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Academic Support AP Statistics Unit 3: Inference for Categorical Data: Proportions

Chi-Square Assumptions, Expected Counts, and Contributions

Learn chi square test assumptions with current AP Statistics scope, proper formulas, worked examples, and original Easy, Tough, and Toughest questions.

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Statistical Procedure

Chi-Square Assumptions, Expected Counts, and Contributions

A decision-and-workflow guide for chi-square conditions and cell contributions, covering method selection, conditions, mathematics, calculator evidence, and contextual reporting.

Course status: Revised 2026-27 course
Updated: July 18, 2026
Practice: Easy, Tough and Toughest

Method at a Glance: Chi-Square Test Assumptions

Chi-square approximation checks use expected rather than observed counts, and each cell contribution reveals where observed and null-model counts differ most.

Reader taskrandomness, independence, expected counts, component diagnostics, and robustness
Planned modules7
Mathematics1 expressions
Worked checks45

Boundary: Check expected rather than observed counts.

Procedure Workflow

  1. Identify the data structure and parameter before selecting chi square test assumptions; the name of a calculator menu is not method evidence.
  2. State the hypotheses or estimation target for chi square test assumptions using population notation and the order defined by the question.
  3. Verify the design, independence, and approximation conditions that specifically justify chi square test assumptions rather than reciting every condition learned in the course.
  4. Compute the statistic, standard error, interval, or p-value for chi square test assumptions with defined symbols, guard digits, and an independent arithmetic check.
  5. Interpret chi square test assumptions in the population and units named by the problem, then limit causation and generalization to what the collection design supports.

Procedure Formulas and Notation

Chi-square cell contribution

cell contribution=(OE)2E

Chi-square cell contribution in Chi-Square Test Assumptions: This expression belongs specifically to chi-square conditions and cell contributions; define every symbol and apply the scope rule for randomness, independence, expected counts, component diagnostics, and robustness before calculation.

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Step 1

Random condition

Decision

For Random condition in chi square test assumptions, For a constructed 2 by 3 table from a battery-life laboratory trial, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for random condition.

Random condition result in chi square test assumptions: The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals.

Eij=(row total)(column total)120,χ2=5.043,df=(21)(31)=2.

Interpretation and validity

Random condition interpretation for chi square test assumptions: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.

Condition evidence for chi square test assumptions and Random condition: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

Procedure error: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.
Step 2

Independence and 10% condition

Decision

For Independence and 10% condition in chi square test assumptions, For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for independence and 10% condition.

Independence and 10% condition result in chi square test assumptions: The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals.

Eij=(row total)(column total)122,χ2=5.782,df=(21)(31)=2.

Interpretation and validity

Independence and 10% condition interpretation for chi square test assumptions: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.

Condition evidence for chi square test assumptions and Independence and 10% condition: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

Procedure error: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.
Step 3

Expected-count condition

Decision

For Expected-count condition in chi square test assumptions, For a constructed 2 by 3 table from a battery-life laboratory trial, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for expected-count condition.

Expected-count condition result in chi square test assumptions: The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals.

Eij=(row total)(column total)124,χ2=6.573,df=(21)(31)=2.

Interpretation and validity

Expected-count condition interpretation for chi square test assumptions: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.

Condition evidence for chi square test assumptions and Expected-count condition: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

Procedure error: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.
Step 4

Calculating expected counts

Decision

For Calculating expected counts in chi square test assumptions, For a constructed 2 by 3 table from an online-course completion sample, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for calculating expected counts.

Calculating expected counts result in chi square test assumptions: The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals.

Eij=(row total)(column total)126,χ2=7.412,df=(21)(31)=2.

Interpretation and validity

Calculating expected counts interpretation for chi square test assumptions: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.

Condition evidence for chi square test assumptions and Calculating expected counts: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

Procedure error: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.
Step 5

Cell contributions

Decision

For Cell contributions in chi square test assumptions, For a constructed 2 by 3 table from a classroom memory study, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for cell contributions.

Cell contributions result in chi square test assumptions: The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals.

