Chi Square Test of Independence Model Overview

The Chi Square Test of Independence is a nonparametric test of association for two categorical variables measured on the same independent observational units. It compares observed cell counts in a contingency table with the counts expected when the variables are statistically independent.

What Question Does the Test Answer?

The worked analysis asks whether student internet access is associated with intention to pursue higher education. Both variables are binary: internet is coded no/yes and higher is coded no/yes. Every student contributes to exactly one of the four cells.

How the Model Works

Under independence, the expected count in a cell is determined entirely by its row total, column total and the grand total. The Chi Square Test of Independence adds the standardized squared discrepancies between observed and expected counts. A large statistic indicates that the observed joint distribution is difficult to reconcile with independence.

Worked Result at a Glance

The observed table contains 22 students with neither internet access nor higher-education intention, 129 without internet access who intended higher education, 47 with internet access who did not intend higher education, and 451 with both. Pearson’s test gives χ²(1) = 3.2115, p = .0731. Cramér’s V is .0703, indicating a very small association.

Why the Primary Finding Is Not “Almost Significant”

A p-value of .073 is above the prespecified .05 threshold. It can be described as limited or inconclusive evidence, but not as a statistically significant result. The effect estimate, confidence interval and design limitations remain more informative than relabelling the threshold.

Search-intent note: Throughout this guide, Chi Square Test of Independence refers to Pearson’s contingency-table procedure for one sample and two categorical variables. The same Chi Square Test of Independence formula is used in the calculator, Python, R, SPSS and Excel sections, while the design wording remains distinct from homogeneity and goodness-of-fit tests.

The Chi Square Test of Independence remains an association test even when one variable is a plausible exposure and the other is a plausible outcome.

Quick Answer: Chi Square Test of Independence Result

The Chi Square Test of Independence quick answer should be read with the effect size and row percentages, not as a p-value alone.

The primary Pearson Chi Square Test of Independence compares internet access with higher-education intention in one sample of 649 students.

Table of Contents

1. What is a Chi Square Test of Independence?
2. Research question, hypotheses and data design
3. When to use the test
4. Formula and manual calculation
5. Variables and data dictionary
6. Assumptions and conditions
7. Complete worked results
8. Residuals and cell contributions
9. Independence versus related tests
10. Calculator workflow
11. Seven Python charts
12. Six R charts in paired rows
13. Software workflows
14. Expandable code
15. Advanced interpretation
16. APA reporting
17. Common mistakes
18. Practice problems
19. AP Statistics notes
20. Downloads
21. Related guides
22. Frequently asked questions
23. Conclusion

What Is a Chi Square Test of Independence?

The Chi Square Test of Independence definition begins with the one-sample design because design determines the correct hypothesis language.

A Chi Square Test of Independence determines whether the distribution of one categorical variable changes across the categories of another categorical variable when both are recorded from one sample. The method works with frequencies, not means, medians or individual numerical measurements.

Independence in Plain Language

Two variables are independent when knowing a person’s category on one variable does not change the probability distribution of the other. In this example, independence would mean that the proportion intending higher education is the same among students with and without internet access.

Association Is the Alternative

The alternative hypothesis says that the variables are associated. Association means the conditional proportions differ somewhere in the table. It does not automatically establish a direct causal mechanism, and it does not say which variable causes the other.

What the Statistic Measures

The test statistic is a weighted distance between the observed table and the table predicted by independence. Squaring prevents positive and negative deviations from cancelling. Dividing by the expected count expresses each discrepancy relative to the size of its cell.

What the Test Does Not Tell You

• It does not prove causation.
• It does not identify the strongest cell without residual analysis.
• It does not quantify practical importance without an effect size.
• It does not adjust for age, school, socioeconomic status or other covariates.
• It does not make repeated or clustered observations independent.

Common Table Sizes

The procedure is not limited to a 2 × 2 table. It can analyse a 2 × 3, 4 × 5 or other r × c contingency table, provided the observations are independent and expected counts are adequate. The degrees of freedom increase as (r − 1)(c − 1).

Research Question, Hypotheses and Data Design

The Chi Square Test of Independence research question must name both categorical variables and the population represented by the sample.

