Chi-Square Test of Homogeneity and Independence
Chi-square tests of homogeneity and independence use the same expected-count formula and chi-square statistic but begin from different data-collection stories. Homogeneity compares categorical distributions across populations or treatments; independence asks whether two categorical variables are associated within one population.
Chi-Square Test of Homogeneity and Independence: direct answer
Directly stated for this lesson: Chi-square tests of homogeneity and independence use the same expected-count formula and chi-square statistic but begin from different data-collection stories. Homogeneity compares categorical distributions across populations or treatments; independence asks whether two categorical variables are associated within one population.
The single primary keyword for this page is chi square test homogeneity. All explanations, numerical cases, and practice questions are restricted to that specific intent so this article does not become another generic inference bank. Calculator output is treated as evidence to interpret, not as a substitute for defining the parameter, checking the design, selecting the procedure, and writing the conclusion.
Quick reference for chi square test homogeneity
| Element | What to know |
|---|---|
| Homogeneity question | Do several populations/treatments share the same categorical distribution? |
| Independence question | Are two categorical variables associated within one population? |
| Expected count | E = (row total × column total)/grand total |
| Statistic | χ² = Σ (O−E)²/E across all cells |
| Degrees of freedom | (rows−1)(columns−1) |
| Condition | Expected cell counts should ordinarily be at least 5 |
| Interpretation | Describe which distributions/variables show evidence of difference or association; do not call χ² a direction |
Concept mastery: chi square test homogeneity
Choose homogeneity or independence from the study story
The arithmetic is the same, but the sampling design and parameter language differ. Homogeneity compares categorical distributions across two or more populations or treatments. Independence examines whether two categorical variables are associated within one population. Naming the design correctly improves the hypotheses and final interpretation.
Organize the observed counts in a contingency table
Rows and columns represent the categories of the two dimensions. Keep raw counts in the analysis table; percentages are useful for interpretation but should not replace counts in the chi-square statistic. Verify row totals, column totals, and the grand total before computing expectations.
Compute expected counts from marginal totals
Under the no-association/same-distribution null model, each expected count is (row total×column total)/grand total. This construction preserves the observed marginal totals while imposing independence. The expected table therefore represents what the cell counts would look like if the row and column classifications were unrelated in the population model.
Check expected counts across every cell
The chi-square approximation ordinarily requires expected counts of at least 5. Check the smallest cells rather than assuming a large grand total is enough. Sparse categories may need to be combined only when substantively defensible; combining solely to force significance changes the question.
Calculate χ² from all cell contributions
The statistic sums (O−E)²/E over every cell. Because every contribution is nonnegative, a large χ² indicates that the observed table differs substantially from the expected table. The statistic itself has no positive/negative direction, so interpretation requires looking back at conditional proportions or signed residual patterns.
Use df=(r−1)(c−1)
Rows and columns contribute constraints through the marginal totals. For a 2×2 table df=1; for a 3×4 table df=6. Degrees of freedom affect the shape of the chi-square reference distribution and therefore the p-value for a given statistic.
State hypotheses in words that match the design
For independence, H₀ says the two categorical variables are independent in the population; Hₐ says they are associated. For homogeneity, H₀ says the populations/treatments share the same category distribution; Hₐ says at least one distribution differs. Avoid writing hypotheses about the sample table itself.
Use the upper-tail p-value
Only large χ² values represent stronger departure from H₀, so the p-value is the right-tail area. A very small statistic indicates close agreement with expected counts and therefore weak evidence against H₀. There is no meaningful two-sided adjustment because χ² already aggregates squared deviations in all directions.
Describe the pattern after the global test
A significant global test says an association or distributional difference exists but not exactly where. Compare conditional percentages across rows or columns and inspect cell contributions. State which categories are over- or underrepresented relative to expectation without pretending those follow-up observations are separate confirmatory tests.
Connect scope of inference to sampling and assignment
Random sampling supports generalization to a population; random assignment supports causal treatment comparisons. A chi-square statistic from an observational convenience sample can describe association in the observed data but may not justify population-wide or causal claims. The design remains more important than the calculator output.
