What is the chi-square test formula?

The chi-square test formula is χ² = Σ[(O − E)² / E], where O is the observed frequency in each cell, E is the expected frequency, and Σ means sum over all cells. For a test of independence, the expected frequency for each cell is E = (Row Total × Column Total) / Grand Total.

How do you calculate expected frequencies for a chi-square test?

For a goodness-of-fit test: E = n × p, where n is the sample size and p is the theoretical proportion. For a test of independence: E = (Row Total × Column Total) / Grand Total — calculated separately for every cell in the contingency table.

How do you read a chi-square table?

To read a chi-square table: (1) Calculate your degrees of freedom — df = (r−1)(c−1) for independence, or df = k−1 for goodness-of-fit. (2) Choose your significance level α (usually 0.05). (3) Find the row matching your df and the column matching your α. The value at their intersection is your critical value. (4) If your calculated χ² exceeds this critical value, reject the null hypothesis.

How do you calculate a chi-square test step by step?

Step 1: State H₀ and H₁. Step 2: Set α (typically 0.05). Step 3: Build the contingency table with observed counts. Step 4: Calculate expected frequencies E = (Row × Col)/N. Step 5: Compute χ² = Σ[(O−E)²/E]. Step 6: Find df = (r−1)(c−1). Step 7: Look up the critical value. Step 8: If χ² > critical value, reject H₀.

What is the chi-square critical value at α = 0.05 with 1 degree of freedom?

The chi-square critical value at α = 0.05 with df = 1 is 3.841. This applies to any 2×2 contingency table or 2-category goodness-of-fit test. If your calculated χ² exceeds 3.841, reject the null hypothesis at the 5% significance level.

Can chi-square be negative?

No. Because the formula squares each difference (O − E)² before dividing, every term is non-negative. The chi-square statistic is always ≥ 0. A value of 0 means observed frequencies perfectly match expected frequencies.

Is chi-square parametric or nonparametric?

Chi-square is a nonparametric test. It does not assume the data follow a normal distribution, making it appropriate for categorical count data.

What is the difference between chi-square goodness-of-fit and chi-square test of independence?

A goodness-of-fit test examines a single categorical variable and tests whether its observed distribution matches a specified theoretical distribution. A test of independence examines two categorical variables simultaneously and tests whether they are statistically associated. Both use χ² = Σ[(O − E)² / E], but differ in how expected frequencies are calculated and in the df formula.

Chi-Square Test: Formula, Table, Examples & Step-by-Step Guide (2026)

Q: What is the minimum sample size for a chi-square test?

There is no single minimum total sample size, but every expected cell frequency must be at least 5. If more than 20% of cells have expected frequencies below 5, use Fisher's exact test instead.

What Is a Chi-Square Test?

Definition — Chi-Square Test (χ² Test)

A chi-square test is a nonparametric statistical test used to determine whether observed categorical data differ significantly from expected frequencies, or whether two categorical variables are independent of each other. It measures how much the data deviate from what chance alone would predict.

χ² = Σ [(O − E)² / E]

Chi-square tests work on count data — the number of people, items, or observations that fall into each category. They do not require the data to follow a normal distribution, which makes them one of the most broadly applicable tools in hypothesis testing. Researchers use chi-square tests in medical trials, ecology, marketing surveys, genetics, machine learning, and any field where outcomes are measured in categories rather than continuous numbers.

The test was developed by Karl Pearson in 1900. According to the NIST/SEMATECH Engineering Statistics Handbook, it is the standard approach for categorical data inference when expected cell counts are adequate. Understanding its assumptions before running the test is as important as the calculation itself — see the assumptions page for the broader picture.

