Introduction
The chi-square test of independence is used to determine whether there is a statistically significant association between two categorical variables. Unlike the t-test or ANOVA, which compare means of continuous variables, the chi-square test works with count data organized in a contingency table.
This test appears frequently in social science research, market research, medical studies, and quality control. If you have ever asked a question like "Is there a relationship between gender and product preference?" or "Does treatment type affect recovery status?", the chi-square test of independence is the analysis you need.
This guide walks you through every step, from setting up your contingency table to interpreting the results, using a concrete worked example.
When to Use the Chi-Square Test of Independence
Use this test when:
- You have two categorical variables (nominal or ordinal).
- You want to test whether the variables are independent or associated.
- Your data consist of counts (frequencies), not means or continuous measurements.
- Each observation falls into exactly one cell of the contingency table.
If you are comparing a categorical variable against a continuous outcome, consider a t-test or ANOVA instead.
Step 1: State Your Hypotheses
Example scenario: A university researcher wants to know whether there is a relationship between study method (Online, Library, Study Group) and exam outcome (Pass, Fail) among 200 students.
- H0 (Null hypothesis): Study method and exam outcome are independent. There is no association between the two variables.
- H1 (Alternative hypothesis): Study method and exam outcome are not independent. There is an association between the two variables.
Step 2: Organize Data in a Contingency Table
Collect the frequency counts and arrange them in a contingency table (also called a cross-tabulation):
| | Pass | Fail | Row Total | |---|------|------|-----------| | Online | 42 | 28 | 70 | | Library | 51 | 14 | 65 | | Study Group | 48 | 17 | 65 | | Column Total | 141 | 59 | 200 |
Each cell contains the count of students who fall into that combination of study method and exam outcome.
Step 3: Calculate Expected Frequencies
Under the null hypothesis (independence), the expected frequency for each cell is:
Expected = (Row Total * Column Total) / Grand Total
| | Pass (Expected) | Fail (Expected) | |---|-----------------|-----------------| | Online | (70 * 141) / 200 = 49.35 | (70 * 59) / 200 = 20.65 | | Library | (65 * 141) / 200 = 45.83 | (65 * 59) / 200 = 19.18 | | Study Group | (65 * 141) / 200 = 45.83 | (65 * 59) / 200 = 19.18 |
Step 4: Check Assumptions
Before proceeding, verify these assumptions:
1. Independence of Observations
Each student is counted only once and appears in exactly one cell. No student used multiple study methods simultaneously in this study.
2. Expected Frequency Rule
The chi-square approximation is valid when no more than 20% of expected frequencies are below 5, and no expected frequency is below 1. Looking at our expected values:
- Smallest expected frequency: 19.18
All expected frequencies are well above 5, so the assumption is met. If you had very small expected counts, consider using Fisher's Exact Test instead.
3. Sample Size
A general guideline is that the total sample size should be at least 5 times the number of cells. With 6 cells and N = 200, this is easily satisfied.
Step 5: Calculate the Chi-Square Statistic
The chi-square statistic is computed as:
chi-square = sum of (Observed - Expected)^2 / Expected
Calculate for each cell:
| Cell | Observed (O) | Expected (E) | (O - E)^2 / E | |------|-------------|-------------|----------------| | Online, Pass | 42 | 49.35 | (42 - 49.35)^2 / 49.35 = 1.095 | | Online, Fail | 28 | 20.65 | (28 - 20.65)^2 / 20.65 = 2.615 | | Library, Pass | 51 | 45.83 | (51 - 45.83)^2 / 45.83 = 0.583 | | Library, Fail | 14 | 19.18 | (14 - 19.18)^2 / 19.18 = 1.398 | | Study Group, Pass | 48 | 45.83 | (48 - 45.83)^2 / 45.83 = 0.103 | | Study Group, Fail | 17 | 19.18 | (17 - 19.18)^2 / 19.18 = 0.248 |
chi-square = 1.095 + 2.615 + 0.583 + 1.398 + 0.103 + 0.248 = 6.042
Step 6: Determine Degrees of Freedom
df = (number of rows - 1) * (number of columns - 1)
df = (3 - 1) * (2 - 1) = 2 * 1 = 2
Step 7: Find the P Value
With chi-square = 6.042 and df = 2, the p value is approximately .049.