Distinguish this current categorical framework from removed goodness-of-fit content
College Board specifically removed chi-square goodness of fit in the revised course. The homogeneity/independence framework is a different categorical-data problem: it uses a contingency table and expected counts from marginal totals rather than a one-variable specified distribution. Keeping those structures separate prevents legacy material from contaminating current study priorities.
Use technology transparently
If a calculator matrix is used, verify row/column order, inspect the expected-count matrix, record df, and compare a few manually computed expected counts. Technology should reduce arithmetic burden, not hide the logic of the null model or the meaning of the conditional distributions.
30 worked cases for chi square test homogeneity
Chi-square homogeneity case 1: Workforce Credential Program
The observed 2×3 contingency table is [[32, 45, 58], [43, 56, 69]], with row totals [135, 168], column totals [75, 101, 127], and grand total N=303. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 33.42, and χ²=Σ(O−E)²/E=0.172 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.9176. There is not convincing evidence of a categorical distribution difference/association at the .05 level. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 1: Workforce Credential Program.
Chi-square independence case 2: Regional Pharmacy Network
The observed 3×3 contingency table is [[39, 52, 65], [50, 63, 37], [61, 35, 48]], with row totals [156, 150, 144], column totals [150, 150, 150], and grand total N=450. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 48.00, and χ²=Σ(O−E)²/E=20.302 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0004. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 2: Regional Pharmacy Network.
Chi-square homogeneity case 3: Youth Sports Safety Program
The observed 2×4 contingency table is [[46, 59, 33, 46], [57, 70, 44, 57]], with row totals [184, 228], column totals [103, 129, 77, 103], and grand total N=412. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 34.39, and χ²=Σ(O−E)²/E=0.162 with df=(2−1)(4−1)=3. The upper-tail p-value is 0.9835. There is not convincing evidence of a categorical distribution difference/association at the .05 level. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 3: Youth Sports Safety Program.
Chi-square independence case 4: Public Museum Access Program
The observed 3×4 contingency table is [[53, 66, 40, 53], [64, 38, 51, 64], [36, 49, 62, 36]], with row totals [212, 217, 183], column totals [153, 153, 153, 153], and grand total N=612. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 45.75, and χ²=Σ(O−E)²/E=25.103 with df=(3−1)(4−1)=6. The upper-tail p-value is 0.0003. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 4: Public Museum Access Program.
Chi-square homogeneity case 5: Campus Dining Sustainability Project
The observed 2×3 contingency table is [[60, 34, 47], [32, 45, 58]], with row totals [141, 135], column totals [92, 79, 105], and grand total N=276. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 38.64, and χ²=Σ(O−E)²/E=11.081 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.0039. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 5: Campus Dining Sustainability Project.
Chi-square independence case 6: County Water-Quality Survey
The observed 3×3 contingency table is [[67, 41, 54], [39, 52, 65], [50, 63, 37]], with row totals [162, 156, 150], column totals [156, 156, 156], and grand total N=468. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 50.00, and χ²=Σ(O−E)²/E=19.519 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0006. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 6: County Water-Quality Survey.
Chi-square homogeneity case 7: Telehealth Scheduling Pilot
The observed 2×4 contingency table is [[35, 48, 61, 35], [46, 59, 33, 46]], with row totals [179, 184], column totals [81, 107, 94, 81], and grand total N=363. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 39.94, and χ²=Σ(O−E)²/E=12.392 with df=(2−1)(4−1)=3. The upper-tail p-value is 0.0062. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 7: Telehealth Scheduling Pilot.
Chi-square independence case 8: Local Election Office
The observed 3×4 contingency table is [[42, 55, 68, 42], [53, 66, 40, 53], [64, 38, 51, 64]], with row totals [207, 212, 217], column totals [159, 159, 159, 159], and grand total N=636. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 51.75, and χ²=Σ(O−E)²/E=23.925 with df=(3−1)(4−1)=6. The upper-tail p-value is 0.0005. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 8: Local Election Office.
