When should you use a chi-square test?

Use a chi-square test when: (1) your outcome variable is categorical (nominal or ordinal counts), (2) observations are independent, (3) each expected cell frequency is at least 5, and (4) you want to test either whether a single variable follows a specified distribution (goodness-of-fit) or whether two variables are associated (test of independence).

What is the chi-square test formula?

The chi-square test statistic 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 degrees of freedom for a chi-square test?

For a goodness-of-fit test: df = k − 1, where k is the number of categories. For a test of independence: df = (r − 1)(c − 1), where r is the number of rows and c is the number of columns in the contingency table.

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 is the most commonly used value in statistics — it applies to any 2×2 contingency table or a 2-category goodness-of-fit test at the 5% significance level. If your calculated χ² exceeds 3.841, reject the null hypothesis.

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 they differ in how expected frequencies are calculated and in the degrees of freedom formula.

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 exactly 0 means observed frequencies perfectly match expected frequencies — a result that would almost never occur with real data.

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

There is no single minimum sample size, but every expected cell frequency should be at least 5. If more than 20% of cells have expected frequencies below 5, the chi-square approximation becomes unreliable. For 2×2 tables with small samples (expected frequencies < 5 in any cell), use Fisher's exact test instead.

How do you interpret a chi-square p-value?

The p-value in a chi-square test is the probability of obtaining a χ² statistic as large as observed, assuming the null hypothesis is true. If p < α (e.g., p < 0.05), reject H₀ and conclude there is a statistically significant association or deviation. If p ≥ α, fail to reject H₀ — the data are consistent with the null hypothesis.

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. The chi-square distribution itself is a family of continuous probability distributions used to approximate the test statistic's sampling distribution when expected frequencies are adequate (≥ 5 per cell).

Chi-Square Test: Complete Guide, Formula & Distribution Table

Q: What is a chi-square test?

A chi-square test is a nonparametric statistical test used to determine whether observed categorical data differ significantly from expected frequencies (goodness-of-fit) or whether two categorical variables are independent of each other (test of independence). The test statistic χ² = Σ[(O − E)² / E] measures total deviation between observed and expected counts.

Q: How do you interpret a chi-square p-value?

The p-value in a chi-square test is the probability of obtaining a χ² statistic as large as observed, assuming the null hypothesis is true. If p < α (e.g., p < 0.05), reject H₀ and conclude there is a statistically significant association or deviation. If p ≥ α, fail to reject H₀ — the data are consistent with the null hypothesis.

Q: 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. The chi-square distribution itself is a family of continuous probability distributions used to approximate the test statistic's sampling distribution when expected frequencies are adequate (≥ 5 per cell).

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 statistics. Researchers use chi-square tests in medical trials, ecology, marketing surveys, genetics, machine learning, and any other field where outcomes are measured in categories rather than continuous numbers.

The test was developed by Karl Pearson in 1900 and remains one of the most cited statistical procedures in scientific literature. According to the NIST/SEMATECH Engineering Statistics Handbook, the chi-square test is the standard approach for categorical data inference when expected cell counts are adequate. At Statistics Fundamentals, chi-square sits within the broader framework of hypothesis testing, which establishes the null and alternative hypothesis structure all these tests share.

⚡ 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)
Two main types: Goodness-of-fit (1 variable) and test of independence (2 variables)
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)])

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)

Two Types of Chi-Square Tests

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

Feature	Goodness-of-Fit Test	Test of Independence
Number of variables	1 categorical variable	2 categorical variables
Data structure	Single frequency table	Contingency table (rows × columns)
Research question	Does this variable follow a specified distribution?	Are these two variables associated?
Null hypothesis (H₀)	Observed = expected distribution	The two variables are independent
Expected frequency	E = n × p (theoretical proportion)	E = (Row Total × Col Total) / N
Degrees of freedom	k − 1	(r − 1)(c − 1)
Classic example	Do the six faces of a die appear equally often?	Is gender associated with product preference?

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 the counts you would expect if a specific null hypothesis were true. A genetics researcher testing whether a cross follows Mendel's 9:3:3:1 ratio, or a quality control engineer testing whether defect types are uniformly distributed, both use this form.

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 — rows for one variable, columns for the other — and the test determines whether the pattern of cell frequencies is consistent with the two variables being unrelated. This is the more commonly encountered form in social science, medical, and marketing research.

Chi-Square Test Formula

All chi-square tests use the same core formula for the test statistic. The distinction between test types lies in how expected frequencies are calculated, not in the formula itself.

The Chi-Square Test Statistic: 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.

Expected Frequency Formulas

Expected Frequency — Goodness-of-Fit

E = n × p

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

Expected Frequency — Test of Independence

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

Calculate for every cell individually using the row and column marginals

Degrees of Freedom Formulas

Degrees of Freedom (df)

Goodness-of-fit: df = k − 1

Test of independence: 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).

Assumptions of the Chi-Square Test

Chi-square test results are only valid when these five conditions hold. Penn State's STAT 500 course — a widely cited statistics curriculum — 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 measured as categories — nominal (unordered labels like colors or countries) or ordinal (ordered categories like rating scales). Chi-square cannot be applied to continuous measurements directly.

