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

Chi-Square Test: Formula, Types, Conditions, and Examples

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

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Statistical Procedure

Chi-Square Test: Formula, Types, Conditions, and Examples

A decision-and-workflow guide for chi-square tests in the revised course, 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

The revised course retains chi-square tests of independence and homogeneity, both using expected counts and cell contributions, while goodness-of-fit is legacy enrichment.

Reader taskindependence and homogeneity, expected counts, contributions, degrees of freedom, and conclusions
Planned modules8
Mathematics1 expressions
Worked checks45

Boundary: Goodness-of-fit is legacy P70, not revised Unit 3 core.

Procedure Workflow

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

χ2=(OE)2E

Chi-square statistic in Chi-Square Test: Compute expected counts from the null model, then retain each cell contribution before summing so the result can be audited.

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Step 1

Chi-square statistic

Decision

For Chi-square statistic in chi square test, For a constructed 2 by 3 table from a water-filtration experiment, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for chi-square statistic.

Chi-square statistic result in chi square test: 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

Chi-square statistic interpretation for chi square test: 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 and Chi-square statistic: 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

Observed and expected counts

Decision

For Observed and expected counts in chi square test, For a constructed 2 by 3 table from a classroom memory study, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for observed and expected counts.

Observed and expected counts result in chi square test: 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

Observed and expected counts interpretation for chi square test: 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 and Observed and 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 3

Types of chi-square tests

Decision

For Types of chi-square tests in chi square test, For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for types of chi-square tests.

Types of chi-square tests result in chi square test: 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

Types of chi-square tests interpretation for chi square test: 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 and Types of chi-square tests: 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

Degrees of freedom

Decision

For Degrees of freedom in chi square test, For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for degrees of freedom.

Degrees of freedom result in chi square test: 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

Degrees of freedom interpretation for chi square test: 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 and Degrees of freedom: 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

P-value

Decision

For P-value in chi square test, For a constructed 2 by 3 table from a quality-control inspection, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for p-value.

P-value result in chi square test: 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

P-value interpretation for chi square test: 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 and P-value: 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

Conditions

Decision

For Conditions in chi square test, For a constructed 2 by 3 table from an online-course completion sample, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for conditions.

Conditions result in chi square test: 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

Conditions interpretation for chi square test: 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 and Conditions: 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

Worked overview

Decision

For Worked overview in chi square test, 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 worked overview.

Worked overview result in chi square test: 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

Worked overview interpretation for chi square test: 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 and Worked overview: 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 8

Procedure chooser

Decision

For Procedure chooser in chi square test, For a constructed 2 by 3 table from a water-filtration experiment, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for procedure chooser.

Procedure chooser result in chi square test: 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 and validity

Procedure chooser interpretation for chi square test: 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 and Procedure chooser: 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 Test: Formula, Types, Conditions, and Examples 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: Chi-square statistic

Question P68-Easy-1. For a constructed 2 by 3 table from a commuter route study, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for chi-square statistic.

Worked solution and validity check

Worked solution P68-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: Observed and expected counts

Question P68-Easy-2. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for observed and expected counts.

Worked solution and validity check

Worked solution P68-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: Types of chi-square tests

Question P68-Easy-3. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for types of chi-square tests.

Worked solution and validity check

Worked solution P68-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: Degrees of freedom

Question P68-Easy-4. For a constructed 2 by 3 table from a school library checkout study, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for degrees of freedom.

Worked solution and validity check

Worked solution P68-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: P-value

Question P68-Easy-5. For a constructed 2 by 3 table from a commuter route study, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for p-value.

Worked solution and validity check

Worked solution P68-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: Conditions

Question P68-Easy-6. For a constructed 2 by 3 table from a school library checkout study, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for conditions.

Worked solution and validity check

Worked solution P68-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: Worked overview

Question P68-Easy-7. For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for worked overview.

Worked solution and validity check

Worked solution P68-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: Procedure chooser

Question P68-Easy-8. For a constructed 2 by 3 table from a website response-time study, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for procedure chooser.

Worked solution and validity check

Worked solution P68-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: Chi-square statistic

Question P68-Easy-9. For a constructed 2 by 3 table from a water-filtration experiment, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for chi-square statistic.