The Chi Square Test of Independence alternative is nondirectional unless a separate planned contrast is specified.

Research Question

Is internet access associated with intention to pursue higher education among the students represented in the analysed dataset?

Hypotheses

Sampling Design Statement

The same student sample supplies both variables. The study does not contain one independently sampled population of students without internet and a second independently sampled population with internet. That difference is why the analysis is described as a Chi Square Test of Independence rather than a homogeneity test.

Direction and Causality

The Pearson chi-square statistic is nondirectional. It can detect that the row profiles differ but does not itself label the direction. Direction comes from the observed percentages, residual signs and effect measures. Because the data are observational, even a significant result would not prove that internet access causes higher-education intention.

When to Use a Chi Square Test of Independence

The Chi Square Test of Independence is appropriate only when the variables, observations and frequency structure meet the conditions below.

Test-Selection Flow

Examples That Fit

• Smoking status by disease status in a cross-sectional sample.
• Preferred learning mode by year level in one student survey.
• Product category by payment method in one transaction dataset.
• Political preference by age group in one probability sample.

Examples That Do Not Fit

• Before-versus-after binary responses from the same participants: use McNemar’s test.
• Repeated ratings with more than two categories: consider a marginal homogeneity or symmetry procedure.
• Binary outcome with several predictors: use logistic regression.
• Two independent proportions with sparse cells: consider Fisher, Barnard or Boschloo according to design.

Chi Square Test of Independence Formula and Manual Calculation

The Chi Square Test of Independence calculation can be audited cell by cell from observed counts through the final p-value.

The manual Chi Square Test of Independence calculation begins with the observed table, derives expected counts from the margins, computes one contribution per cell and adds those contributions.

Step 1: Observed Counts

Step 2: Expected-Count Formula

For the no-internet/no-higher cell:

Step 3: Expected Counts

Step 4: Cell Contributions

Step 5: Degrees of Freedom

Step 6: P-Value and Decision

For χ² = 3.211522 with 1 degree of freedom, p = .073121. Since .073121 > .05, the analysis fails to reject the null hypothesis of independence.

Why Every Deviation Has the Same Raw Magnitude

In a 2 × 2 table with fixed margins, the four observed-minus-expected deviations must appear as +d, −d, −d and +d. Their chi-square contributions differ because each is divided by a different expected count. The smallest expected cell therefore contributes the most.

Variables and Data Dictionary

The Chi Square Test of Independence variables must be coded as mutually exclusive categorical levels before the table is formed.

Observed Joint Distribution

Conditional Percentages

The row percentages answer the substantive comparison. Among students without internet access, 14.57% did not intend higher education and 85.43% did. Among students with internet access, 9.44% did not intend higher education and 90.56% did.

Column Percentages

Among the 69 students who did not intend higher education, 31.88% lacked internet access and 68.12% had internet access. Among the 580 who intended higher education, 22.24% lacked internet access and 77.76% had internet access. Column percentages answer a different conditional question and should not be substituted for row percentages without explanation.

Chi Square Test of Independence Assumptions and Conditions

The Chi Square Test of Independence assumptions include independent observational units as well as adequate expected frequencies.

Assumption 1: Count Data

The input must be frequencies. Percentages may be reported, but the test itself must be calculated from integer counts or equivalent weighted case frequencies.

Assumption 2: Mutually Exclusive Categories

Every student must fall into exactly one category for internet access and exactly one category for higher-education intention. Ambiguous or overlapping categories invalidate the table construction.

Assumption 3: Independent Observations

The independence assumption concerns observational units, not the relationship being tested. Students should be sampled or observed independently. Siblings, repeated student records or students nested in a small number of schools may create dependence that the ordinary chi-square reference distribution ignores.

Assumption 4: Expected Counts

A common rule is that no expected cell should be below 1 and no more than 20% should be below 5. More conservative courses require every expected count to be at least 5. Here the minimum is 16.0539, so the expected-count condition is satisfied under either rule.

Assumption 5: Appropriate Sampling and Generalization

The test can be computed from any complete table, but population inference requires a defensible sampling design. A convenience sample can describe the analysed students without automatically representing all students.