Eij=(row total)(column total)124,χ2=6.844,df=(21)(31)=2.

Interpretation and validity

Cell contributions interpretation for chi square test assumptions: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.

Condition evidence for chi square test assumptions and Cell contributions: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

Procedure error: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.
Step 6

Standardized residuals

Decision

For Standardized residuals in chi square test assumptions, For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for standardized residuals.

Standardized residuals result in chi square test assumptions: The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals.

Eij=(row total)(column total)121,χ2=5.325,df=(21)(31)=2.

Interpretation and validity

Standardized residuals interpretation for chi square test assumptions: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.

Condition evidence for chi square test assumptions and Standardized residuals: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

Procedure error: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.
Step 7

What to do when assumptions fail

Decision

For What to do when assumptions fail in chi square test assumptions, For a constructed 2 by 3 table from a battery-life laboratory trial, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

What to do when assumptions fail result in chi square test assumptions: The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals.

Eij=(row total)(column total)123,χ2=6.094,df=(21)(31)=2.

Interpretation and validity

What to do when assumptions fail interpretation for chi square test assumptions: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.

Condition evidence for chi square test assumptions and What to do when assumptions fail: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

Procedure error: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Procedure Practice and Full Solutions

Every question in Chi-Square Assumptions, Expected Counts, and Contributions is newly written from the revised framework and the logic visible in public College Board materials. Constructed numerical settings are identified as instructional scenarios and are never represented as measurements from a real population. No released or secure question wording is reproduced.

Easy Practice

Easy 1: What to do when assumptions fail

Question P73-Easy-1. For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

Worked solution and validity check

Worked solution P73-Easy-1. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. Eij=(row total)(column total)120,χ2=5.043,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 2: Random condition

Question P73-Easy-2. For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for random condition.

Worked solution and validity check

Worked solution P73-Easy-2. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.782,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 3: Independence and 10% condition

Question P73-Easy-3. For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for independence and 10% condition.

Worked solution and validity check

Worked solution P73-Easy-3. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.573,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 4: Expected-count condition

Question P73-Easy-4. For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for expected-count condition.

Worked solution and validity check

Worked solution P73-Easy-4. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. Eij=(row total)(column total)126,χ2=7.412,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 5: Calculating expected counts

Question P73-Easy-5. For a constructed 2 by 3 table from a greenhouse germination experiment, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for calculating expected counts.

Worked solution and validity check

Worked solution P73-Easy-5. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.844,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 6: Cell contributions

Question P73-Easy-6. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for cell contributions.

Worked solution and validity check

Worked solution P73-Easy-6. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. Eij=(row total)(column total)121,χ2=5.325,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 7: Standardized residuals

Question P73-Easy-7. For a constructed 2 by 3 table from a classroom memory study, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for standardized residuals.

Worked solution and validity check

Worked solution P73-Easy-7. The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals. Eij=(row total)(column total)123,χ2=6.094,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 8: What to do when assumptions fail

Question P73-Easy-8. For a constructed 2 by 3 table from a school library checkout study, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

Worked solution and validity check

Worked solution P73-Easy-8. The statistic is 6.912, df=2, and p=0.0315; expected counts are computed from row and column totals. Eij=(row total)(column total)125,χ2=6.912,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 9: Random condition

Question P73-Easy-9. For a constructed 2 by 3 table from a greenhouse germination experiment, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for random condition.

Worked solution and validity check

Worked solution P73-Easy-9. The statistic is 6.385, df=2, and p=0.0411; expected counts are computed from row and column totals. Eij=(row total)(column total)123,χ2=6.385,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 10: Independence and 10% condition

Question P73-Easy-10. For a constructed 2 by 3 table from a campus dining survey, observed rows are [34, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for independence and 10% condition.

Worked solution and validity check

Worked solution P73-Easy-10. The statistic is 7.189, df=2, and p=0.0275; expected counts are computed from row and column totals. Eij=(row total)(column total)125,χ2=7.189,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 11: Expected-count condition

Question P73-Easy-11. For a constructed 2 by 3 table from a city bus arrival investigation, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for expected-count condition.