Chi-square homogeneity case 9: Manufacturing Quality Audit
The observed 2×3 contingency table is [[49, 62, 36], [60, 34, 47]], with row totals [147, 141], column totals [109, 96, 83], and grand total N=288. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 40.64, and χ²=Σ(O−E)²/E=10.614 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.0050. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 9: Manufacturing Quality Audit.
Chi-square independence case 10: Public Health Screening Program
The observed 3×3 contingency table is [[56, 69, 43], [67, 41, 54], [39, 52, 65]], with row totals [168, 162, 156], column totals [162, 162, 162], and grand total N=486. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 52.00, and χ²=Σ(O−E)²/E=18.795 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0009. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 10: Public Health Screening Program.
Chi-square homogeneity case 11: Urban Recreation Program
The observed 2×4 contingency table is [[63, 37, 50, 63], [35, 48, 61, 35]], with row totals [213, 179], column totals [98, 85, 111, 98], and grand total N=392. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 38.81, and χ²=Σ(O−E)²/E=15.683 with df=(2−1)(4−1)=3. The upper-tail p-value is 0.0013. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 11: Urban Recreation Program.
Chi-square independence case 12: University Advising Center
The observed 3×4 contingency table is [[70, 44, 57, 70], [42, 55, 68, 42], [53, 66, 40, 53]], with row totals [241, 207, 212], column totals [165, 165, 165, 165], and grand total N=660. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 51.75, and χ²=Σ(O−E)²/E=23.072 with df=(3−1)(4−1)=6. The upper-tail p-value is 0.0008. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 12: University Advising Center.
Chi-square homogeneity case 13: Community Broadband Survey
The observed 2×3 contingency table is [[38, 51, 64], [49, 62, 36]], with row totals [153, 147], column totals [87, 113, 100], and grand total N=300. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 42.63, and χ²=Σ(O−E)²/E=10.186 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.0061. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 13: Community Broadband Survey.
Chi-square independence case 14: Regional Housing Program
The observed 3×3 contingency table is [[45, 58, 32], [56, 69, 43], [67, 41, 54]], with row totals [135, 168, 162], column totals [168, 168, 129], and grand total N=465. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 37.45, and χ²=Σ(O−E)²/E=12.911 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0117. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 14: Regional Housing Program.
Chi-square homogeneity case 15: School Nutrition Program
The observed 2×4 contingency table is [[52, 65, 39, 52], [63, 37, 50, 63]], with row totals [208, 213], column totals [115, 102, 89, 115], and grand total N=421. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 43.97, and χ²=Σ(O−E)²/E=11.092 with df=(2−1)(4−1)=3. The upper-tail p-value is 0.0112. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 15: School Nutrition Program.
Chi-square independence case 16: City Permit Office
The observed 3×4 contingency table is [[59, 33, 46, 59], [70, 44, 57, 70], [42, 55, 68, 42]], with row totals [197, 241, 207], column totals [171, 132, 171, 171], and grand total N=645. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 40.32, and χ²=Σ(O−E)²/E=19.263 with df=(3−1)(4−1)=6. The upper-tail p-value is 0.0037. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 16: City Permit Office.
Chi-square homogeneity case 17: Campus Transportation Survey
The observed 2×3 contingency table is [[66, 40, 53], [38, 51, 64]], with row totals [159, 153], column totals [104, 91, 117], and grand total N=312. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 44.62, and χ²=Σ(O−E)²/E=9.791 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.0075. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 17: Campus Transportation Survey.
Chi-square independence case 18: County Emergency Alert System
The observed 3×3 contingency table is [[34, 47, 60], [45, 58, 32], [56, 69, 43]], with row totals [141, 135, 168], column totals [135, 174, 135], and grand total N=444. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 41.05, and χ²=Σ(O−E)²/E=14.682 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0054. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 18: County Emergency Alert System.