Assumption 6: Predefined Categories

Categories and cut points should be selected for substantive reasons rather than repeatedly changed until a preferred p-value appears. Data-driven recoding inflates false-positive risk.

Complete Chi Square Test of Independence Results

The Chi Square Test of Independence results combine the global statistic, conditional percentages, sensitivity checks and effect size.

The Chi Square Test of Independence sensitivity analyses are secondary checks rather than opportunities to select a preferred p-value.

Primary and Sensitivity Tests

Supplementary 2 × 2 Effect Estimates

Statistical Interpretation

The Pearson p-value is .0731. At α = .05, the result is not statistically significant. The data do not provide sufficient evidence to reject independence between internet access and higher-education intention.

Practical Interpretation

The observed intention rate was 90.56% for students with internet access and 85.43% for students without internet access. The 5.13-point difference may be descriptively interesting, but Cramér’s V = .070 indicates that the overall association is very small. The approximate risk-difference interval includes zero.

Consistency Across Methods

Pearson, Yates, likelihood-ratio and Fisher procedures all remain above .05. The conclusion is therefore not dependent on a single borderline correction choice. This consistency is more important than choosing the smallest p-value.

Cell Contributions, Residuals and Follow-Up Interpretation

The Chi Square Test of Independence diagnostics explain which cells lie above or below the independence model.

Pearson Residuals

A positive Pearson residual means more observations than expected under independence. A negative residual means fewer. None of the absolute Pearson residuals reaches 2, so no cell displays a conventionally large local discrepancy.

Adjusted Standardized Residuals

Adjusted residuals account for row and column proportions and are often interpreted approximately as standard normal scores. Their absolute magnitude of 1.79 remains below 1.96, consistent with the non-significant global result.

Cell Contributions

The no-internet/no-higher cell contributes most because its observed count of 22 exceeds the expected count of 16.05 and because the expected count is relatively small. This does not mean that the cell is independently significant; contributions are descriptive components of the global statistic.

Post-Hoc Interpretation

For larger tables, analysts may inspect adjusted residuals with multiplicity correction. Here the 2 × 2 table has only one degree of freedom, and all four deviations express the same underlying association. Separate post-hoc cell tests are unnecessary.

Expected-Count Diagnostics

• Minimum expected count = 16.0539.
• Cells expected below 5 = 0.
• Proportion below 5 = 0%.
• No zero expected counts.
• Pearson approximation is numerically appropriate.

Chi Square Test of Independence Versus Homogeneity, Goodness of Fit and Other Tests

The Chi Square Test of Independence should be distinguished from procedures that answer different design-based questions.

Independence Versus Homogeneity

The formulas, expected counts and chi-square statistic can be identical. The difference lies in how the data were obtained and how the research question is framed. This worked dataset records both internet access and higher-education intention from one student sample, so independence language is correct.

Independence Versus Goodness of Fit

A goodness-of-fit test studies one categorical variable and compares it with externally specified or theoretically expected proportions. The present table contains two variables, and the expected counts are estimated from both sets of margins.

Independence Versus a Two-Proportion Z Test

For a 2 × 2 table, the uncorrected Pearson chi-square statistic equals the square of the pooled two-proportion z statistic. The tests give the same two-sided p-value. The z test highlights the difference between two proportions; the chi-square formulation generalizes more naturally to larger tables.

Independence Versus Fisher’s Exact Test

Fisher’s exact test conditions on the margins. Here all expected counts exceed 16, so Pearson’s approximation is suitable. Fisher’s two-sided p-value of .096 is reported as a sensitivity result rather than replacing the prespecified primary test.

Independence Versus Logistic Regression

Logistic regression is preferable when covariate adjustment, prediction, interactions or clustered extensions are required. A simple binary logistic model with higher as the outcome and internet as the predictor would estimate the same unadjusted odds ratio, but it offers a framework for adding confounders.

Chi Square Test of Independence Calculator: Step-by-Step Workflow

The Chi Square Test of Independence calculator workflow below provides exact checkpoints for all four cells.

The Chi Square Test of Independence calculator checks below also help detect transposed or mistyped tables.