Worked solution and validity check

Worked solution P73-Easy-11. The statistic is 5.622, df=2, and p=0.0602; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.622,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 12: Calculating expected counts

Question P73-Easy-12. For a constructed 2 by 3 table from a campus dining survey, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for calculating expected counts.

Worked solution and validity check

Worked solution P73-Easy-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.418,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 13: Cell contributions

Question P73-Easy-13. For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for cell contributions.

Worked solution and validity check

Worked solution P73-Easy-13. The statistic is 5.931, df=2, and p=0.0515; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.931,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 14: Standardized residuals

Question P73-Easy-14. For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [33, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for standardized residuals.

Worked solution and validity check

Worked solution P73-Easy-14. The statistic is 6.715, df=2, and p=0.0348; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.715,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Easy 15: What to do when assumptions fail

Question P73-Easy-15. For a constructed 2 by 3 table from a city bus arrival investigation, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

Worked solution and validity check

Worked solution P73-Easy-15. The statistic is 7.546, df=2, and p=0.0230; expected counts are computed from row and column totals. Eij=(row total)(column total)126,χ2=7.546,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough Practice

Tough 1: Cell contributions

Question P73-Tough-1. For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for cell contributions.

Worked solution and validity check

Worked solution P73-Tough-1. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. Eij=(row total)(column total)120,χ2=5.043,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 2: Standardized residuals

Question P73-Tough-2. For a constructed 2 by 3 table from a school library checkout study, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for standardized residuals.

Worked solution and validity check

Worked solution P73-Tough-2. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.782,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 3: What to do when assumptions fail

Question P73-Tough-3. For a constructed 2 by 3 table from a commuter route study, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

Worked solution and validity check

Worked solution P73-Tough-3. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.573,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 4: Random condition

Question P73-Tough-4. For a constructed 2 by 3 table from a battery-life laboratory trial, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for random condition.

Worked solution and validity check

Worked solution P73-Tough-4. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. Eij=(row total)(column total)126,χ2=7.412,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 5: Independence and 10% condition

Question P73-Tough-5. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for independence and 10% condition.

Worked solution and validity check

Worked solution P73-Tough-5. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.844,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 6: Expected-count condition

Question P73-Tough-6. For a constructed 2 by 3 table from a classroom memory study, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for expected-count condition.

Worked solution and validity check

Worked solution P73-Tough-6. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. Eij=(row total)(column total)121,χ2=5.325,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 7: Calculating expected counts

Question P73-Tough-7. For a constructed 2 by 3 table from a commuter route study, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for calculating expected counts.

Worked solution and validity check

Worked solution P73-Tough-7. The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals. Eij=(row total)(column total)123,χ2=6.094,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 8: Cell contributions

Question P73-Tough-8. For a constructed 2 by 3 table from a city bus arrival investigation, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for cell contributions.

Worked solution and validity check

Worked solution P73-Tough-8. The statistic is 6.912, df=2, and p=0.0315; expected counts are computed from row and column totals. Eij=(row total)(column total)125,χ2=6.912,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 9: Standardized residuals

Question P73-Tough-9. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for standardized residuals.

Worked solution and validity check

Worked solution P73-Tough-9. The statistic is 6.385, df=2, and p=0.0411; expected counts are computed from row and column totals. Eij=(row total)(column total)123,χ2=6.385,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 10: What to do when assumptions fail

Question P73-Tough-10. For a constructed 2 by 3 table from a school library checkout study, observed rows are [34, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

Worked solution and validity check

Worked solution P73-Tough-10. The statistic is 7.189, df=2, and p=0.0275; expected counts are computed from row and column totals. Eij=(row total)(column total)125,χ2=7.189,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 11: Random condition

Question P73-Tough-11. For a constructed 2 by 3 table from a classroom memory study, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for random condition.

Worked solution and validity check

Worked solution P73-Tough-11. The statistic is 5.622, df=2, and p=0.0602; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.622,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 12: Independence and 10% condition

Question P73-Tough-12. For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for independence and 10% condition.

