, p = c(1/3, 1/3, 1/3))

3Interpret the Output

Chi-squared test for given probabilities

data:  observed
X-squared = 4.3, df = 2, p-value = 0.1165

What this output tells you:

X-squared = 4.3: The Chi-square statistic (χ² = 4.3)
df = 2: Degrees of freedom (number of categories - 1)
p-value = 0.1165: Probability of seeing this difference by chance

Interpretation: p = 0.117 > 0.05, so we fail to reject H₀. The observed distribution doesn't significantly differ from equal proportions. Rats don't show a clear preference.

🔢 Understanding Degrees of Freedom (df)

df = k - 1

k = number of categories

Why k - 1? Once you know the frequencies in (k-1) categories and the total n, the last category is determined. That's one constraint, so we lose one degree of freedom.

Examples:

3 maze arms: df = 3 - 1 = 2
2 categories (e.g., yes/no): df = 2 - 1 = 1
6 categories: df = 6 - 1 = 5

📊 Visualizing Your Results

Always create a bar plot to visualize observed vs. expected frequencies!

# Create a bar plot comparing observed and expected
observed <- c(15, 18, 27)
expected <- c(20, 20, 20)
categories <- c("Left", "Center", "Right")

# Combine into a matrix for plotting
data <- rbind(observed, expected)
colnames(data) <- categories

# Create grouped bar plot
barplot(data, beside = TRUE, 
        col = c("#f5576c", "#9c27b0"),
        legend = c("Observed", "Expected"),
        main = "Rat Maze Arm Preferences",
        xlab = "Maze Arm",
        ylab = "Frequency",
        ylim = c(0, 30))

🎯 Practice Problem 1: Birth Days

Research Question: Are births equally distributed across days of the week?

Data: Hospital records show the following births:

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun
Births	28	32	29	31	35	18	17

Total births: 190

Expected per day (if equal): 190 ÷ 7 = 27.14

# Your turn! Complete this R code:

# Step 1: Enter the observed frequencies
births <- c(28, 32, 29, 31, 35, 18, 17)
names(births) <- c("Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun")

# Step 2: Run Chi-square test
result <- chisq.test(births)
result

# Step 3: Visualize
barplot(births, col = "#f5576c",
        main = "Births by Day of Week",
        ylab = "Number of Births")

Chi-squared test for given probabilities

data:  births
X-squared = 14.63, df = 6, p-value = 0.023

Interpretation: χ² = 14.63, p = 0.023 < 0.05. We reject H₀. Births are NOT equally distributed across days. Looking at the data, weekends (Sat/Sun) have notably fewer births - likely due to scheduled C-sections and inductions happening on weekdays!

🎯 Practice Problem 2: Mendelian Genetics

Research Question: Do flower colors follow Mendelian 3:1 ratio?

Background: According to Mendelian genetics, we expect 75% purple flowers and 25% white flowers.

Observed data: 152 purple, 48 white (n = 200 total)

# Test against 3:1 ratio (0.75 and 0.25 proportions)
flowers <- c(152, 48)
names(flowers) <- c("Purple", "White")

# Specify expected proportions
chisq.test(flowers, p = c(0.75, 0.25))

# Calculate expected frequencies manually
expected <- 200 * c(0.75, 0.25)
expected  # Should be 150 and 50

Chi-squared test for given probabilities

data:  flowers
X-squared = 0.107, df = 1, p-value = 0.744

Interpretation: χ² = 0.107, p = 0.744 >> 0.05. We fail to reject H₀. The observed data (152:48) fits the expected 3:1 Mendelian ratio very well. This supports Mendelian inheritance!

🤔 Check Your Understanding

Question 1: You test 100 animals for color preference. Observed: Red=35, Blue=40, Green=25. You expect equal preferences. What are the expected frequencies?

A) Red=35, Blue=40, Green=25 (same as observed)

B) Red=33.33, Blue=33.33, Green=33.33 (equal split)

C) Red=50, Blue=30, Green=20

Correct! With 100 animals and 3 categories, if we expect equal preferences, each category should have 100÷3 = 33.33 animals. Expected frequencies always reflect your null hypothesis!

Question 2: You get χ² = 8.5, df = 3, p = 0.037. What do you conclude?

A) Reject H₀: Observed distribution differs from expected (p < 0.05)

B) Fail to reject H₀: No significant difference

C) Cannot determine without seeing the data

Correct! p = 0.037 < 0.05, so we reject the null hypothesis. The observed distribution significantly differs from the expected distribution. The chi-square statistic (8.5) exceeded the critical value for df=3.

Question 3: A study has 4 categories. What are the degrees of freedom?

A) df = 4

B) df = 3 (k - 1 = 4 - 1)

C) df = 5

Correct! Degrees of freedom = number of categories minus 1. With 4 categories: df = 4 - 1 = 3. Once you know the frequencies in 3 categories and the total, the 4th is determined.

