Chi-Square Calculator, Critical Values, and Degrees of Freedom
Chi square calculator table critical values work is easiest when you separate four jobs: build or verify expected counts, calculate the chi-square statistic, determine degrees of freedom, and then use a p-value or critical value to make the decision. This page is a calculation-first guide with an interactive engine, hand-worked checks, and interpretation practice.
Quick answer: how to calculate chi-square
The core statistic is χ2 = Σ (O−E)2/E. For a contingency table, first compute every expected count from row total × column total ÷ grand total. Then add the cell contributions, determine degrees of freedom with (rows − 1)(columns − 1), and compare the resulting tail probability with the chosen significance level. A larger chi-square statistic means the observed counts are farther from what the null model predicts.
A calculator is useful for arithmetic, but it does not decide whether the design justifies inference or whether the null and alternative hypotheses match the question. Always identify the variables, the sampling or treatment design, and the correct expected-count model before entering numbers. The numerical output only becomes statistical evidence after those pieces are established.
Chi square calculator table critical values questions often mix calculation with interpretation. A complete solution states the hypotheses in context, checks conditions, reports χ² and df, gives the p-value or critical-value comparison, and finishes with a conclusion about the population or treatment distributions—not merely a statement that one number is larger than another.
A reliable chi-square calculator workflow
Start by organizing the counts exactly as they were collected. In a test of homogeneity, the rows can represent independently sampled populations and the columns can represent response categories. In a test of independence, both variables are measured on one random sample and the table records combinations of their categories. The arithmetic may look identical, but the sampling story changes the hypotheses and conclusion.
If expected counts are already provided, verify that their total matches the total observed count. If they are not provided, compute them from the null model. For a contingency table the expected count in cell (i,j) equals row i total times column j total divided by the grand total. This preserves both margins while representing the null claim of identical distributions or independence.
After calculating χ², remember that chi-square tests use the right tail. The statistic cannot be negative because every squared deviation is nonnegative. A value near zero means the observed table is close to the null expectation; larger values indicate more discrepancy. Whether a particular value is surprising depends on degrees of freedom, which is why a statistic without df is incomplete.
| Stage | What to enter or compute | What to verify |
|---|---|---|
| 1. Observed counts | Counts in mutually exclusive categories | Raw counts, not percentages |
| 2. Expected counts | Null-model counts for each category/cell | Every expected count is positive and appropriately large |
| 3. Statistic | Σ(O−E)²/E | Cell contributions add to the reported χ² |
| 4. Degrees of freedom | Model-specific df | Contingency table: (r−1)(c−1) |
| 5. Evidence | Right-tail p-value or critical value | Use the same df and significance level |
| 6. Conclusion | Decision plus context | Do not say the null is proved |
Interactive AP Statistics calculator
Interactive AP Statistics Formula Engine
Z-scoreBinomial probabilityOne-proportion z intervalTwo-proportion z intervalOne-proportion z testTwo-proportion z testMean and sample standard deviationOne-sample t statisticTwo-sample t statisticLeast-squares lineResidualChi-square statisticNormal probability between two valuesNormal percentile / inverse normalDiscrete expected value and standard deviationSampling distribution of x̄ probabilitySampling distribution of p̂ probabilityOne-sample t interval (enter t*)Two-sample t interval (enter t*)One-proportion sample size for margin of error
Select a method and enter data.
What the engine can calculate
- Z-score: Value x, mean, and standard deviation
- Binomial probability: n trials, x successes, and success probability p
- One-proportion z interval: successes x, sample size n, and confidence level
- Two-proportion z interval: two success counts, two sample sizes, and confidence level
- One-proportion z test: x, n, null p0, and alternative
- Two-proportion z test: x1, n1, x2, n2, and alternative
- Mean and sample standard deviation: comma-separated quantitative data
- One-sample t statistic: sample mean, null mean, sample SD, and n
- Two-sample t statistic: two means, SDs, and sample sizes
- Least-squares line: comma-separated x values and y values
- Residual: observed y and predicted y
- Chi-square statistic: comma-separated observed and expected counts
- Normal probability: Mean, standard deviation, lower bound, and upper bound
- Normal percentile: Mean, standard deviation, and cumulative area to the left
- Expected value and SD: Comma-separated outcomes and matching probabilities
- Sampling distribution of x̄: Population mean, population SD, n, and interval bounds
- Sampling distribution of p̂: Population proportion, n, and interval bounds
- One-sample t interval: Sample mean, sample SD, n, and the selected t* critical value
- Two-sample t interval: Two means, SDs, sample sizes, and the selected t* critical value
- Proportion sample size: Confidence level, target margin of error, and planning value of p
Expected counts are the foundation of the calculation
Expected counts answer a counterfactual question: what cell counts would be anticipated if the null relationship were exactly true while the observed margins stayed fixed? They are not forecasts from personal judgment and they are not the average of neighboring cells. Their purpose is to create the reference pattern against which the observed table is compared.