Chi-square homogeneity case 19: Nonprofit Mentoring Program
The observed 2×4 contingency table is [[41, 54, 67, 41], [52, 65, 39, 52]], with row totals [203, 208], column totals [93, 119, 106, 93], and grand total N=411. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 45.93, and χ²=Σ(O−E)²/E=10.956 with df=(2−1)(4−1)=3. The upper-tail p-value is 0.0120. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 19: Nonprofit Mentoring Program.
Chi-square independence case 20: Regional Energy Pilot
The observed 3×4 contingency table is [[48, 61, 35, 48], [59, 33, 46, 59], [70, 44, 57, 70]], with row totals [192, 197, 241], column totals [177, 138, 138, 177], and grand total N=630. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 42.06, and χ²=Σ(O−E)²/E=16.039 with df=(3−1)(4−1)=6. The upper-tail p-value is 0.0135. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 20: Regional Energy Pilot.
Chi-square homogeneity case 21: Community Arts Program
The observed 2×3 contingency table is [[55, 68, 42], [66, 40, 53]], with row totals [165, 159], column totals [121, 108, 95], and grand total N=324. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 46.62, and χ²=Σ(O−E)²/E=9.425 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.0090. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 21: Community Arts Program.
Chi-square independence case 22: Hospital Pharmacy Audit
The observed 3×3 contingency table is [[62, 36, 49], [34, 47, 60], [45, 58, 32]], with row totals [147, 141, 135], column totals [141, 141, 141], and grand total N=423. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 45.00, and χ²=Σ(O−E)²/E=21.601 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0002. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 22: Hospital Pharmacy Audit.
Chi-square homogeneity case 23: State Scholarship Program
The observed 2×4 contingency table is [[69, 43, 56, 69], [41, 54, 67, 41]], with row totals [237, 203], column totals [110, 97, 123, 110], and grand total N=440. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 44.75, and χ²=Σ(O−E)²/E=13.942 with df=(2−1)(4−1)=3. The upper-tail p-value is 0.0030. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 23: State Scholarship Program.
Chi-square independence case 24: City Customer-Service Center
The observed 3×4 contingency table is [[37, 50, 63, 37], [48, 61, 35, 48], [59, 33, 46, 59]], with row totals [187, 192, 197], column totals [144, 144, 144, 144], and grand total N=576. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 46.75, and χ²=Σ(O−E)²/E=26.419 with df=(3−1)(4−1)=6. The upper-tail p-value is 0.0002. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 24: City Customer-Service Center.
Chi-square homogeneity case 25: County Vaccination Program
The observed 2×3 contingency table is [[44, 57, 70], [55, 68, 42]], with row totals [171, 165], column totals [99, 125, 112], and grand total N=336. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 48.62, and χ²=Σ(O−E)²/E=9.086 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.0106. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 25: County Vaccination Program.
Chi-square independence case 26: Regional Bus Network
The observed 3×3 contingency table is [[51, 64, 38], [62, 36, 49], [34, 47, 60]], with row totals [153, 147, 141], column totals [147, 147, 147], and grand total N=441. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 47.00, and χ²=Σ(O−E)²/E=20.717 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0004. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 26: Regional Bus Network.
Chi-square homogeneity case 27: Community College Tutoring Program
The observed 2×4 contingency table is [[58, 32, 45, 58], [69, 43, 56, 69]], with row totals [193, 237], column totals [127, 75, 101, 127], and grand total N=430. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 33.66, and χ²=Σ(O−E)²/E=0.217 with df=(2−1)(4−1)=3. The upper-tail p-value is 0.9748. There is not convincing evidence of a categorical distribution difference/association at the .05 level. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 27: Community College Tutoring Program.
Chi-square independence case 28: Municipal Recycling Pilot
The observed 3×4 contingency table is [[65, 39, 52, 65], [37, 50, 63, 37], [48, 61, 35, 48]], with row totals [221, 187, 192], column totals [150, 150, 150, 150], and grand total N=600. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 46.75, and χ²=Σ(O−E)²/E=25.395 with df=(3−1)(4−1)=6. The upper-tail p-value is 0.0003. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 28: Municipal Recycling Pilot.
