Chi-Square Calculator, Critical Value Table, and Degrees of Freedom
A lookup-first reference for chi-square calculator output and critical-value tables, organized around notation, formulas or controls, correct selection, and worked verification.
Reference at a Glance: Chi-Square Test Calculator
Calculator chi-square output is valid only after the observed matrix, expected matrix, degrees of freedom, and expected-count conditions have been checked independently.
Reference Formula Index
Chi-square statistic
Chi-square statistic in Chi-Square Test Calculator: Compute expected counts from the null model, then retain each cell contribution before summing so the result can be audited.
Chi-square degrees of freedom
Chi-square degrees of freedom in Chi-Square Test Calculator: This expression belongs specifically to chi-square calculator output and critical-value tables; define every symbol and apply the scope rule for matrix entry, expected counts, statistic, df, p-value, and table bounds before calculation.
Calculator
Lookup decision
Calculator: 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.
Selection check for chi square test calculator and Calculator: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Chi-square table
Lookup decision
Chi-square table: 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.
Selection check for chi square test calculator and Chi-square table: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Critical values
Lookup decision
Critical values: 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.
Selection check for chi square test calculator and Critical values: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Degrees of freedom
Lookup decision
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.
Selection check for chi square test calculator and Degrees of freedom: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Right-tail probability
Lookup decision
Right-tail probability: 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.
Selection check for chi square test calculator and Right-tail probability: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Expected-count calculator
Lookup decision
Expected-count calculator: 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.
Selection check for chi square test calculator and Expected-count calculator: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Worked lookups
Lookup decision
Worked lookups: 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.
Selection check for chi square test calculator and Worked lookups: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Downloadable reference
Lookup decision
Downloadable reference: The statistic is 6.912, df=2, and p=0.0315; 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.
Selection check for chi square test calculator and Downloadable reference: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts.
Reference Drills with Worked Answers
Every question in Chi-Square Calculator, Critical Value Table, and Degrees of Freedom 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: Worked lookups
Question P72-Easy-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 worked lookups.
Worked solution and validity check
Worked solution P72-Easy-1. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. 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: Downloadable reference
Question P72-Easy-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 downloadable reference.
Worked solution and validity check
Worked solution P72-Easy-2. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. 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: Calculator
Question P72-Easy-3. For a constructed 2 by 3 table from a school library checkout study, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for calculator.
Worked solution and validity check
Worked solution P72-Easy-3. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. 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: Chi-square table
Question P72-Easy-4. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for chi-square table.
Worked solution and validity check
Worked solution P72-Easy-4. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. 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: Critical values
Question P72-Easy-5. For a constructed 2 by 3 table from a reading-speed investigation, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for critical values.
Worked solution and validity check
Worked solution P72-Easy-5. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. 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: Degrees of freedom
Question P72-Easy-6. 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 degrees of freedom.
Worked solution and validity check
Worked solution P72-Easy-6. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. 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: Right-tail probability
Question P72-Easy-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 right-tail probability.
Worked solution and validity check
Worked solution P72-Easy-7. The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals. 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: Expected-count calculator
Question P72-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 expected-count calculator.
Worked solution and validity check
Worked solution P72-Easy-8. The statistic is 6.912, df=2, and p=0.0315; expected counts are computed from row and column totals. 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: Worked lookups
Question P72-Easy-9. For a constructed 2 by 3 table from an online-course completion sample, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for worked lookups.
Worked solution and validity check
Worked solution P72-Easy-9. The statistic is 6.385, df=2, and p=0.0411; expected counts are computed from row and column totals. 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: Downloadable reference
Question P72-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 downloadable reference.
Worked solution and validity check
Worked solution P72-Easy-10. The statistic is 7.189, df=2, and p=0.0275; expected counts are computed from row and column totals. 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: Calculator
Question P72-Easy-11. For a constructed 2 by 3 table from a seedling-growth comparison, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for calculator.