Worked solution and validity check

Worked solution P73-Tough-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.418,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 13: Expected-count condition

Question P73-Tough-13. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for expected-count condition.

Worked solution and validity check

Worked solution P73-Tough-13. The statistic is 5.931, df=2, and p=0.0515; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.931,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 14: Calculating expected counts

Question P73-Tough-14. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [33, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for calculating expected counts.

Worked solution and validity check

Worked solution P73-Tough-14. The statistic is 6.715, df=2, and p=0.0348; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.715,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Tough 15: Cell contributions

Question P73-Tough-15. For a constructed 2 by 3 table from a battery-life laboratory trial, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for cell contributions.

Worked solution and validity check

Worked solution P73-Tough-15. The statistic is 7.546, df=2, and p=0.0230; expected counts are computed from row and column totals. Eij=(row total)(column total)126,χ2=7.546,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest Practice

Toughest 1: Expected-count condition

Question P73-Toughest-1. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for expected-count condition.

Worked solution and validity check

Worked solution P73-Toughest-1. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. Eij=(row total)(column total)120,χ2=5.043,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 2: Calculating expected counts

Question P73-Toughest-2. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for calculating expected counts.

Worked solution and validity check

Worked solution P73-Toughest-2. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.782,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 3: Cell contributions

Question P73-Toughest-3. For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for cell contributions.

Worked solution and validity check

Worked solution P73-Toughest-3. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.573,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 4: Standardized residuals

Question P73-Toughest-4. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for standardized residuals.

Worked solution and validity check

Worked solution P73-Toughest-4. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. Eij=(row total)(column total)126,χ2=7.412,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 5: What to do when assumptions fail

Question P73-Toughest-5. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

Worked solution and validity check

Worked solution P73-Toughest-5. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.844,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 6: Random condition

Question P73-Toughest-6. For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for random condition.

Worked solution and validity check

Worked solution P73-Toughest-6. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. Eij=(row total)(column total)121,χ2=5.325,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 7: Independence and 10% condition

Question P73-Toughest-7. For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for independence and 10% condition.

Worked solution and validity check

Worked solution P73-Toughest-7. The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals. Eij=(row total)(column total)123,χ2=6.094,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 8: Expected-count condition

Question P73-Toughest-8. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for expected-count condition.

Worked solution and validity check

Worked solution P73-Toughest-8. The statistic is 6.912, df=2, and p=0.0315; expected counts are computed from row and column totals. Eij=(row total)(column total)125,χ2=6.912,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 9: Calculating expected counts

Question P73-Toughest-9. For a constructed 2 by 3 table from a classroom memory study, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for calculating expected counts.

Worked solution and validity check

Worked solution P73-Toughest-9. The statistic is 6.385, df=2, and p=0.0411; expected counts are computed from row and column totals. Eij=(row total)(column total)123,χ2=6.385,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 10: Cell contributions

Question P73-Toughest-10. For a constructed 2 by 3 table from a city bus arrival investigation, observed rows are [34, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for cell contributions.

Worked solution and validity check

Worked solution P73-Toughest-10. The statistic is 7.189, df=2, and p=0.0275; expected counts are computed from row and column totals. Eij=(row total)(column total)125,χ2=7.189,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 11: Standardized residuals

Question P73-Toughest-11. For a constructed 2 by 3 table from a battery-life laboratory trial, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for standardized residuals.

Worked solution and validity check

Worked solution P73-Toughest-11. The statistic is 5.622, df=2, and p=0.0602; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.622,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 12: What to do when assumptions fail

Question P73-Toughest-12. For a constructed 2 by 3 table from a greenhouse germination experiment, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for what to do when assumptions fail.

Worked solution and validity check

Worked solution P73-Toughest-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.418,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 13: Random condition

Question P73-Toughest-13. For a constructed 2 by 3 table from a school library checkout study, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for random condition.

Worked solution and validity check

Worked solution P73-Toughest-13. The statistic is 5.931, df=2, and p=0.0515; expected counts are computed from row and column totals. Eij=(row total)(column total)122,χ2=5.931,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 14: Independence and 10% condition

Question P73-Toughest-14. For a constructed 2 by 3 table from a commuter route study, observed rows are [33, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for independence and 10% condition.