Worked solution and validity check

Worked solution P68-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: Observed and expected counts

Question P68-Easy-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 observed and expected counts.

Worked solution and validity check

Worked solution P68-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: Types of chi-square tests

Question P68-Easy-11. For a constructed 2 by 3 table from a reading-speed investigation, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for types of chi-square tests.

Worked solution and validity check

Worked solution P68-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: Degrees of freedom

Question P68-Easy-12. For a constructed 2 by 3 table from a school library checkout study, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for degrees of freedom.

Worked solution and validity check

Worked solution P68-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: P-value

Question P68-Easy-13. For a constructed 2 by 3 table from a commuter route study, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for p-value.

Worked solution and validity check

Worked solution P68-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: Conditions

Question P68-Easy-14. For a constructed 2 by 3 table from a tutoring-program evaluation, observed rows are [33, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for conditions.

Worked solution and validity check

Worked solution P68-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: Worked overview

Question P68-Easy-15. For a constructed 2 by 3 table from a reading-speed investigation, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for worked overview.

Worked solution and validity check

Worked solution P68-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: Types of chi-square tests

Question P68-Tough-1. For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for types of chi-square tests.

Worked solution and validity check

Worked solution P68-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: Degrees of freedom

Question P68-Tough-2. For a constructed 2 by 3 table from a tutoring-program evaluation, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for degrees of freedom.

Worked solution and validity check

Worked solution P68-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: P-value

Question P68-Tough-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 p-value.

Worked solution and validity check

Worked solution P68-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: Conditions

Question P68-Tough-4. For a constructed 2 by 3 table from a campus dining survey, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for conditions.

Worked solution and validity check

Worked solution P68-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: Worked overview

Question P68-Tough-5. For a constructed 2 by 3 table from a tutoring-program evaluation, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for worked overview.

Worked solution and validity check

Worked solution P68-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: Procedure chooser

Question P68-Tough-6. For a constructed 2 by 3 table from a website response-time study, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for procedure chooser.

Worked solution and validity check

Worked solution P68-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: Chi-square statistic

Question P68-Tough-7. For a constructed 2 by 3 table from a water-filtration experiment, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for chi-square statistic.

Worked solution and validity check

Worked solution P68-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: Observed and expected counts

Question P68-Tough-8. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for observed and expected counts.

Worked solution and validity check

Worked solution P68-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: Types of chi-square tests

Question P68-Tough-9. For a constructed 2 by 3 table from a commuter route study, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for types of chi-square tests.

Worked solution and validity check

Worked solution P68-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: Degrees of freedom

Question P68-Tough-10. For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [34, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for degrees of freedom.

Worked solution and validity check

Worked solution P68-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: P-value

Question P68-Tough-11. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for p-value.

Worked solution and validity check

Worked solution P68-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: Conditions

Question P68-Tough-12. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for conditions.

Worked solution and validity check

Worked solution P68-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: Worked overview

Question P68-Tough-13. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for worked overview.

Worked solution and validity check

Worked solution P68-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: Procedure chooser

Question P68-Tough-14. For a constructed 2 by 3 table from a greenhouse germination experiment, observed rows are [33, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for procedure chooser.

Worked solution and validity check

Worked solution P68-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: Chi-square statistic

Question P68-Tough-15. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for chi-square statistic.

Worked solution and validity check

Worked solution P68-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: Worked overview

Question P68-Toughest-1. For a constructed 2 by 3 table from a manufacturing fill-volume check, observed rows are [30, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for worked overview.

Worked solution and validity check

Worked solution P68-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: Procedure chooser

Question P68-Toughest-2. For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for procedure chooser.

Worked solution and validity check

Worked solution P68-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: Chi-square statistic

Question P68-Toughest-3. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for chi-square statistic.

Worked solution and validity check

Worked solution P68-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: Observed and expected counts

Question P68-Toughest-4. 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 observed and expected counts.

Worked solution and validity check

Worked solution P68-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: Types of chi-square tests

Question P68-Toughest-5. For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for types of chi-square tests.

Worked solution and validity check

Worked solution P68-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: Degrees of freedom

Question P68-Toughest-6. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [30, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for degrees of freedom.