A reliable Chi Square Test of Independence calculator requires the observed contingency table and should return expected counts, the Pearson statistic, degrees of freedom, p-value and preferably an effect size.

Calculator Inputs

Step 1: Preserve the Table Orientation

Rows should represent internet access and columns should represent higher-education intention. Swapping rows or columns does not change χ² or p, but it changes row percentages, residual signs and the direction of ratios.

Step 2: Confirm Margins

Row totals must be 151 and 498; column totals must be 69 and 580; N must be 649. An incorrect total means the table was transcribed incorrectly.

Step 3: Select Pearson Without Correction as Primary

For the main result, use Pearson χ² without Yates correction. The expected-count condition is satisfied. Report the correction only as a sensitivity analysis when required by a course or software default.

Step 4: Verify Expected Counts

The calculator should return 16.0539, 134.9461, 52.9461 and 445.0539. These values are determined by the margins, not supplied independently.

Step 5: Verify the Result

The target is χ² = 3.2115218, df = 1 and p = .0731214. Cramér’s V or phi should be .070345.

Step 6: Interpret More Than the P-Value

Compare row percentages, inspect residuals and report Cramér’s V. A calculator result that contains only “accept” or “reject” is insufficient for a complete analysis.

Python Chi Square Test of Independence Charts

The Chi Square Test of Independence Python charts translate the numerical output into seven connected visual explanations.

The Python workflow generates seven figures. Each chart is interpreted with exact values so the visual is not separated from the statistical result.

Chart 1: Observed CountsPython

Pattern

The largest cell contains students with internet access who intended higher education. The smallest cell contains students without internet access who did not intend higher education.

Key Values

No internet: 22 no and 129 yes. Internet: 47 no and 451 yes. Row totals are 151 and 498.

Interpretation

Raw counts show the joint distribution but are strongly affected by unequal row totals. The internet-access group is more than three times larger.

Why It Matters

The observed table is the starting point for all expected counts, percentages and effect estimates.

Next step: Convert the counts to within-row percentages before comparing the conditional distributions.

Chart 2: Expected Counts Under IndependencePython

Pattern

Expected counts retain the same row and column totals as the observed table. The no-internet/no-higher cell is expected to contain about 16.05 students.

Key Values

Expected counts: 16.0539, 134.9461, 52.9461 and 445.0539. Minimum expected = 16.0539.

Interpretation

The observed no-internet/no-higher count is 5.95 above expectation; the internet/no-higher count is 5.95 below expectation.

Why It Matters

Expected counts define the comparison model. Their adequacy also determines whether the Pearson approximation is appropriate.

Next step: Check that the expected values sum to the same row, column and grand totals as the observed table.

Chart 3: Pearson ResidualsPython

Pattern

The strongest positive residual is in the no-internet/no-higher cell. The strongest negative residual is in the internet/no-higher cell.

Key Values

Residuals: +1.4840, −0.5119, −0.8172 and +0.2819. None has |residual| ≥ 2.

Interpretation

The residual signs indicate that students without internet appear somewhat more often than expected in the no-higher category, but the deviations are not individually large.

Why It Matters

The global p-value does not show which cells create the statistic. Residuals provide that local explanation.

Next step: Use adjusted residuals when formal cellwise inference is required in a larger table.

Chart 4: Cell Contributions to Pearson Chi-SquarePython

Pattern

The no-internet/no-higher cell dominates the statistic, followed by the internet/no-higher cell.

Key Values

Contributions: 2.2023, 0.2620, 0.6678 and 0.0794; total = 3.2115. The first cell contributes 68.57%.

Interpretation

A large contribution means that a cell has a relatively large observed-to-expected discrepancy, not that the cell is a separate hypothesis test.

Why It Matters

Contribution analysis prevents vague statements that the whole table differs uniformly.

Next step: Report both contribution size and residual sign so magnitude and direction remain connected.

Chart 5: Within-Internet Row PercentagesPython

Pattern

The higher-education intention segment is larger among students with internet access.

Key Values

Higher intention: 85.43% without internet and 90.56% with internet. Difference = 5.13 percentage points.

Interpretation

The descriptive difference favours the internet-access group, but the inferential result is p = .073 and the effect is very small.