For a 2×3 table with row totals 90 and 110, column totals 70, 60, and 70, and grand total 200, the expected counts in the first row are 31.5, 27, and 31.5. The second row expectations are 38.5, 33, and 38.5. Notice that each expected row sums to its observed row total and each expected column sums to its observed column total. Those margin checks catch many data-entry errors.
Cell contributions reveal which parts of a table drive the overall statistic. A cell that differs from expectation by 8 does not automatically contribute more than a cell that differs by 5, because the squared deviation is divided by the expected count. Comparing contributions is useful for diagnosis, but the formal test still uses their sum.
| Observed O | Expected E | Contribution (O−E)²/E |
|---|---|---|
| 42 | 36 | 1.000 |
| 28 | 34 | 1.059 |
| 30 | 30 | 0.000 |
| 24 | 30 | 1.200 |
| 36 | 30 | 1.200 |
| 40 | 40 | 0.000 |
Degrees of freedom: choose the formula that matches the model
For an r×c contingency table, df=(r−1)(c−1). A 2×2 table has 1 degree of freedom, a 2×4 table has 3, and a 3×4 table has 6. The formula reflects how many cell counts can vary once the row and column margins constrain the table.
A one-variable goodness-of-fit test uses a different df rule, commonly categories minus one when no parameters are estimated from the same data. However, goodness-of-fit was removed from the revised AP Statistics course effective 2026–27, so on this site it is treated as legacy/enrichment. The current-core use of chi-square emphasizes categorical association/comparison in contingency tables.
Critical values change with df. Reusing a memorized cutoff for every table is unsafe. When technology gives a p-value directly, df is still part of the test because the tail area comes from the chi-square distribution with that specific df.
| Table shape | df |
|---|---|
| 2×2 | 1 |
| 2×3 | 2 |
| 2×5 | 4 |
| 3×3 | 4 |
| 3×4 | 6 |
| 4×5 | 12 |
Critical values, p-values, and the right-tail decision
A critical value is the chi-square value that leaves a specified upper-tail area α. If the calculated statistic exceeds the critical value, the p-value is smaller than α and the null hypothesis is rejected. If it does not exceed the cutoff, the evidence is insufficient to reject. The p-value approach and the critical-value approach are equivalent when they use the same df and α.
Do not compare χ² directly with α. A statistic such as 7.81 and a significance level such as 0.05 are on different scales. Compare the p-value with α, or compare χ² with the correct critical value. A common exam error is writing “7.81 > .05, so reject,” which reaches a decision through an invalid comparison even if the conclusion happens to be correct.
A p-value is the probability, assuming the null model is true, of obtaining a chi-square statistic at least as large as the observed statistic. It is not the probability that the null hypothesis is true. It also is not the probability that the observed table occurred by random chance. The conditional wording matters because it defines what the tail area represents.
Worked chi-square calculation cases
Worked case 1: School lunch preference
The observed 2×3 table is [[38, 27, 15], [31, 34, 25]]. Row totals are [80, 90], column totals are [69, 61, 40], and n=170. Under the null model, the expected table is [[32.47, 28.71, 18.82], [36.53, 32.29, 21.18]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈3.437 with df=2. The largest contribution is about 0.942, from row 1, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about school lunch preference, not merely to the phrase “statistically significant.”
Worked case 2: Transit payment method
The observed 2×3 table is [[52, 33, 25], [43, 39, 28]]. Row totals are [110, 110], column totals are [95, 72, 53], and n=220. Under the null model, the expected table is [[47.5, 36.0, 26.5], [47.5, 36.0, 26.5]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈1.522 with df=2. The largest contribution is about 0.426, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about transit payment method, not merely to the phrase “statistically significant.”