Worked solution and validity check

Worked solution P73-Toughest-14. The statistic is 6.715, df=2, and p=0.0348; expected counts are computed from row and column totals. Eij=(row total)(column total)124,χ2=6.715,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Toughest 15: Expected-count condition

Question P73-Toughest-15. For a constructed 2 by 3 table from a greenhouse germination experiment, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for expected-count condition.

Worked solution and validity check

Worked solution P73-Toughest-15. The statistic is 7.546, df=2, and p=0.0230; expected counts are computed from row and column totals. Eij=(row total)(column total)126,χ2=7.546,df=(21)(31)=2. Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

AP Response and Publication Checklist

Audit pointRequired evidence for chi square test assumptions
ScopeCheck expected rather than observed counts.
Method or sourceChi-square approximation checks use expected rather than observed counts, and each cell contribution reveals where observed and null-model counts differ most.
CalculationEij=(row total)(column total)120,χ2=5.043,df=(21)(31)=2.
InterpretationA small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects.
ValidityUse random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
CorrectionCheck expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Frequently Asked Questions

How does random condition work in chi square test assumptions?

Answer for chi square test assumptions and Random condition. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. The required validity evidence is: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

How does independence and 10% condition work in chi square test assumptions?

Answer for chi square test assumptions and Independence and 10% condition. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. The required validity evidence is: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

How does expected-count condition work in chi square test assumptions?

Answer for chi square test assumptions and Expected-count condition. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. The required validity evidence is: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

How does calculating expected counts work in chi square test assumptions?

Answer for chi square test assumptions and Calculating expected counts. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. The required validity evidence is: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

How does cell contributions work in chi square test assumptions?

Answer for chi square test assumptions and Cell contributions. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. The required validity evidence is: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

How does standardized residuals work in chi square test assumptions?

Answer for chi square test assumptions and Standardized residuals. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. The required validity evidence is: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.

How does assumptions in chi square test connect to Chi-Square Test Assumptions?

assumptions in chi square test within chi square test assumptions. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. For Random condition, the controlling scope is: Check expected rather than observed counts.

How does conditions for chi square test connect to Chi-Square Test Assumptions?

conditions for chi square test within chi square test assumptions. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. For Independence and 10% condition, the controlling scope is: Check expected rather than observed counts.

How does assumptions for a chi square test connect to Chi-Square Test Assumptions?

assumptions for a chi square test within chi square test assumptions. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. For Expected-count condition, the controlling scope is: Check expected rather than observed counts.

How does conditions for a chi square test connect to Chi-Square Test Assumptions?

conditions for a chi square test within chi square test assumptions. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. For Calculating expected counts, the controlling scope is: Check expected rather than observed counts.

How does chi square test conditions connect to Chi-Square Test Assumptions?

chi square test conditions within chi square test assumptions. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. For Cell contributions, the controlling scope is: Check expected rather than observed counts.

How does assumptions for chi square test connect to Chi-Square Test Assumptions?

assumptions for chi square test within chi square test assumptions. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. For Standardized residuals, the controlling scope is: Check expected rather than observed counts.

How does assumptions of chi square test connect to Chi-Square Test Assumptions?

assumptions of chi square test within chi square test assumptions. The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals. A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. For What to do when assumptions fail, the controlling scope is: Check expected rather than observed counts.

Sources

Administrative and curricular statements in Chi-Square Assumptions, Expected Counts, and Contributions were checked on July 18, 2026. The linked College Board pages control any later policy change; all instructional datasets in original questions are explicitly constructed rather than attributed to a real study.

Chi-Square Test Assumptions Conclusion

Chi-square approximation checks use expected rather than observed counts, and each cell contribution reveals where observed and null-model counts differ most. Mastery of chi square test assumptions therefore requires the exact evidence, mathematics, interpretation, and scope developed in this guide, while preserving this boundary: Check expected rather than observed counts.

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