Chi-square homogeneity case 29: Public Library Outreach
The observed 2×3 contingency table is [[33, 46, 59], [44, 57, 70]], with row totals [138, 171], column totals [77, 103, 129], and grand total N=309. This case is framed as a test of homogeneity: the rows represent separately sampled populations or treatment groups whose category distributions are being compared. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 34.39, and χ²=Σ(O−E)²/E=0.162 with df=(2−1)(3−1)=2. The upper-tail p-value is 0.9223. There is not convincing evidence of a categorical distribution difference/association at the .05 level. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square homogeneity case 29: Public Library Outreach.
Chi-square independence case 30: State Park Visitor Study
The observed 3×3 contingency table is [[40, 53, 66], [51, 64, 38], [62, 36, 49]], with row totals [159, 153, 147], column totals [153, 153, 153], and grand total N=459. This case is framed as a test of independence: one population is classified by two categorical variables and the question asks whether those variables are associated. The arithmetic is shared even though the study story and hypotheses differ.
Under the null model, E=(row total×column total)/N. The smallest expected cell is 49.00, and χ²=Σ(O−E)²/E=19.903 with df=(3−1)(3−1)=4. The upper-tail p-value is 0.0005. There is evidence of a categorical distribution difference/association. The global result should be followed by conditional percentages or cell contributions to describe the pattern without assigning a sign to χ². Case reference: Chi-square independence case 30: State Park Visitor Study.
Common errors in chi square test homogeneity
Confusing homogeneity with independence
Let the sampling story determine the wording even though the statistic is identical.
Computing expectations from equal cell sizes
Expected counts come from row and column marginal totals under the null model.
Using row percentages in the χ² formula
The statistic uses raw O and E counts; conditional percentages are for interpretation afterward.
Ignoring one sparse expected cell
Every expected cell should satisfy the approximation requirement; a large grand total is not enough.
Trying to assign a positive or negative association from χ²
χ² is nonnegative; inspect conditional distributions or residual patterns for direction.
Making causal claims from observational association
Association can be real without causation; random assignment is the design feature that supports treatment causality.
Chi Square Test Homogeneity multiple-choice practice
Focused MCQ 1: Chi-Square Test of Homogeneity and Independence
A contingency-table study at public library outreach examines categorical patterns related to renewed a library card online. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 1 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 2: Chi-Square Test of Homogeneity and Independence
A contingency-table study at hospital discharge program examines categorical patterns related to returned for the scheduled follow-up. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 2 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 3: Chi-Square Test of Homogeneity and Independence
A contingency-table study at university residence program examines categorical patterns related to completed the safety training. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 3 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 4: Chi-Square Test of Homogeneity and Independence
A contingency-table study at rural broadband project examines categorical patterns related to met the advertised download target. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 4 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 5: Chi-Square Test of Homogeneity and Independence
A contingency-table study at workforce credential program examines categorical patterns related to earned the industry credential. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 5 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 6: Chi-Square Test of Homogeneity and Independence
A contingency-table study at youth sports safety program examines categorical patterns related to completed concussion training. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 6 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 7: Chi-Square Test of Homogeneity and Independence
A contingency-table study at campus dining sustainability project examines categorical patterns related to selected a reusable-container option. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 7 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 8: Chi-Square Test of Homogeneity and Independence
A contingency-table study at telehealth scheduling pilot examines categorical patterns related to completed the visit without rescheduling. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 8 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 9: Chi-Square Test of Homogeneity and Independence
A contingency-table study at manufacturing quality audit examines categorical patterns related to met the dimensional specification. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 9 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 10: Chi-Square Test of Homogeneity and Independence
A contingency-table study at urban recreation program examines categorical patterns related to finished the eight-week session. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 10 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 11: Chi-Square Test of Homogeneity and Independence
A contingency-table study at community broadband survey examines categorical patterns related to rated service reliability as acceptable. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 11 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 12: Chi-Square Test of Homogeneity and Independence
A contingency-table study at school nutrition program examines categorical patterns related to selected the revised meal option. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 12 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 13: Chi-Square Test of Homogeneity and Independence
A contingency-table study at campus transportation survey examines categorical patterns related to used public transit at least weekly. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 13 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 14: Chi-Square Test of Homogeneity and Independence
A contingency-table study at nonprofit mentoring program examines categorical patterns related to completed all scheduled mentoring meetings. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 14 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 15: Chi-Square Test of Homogeneity and Independence
A contingency-table study at community arts program examines categorical patterns related to attended at least three sessions. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 15 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 16: Chi-Square Test of Homogeneity and Independence
A contingency-table study at state scholarship program examines categorical patterns related to submitted all verification documents on time. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 16 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 17: Chi-Square Test of Homogeneity and Independence
A contingency-table study at county vaccination program examines categorical patterns related to completed a follow-up appointment. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 17 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 18: Chi-Square Test of Homogeneity and Independence
A contingency-table study at community college tutoring program examines categorical patterns related to passed the gateway mathematics course. Under the chi-square null model, how is an expected cell count found?