Worked case 3: Clinic appointment channel
The observed 2×3 table is [[44, 36, 20], [35, 40, 25]]. Row totals are [100, 100], column totals are [79, 76, 45], and n=200. Under the null model, the expected table is [[39.5, 38.0, 22.5], [39.5, 38.0, 22.5]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈1.791 with df=2. The largest contribution is about 0.513, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about clinic appointment channel, not merely to the phrase “statistically significant.”
Worked case 4: Library resource format
The observed 2×3 table is [[29, 41, 30], [36, 34, 30]]. Row totals are [100, 100], column totals are [65, 75, 60], and n=200. Under the null model, the expected table is [[32.5, 37.5, 30.0], [32.5, 37.5, 30.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈1.407 with df=2. The largest contribution is about 0.377, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about library resource format, not merely to the phrase “statistically significant.”
Worked case 5: Community survey response
The observed 2×3 table is [[48, 32, 20], [39, 36, 25]]. Row totals are [100, 100], column totals are [87, 68, 45], and n=200. Under the null model, the expected table is [[43.5, 34.0, 22.5], [43.5, 34.0, 22.5]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈1.722 with df=2. The largest contribution is about 0.466, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about community survey response, not merely to the phrase “statistically significant.”
Worked case 6: Device type by program
The observed 2×3 table is [[33, 47, 20], [42, 38, 20]]. Row totals are [100, 100], column totals are [75, 85, 40], and n=200. Under the null model, the expected table is [[37.5, 42.5, 20.0], [37.5, 42.5, 20.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.033 with df=2. The largest contribution is about 0.540, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about device type by program, not merely to the phrase “statistically significant.”
Worked case 7: Recycling category choice
The observed 2×3 table is [[50, 30, 20], [40, 35, 25]]. Row totals are [100, 100], column totals are [90, 65, 45], and n=200. Under the null model, the expected table is [[45.0, 32.5, 22.5], [45.0, 32.5, 22.5]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.051 with df=2. The largest contribution is about 0.556, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about recycling category choice, not merely to the phrase “statistically significant.”
Worked case 8: Training completion mode
The observed 2×3 table is [[46, 29, 25], [35, 41, 24]]. Row totals are [100, 100], column totals are [81, 70, 49], and n=200. Under the null model, the expected table is [[40.5, 35.0, 24.5], [40.5, 35.0, 24.5]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈3.571 with df=2. The largest contribution is about 1.029, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about training completion mode, not merely to the phrase “statistically significant.”
Worked case 9: Museum ticket source
The observed 2×3 table is [[37, 43, 20], [44, 31, 25]]. Row totals are [100, 100], column totals are [81, 74, 45], and n=200. Under the null model, the expected table is [[40.5, 37.0, 22.5], [40.5, 37.0, 22.5]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈3.106 with df=2. The largest contribution is about 0.973, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about museum ticket source, not merely to the phrase “statistically significant.”
Worked case 10: Course resource use
The observed 2×3 table is [[41, 35, 24], [32, 44, 24]]. Row totals are [100, 100], column totals are [73, 79, 48], and n=200. Under the null model, the expected table is [[36.5, 39.5, 24.0], [36.5, 39.5, 24.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.135 with df=2. The largest contribution is about 0.555, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about course resource use, not merely to the phrase “statistically significant.”
Worked case 11: Emergency alert channel
The observed 2×3 table is [[55, 25, 20], [45, 35, 20]]. Row totals are [100, 100], column totals are [100, 60, 40], and n=200. Under the null model, the expected table is [[50.0, 30.0, 20.0], [50.0, 30.0, 20.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.667 with df=2. The largest contribution is about 0.833, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about emergency alert channel, not merely to the phrase “statistically significant.”
Worked case 12: Sports registration route
The observed 2×3 table is [[34, 46, 20], [43, 37, 20]]. Row totals are [100, 100], column totals are [77, 83, 40], and n=200. Under the null model, the expected table is [[38.5, 41.5, 20.0], [38.5, 41.5, 20.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.028 with df=2. The largest contribution is about 0.526, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about sports registration route, not merely to the phrase “statistically significant.”