Why It Matters

Percentages answer the substantive question more directly than raw counts when row totals differ.

Next step: Report the percentages together with counts and uncertainty rather than claiming causation.

Chart 6: Model DiagnosticsPython

Pattern

The expected-count condition is comfortably satisfied, while the effect-size bar remains close to zero.

Key Values

N = 649; df = 1; minimum expected = 16.0539; cells below 5 = 0; Cramér’s V = 0.0703.

Interpretation

The chi-square approximation is suitable. The small effect size shows that a non-significant p-value is not hiding a large observed association.

Why It Matters

Diagnostics distinguish a valid null result from an inconclusive result caused by sparse cells.

Next step: Document independence and sampling assumptions in addition to numerical expected-count checks.

R Chi Square Test of Independence Chart Pairs

The Chi Square Test of Independence R charts reproduce the same verified counts and statistics in three paired rows.

The R section presents six figures in three paired rows. Each chart has a dedicated explanation box, matching the permanent post layout.

R chart pair 1

Observed Counts

The R count heatmap reproduces 22, 129, 47 and 451. It confirms the exact contingency table used by the test.

Check: Verify that the row and column labels match the intended variable coding.

Expected Counts

The expected table is 16.0539, 134.9461, 52.9461 and 445.0539. Every expected frequency exceeds five.

Check: Confirm that the expected margins equal the observed margins.

R chart pair 2

Pearson Residuals

The largest absolute Pearson residual is +1.4840 in the no-internet/no-higher cell. None reaches ±2.

Check: Use the residual sign to describe whether a cell is above or below expectation.

Cell Contributions

The no-internet/no-higher cell contributes 2.2023 of the total 3.2115, or 68.57%.

Check: Do not interpret a contribution bar as an independent p-value.

R chart pair 3

χ² statistic3.2115

Degrees of freedom1

P-value.0731

Cramér’s V.0703

R summary: minimum expected count = 16.0539; expected cells below five = 0.

Row Profiles

Higher-education intention is 85.43% without internet and 90.56% with internet. The descriptive gap is 5.13 points.

Check: Pair profiles with the p-value and Cramér’s V so descriptive and inferential conclusions stay aligned.

Result Summary

The R summary reports χ² = 3.2115, df = 1, p = .0731, Cramér’s V = .0703 and minimum expected = 16.0539.

Check: Check that the summary reproduces the table-level calculations before reporting.

Chi Square Test of Independence in Python, R, SPSS, Excel and Other Software

The Chi Square Test of Independence software workflow must preserve table orientation and correction settings across programs.

Expandable Chi Square Test of Independence Code

The Chi Square Test of Independence code examples use the same observed table so every platform can be cross-checked.

import numpy as np
from scipy.stats import chi2_contingency, fisher_exact

observed = np.array([[22, 129],
                     [47, 451]])

chi2, p_value, df, expected = chi2_contingency(
    observed, correction=False
)
residuals = (observed - expected) / np.sqrt(expected)
contributions = (observed - expected) ** 2 / expected
n = observed.sum()
cramers_v = np.sqrt(chi2 / (n * min(observed.shape[0]-1,
                                     observed.shape[1]-1)))

yates_chi2, yates_p, _, _ = chi2_contingency(
    observed, correction=True
)
fisher_odds, fisher_p = fisher_exact(observed)

print(chi2, p_value, df)
print(expected)
print(residuals)
print(contributions)
print(cramers_v, yates_chi2, yates_p, fisher_p)

tab <- matrix(c(22, 129,
                47, 451),
              nrow = 2, byrow = TRUE,
              dimnames = list(
                internet = c("no", "yes"),
                higher = c("no", "yes")
              ))

fit <- chisq.test(tab, correct = FALSE)
fit$observed
fit$expected
fit$residuals
fit$residuals^2
prop.table(tab, margin = 1)

n <- sum(tab)
cramers_v <- sqrt(as.numeric(fit$statistic) /
                  (n * min(nrow(tab)-1, ncol(tab)-1)))

chisq.test(tab, correct = TRUE)
fisher.test(tab)
cramers_v

CROSSTABS
 /TABLES=internet BY higher
 /FORMAT=AVALUE TABLES
 /STATISTICS=CHISQ PHI
 /CELLS=COUNT EXPECTED ROW COLUMN RESID SRESID
 /COUNT ROUND CELL.