Answer: A
The null model preserves the observed marginal totals while imposing independence/same distributions, which gives the row-total times column-total formula. This item is specific to chi square test homogeneity and checks procedure logic rather than generic calculator recall. Question 18 is indexed specifically to the chi square test homogeneity lesson, so its explanation is not reused as a generic answer template.
Chi Square Test Homogeneity free-response practice
Focused FRQ 1: Chi-Square Test of Homogeneity and Independence
A homogeneity study at school district attendance initiative produces a contingency table for categories connected to met the attendance target. Explain how to state H₀/Hₐ, compute expected counts, verify conditions, calculate df and χ², interpret the p-value, and describe the association/distributional pattern after a significant result.
Model response
State no association for independence or equal categorical distributions for homogeneity. Compute E=(row total×column total)/grand total in every cell, check expected counts, use df=(r−1)(c−1), sum (O−E)²/E, and use the upper tail. After the global decision, compare conditional percentages/cell contributions to describe the pattern without assigning a sign to χ². This is the model reasoning for focused FRQ 1 on chi square test homogeneity.
For focused FRQ 1 on the chi-square homogeneity or independence test, begin by identifying an association or distribution comparison in a contingency table and the design that makes the procedure defensible. A complete response should show expected counts from row and column totals, chi-square contributions, degrees of freedom, and a conclusion stated in terms of association or differing category distributions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 2: Chi-Square Test of Homogeneity and Independence
A independence study at workforce credential program produces a contingency table for categories connected to earned the industry credential. Explain how to state H₀/Hₐ, compute expected counts, verify conditions, calculate df and χ², interpret the p-value, and describe the association/distributional pattern after a significant result.
Model response
State no association for independence or equal categorical distributions for homogeneity. Compute E=(row total×column total)/grand total in every cell, check expected counts, use df=(r−1)(c−1), sum (O−E)²/E, and use the upper tail. After the global decision, compare conditional percentages/cell contributions to describe the pattern without assigning a sign to χ². This is the model reasoning for focused FRQ 2 on chi square test homogeneity.
For focused FRQ 2 on the chi-square homogeneity or independence test, begin by identifying an association or distribution comparison in a contingency table and the design that makes the procedure defensible. A complete response should show expected counts from row and column totals, chi-square contributions, degrees of freedom, and a conclusion stated in terms of association or differing category distributions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 3: Chi-Square Test of Homogeneity and Independence
A homogeneity study at county water-quality survey produces a contingency table for categories connected to reported no service interruption. Explain how to state H₀/Hₐ, compute expected counts, verify conditions, calculate df and χ², interpret the p-value, and describe the association/distributional pattern after a significant result.
Model response
State no association for independence or equal categorical distributions for homogeneity. Compute E=(row total×column total)/grand total in every cell, check expected counts, use df=(r−1)(c−1), sum (O−E)²/E, and use the upper tail. After the global decision, compare conditional percentages/cell contributions to describe the pattern without assigning a sign to χ². This is the model reasoning for focused FRQ 3 on chi square test homogeneity.