Worked case 13: Campus dining location
The observed 2×3 table is [[39, 31, 30], [31, 39, 30]]. Row totals are [100, 100], column totals are [70, 70, 60], and n=200. Under the null model, the expected table is [[35.0, 35.0, 30.0], [35.0, 35.0, 30.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈1.829 with df=2. The largest contribution is about 0.457, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about campus dining location, not merely to the phrase “statistically significant.”
Worked case 14: Household internet plan
The observed 2×3 table is [[42, 28, 30], [34, 36, 30]]. Row totals are [100, 100], column totals are [76, 64, 60], and n=200. Under the null model, the expected table is [[38.0, 32.0, 30.0], [38.0, 32.0, 30.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈1.842 with df=2. The largest contribution is about 0.500, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about household internet plan, not merely to the phrase “statistically significant.”
Worked case 15: Volunteer shift choice
The observed 2×3 table is [[47, 33, 20], [38, 42, 20]]. Row totals are [100, 100], column totals are [85, 75, 40], and n=200. Under the null model, the expected table is [[42.5, 37.5, 20.0], [42.5, 37.5, 20.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.033 with df=2. The largest contribution is about 0.540, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about volunteer shift choice, not merely to the phrase “statistically significant.”
Worked case 16: Customer support channel
The observed 2×3 table is [[51, 29, 20], [42, 38, 20]]. Row totals are [100, 100], column totals are [93, 67, 40], and n=200. Under the null model, the expected table is [[46.5, 33.5, 20.0], [46.5, 33.5, 20.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.080 with df=2. The largest contribution is about 0.604, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about customer support channel, not merely to the phrase “statistically significant.”
Worked case 17: Park activity choice
The observed 2×3 table is [[36, 39, 25], [45, 30, 25]]. Row totals are [100, 100], column totals are [81, 69, 50], and n=200. Under the null model, the expected table is [[40.5, 34.5, 25.0], [40.5, 34.5, 25.0]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈2.174 with df=2. The largest contribution is about 0.587, from row 2, column 2; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about park activity choice, not merely to the phrase “statistically significant.”
Worked case 18: Workshop attendance mode
The observed 2×3 table is [[40, 40, 20], [32, 43, 25]]. Row totals are [100, 100], column totals are [72, 83, 45], and n=200. Under the null model, the expected table is [[36.0, 41.5, 22.5], [36.0, 41.5, 22.5]]. These expected counts preserve the observed margins while removing the association or distribution difference described by the alternative.
Summing all six contributions gives χ²≈1.553 with df=2. The largest contribution is about 0.444, from row 2, column 1; that cell is useful diagnostically because it contributes most to the overall discrepancy. A formal conclusion would still use the right-tail p-value for df=2 rather than treating the largest cell as a separate hypothesis test.
Before relying on software, reproduce at least one expected count and one contribution by hand. That cross-check catches transposed rows, percentages entered as counts, and expected values copied into the observed field. In a written response, report the null relationship in context and connect the p-value to evidence about workshop attendance mode, not merely to the phrase “statistically significant.”
Chi-square calculator multiple-choice practice
Question 1. Chi-square calculation
A 3×4 table is analyzed. What are the degrees of freedom?
Answer: B
For a contingency table, df=(3−1)(4−1)=6.
Question 2. Chi-square calculation
Which expression gives an expected count in a contingency-table cell?
Answer: B
The null-model expectation is row total × column total divided by the grand total.
Question 3. Chi-square calculation
A chi-square statistic equals 0. What does that mean numerically?
Answer: A
χ² is a sum of nonnegative cell contributions, so it can be zero only when every observed count matches expectation.
Question 4. Chi-square calculation
For df=4, software reports p=.018. At α=.05, what is the appropriate decision?
Answer: A
Because .018 < .05, the result is in the rejection region.
Question 5. Chi-square calculation
Which comparison is valid in a critical-value approach?
Answer: C
The test statistic and critical value are on the same chi-square scale.
Question 6. Chi-square calculation
Why are chi-square tests right-tailed?
Answer: B
χ² is nonnegative and increasingly large values indicate greater discrepancy from the null expectation.
Question 7. Chi-square calculation
A student enters row percentages instead of counts into a chi-square calculator. What is the main problem?