⚡ Quick Reference — Chi-Square Key Facts

Formula: χ² = Σ[(O − E)² / E], where O = observed frequency, E = expected frequency
Data requirement: Categorical variables only (counts/frequencies, not means)
Three main types: Goodness-of-fit, test of independence, test of homogeneity
Key assumption: Every expected cell frequency must be ≥ 5
Most common critical value: df = 1, α = 0.05 → χ² = 3.841
Effect size: Use Cramér's V — V = √(χ² / [n × min(r−1, c−1)])
Small samples: When any E < 5, use Fisher's exact test instead

3.841

Critical value
df=1, α=0.05

≥ 5

Min. expected
cell frequency

1900

Year Pearson
developed it

χ²

Symbol (Greek
letter chi-squared)

Three Types of Chi-Square Tests

Chi-square tests take three distinct forms. Choosing the wrong one is a common mistake that invalidates your analysis, so it is worth being precise about what each one does.

Feature	Goodness-of-Fit Test	Test of Independence	Test of Homogeneity
Number of variables	1 categorical variable	2 categorical variables	1 variable across 2+ populations
Data structure	Single frequency table	Contingency table (rows × columns)	Separate samples from each population
Research question	Does this variable follow a specified distribution?	Are these two variables associated?	Do multiple populations have the same distribution?
Null hypothesis (H₀)	Observed = expected distribution	The two variables are independent	All populations have equal proportions
Expected frequency	E = n × p (theoretical proportion)	E = (Row Total × Col Total) / N	E = (Row Total × Col Total) / N (same formula)
Degrees of freedom	k − 1	(r − 1)(c − 1)	(r − 1)(c − 1)
Classic example	Do die faces appear equally often?	Is gender associated with product preference?	Do smoking rates differ across three cities?

Chi-Square Goodness-of-Fit Test

The goodness-of-fit test asks: does a single categorical variable match a theoretical distribution? You collect counts across k categories and compare them to expected counts under a specified null hypothesis. A genetics researcher testing whether a cross follows Mendel's 9:3:3:1 ratio, or a quality engineer testing whether defect types are uniformly distributed, both use this form. Degrees of freedom: df = k − 1.

Chi-Square Test of Independence

The test of independence asks: are two categorical variables related? Both variables are measured on the same sample. Their joint counts are arranged in a contingency table and the test determines whether the pattern is consistent with independence. This is the most common form in social science, medical, and marketing research. Degrees of freedom: df = (r − 1)(c − 1).

Chi-Square Test of Homogeneity

The test of homogeneity asks: do multiple independent populations have the same distribution of a categorical variable? Unlike the test of independence (one sample, two measured variables), homogeneity uses separate random samples from each population. The calculation is mechanically identical — same formula, same df formula — but the research design and interpretation differ. For example: "Do students in three different universities prefer the same study methods?" uses homogeneity; "Are gender and study method preference related in one sample?" uses independence.

Chi-Square Test Formula Explained

The Chi-Square Formula: χ² = Σ[(O − E)² / E]

Chi-Square Test Statistic — Pearson (1900)

χ² = Σ [(O − E)² / E]

Sum across all categories (goodness-of-fit) or all cells (test of independence)

χ² = chi-square test statistic O = observed frequency (actual count) E = expected frequency (under H₀) Σ = sum over all categories or cells

In plain terms: For every cell or category, subtract the expected count from the observed count, square the result (to eliminate negative values), divide by the expected count (to standardize for scale), then add all those values together. A larger χ² means the data deviate more from the null hypothesis. Because every term is squared, χ² is always ≥ 0 — it can never be negative.

How to Calculate Expected Frequencies

Expected Frequency — Goodness-of-Fit

E = n × p

n = total sample size, p = theoretical expected proportion for that category

Expected Frequency — Test of Independence / Homogeneity

E = (Row Total × Column Total) / Grand Total

Calculate for every cell individually using the marginal totals

Degrees of Freedom Formula

Degrees of Freedom (df)

Goodness-of-fit: df = k − 1

Independence / Homogeneity: df = (r − 1)(c − 1)

k = number of categories; r = rows; c = columns in the contingency table

The degrees of freedom determine which chi-square distribution to use when finding the p-value or critical value. Higher df shifts the distribution to the right — the same χ² value corresponds to a larger p-value when df is larger. This is why the critical value at df = 1 (3.841 at α = 0.05) is much smaller than the critical value at df = 9 (16.919 at α = 0.05). Understanding degrees of freedom is key to correctly applying this test.