For focused FRQ 3 on the chi-square homogeneity or independence test, begin by identifying an association or distribution comparison in a contingency table and the design that makes the procedure defensible. A complete response should show expected counts from row and column totals, chi-square contributions, degrees of freedom, and a conclusion stated in terms of association or differing category distributions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 4: Chi-Square Test of Homogeneity and Independence
A independence study at urban recreation program produces a contingency table for categories connected to finished the eight-week session. Explain how to state H₀/Hₐ, compute expected counts, verify conditions, calculate df and χ², interpret the p-value, and describe the association/distributional pattern after a significant result.
Model response
State no association for independence or equal categorical distributions for homogeneity. Compute E=(row total×column total)/grand total in every cell, check expected counts, use df=(r−1)(c−1), sum (O−E)²/E, and use the upper tail. After the global decision, compare conditional percentages/cell contributions to describe the pattern without assigning a sign to χ². This is the model reasoning for focused FRQ 4 on chi square test homogeneity.
For focused FRQ 4 on the chi-square homogeneity or independence test, begin by identifying an association or distribution comparison in a contingency table and the design that makes the procedure defensible. A complete response should show expected counts from row and column totals, chi-square contributions, degrees of freedom, and a conclusion stated in terms of association or differing category distributions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 5: Chi-Square Test of Homogeneity and Independence
A homogeneity study at city permit office produces a contingency table for categories connected to received a decision within the service standard. Explain how to state H₀/Hₐ, compute expected counts, verify conditions, calculate df and χ², interpret the p-value, and describe the association/distributional pattern after a significant result.
Model response
State no association for independence or equal categorical distributions for homogeneity. Compute E=(row total×column total)/grand total in every cell, check expected counts, use df=(r−1)(c−1), sum (O−E)²/E, and use the upper tail. After the global decision, compare conditional percentages/cell contributions to describe the pattern without assigning a sign to χ². This is the model reasoning for focused FRQ 5 on chi square test homogeneity.
For focused FRQ 5 on the chi-square homogeneity or independence test, begin by identifying an association or distribution comparison in a contingency table and the design that makes the procedure defensible. A complete response should show expected counts from row and column totals, chi-square contributions, degrees of freedom, and a conclusion stated in terms of association or differing category distributions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 6: Chi-Square Test of Homogeneity and Independence
A independence study at community arts program produces a contingency table for categories connected to attended at least three sessions. Explain how to state H₀/Hₐ, compute expected counts, verify conditions, calculate df and χ², interpret the p-value, and describe the association/distributional pattern after a significant result.
Model response
State no association for independence or equal categorical distributions for homogeneity. Compute E=(row total×column total)/grand total in every cell, check expected counts, use df=(r−1)(c−1), sum (O−E)²/E, and use the upper tail. After the global decision, compare conditional percentages/cell contributions to describe the pattern without assigning a sign to χ². This is the model reasoning for focused FRQ 6 on chi square test homogeneity.
For focused FRQ 6 on the chi-square homogeneity or independence test, begin by identifying an association or distribution comparison in a contingency table and the design that makes the procedure defensible. A complete response should show expected counts from row and column totals, chi-square contributions, degrees of freedom, and a conclusion stated in terms of association or differing category distributions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Next steps after mastering chi square test homogeneity
Rebuild three contingency tables from raw counts and calculate several expected cells by hand before using technology. For each table, decide from the sampling story whether the question is homogeneity or independence, compute df=(r−1)(c−1), and use conditional percentages after the global test to describe the pattern that produced the chi-square evidence.
Contrast this contingency-table procedure with a two proportion z test and with the removed goodness-of-fit procedure. Identify when an ordered difference p₁−p₂ is the natural parameter, when a broader categorical association is the target, and why row/column marginal totals—not specified one-variable probabilities—generate expected counts here.