Since p = .049 < .05, we reject the null hypothesis at the .05 significance level. There is a statistically significant association between study method and exam outcome.
Step 8: Calculate Effect Size
For chi-square tests, Cramer's V is the standard effect size measure:
V = sqrt(chi-square / (N * min(r-1, c-1)))
V = sqrt(6.042 / (200 * 1)) = sqrt(0.0302) = 0.174
Cramer's V interpretation guidelines:
| df* | Small | Medium | Large | |-----|-------|--------|-------| | 1 | 0.10 | 0.30 | 0.50 | | 2 | 0.07 | 0.21 | 0.35 | | 3 | 0.06 | 0.17 | 0.29 |
df here refers to min(r-1, c-1). With df* = 1, our V = 0.174 indicates a small-to-medium effect.
Step 9: Examine the Pattern
To understand the nature of the association, compare observed and expected values or look at column percentages:
| | Pass Rate (%) | |---|--------------| | Online | 42/70 = 60.0% | | Library | 51/65 = 78.5% | | Study Group | 48/65 = 73.8% |
The Library group has the highest pass rate (78.5%), followed by Study Group (73.8%) and Online (60.0%). The significant chi-square result appears to be driven primarily by the lower pass rate in the Online group.
Step 10: Report the Results
A chi-square test of independence was performed to examine the association between study method and exam outcome. The analysis revealed a statistically significant association, chi-square(2, N = 200) = 6.04, p = .049, V = .17. Students who studied in the library had the highest pass rate (78.5%), compared to study groups (73.8%) and online study (60.0%).
2x2 Tables: A Special Case
For 2x2 contingency tables, the procedure is the same, but you have additional options:
- Yates' continuity correction: A small correction that makes the chi-square test more conservative for 2x2 tables.
- Fisher's Exact Test: Preferred when any expected frequency is below 5.
- Odds Ratio: A useful effect size measure specific to 2x2 tables.
Common Mistakes to Avoid
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Using percentages instead of counts: The chi-square test requires raw frequency counts, not percentages or proportions. Convert back to counts before analysis.
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Violating the independence assumption: Each observation must be independent. If the same person could appear in multiple cells (e.g., repeated measures), the chi-square test is inappropriate.
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Ignoring small expected frequencies: When expected counts are below 5 in more than 20% of cells, the chi-square approximation becomes unreliable. Use Fisher's Exact Test in such cases.
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Confusing statistical and practical significance: A significant chi-square with a tiny Cramer's V may not be practically meaningful. Always report effect size.
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Over-interpreting the direction: The chi-square test tells you an association exists but does not indicate causation. Study method and exam outcome may both be influenced by unmeasured variables.
Frequently Asked Questions
What is the difference between chi-square goodness-of-fit and test of independence?
The goodness-of-fit test compares observed frequencies of a single variable to expected frequencies (e.g., testing whether a die is fair). The test of independence examines the association between two categorical variables. This guide focuses on the test of independence.
Can I use the chi-square test with ordinal variables?
Yes, but the chi-square test treats all categories as nominal and ignores the ordering. If you want to account for the ordinal nature, consider a Mantel-Haenszel test for trend or Spearman's correlation on ranked data.
What sample size do I need?
A common rule of thumb is a minimum of 5 expected observations per cell. For a 3x2 table, this means you need at least 30 total observations. Larger samples provide more statistical power.
What if I have more than two variables?
For three or more categorical variables, you can use log-linear analysis or run separate chi-square tests. However, running multiple tests increases the risk of Type I error, so apply a correction such as Bonferroni.
Can I use chi-square with continuous data?
Not directly. You would first need to categorize the continuous variable into groups (e.g., low/medium/high). However, categorizing continuous variables loses information, so consider using correlation or regression analyses designed for continuous data.
Run Your Chi-Square Test with StatMate
StatMate's chi-square calculator makes this process effortless. Enter your contingency table data, and StatMate will compute the chi-square statistic, degrees of freedom, p value, Cramer's V, expected frequencies, and standardized residuals automatically. If your table is 2x2, it also provides Yates' correction and Fisher's Exact Test results.