Worked solution and validity check
Worked solution P72-Easy-11. The statistic is 5.622, df=2, and p=0.0602; expected counts are computed from row and column totals. 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: Chi-square table
Question P72-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 chi-square table.
Worked solution and validity check
Worked solution P72-Easy-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. 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: Critical values
Question P72-Easy-13. For a constructed 2 by 3 table from an online-course completion sample, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for critical values.
Worked solution and validity check
Worked solution P72-Easy-13. The statistic is 5.931, df=2, and p=0.0515; expected counts are computed from row and column totals. 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: Degrees of freedom
Question P72-Easy-14. For a constructed 2 by 3 table from a water-filtration experiment, 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 P72-Easy-14. The statistic is 6.715, df=2, and p=0.0348; expected counts are computed from row and column totals. 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: Right-tail probability
Question P72-Easy-15. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for right-tail probability.
Worked solution and validity check
Worked solution P72-Easy-15. The statistic is 7.546, df=2, and p=0.0230; expected counts are computed from row and column totals. 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 16: Expected-count calculator
Question P72-Easy-16. For a constructed 2 by 3 table from an online-course completion sample, observed rows are [30, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for expected-count calculator.
Worked solution and validity check
Worked solution P72-Easy-16. The statistic is 5.930, df=2, and p=0.0516; expected counts are computed from row and column totals. 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: Worked lookups
Question P72-Tough-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 worked lookups.
Worked solution and validity check
Worked solution P72-Tough-1. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. 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: Downloadable reference
Question P72-Tough-2. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [31, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for downloadable reference.
Worked solution and validity check
Worked solution P72-Tough-2. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. 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: Calculator
Question P72-Tough-3. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [32, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for calculator.
Worked solution and validity check
Worked solution P72-Tough-3. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. 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: Chi-square table
Question P72-Tough-4. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for chi-square table.
Worked solution and validity check
Worked solution P72-Tough-4. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. 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: Critical values
Question P72-Tough-5. For a constructed 2 by 3 table from a school library checkout study, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for critical values.
Worked solution and validity check
Worked solution P72-Tough-5. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. 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: Degrees of freedom
Question P72-Tough-6. 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 degrees of freedom.
Worked solution and validity check
Worked solution P72-Tough-6. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. 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: Right-tail probability
Question P72-Tough-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 right-tail probability.
Worked solution and validity check
Worked solution P72-Tough-7. The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals. 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: Expected-count calculator
Question P72-Tough-8. For a constructed 2 by 3 table from a reading-speed investigation, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for expected-count calculator.
Worked solution and validity check
Worked solution P72-Tough-8. The statistic is 6.912, df=2, and p=0.0315; expected counts are computed from row and column totals. 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: Worked lookups
Question P72-Tough-9. For a constructed 2 by 3 table from a school library checkout study, observed rows are [33, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for worked lookups.
Worked solution and validity check
Worked solution P72-Tough-9. The statistic is 6.385, df=2, and p=0.0411; expected counts are computed from row and column totals. 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: Downloadable reference
Question P72-Tough-10. For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [34, 20, 10] and [18, 28, 15]. Carry out the chi-square calculation for downloadable reference.
Worked solution and validity check
Worked solution P72-Tough-10. The statistic is 7.189, df=2, and p=0.0275; expected counts are computed from row and column totals. 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: Calculator
Question P72-Tough-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 calculator.
Worked solution and validity check
Worked solution P72-Tough-11. The statistic is 5.622, df=2, and p=0.0602; expected counts are computed from row and column totals. 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: Chi-square table
Question P72-Tough-12. For a constructed 2 by 3 table from a recycling-behavior survey, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for chi-square table.
Worked solution and validity check
Worked solution P72-Tough-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. 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: Critical values
Question P72-Tough-13. For a constructed 2 by 3 table from a tutoring-program evaluation, observed rows are [32, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for critical values.