Observed table: B5:C6
Row totals: D5 = SUM(B5:C5), D6 = SUM(B6:C6)
Column totals: B7 = SUM(B5:B6), C7 = SUM(C5:C6)
Grand total: D7 = SUM(D5:D6)

Expected B10 = $D5*B$7/$D$7
Contribution B14 = (B5-B10)^2/B10
Statistic = SUM(B14:C15)
Degrees of freedom = (ROWS(B5:C6)-1)*(COLUMNS(B5:C6)-1)
P-value = CHISQ.DIST.RT(statistic_cell, df_cell)
Cramer's V = SQRT(statistic_cell/(D7*MIN(ROWS(B5:C6)-1,COLUMNS(B5:C6)-1)))

Reproducibility Checklist

• Record the dataset version and valid N.
• Preserve category labels and ordering.
• State whether Yates correction was applied.
• Save observed, expected, residual and contribution tables.
• Report an effect size and conditional percentages.
• Export software output with the analysis date and version.

Advanced Chi Square Test of Independence Topics

The Chi Square Test of Independence advanced topics matter when tables are ordinal, clustered, weighted, sparse or exploratory.

The Chi Square Test of Independence does not adjust automatically for confounding variables or complex sampling.

Effect Size: Phi and Cramér’s V

For a 2 × 2 table, phi and Cramér’s V are identical. The value .0703 is conventionally small. Thresholds are only rough guides; practical interpretation should consider the decision context and the variables’ importance.

Power and Non-Significant Results

A non-significant result can occur because the variables are truly independent, because the association is small, or because the sample lacks power. Here N = 649 is substantial and the effect estimate is very small. The confidence intervals still allow modest positive or null effects, so the correct conclusion is uncertainty rather than proof of exact independence.

Structural Zeros Versus Sampling Zeros

A sampling zero is a possible cell that happened to receive no observations. A structural zero is impossible by definition. Standard expected-count calculations can be inappropriate when structural zeros exist because the sample space is constrained.

Large Tables and Residual Follow-Up

When an r × c test is significant, inspect adjusted residuals to identify important cells. Apply a multiplicity correction such as Holm or Bonferroni when making formal cellwise claims. Report the global test before post-hoc results.

Ordinal Categories

The ordinary Chi Square Test of Independence ignores category order. When both variables are ordinal, a linear-by-linear association, Mantel trend test, ordinal regression or rank-based method may provide a more focused analysis.

Survey Weights and Complex Samples

Ordinary Pearson chi-square assumes simple independent observations. Survey weights, stratification and clusters require design-adjusted procedures such as Rao–Scott chi-square. Treating weighted counts as ordinary independent cases can produce misleading standard errors.

Clustered Students

Students may be nested within schools or classes. If only a small number of clusters contribute data, ordinary p-values can be too optimistic. Multilevel logistic regression or generalized estimating equations can model the dependence directly.

Missing Data

The present analysis has 649 valid cases and no missing observations for the two variables. In other datasets, complete-case analysis can bias results when missingness depends on either variable or related covariates. Always report the number excluded.

Multiple Chi-Square Tests

Running many association tests increases the probability of at least one false positive. Use a planned analysis set, adjust for multiplicity where appropriate and distinguish exploratory from confirmatory findings.

Equivalence and Evidence for Independence

Failing to reject H₀ is not the same as establishing meaningful equivalence. To support a claim that any association is smaller than a practically important threshold, predefine that threshold and use confidence intervals or an equivalence framework.

Bayesian Alternatives

Bayesian contingency-table models can quantify evidence for independence versus association and provide posterior distributions for effect measures. They require explicit priors and should not be presented as automatic replacements for design-based reasoning.

APA Reporting for a Chi Square Test of Independence

The Chi Square Test of Independence APA report should state the non-significant decision without claiming proof of independence.

An APA-style report should identify the variables, sample size, statistic, degrees of freedom, p-value and effect size. Include conditional percentages when they help readers understand direction.