Chi-Square Test Assumptions (Conditions)

Chi-square test results are only valid when these five conditions hold. Penn State's STAT 500 course lists adequate expected cell frequency as the most commonly violated assumption in practice (Penn State STAT 500, Lesson 8).

Categorical data

Both variables must be categorical — nominal (unordered labels like colors or countries) or ordinal (ordered categories like rating scales). Chi-square cannot be directly applied to continuous measurements. See types of data for the distinction.

Independent observations

Each subject or observation contributes to exactly one cell. Observations must not be paired, matched, or repeated. For paired categorical data, use McNemar's test instead.

Adequate expected frequencies (E ≥ 5 per cell)

Every expected cell frequency must be at least 5. If more than 20% of cells have E < 5, the chi-square approximation is unreliable. For 2×2 tables with small expected counts, use Fisher's exact test (see Fisher's exact test examples).

Random or representative sampling

Data must come from a random sample or a representative sampling design. Non-random convenience samples limit the generalizability of the result. See study design for sampling principles.

Mutually exclusive categories

Each observation must fall into one and only one category. Overlapping categories (where an observation could be counted in two cells) violate the independence assumption and invalidate the test.

⚠️

Most Violated Assumption

Expected frequencies below 5 account for the majority of chi-square misapplications in published research. Always check E values before reporting results. The SPSS output footnote and R's chisq.test()$expected both flag this automatically. When the violation occurs, Fisher's exact test is your primary alternative.

Chi-Square Distribution Table (Critical Values)

Use this table to find the critical value for your test. Locate your degrees of freedom (df) in the left column, find the column matching your significance level (α), and read the critical value at their intersection. If your calculated χ² exceeds the critical value, reject the null hypothesis.

How to Read the Chi-Square Table

📖 4-Step Guide — How to Read a Chi-Square Critical Value Table

Calculate your degrees of freedom (df). For a test of independence or homogeneity: df = (rows − 1)(columns − 1). For goodness-of-fit: df = k − 1, where k is the number of categories. Find this number in the leftmost column of the table.
Choose your significance level (α). The most common choice is α = 0.05 (5%). Identify the column header that matches your α level — the table below shows 0.10, 0.05, 0.025, 0.01, and 0.001.
Read the critical value at the intersection. Where your df row meets your α column is your critical value. Example: df = 1 and α = 0.05 gives the critical value 3.841 (highlighted in the table below).
Compare your test statistic to the critical value. If your calculated χ² is greater than the critical value → reject H₀ (statistically significant). If χ² is less than or equal to the critical value → fail to reject H₀.

df	α = 0.10	α = 0.05	α = 0.025	α = 0.01	α = 0.001
1	2.706	3.841	5.024	6.635	10.828
2	4.605	5.991	7.378	9.210	13.816
3	6.251	7.815	9.348	11.345	16.266
4	7.779	9.488	11.143	13.277	18.467
5	9.236	11.070	12.833	15.086	20.515
6	10.645	12.592	14.449	16.812	22.458
7	12.017	14.067	16.013	18.475	24.322
8	13.362	15.507	17.535	20.090	26.125
9	14.684	16.919	19.023	21.666	27.877
10	15.987	18.307	20.483	23.209	29.588
12	18.549	21.026	23.337	26.217	32.910
15	22.307	24.996	27.488	30.578	37.697
20	28.412	31.410	34.170	37.566	45.315
25	34.382	37.652	40.646	44.314	52.620
30	40.256	43.773	46.979	50.892	59.703
40	51.805	55.758	59.342	63.691	73.402
50	63.167	67.505	71.420	76.154	86.661
60	74.397	79.082	83.298	88.379	99.607
80	96.578	101.879	106.629	112.329	124.839
100	118.498	124.342	129.561	135.807	149.449

Highlighted: χ² = 3.841 at df = 1, α = 0.05 — the most commonly referenced critical value. Source: tabulated from the chi-square CDF. Values match the NIST/SEMATECH Statistics Handbook Table. For the extended downloadable version, visit the Chi-Square Table reference page or the how to read the chi-square table guide.