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.

Random or representative sampling

Data must come from a random sample or a sampling design that is representative of the population being studied. Non-random convenience samples limit the generalizability of the result.

Mutually exclusive categories

Each observation must fall into one and only one category. Overlapping categories (where an observation could be counted twice) violate the independence of cells 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.

Chi-Square Distribution Table (Critical Values)

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

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 cumulative distribution function. Values match those in the NIST/SEMATECH Statistics Handbook Table. For the extended downloadable version, visit the Chi-Square Table reference page.

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 Perform a Chi-Square Test: 8-Step Method

📋

Featured Snippet — 8 Steps

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. Step 7: Look up the critical value. Step 8: Compare χ² to the critical value and state the conclusion.

Worked Example — Full 8-Step Walkthrough

Research question: Is there a statistically significant association between gender (Male/Female) and 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:

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, critical value = 3.841 (from the distribution table above)

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

🧮 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.

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.

Worked Examples Across Three Fields

Example 1 — Medical Research: Smoking and Lung Disease

This example follows the design used in classic epidemiology studies on smoking. The data below are representative of the proportions established in large-sample clinical research, consistent with findings documented by the CDC Tobacco Statistics and Data program.

Worked Example 2 — 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 = (2−1)(2−1) = 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, indicating a medium-to-large effect.

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

Worked Example 3 — 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 = 4 − 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). The data do not provide evidence against the Mendelian model.

How to Report Chi-Square Results (APA Format)

When writing up chi-square test results, follow the APA 7th edition format. This format is required by most journals and expected in graduate-level coursework. The Publication Manual of the American Psychological Association (7th ed.) specifies this structure for nonparametric test reporting.

APA 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."

✅

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 (φ) for 2×2 tables.

Effect Size: Cramér's V

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 more recent work by Lakens (2013) at Eindhoven University of Technology notes that these benchmarks 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)] )

Where 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 there is one degree of freedom.

Chi-Square Test in SPSS

Test of Independence in SPSS

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

Reading SPSS output:

In the Chi-Square Tests table, read the Pearson Chi-Square row
The Asymptotic Significance (2-sided) column gives your p-value
Check the footnote for "X cells have expected count less than 5" — if this appears, consider Fisher's exact test
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 test
result <- chisq.test(data_matrix)
print(result)

# Output: X-squared = 16.667, df = 1, p-value = 4.46e-05

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

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

⚠️

Yates' Continuity Correction

R applies Yates' continuity correction by default for 2×2 tables — this slightly reduces χ². To match hand calculations or SPSS output, disable it: chisq.test(data_matrix, correct = FALSE)

Goodness-of-Fit in R

# Observed phenotype counts
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 test
chisq.test(observed, p = expected_probs)

# Output: X-squared = 0.267, df = 3, p-value = 0.966
# Interpretation: Data consistent with 9:3:3:1 ratio
        

Chi-Square vs. Other Statistical Tests

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	Checkbox in SPSS; `fisher.test()` in R

Chi-Square vs. t-Test

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. 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 on this site.

Where Chi-Square Tests Are Used

🏥

Medical Research

Testing whether treatment outcomes (recovered/not recovered) differ by treatment group. Used routinely in randomized controlled trials with binary endpoints.

🧬

Genetics & Biology

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

📊

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.

🤖

Machine Learning

Feature selection: identifying which categorical features are statistically associated with the target variable before model training.

🏛️

Social Sciences

Analyzing voting patterns, criminal justice outcomes, educational attainment, and social mobility data at the population level.

🌿

Ecology

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

🏭

Quality Control

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

When Chi-Square Fails: Alternatives

Problem	Recommended Alternative	Why
Expected frequencies < 5 in any cell (small sample)	Fisher's exact test	Exact computation; no large-sample approximation needed
Paired or matched categorical data	McNemar's test	Accounts for the 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

Frequently Asked Questions

Chi-Square Quick Reference Cheat Sheet

The table below consolidates every key term, formula, and decision rule covered in this guide. It is structured for maximum LLM parsability and serves as a one-page study reference.

Term / Entity	Formula / Value	When to Use	Plain Interpretation
Chi-square statistic (χ²)	`χ² = Σ[(O − E)² / E]`	All chi-square tests	Total deviation of observed from expected counts
Expected frequency — goodness-of-fit	`E = n × p`	Single-variable test	Count predicted by theoretical proportion p
Expected frequency — independence	`E = (Row × Col) / N`	Two-variable test	Count predicted if variables were unrelated
Degrees of freedom (goodness-of-fit)	`df = k − 1`	k = number of categories	Number of free cells after constraints applied
Degrees of freedom (independence)	`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	Effect size: 0.10=small, 0.30=medium, 0.50=large
p-value	P(χ² ≥ observed \| H₀ true)	Always report	p < 0.05 → reject H₀ (at conventional α)
Null hypothesis (independence)	H₀: Variables are independent	Test of independence	No association between the two categorical variables
Null hypothesis (goodness-of-fit)	H₀: Observed = expected distribution	Goodness-of-fit	Data match the theoretical distribution
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