Worked solution and validity check
Worked solution P72-Tough-13. The statistic is 5.931, df=2, and p=0.0515; expected counts are computed from row and column totals. 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: Degrees of freedom
Question P72-Tough-14. For a constructed 2 by 3 table from a reading-speed investigation, 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 P72-Tough-14. The statistic is 6.715, df=2, and p=0.0348; expected counts are computed from row and column totals. 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: Right-tail probability
Question P72-Tough-15. For a constructed 2 by 3 table from a public-parks visitor survey, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for right-tail probability.
Worked solution and validity check
Worked solution P72-Tough-15. The statistic is 7.546, df=2, and p=0.0230; expected counts are computed from row and column totals. 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 16: Expected-count calculator
Question P72-Tough-16. For a constructed 2 by 3 table from a school library checkout study, observed rows are [30, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for expected-count calculator.
Worked solution and validity check
Worked solution P72-Tough-16. The statistic is 5.930, df=2, and p=0.0516; expected counts are computed from row and column totals. 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 calculator
Question P72-Toughest-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 expected-count calculator.
Worked solution and validity check
Worked solution P72-Toughest-1. The statistic is 5.043, df=2, and p=0.0804; expected counts are computed from row and column totals. 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: Worked lookups
Question P72-Toughest-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 worked lookups.
Worked solution and validity check
Worked solution P72-Toughest-2. The statistic is 5.782, df=2, and p=0.0555; expected counts are computed from row and column totals. 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: Downloadable reference
Question P72-Toughest-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 downloadable reference.
Worked solution and validity check
Worked solution P72-Toughest-3. The statistic is 6.573, df=2, and p=0.0374; expected counts are computed from row and column totals. 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: Calculator
Question P72-Toughest-4. For a constructed 2 by 3 table from a quality-control inspection, observed rows are [33, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for calculator.
Worked solution and validity check
Worked solution P72-Toughest-4. The statistic is 7.412, df=2, and p=0.0246; expected counts are computed from row and column totals. 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: Chi-square table
Question P72-Toughest-5. For a constructed 2 by 3 table from a reading-speed investigation, observed rows are [34, 20, 10] and [18, 27, 15]. Carry out the chi-square calculation for chi-square table.
Worked solution and validity check
Worked solution P72-Toughest-5. The statistic is 6.844, df=2, and p=0.0327; expected counts are computed from row and column totals. 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: Critical values
Question P72-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 critical values.
Worked solution and validity check
Worked solution P72-Toughest-6. The statistic is 5.325, df=2, and p=0.0698; expected counts are computed from row and column totals. 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: Degrees of freedom
Question P72-Toughest-7. For a constructed 2 by 3 table from a package-delivery sample, observed rows are [31, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for degrees of freedom.
Worked solution and validity check
Worked solution P72-Toughest-7. The statistic is 6.094, df=2, and p=0.0475; expected counts are computed from row and column totals. 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: Right-tail probability
Question P72-Toughest-8. For a constructed 2 by 3 table from a reading-speed investigation, observed rows are [32, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for right-tail probability.
Worked solution and validity check
Worked solution P72-Toughest-8. The statistic is 6.912, df=2, and p=0.0315; expected counts are computed from row and column totals. 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: Expected-count calculator
Question P72-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 expected-count calculator.
Worked solution and validity check
Worked solution P72-Toughest-9. The statistic is 6.385, df=2, and p=0.0411; expected counts are computed from row and column totals. 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: Worked lookups
Question P72-Toughest-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 worked lookups.
Worked solution and validity check
Worked solution P72-Toughest-10. The statistic is 7.189, df=2, and p=0.0275; expected counts are computed from row and column totals. 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: Downloadable reference
Question P72-Toughest-11. For a constructed 2 by 3 table from a school library checkout study, observed rows are [30, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for downloadable reference.
Worked solution and validity check
Worked solution P72-Toughest-11. The statistic is 5.622, df=2, and p=0.0602; expected counts are computed from row and column totals. 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: Calculator
Question P72-Toughest-12. For a constructed 2 by 3 table from an online-course completion sample, observed rows are [31, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for calculator.