Quick Lookup — Common Scenarios

Scenario	df	Critical value α = 0.05	Critical value α = 0.01
2-category goodness-of-fit	1	3.841	6.635
3-category goodness-of-fit	2	5.991	9.210
4-category goodness-of-fit	3	7.815	11.345
2×2 contingency table	1	3.841	6.635
2×3 contingency table	2	5.991	9.210
3×3 contingency table	4	9.488	13.277
3×4 contingency table	6	12.592	16.812
4×4 contingency table	9	16.919	21.666
Mendel's 9:3:3:1 ratio (4 categories)	3	7.815	11.345

How to Calculate a Chi-Square Test: 8-Step Method

📋

Chi-Square Test — 8 Steps (Featured Snippet)

Step 1: State H₀ and H₁. Step 2: Set α. Step 3: Build the contingency table with observed counts. Step 4: Calculate expected frequencies using E = (Row × Col) / N. Step 5: Compute χ² = Σ[(O − E)² / E]. Step 6: Find df = (r−1)(c−1). Step 7: Look up the critical value in the chi-square table. Step 8: Compare χ² to the critical value and state the conclusion.

Worked Example — Full 8-Step Calculation

Research question: Is there a statistically significant association between gender (Male/Female) and brand preference (Brand A / Brand B) in a sample of 100 consumers?

State the hypotheses:
H₀: Gender and brand preference are independent (no association)
H₁: Gender and brand preference are associated

Set the significance level: α = 0.05

Build the observed contingency table:

	Brand A	Brand B	Row Total
Male	30	20	50
Female	10	40	50
Col Total	40	60	100

Calculate expected frequencies using E = (Row Total × Column Total) / Grand Total:

Cell	Calculation	Expected (E)
Male / Brand A	(50 × 40) / 100	20.0
Male / Brand B	(50 × 60) / 100	30.0
Female / Brand A	(50 × 40) / 100	20.0
Female / Brand B	(50 × 60) / 100	30.0

✓ All expected frequencies ≥ 5. Assumption satisfied.

Calculate the chi-square statistic using χ² = Σ[(O − E)² / E]:

Cell	O	E	(O − E)²	(O − E)² / E
Male / Brand A	30	20	100	5.000
Male / Brand B	20	30	100	3.333
Female / Brand A	10	20	100	5.000
Female / Brand B	40	30	100	3.333
Total				χ² = 16.667

Degrees of freedom: df = (r − 1)(c − 1) = (2 − 1)(2 − 1) = 1

Critical value: At df = 1 and α = 0.05 → from the chi-square table above: critical value = 3.841

Decision: χ² = 16.667 > critical value 3.841 → Reject H₀

✓ Conclusion: There is a statistically significant association between gender and brand preference (χ²(1, N = 100) = 16.67, p < .001). Men and women differ in their brand preferences beyond what chance alone would predict.

Chi-Square Calculator (Interactive)

🧮 Chi-Square Test of Independence Calculator

Enter observed counts for a 2×2 contingency table. The calculator computes χ², degrees of freedom, the p-value approximation, and Cramér's V effect size. For larger tables, use the full chi-square calculator.

Observed Counts (O)

Row 1, Col 1

Row 1, Col 2

Row 2, Col 1

Row 2, Col 2

📊

Enter counts in the 2×2 table to the left and click Calculate.