Worked solution and validity check

Worked solution P68-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: P-value

Question P68-Toughest-7. For a constructed 2 by 3 table from a city bus arrival investigation, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for p-value.

Worked solution and validity check

Worked solution P68-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: Conditions

Question P68-Toughest-8. For a constructed 2 by 3 table from a water-filtration experiment, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for conditions.

Worked solution and validity check

Worked solution P68-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: Worked overview

Question P68-Toughest-9. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for worked overview.

Worked solution and validity check

Worked solution P68-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: Procedure chooser

Question P68-Toughest-10. For a constructed 2 by 3 table from an online-course completion sample, observed rows are [34, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for procedure chooser.

Worked solution and validity check

Worked solution P68-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: Chi-square statistic

Question P68-Toughest-11. For a constructed 2 by 3 table from a website response-time study, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for chi-square statistic.

Worked solution and validity check

Worked solution P68-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: Observed and expected counts

Question P68-Toughest-12. For a constructed 2 by 3 table from a website response-time study, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for observed and expected counts.

Worked solution and validity check

Worked solution P68-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: Types of chi-square tests

Question P68-Toughest-13. For a constructed 2 by 3 table from a classroom memory study, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for types of chi-square tests.

Worked solution and validity check

Worked solution P68-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: Degrees of freedom

Question P68-Toughest-14. For a constructed 2 by 3 table from a battery-life laboratory trial, observed rows are [33, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for degrees of freedom.

Worked solution and validity check

Worked solution P68-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: P-value

Question P68-Toughest-15. For a constructed 2 by 3 table from a website response-time study, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for p-value.

Worked solution and validity check

Worked solution P68-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
ScopeGoodness-of-fit is legacy P70, not revised Unit 3 core.
Method or sourceThe revised course retains chi-square tests of independence and homogeneity, both using expected counts and cell contributions, while goodness-of-fit is legacy enrichment.
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 chi-square statistic work in chi square test?

Answer for chi square test and Chi-square statistic. 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 observed and expected counts work in chi square test?

Answer for chi square test and Observed and expected counts. 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 types of chi-square tests work in chi square test?

Answer for chi square test and Types of chi-square tests. 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 degrees of freedom work in chi square test?

Answer for chi square test and Degrees of freedom. 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 p-value work in chi square test?

Answer for chi square test and P-value. 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 conditions work in chi square test?

Answer for chi square test and Conditions. 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 chi-square test connect to Chi-Square Test?

chi-square test within chi square test. 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 Chi-square statistic, the controlling scope is: Goodness-of-fit is legacy P70, not revised Unit 3 core.

How does chi squared test connect to Chi-Square Test?

chi squared test within chi square test. 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 Observed and expected counts, the controlling scope is: Goodness-of-fit is legacy P70, not revised Unit 3 core.

How does chi-squared test connect to Chi-Square Test?

chi-squared test within chi square test. 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 Types of chi-square tests, the controlling scope is: Goodness-of-fit is legacy P70, not revised Unit 3 core.

How does what is a chi square test connect to Chi-Square Test?

what is a chi square test within chi square test. 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 Degrees of freedom, the controlling scope is: Goodness-of-fit is legacy P70, not revised Unit 3 core.

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

chi square test equation within chi square test. 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 P-value, the controlling scope is: Goodness-of-fit is legacy P70, not revised Unit 3 core.

How does how to do chi square test connect to Chi-Square Test?

how to do chi square test within chi square test. 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 Conditions, the controlling scope is: Goodness-of-fit is legacy P70, not revised Unit 3 core.

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

analysis of chi square test within chi square test. 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 Worked overview, the controlling scope is: Goodness-of-fit is legacy P70, not revised Unit 3 core.

Sources

Administrative and curricular statements in Chi-Square Test: Formula, Types, Conditions, and Examples 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 Conclusion

The revised course retains chi-square tests of independence and homogeneity, both using expected counts and cell contributions, while goodness-of-fit is legacy enrichment. Mastery of chi square test therefore requires the exact evidence, mathematics, interpretation, and scope developed in this guide, while preserving this boundary: Goodness-of-fit is legacy P70, not revised Unit 3 core.

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Engr. Muhammad Yar Saqib

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