Answer: B
Chi-square expected counts and contributions are based on frequencies, not an arbitrary percentage rescaling by row.
Question 8. Chi-square calculation
In a 2×5 contingency table, df equals
Answer: A
df=(2−1)(5−1)=4.
Question 9. Chi-square calculation
Which statement correctly interprets p=.003 in a chi-square test?
Answer: B
A p-value is conditional on the null model; it is not a probability assigned to either hypothesis.
Question 10. Chi-square calculation
A cell has O=30 and E=20. Its contribution to χ² is
Answer: C
(30−20)²/20=100/20=5.
Question 11. Chi-square calculation
If every observed and expected count is doubled while proportions stay the same, what generally happens to χ²?
Answer: C
Each squared difference grows by a factor of four while E doubles, so each contribution doubles.
Question 12. Chi-square calculation
Why should the total expected count equal the total observed count?
Answer: A
Expected counts redistribute the same grand total under the null relationship.
Question 13. Chi-square calculation
Which output is incomplete for reporting a chi-square test?
Answer: B
The statistic alone lacks df, evidence, and a contextual decision.
Question 14. Chi-square calculation
A p-value of .22 at α=.05 means
Answer: B
The observed discrepancy is not sufficiently unusual at the .05 significance level.
Question 15. Chi-square calculation
Which error is most likely to change the computed chi-square statistic?
Answer: B
Cell-specific expected values must align with the corresponding observed cells.
Question 16. Chi-square calculation
For a 4×3 table, which df should be sent to a critical-value table?
Answer: C
(4−1)(3−1)=6.
Question 17. Chi-square calculation
If one cell contributes 8.0 and every other cell contributes less than 0.5, what is a reasonable diagnostic statement?
Answer: A
Cell contributions can identify where discrepancy is concentrated, but the inferential test uses the full table.
Question 18. Chi-square calculation
What is the best final sentence after rejecting a homogeneity null?
Answer: B
The conclusion should match the population-distribution alternative without overclaiming specific cells or causation.
Chi-square calculator free-response practice
FRQ 1. Contingency-table calculation
A community survey records transportation preference in two neighborhoods: [[48,32,20],[36,39,25]]. Calculate the expected counts, χ², and df, and explain what a small p-value would mean.
Model response
The row totals are both 100 and the column totals are 84, 71, and 45, so the first-row expected counts are 42, 35.5, and 22.5; the second row has the same expectations.
The six contributions sum to χ²≈2.584 and df=(2−1)(3−1)=2. A small right-tail p-value would mean that a discrepancy at least this large would be unusual if the two population distributions were actually the same.
FRQ 2. Critical-value method
A 3×3 table produces χ²=11.2 with df=4. Describe how to use a chi-square critical-value table at α=.05 without confusing α with the statistic.
Model response
Locate the upper-tail .05 critical value in the df=4 row. Compare 11.2 with that chi-square cutoff; do not compare 11.2 directly with .05.
If 11.2 exceeds the cutoff, reject the null at the .05 level; otherwise fail to reject. The conclusion must be written in the context of the categorical relationship being tested.
FRQ 3. Contribution diagnosis
A test has χ²=14.6. One cell contributes 7.9, while no other cell contributes more than 1.4. Explain what the contribution tells you and what it does not prove.
Model response
The cell is the largest numerical source of discrepancy between observed and expected counts, so it is a sensible place to inspect when describing the table.
The cell contribution is not a separate inferential test and does not by itself prove which population category caused the association. The formal evidence comes from the overall statistic and its reference distribution.
FRQ 4. Data-entry audit
A student gets χ²=0 after entering a table whose observed counts visibly differ across groups. Give two checks that should be performed before accepting the output.
Model response
Verify that the observed and expected lists were not accidentally made identical and that counts rather than percentages were entered.
Recompute one expected count from the margins and one (O−E)²/E contribution by hand. Any nonzero discrepancy should generate a positive contribution.
FRQ 5. Degrees-of-freedom explanation
Explain why a 3×5 contingency table has 8 degrees of freedom, not 15.
Model response
Once the row and column margins are fixed, all 15 cell counts cannot vary independently. The contingency-table rule captures those constraints.
Using (r−1)(c−1) gives (3−1)(5−1)=8 degrees of freedom.