Chi-Square Test Examples (Three Fields)

Example 1 — Medical Research: Smoking and Lung Disease

Worked Example — Test of Independence (Medicine)

Research question: Is smoking status (Smoker / Non-Smoker) associated with lung disease diagnosis (Yes / No) in a sample of 300 patients?

Observed contingency table:

	Lung Disease: Yes	Lung Disease: No	Row Total
Smoker	90	60	150
Non-Smoker	30	120	150
Col Total	120	180	300

Expected frequencies:
Smoker/Yes: (150 × 120) / 300 = 60 | Smoker/No: (150 × 180) / 300 = 90
Non-Smoker/Yes: (150 × 120) / 300 = 60 | Non-Smoker/No: (150 × 180) / 300 = 90

Chi-square statistic:
(90−60)²/60 + (60−90)²/90 + (30−60)²/60 + (120−90)²/90
= 900/60 + 900/90 + 900/60 + 900/90 = 15 + 10 + 15 + 10 = χ² = 50.00

df = 1. Critical value at α = 0.05, df = 1: 3.841. Since 50.00 ≫ 3.841, reject H₀.

✓ Conclusion: Smoking status and lung disease are significantly associated (χ²(1, N = 300) = 50.00, p < .001). Cramér's V = √(50/300) = 0.408 — a medium-to-large effect.

Example 2 — Genetics: Mendel's Goodness-of-Fit Test

Worked Example — Goodness-of-Fit (Biology / Genetics)

Research question: Does a dihybrid pea plant cross produce phenotype ratios consistent with Mendel's predicted 9:3:3:1 ratio in a sample of 160 offspring?

Observed vs. expected counts:

Phenotype	Observed (O)	Ratio	Expected (E = n × p)	(O−E)²/E
Round/Yellow	90	9/16	90.0	0.000
Round/Green	28	3/16	30.0	0.133
Wrinkled/Yellow	32	3/16	30.0	0.133
Wrinkled/Green	10	1/16	10.0	0.000
Total	160		160	χ² = 0.267

df = k − 1 = 3. Critical value at α = 0.05, df = 3: 7.815. Since 0.267 ≪ 7.815, fail to reject H₀.

✓ Conclusion: The observed phenotype ratios are consistent with Mendel's 9:3:3:1 prediction (χ²(3, N = 160) = 0.27, p = .966).

Example 3 — Market Research: Chi-Square Test of Homogeneity

Worked Example — Test of Homogeneity (Market Research)

Research question: Do customers in three cities (London, Manchester, Leeds) have the same distribution of subscription tier preferences (Basic / Standard / Premium)?

Observed contingency table (separate samples from each city):

	Basic	Standard	Premium	Row Total
London	40	80	30	150
Manchester	50	60	40	150
Leeds	30	70	50	150
Col Total	120	210	120	450

Expected frequencies: E = (Row Total × Col Total) / Grand Total
Example: London/Basic: (150 × 120) / 450 = 40. London/Standard: (150 × 210) / 450 = 70. London/Premium: (150 × 120) / 450 = 40. (Repeat for each city.)

Chi-square calculation:
χ² = (40−40)²/40 + (80−70)²/70 + (30−40)²/40 + (50−40)²/40 + (60−70)²/70 + (40−40)²/40 + (30−40)²/40 + (70−70)²/70 + (50−40)²/40
= 0 + 1.429 + 2.500 + 2.500 + 1.429 + 0 + 2.500 + 0 + 2.500 = χ² = 12.857

df = (3−1)(3−1) = 4. Critical value at α = 0.05, df = 4: 9.488. Since 12.857 > 9.488, reject H₀.

✓ Conclusion: Subscription tier preferences differ significantly across the three cities (χ²(4, N = 450) = 12.86, p = .012). The three populations do not share the same distribution of subscription tiers. For more solved examples, see chi-square test examples.

How to Report Chi-Square Results (APA Format)

When writing up chi-square test results, follow the APA 7th edition format. This is required by most journals and expected in graduate-level coursework.