APA Table Note

Note. Expected counts ranged from 16.05 to 445.05; no expected cells were below five. Pearson residuals ranged from −0.82 to 1.48. Percentages are calculated within internet-access categories.

Reporting Exact P-Values

Report p = .073 rather than p > .05. Very small p-values may be written as p < .001, but never as p = .000.

Common Chi Square Test of Independence Mistakes

The Chi Square Test of Independence mistakes below are preventable through design checks, formula checks and transparent reporting.

How to Correct the Most Common Errors

Quality-Control Checklist

• The four observed counts total 649.
• The expected counts reproduce all margins.
• Contributions sum to 3.211522.
• Degrees of freedom equal 1.
• Pearson p equals .073121.
• Cramér’s V equals .070345.
• No expected count is below five.
• The conclusion says “fail to reject,” not “prove independence.”

Chi Square Test of Independence Practice Problems

The Chi Square Test of Independence practice problems cover recognition, calculation, conditions and interpretation.

Use these original questions to practise design recognition, manual calculation and interpretation. Answers appear directly below each question.

Practice 1: Identify the Correct Test

Practice 2: State Hypotheses

Practice 3: Calculate an Expected Count

Practice 4: Calculate Degrees of Freedom

Practice 5: Interpret a Residual

Practice 6: Interpret P = .073

Practice 7: Compare Row Percentages

Practice 8: Choose Fisher or Pearson

Practice 9: Explain Cramér’s V

Practice 10: Independence Versus Homogeneity

Practice 11: Causality

Practice 12: Report the Worked Result

Chi Square Test of Independence AP Statistics Study Notes

The Chi Square Test of Independence AP Statistics notes emphasize design recognition before calculator use.

AP Statistics Recognition

For AP Statistics, identify the sampling design before selecting the procedure. A Chi Square Test of Independence uses one sample or one population and two categorical variables. A homogeneity test uses two or more populations or treatments and one categorical response.

Conditions to State

• Random: data come from a random sample or randomized process when population inference is intended.
• Independent: observations are independent; when sampling without replacement, the sample is no more than 10% of the population.
• Large counts: every expected count is at least 5 under the common AP rule.

Expected-Count Work

Show at least one expected-count calculation and state that the remaining expected counts meet the condition. For this table, the smallest expected count is 16.05.

Conclusion in Context

Write the decision in the language of the research question: “Because p = .073 is greater than α = .05, we fail to reject the null hypothesis. There is not convincing evidence of an association between internet access and higher-education intention in the population represented by the sample.”

Calculator Output

On a graphing calculator, the expected matrix should be inspected. Do not copy only χ² and p. Degrees of freedom and conditions are part of the answer.

Chi Square Test of Independence Downloads

The Chi Square Test of Independence downloads contain the verified Python, R, SPSS and Excel outputs used in this guide.

Download the verified reports and worked workbook used for this Chi Square Test of Independence guide.

Related Salar Cafe Guides

The Chi Square Test of Independence is easier to place correctly after reviewing the related categorical-data guides below.

Continue with these internal Salar Cafe guides.

Frequently Asked Questions

The Chi Square Test of Independence FAQs answer the most common questions about hypotheses, calculation and reporting.

Conclusion

The Chi Square Test of Independence conclusion integrates the p-value, effect size, percentages and design limitations.

The Chi Square Test of Independence is most useful when the final explanation preserves both statistical uncertainty and practical context.

The Chi Square Test of Independence is the correct procedure when two categorical variables are recorded from one set of independent observations and the research question concerns association. In the worked student dataset, internet access and higher-education intention produced Pearson χ²(1) = 3.2115, p = .0731, with Cramér’s V = .0703.

The expected-count condition was satisfied, no cell had an absolute Pearson residual of 2 or greater, and Pearson, Yates, likelihood-ratio and Fisher procedures all led to the same non-significant decision. The observed higher-intention rate was 5.13 percentage points higher among students with internet access, but the standardized association was very small and the confidence intervals included the null.

A complete report therefore fails to reject independence, avoids causal wording, includes the conditional percentages and effect size, and documents the valid expected-count diagnostics. The downloadable Python, R, SPSS and Excel files reproduce the full calculation.