⚠️ Common Mistakes to Avoid

Mistake 1: Using raw data instead of frequencies

# WRONG - Don't use individual observations
responses <- c("A", "B", "A", "C", "B", "A")
chisq.test(responses)  # Error!

# CORRECT - Use frequency counts
freq <- table(responses)  # Converts to frequencies
chisq.test(freq)

Mistake 2: Forgetting that totals must match

Your expected frequencies must sum to the same total as your observed frequencies!

❌ Observed total = 100, Expected total = 120 → ERROR

✅ Observed total = 100, Expected total = 100 → CORRECT

Mistake 3: Misinterpreting non-significant results

If p > 0.05, we DON'T conclude "the distributions are the same." We say "we don't have evidence they differ" - absence of evidence isn't evidence of absence!

Mistake 4: Ignoring the assumption about expected frequencies

Chi-square requires expected frequencies ≥ 5 in all categories. We'll cover this more in Module 4!

✍️ Writing Up Your Results

Here's how to report Chi-square goodness-of-fit results in APA style:

Example Write-Up

Study: Rats choosing among three maze arms (n=60)

Results:

"A chi-square goodness-of-fit test was conducted to determine whether rats showed equal preference for three maze arms. The distribution of choices did not significantly differ from equal proportions, χ²(2) = 4.30, p = .117. Rats selected the left arm 15 times (25%), center arm 18 times (30%), and right arm 27 times (45%). Although there was a numerical preference for the right arm, this pattern could be due to chance variation."

Essential elements to include:

✓ Test name (chi-square goodness-of-fit)
✓ Chi-square statistic and df: χ²(df) = value
✓ p-value
✓ Observed frequencies or percentages
✓ Clear conclusion about the hypothesis

🎯 Comprehensive Practice Problem

Research Question: Do neurons fire at uniform rates across four time periods?

Background: You record neural activity in 4 consecutive time periods and count spikes.

Observed spike counts:

Time Period	0-250ms	250-500ms	500-750ms	750-1000ms
Spikes	45	38	52	45

Your tasks:

Calculate the total spikes and expected frequency per period
Write the R code to run the test
Interpret the results
Write up the findings in a sentence

# Complete this analysis:

# Step 1: Enter data
spikes <- c(45, 38, 52, 45)
names(spikes) <- c("0-250ms", "250-500ms", "500-750ms", "750-1000ms")

# Step 2: Calculate totals
total <- sum(spikes)
total  # Should be 180

# Step 3: Expected frequency (if uniform)
expected_each <- total / 4
expected_each  # Should be 45

# Step 4: Run chi-square
result <- chisq.test(spikes)
result

# Step 5: Visualize
barplot(spikes, col = c("#f093fb", "#f5576c", "#9c27b0", "#e91e63"),
        main = "Neural Spike Counts by Time Period",
        ylab = "Number of Spikes",
        las = 2)  # Rotates x-axis labels
abline(h = expected_each, lty = 2, col = "blue", lwd = 2)
legend("topright", legend = "Expected", lty = 2, col = "blue", lwd = 2)

Chi-squared test for given probabilities

data:  spikes
X-squared = 2.622, df = 3, p-value = 0.4537

Calculations:

Total spikes = 180
Expected per period = 180 ÷ 4 = 45
df = 4 - 1 = 3

Interpretation: χ²(3) = 2.62, p = .454. We fail to reject H₀. The neural firing rate does not significantly differ from uniform across the four time periods. Although there is some variation (38 to 52 spikes), this could easily occur by chance.

Write-up:

"A chi-square goodness-of-fit test examined whether neural spike counts were uniformly distributed across four consecutive 250ms time periods. The distribution did not significantly differ from uniform, χ²(3) = 2.62, p = .454, suggesting consistent firing rates over time."

📋 Chi-Square Goodness-of-Fit Workflow

Follow these steps every time:

1State hypotheses

H₀: Distribution matches expected; H₁: Distribution differs from expected

2Check assumptions

• Categorical variable
• Independent observations
• Expected frequencies ≥ 5

3Calculate expected frequencies

Based on your null hypothesis (equal proportions, specific ratio, etc.)

4Run test in R

chisq.test(observed, p = c(...))

5Interpret results

Look at χ², df, and p-value. If p < .05, reject H₀

6Visualize and report

Create bar plot, write up results with all key statistics

📝 Module 2 Summary

Key Takeaways:

Goodness-of-fit tests compare observed frequencies to expected distributions
Formula: χ² = Σ[(O-E)²/E] measures deviation from expectations
In R: Use chisq.test(observed, p = c(...))
df = k - 1 where k = number of categories
Larger χ² = more evidence against H₀
Always visualize your data with bar plots
Report χ²(df) = value, p = .xxx in write-ups

Next up: Module 3 covers Chi-square Test of Independence - testing relationships between TWO categorical variables!