APA 7th Edition Reporting Format

χ²(df, N = sample size) = value, p = p-value, V = Cramér's V

Full sentence examples:

For independence: "A chi-square test of independence found a significant relationship between gender and brand preference, χ²(1, N = 100) = 16.67, p < .001, Cramér's V = 0.41."
For goodness-of-fit: "A chi-square goodness-of-fit test indicated the observed phenotype distribution was consistent with the 9:3:3:1 Mendelian ratio, χ²(3, N = 160) = 0.27, p = .97."
For homogeneity: "A chi-square test of homogeneity revealed significant differences in subscription tier preferences across the three cities, χ²(4, N = 450) = 12.86, p = .012, Cramér's V = 0.17."

✅

Always report effect size

Statistical significance alone does not tell the reader how large the association is. With large samples, a trivially small association can produce p < .001. Always accompany a significant chi-square result with Cramér's V (or Phi for 2×2 tables). See the effect size guide for full details.

Effect Size: Cramér's V and Phi (φ)

A significant chi-square result tells you the association is real; Cramér's V tells you how strong it is. Cohen (1988) established the benchmark thresholds below, though Lakens (2013) notes they should be treated as contextual guides rather than rigid rules.

Cramér's V — Effect Size for Chi-Square

V = √( χ² / [n × min(r − 1, c − 1)] )

n = total sample size, r = rows, c = columns. V ranges from 0 (no association) to 1 (perfect association).

Cramér's V	Effect Size	df = 1 (2×2)	df = 2 (2×3)	df = 3 (2×4)
0.10	Small	Weak association	Weak association	Weak association
0.30	Medium	Moderate association	Moderate association	Moderate association
0.50	Large	Strong association	Strong association	Strong association

For 2×2 tables specifically, Phi (φ) is equivalent to Cramér's V: φ = √(χ²/n). Both yield the same value when df = 1. Use the effect size calculator to compute Cramér's V automatically.

Chi-Square Test in SPSS

Test of Independence in SPSS (Step by Step)

SPSS — Step-by-Step Menu Navigation

Running a chi-square test of independence in IBM SPSS Statistics

Go to Analyze → Descriptive Statistics → Crosstabs

Move your first variable into the Row(s) box and your second variable into the Column(s) box

Click Statistics → check Chi-square → also check Phi and Cramer's V for effect size → click Continue

Click Cells → check Observed and Expected under Counts → check Row under Percentages → click Continue → OK

Reading SPSS output: In the Chi-Square Tests table, read the Pearson Chi-Square row. The Asymptotic Significance (2-sided) column is your p-value. Check the footnote for "X cells have expected count less than 5" — if this appears, consider Fisher's Exact Test (also reported in the same table). The Symmetric Measures table gives Cramér's V.

Goodness-of-Fit in SPSS

Navigate to Analyze → Nonparametric Tests → Legacy Dialogs → Chi-Square. Move your variable into the Test Variable List. Under Expected Values, choose "All categories equal" for a uniform distribution, or enter custom expected proportions. Click OK.

Chi-Square Test in R

Test of Independence in R

# Create the observed contingency table
data_matrix <- matrix(c(30, 20, 10, 40), nrow = 2,
                      dimnames = list(
                        Gender = c("Male", "Female"),
                        Preference = c("Brand A", "Brand B")
                      ))

# Run the chi-square test
result <- chisq.test(data_matrix, correct = FALSE)  # correct=FALSE matches hand calc
print(result)
# X-squared = 16.667, df = 1, p-value = 4.46e-05

# Verify expected frequencies (must all be ≥ 5)
result$expected

# Calculate Cramér's V effect size
library(rstatix)
cramer_v(data_matrix)
        

Goodness-of-Fit in R

# Observed phenotype counts (Mendel's cross)
observed <- c(90, 28, 32, 10)

# Expected proportions from Mendel's 9:3:3:1 ratio
expected_probs <- c(9/16, 3/16, 3/16, 1/16)

# Run the goodness-of-fit test
chisq.test(observed, p = expected_probs)
# X-squared = 0.267, df = 3, p-value = 0.966
        

⚠️

Yates' Continuity Correction

R applies Yates' correction by default for 2×2 tables (subtract 0.5 before squaring), which reduces the χ² slightly. To match hand calculations, use chisq.test(data, correct = FALSE). For small samples where any E < 5, use fisher.test() instead.