Worked solution and validity check
Worked solution P72-Toughest-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. 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: Chi-square table
Question P72-Toughest-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 chi-square table.
Worked solution and validity check
Worked solution P72-Toughest-13. The statistic is 5.931, df=2, and p=0.0515; expected counts are computed from row and column totals. 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: Critical values
Question P72-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 critical values.
Worked solution and validity check
Worked solution P72-Toughest-14. The statistic is 6.715, df=2, and p=0.0348; expected counts are computed from row and column totals. 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: Degrees of freedom
Question P72-Toughest-15. For a constructed 2 by 3 table from a tutoring-program evaluation, observed rows are [34, 20, 10] and [18, 29, 15]. Carry out the chi-square calculation for degrees of freedom.
Worked solution and validity check
Worked solution P72-Toughest-15. The statistic is 7.546, df=2, and p=0.0230; expected counts are computed from row and column totals. 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 16: Right-tail probability
Question P72-Toughest-16. For a constructed 2 by 3 table from a school library checkout study, observed rows are [30, 20, 10] and [18, 30, 15]. Carry out the chi-square calculation for right-tail probability.
Worked solution and validity check
Worked solution P72-Toughest-16. The statistic is 5.930, df=2, and p=0.0516; expected counts are computed from row and column totals. 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 point | Required evidence for chi square test calculator |
|---|---|
| Scope | Calculator output does not verify randomization or expected-count conditions. |
| Method or source | Calculator chi-square output is valid only after the observed matrix, expected matrix, degrees of freedom, and expected-count conditions have been checked independently. |
| Calculation | |
| 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. |
| Correction | Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most. |
Frequently Asked Questions
How does calculator work in chi square test calculator?
Answer for chi square test calculator and Calculator. 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 chi-square table work in chi square test calculator?
Answer for chi square test calculator and Chi-square table. 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 critical values work in chi square test calculator?
Answer for chi square test calculator and Critical values. 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 calculator?
Answer for chi square test calculator 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 right-tail probability work in chi square test calculator?
Answer for chi square test calculator and Right-tail probability. 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 expected-count calculator work in chi square test calculator?
Answer for chi square test calculator and Expected-count calculator. 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 table connect to Chi-Square Test Calculator?
chi square test table within chi square test calculator. 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 Calculator, the controlling scope is: Calculator output does not verify randomization or expected-count conditions.
How does chi-square test calculator connect to Chi-Square Test Calculator?
chi-square test calculator within chi square test calculator. 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 Chi-square table, the controlling scope is: Calculator output does not verify randomization or expected-count conditions.
How does critical value chi square test connect to Chi-Square Test Calculator?
critical value chi square test within chi square test calculator. 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 Critical values, the controlling scope is: Calculator output does not verify randomization or expected-count conditions.
How does chi square test critical value connect to Chi-Square Test Calculator?
chi square test critical value within chi square test calculator. 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: Calculator output does not verify randomization or expected-count conditions.
How does calculator chi square test connect to Chi-Square Test Calculator?
calculator chi square test within chi square test calculator. 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 Right-tail probability, the controlling scope is: Calculator output does not verify randomization or expected-count conditions.
How does calculator for chi square test connect to Chi-Square Test Calculator?
calculator for chi square test within chi square test calculator. 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 Expected-count calculator, the controlling scope is: Calculator output does not verify randomization or expected-count conditions.
How does chi square test degrees of freedom connect to Chi-Square Test Calculator?
chi square test degrees of freedom within chi square test calculator. 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 lookups, the controlling scope is: Calculator output does not verify randomization or expected-count conditions.
Sources
Administrative and curricular statements in Chi-Square Calculator, Critical Value Table, and Degrees of Freedom 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 Calculator Conclusion
Calculator chi-square output is valid only after the observed matrix, expected matrix, degrees of freedom, and expected-count conditions have been checked independently. Mastery of chi square test calculator therefore requires the exact evidence, mathematics, interpretation, and scope developed in this guide, while preserving this boundary: Calculator output does not verify randomization or expected-count conditions.