FRQ 6. Complete report
Software reports χ²=9.72, df=3, p=.021 for a random-sample contingency table. Write a complete evidence statement at α=.05.
Model response
Because p=.021 is less than .05, reject the null relationship at the stated significance level.
There is statistically convincing evidence, in the population represented by the random sample, of a categorical association or distribution difference corresponding to the study’s alternative. The result does not by itself establish causation.
Extended topic-specific mastery cases
Expected-Count Arithmetic mastery extension 1
Mastery case 1 focuses on expected-count arithmetic. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes expected-count arithmetic less mechanical and helps expose data-entry errors.
Cell Contribution Audit mastery extension 2
Mastery case 2 focuses on cell contribution audit. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes cell contribution audit less mechanical and helps expose data-entry errors.
Degrees-Of-Freedom Choice mastery extension 3
Mastery case 3 focuses on degrees-of-freedom choice. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes degrees-of-freedom choice less mechanical and helps expose data-entry errors.
Right-Tail Evidence mastery extension 4
Mastery case 4 focuses on right-tail evidence. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes right-tail evidence less mechanical and helps expose data-entry errors.
Critical-Value Comparison mastery extension 5
Mastery case 5 focuses on critical-value comparison. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes critical-value comparison less mechanical and helps expose data-entry errors.
Software Cross-Check mastery extension 6
Mastery case 6 focuses on software cross-check. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes software cross-check less mechanical and helps expose data-entry errors.
Expected-Count Arithmetic mastery extension 7
Mastery case 7 focuses on expected-count arithmetic. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes expected-count arithmetic less mechanical and helps expose data-entry errors.
Cell Contribution Audit mastery extension 8
Mastery case 8 focuses on cell contribution audit. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes cell contribution audit less mechanical and helps expose data-entry errors.
Degrees-Of-Freedom Choice mastery extension 9
Mastery case 9 focuses on degrees-of-freedom choice. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes degrees-of-freedom choice less mechanical and helps expose data-entry errors.
Right-Tail Evidence mastery extension 10
Mastery case 10 focuses on right-tail evidence. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes right-tail evidence less mechanical and helps expose data-entry errors.
Critical-Value Comparison mastery extension 11
Mastery case 11 focuses on critical-value comparison. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes critical-value comparison less mechanical and helps expose data-entry errors.
Software Cross-Check mastery extension 12
Mastery case 12 focuses on software cross-check. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes software cross-check less mechanical and helps expose data-entry errors.
Expected-Count Arithmetic mastery extension 13
Mastery case 13 focuses on expected-count arithmetic. Begin from a fresh categorical table rather than copying a previous result. Recompute the margins, generate one expected count independently, and verify that expected row and column totals reproduce the observed margins. Next calculate at least two cell contributions and explain why their sum is nonnegative. State the table dimensions and determine df before reading a tail probability or critical value. Finish by distinguishing numerical evidence from substantive interpretation: a calculator can quantify disagreement with the null model, but the study design determines what population or treatment conclusion is justified. As a final audit, change one observed count while preserving the grand total and predict qualitatively how the affected contributions should respond before recalculating. This prediction-before-computation habit makes expected-count arithmetic less mechanical and helps expose data-entry errors.
Chi-square calculator FAQs
Does a calculator replace expected-count work? No. Even when software computes expectations automatically, you should understand their null-model meaning and be able to verify at least one cell.
Can χ² be negative? No. It is a sum of squared deviations divided by positive expected counts.
Should I use a p-value or a critical value? Either is valid when the df and α are handled consistently. Technology usually makes the p-value approach more direct.
Why does df matter? The shape of the chi-square reference distribution changes with df, so the same statistic can represent different tail areas for different table sizes.
Is chi-square goodness of fit current AP Statistics content? No. It is legacy/enrichment under the revised 2026–27 course; contingency-table chi-square inference remains the more relevant current categorical setting.
Continue with related AP Statistics resources
Chi Square Calculator Table Critical Values review focus
The phrase chi square calculator table critical values names this page’s specific purpose. Use chi square calculator table critical values as the focus when deciding which workflow, conditions, interpretation, or legacy-status guidance belongs here rather than on a neighboring AP Statistics page.
For final review, return to the opening explanation and verify that you can explain chi square calculator table critical values in context without relying on a memorized label alone.