Chi-Square Test in Python

Test of Independence Using SciPy

Python

from scipy.stats import chi2_contingency
import numpy as np

# Observed contingency table (rows=Gender, cols=Preference)
observed = np.array([[30, 20],   # Male: Brand A, Brand B
                        [10, 40]])  # Female: Brand A, Brand B

# Run the chi-square test (correction=False to match hand calculation)
chi2, p_value, df, expected = chi2_contingency(observed, correction=False)

print(f"Chi-square statistic: {chi2:.4f}")  # 16.6667
print(f"p-value: {p_value:.6f}")           # 0.000045
print(f"Degrees of freedom: {df}")          # 1
print(f"Expected frequencies:\n{expected}")

# Calculate Cramér's V manually
n = observed.sum()
cramers_v = np.sqrt(chi2 / (n * min(observed.shape[0]-1, observed.shape[1]-1)))
print(f"Cramér's V: {cramers_v:.4f}")       # 0.4082
        

Goodness-of-Fit in Python

Python

from scipy.stats import chisquare

# Observed and expected counts (Mendel's cross)
observed = [90, 28, 32, 10]
expected = [90.0, 30.0, 30.0, 10.0]  # n × p for 9:3:3:1 ratio, n=160

chi2, p_value = chisquare(f_obs=observed, f_exp=expected)

print(f"Chi-square: {chi2:.4f}")   # 0.2667
print(f"p-value: {p_value:.4f}")   # 0.9659
# Interpretation: data consistent with 9:3:3:1 ratio
        

Chi-Square vs. Other Statistical Tests

Chi-Square vs. t-Test

The choice between a chi-square test and a t-test comes down to the type of outcome variable. For a full comparison, see the dedicated chi-square vs. t-test guide.

Feature	Chi-Square Test	t-Test
Outcome variable type	Categorical (counts)	Continuous (means)
Normality required	No	Yes (or large n)
What it tests	Association or distribution	Difference between means
Example question	"Is political party related to voting behavior?"	"Is the mean exam score higher in Group A vs B?"
Effect size	Cramér's V	Cohen's d

Chi-Square vs. Fisher's Exact Test

Feature	Chi-Square Test	Fisher's Exact Test
Best for	Large samples (all E ≥ 5)	Small samples (any E < 5)
Calculation	Approximate (asymptotic)	Exact probability
Table size	Any r × c	Most commonly 2×2
Sample size guidance	n > 40 (with all E ≥ 5)	n < 20, or any E < 5
Software	Default in SPSS, R, Python	Checkbox in SPSS; `fisher.test()` in R; `scipy.stats.fisher_exact()` in Python

Chi-Square vs. ANOVA

Feature	Chi-Square Test	ANOVA
Outcome variable	Categorical (counts)	Continuous (means)
Independent variable	Categorical groups	Categorical groups
Tests	Association between categories	Differences in group means
Effect size	Cramér's V	η² (eta squared)

For a structured decision guide covering all major statistical tests, see the Statistical Test Selector tool, or the broader parametric vs. nonparametric tests guide.

Where Chi-Square Tests Are Used

🏥

Medical Research

Testing whether treatment outcomes (recovered/not recovered) differ by treatment group. Routine in randomized controlled trials with binary endpoints. See hypothesis testing in clinical trials.

🧬

Genetics & Biology

Verifying whether observed genotype or phenotype ratios match Mendelian or Hardy-Weinberg predictions in population genetics studies.

📊

Survey Analysis

Determining whether survey responses differ by demographic groups such as age, gender, or education level. Standard in social science research.

🛒

Market Research

Testing whether brand preferences, product choices, or consumer behaviors differ across customer segments or regions. See customer segmentation statistics.

🤖

Machine Learning

Feature selection: identifying which categorical features are statistically associated with the target variable before model training. See statistics for machine learning.

🔬

A/B Testing

Comparing conversion rates, click-through rates, or other binary outcomes between two variants. See how statistics powers A/B testing.

🌿

Ecology

Comparing species distribution across habitat types, or testing whether species are associated with particular environmental conditions.

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Quality Control

Testing whether defect rates or product categories are uniformly distributed across production lines, batches, or suppliers.

When Chi-Square Fails: Alternative Tests

Problem / Condition	Recommended Alternative	Why
Expected frequencies < 5 in any cell (small sample)	Fisher's exact test	Exact computation; no large-sample approximation
Paired or matched categorical data	McNemar's test	Accounts for non-independence of matched pairs
Ordered categories (ordinal data)	Cochran-Armitage trend test	Detects monotonic trend rather than general association
Very large N (χ² inflated trivially)	Report Cramér's V; interpret practically	With huge n, even negligible associations become significant
Three or more repeated measures	Cochran's Q test	Extension of McNemar for k > 2 related groups
Three or more independent groups	Kruskal-Wallis test (if ordinal)	Nonparametric test for comparing k ≥ 3 groups on ordinal outcomes

Frequently Asked Questions

Yates' correction modifies the chi-square formula for 2×2 tables: χ² = Σ[(|O − E| − 0.5)² / E]. Subtracting 0.5 before squaring reduces the test statistic slightly, producing a more conservative p-value for small samples. R applies it by default for 2×2 tables (disable with correct = FALSE). Many statisticians argue it overcorrects — for small samples, Fisher's exact test is generally more reliable.

Chi-Square Test Cheat Sheet (Quick Reference)

Every key term, formula, and decision rule from this guide in one structured table.

Term / Entity	Formula / Value	When to Use	Interpretation
Chi-square statistic (χ²)	`χ² = Σ[(O − E)² / E]`	All chi-square tests	Total deviation of observed from expected counts
Expected freq — goodness-of-fit	`E = n × p`	Single-variable test	Count predicted by theoretical proportion p
Expected freq — independence	`E = (Row × Col) / N`	Two-variable test	Count predicted if variables were unrelated
df — goodness-of-fit	`df = k − 1`	k = number of categories	Number of free cells after constraints applied
df — independence / homogeneity	`df = (r−1)(c−1)`	r × c contingency table	Same logic; both row and column margins constrained
Critical value (df=1, α=0.05)	3.841	Most 2×2 tables	Exceed this → reject H₀ at 5% level
Critical value (df=1, α=0.01)	6.635	2×2 with stricter threshold	Exceed this → reject H₀ at 1% level
Cramér's V	`V = √(χ²/[n × min(r−1,c−1)])`	After significant χ²	0.10=small, 0.30=medium, 0.50=large
Phi (φ) — 2×2 tables only	`φ = √(χ²/n)`	2×2 contingency table	Equivalent to Cramér's V when df = 1
p-value interpretation	P(χ² ≥ observed \| H₀ true)	Always report	p < 0.05 → reject H₀ (at conventional α)
Assumption: expected frequency	E ≥ 5 per cell	Every chi-square test	If violated, use Fisher's exact test instead
APA reporting format	χ²(df, N = n) = value, p = .xxx	All publications	Include df, sample size, statistic, p-value, V
Small sample alternative	Fisher's exact test	Any E < 5	Exact p-value; no approximation needed
Paired data alternative	McNemar's test	Before/after or matched pairs	Same χ² statistic but accounts for dependency