Chi-Square Tests and Contingency Tables MCQs 2026
40 questions with detailed answers · 30 from past papers · 4 quiz batches available
Choose a Quiz Batch. Each batch has 10 questions from this topic, in order. Take them one by one to work through all 40 MCQs. Login to save your scores and see your best per batch.
Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.
- Q1hard
Phi coefficient is special case of association measure for
💡 Explanation:φ = √(χ²/n) for fourfold tables.
- Q2hard
Cramér's V measures
💡 Explanation:V = √[χ²/(n·min(r−1,c−1))].
- Q3hard
Simpson's paradox can appear in
💡 Explanation:Aggregated and stratified tables may show opposite associations.
- Q4hard
Likelihood-ratio chi-square G² is
💡 Explanation:G² = 2ΣOi ln(Oi/Ei) parallels Pearson χ².
- Q5Past Paper · PPSC/FPSC/CSSeasy
If all Oi equal Ei exactly, χ² equals
💡 Explanation:Perfect agreement yields zero statistic.
- Q6Past Paper · PPSC/FPSC/CSSmedium
For 4×3 contingency table, minimum df is
💡 Explanation:(4−1)(3−1) = 6.
- Q7Past Paper · PPSC/FPSC/CSSeasy
Chi-square test is classified as
💡 Explanation:χ² methods apply to frequencies and contingency tables.
- Q8Past Paper · PPSC/FPSC/CSSeasy
Expected counts under independence sum to
💡 Explanation:ΣΣEij = N.
- Q9Past Paper · PPSC/FPSC/CSSmedium
Rejecting independence implies
💡 Explanation:Association does not alone establish causality.
- Q10hard
Goodness-of-fit to binomial with unknown p estimates p from data, reducing df by
💡 Explanation:Estimating p uses one degree of freedom.
- Q11medium
When r = 1 or c = 1, independence test is
💡 Explanation:Both dimensions need at least two categories.
- Q12Past Paper · PPSC/FPSC/CSSmedium
Contribution of one cell to χ² is
💡 Explanation:Each cell adds a squared standardized component.
- Q13Past Paper · PPSC/FPSC/CSSmedium
Pearson chi-square requires
💡 Explanation:Multinomial/count model assumptions apply.
- Q14Past Paper · PPSC/FPSC/CSSmedium
In 5×2 table, df for independence equals
💡 Explanation:(5−1)(2−1) = 4.
- Q15Past Paper · PPSC/FPSC/CSSeasy
Larger discrepancy between O and E produces
💡 Explanation:Big relative errors inflate the statistic.
- Q16Past Paper · PPSC/FPSC/CSSeasy
χ² statistic cannot be negative because
💡 Explanation:Squared terms make χ² ≥ 0.
- Q17medium
Pooling adjacent categories may be done when
💡 Explanation:Combining categories raises Ei at cost of detail.
- Q18Past Paper · PPSC/FPSC/CSSmedium
Chi-square homogeneity test compares
💡 Explanation:Homogeneity: same categories, different groups — similar mechanics to independence.
- Q19hard
McNemar's test applies to
💡 Explanation:McNemar uses discordant pairs on diagonal-off cells.
- Q20Past Paper · PPSC/FPSC/CSShard
Fisher's exact test is used for
💡 Explanation:Exact test avoids chi-square approximation issues.
- Q21hard
Standardized residual for cell (i,j) is approximately
💡 Explanation:Large standardized residuals flag cells contributing to χ².
- Q22Past Paper · PPSC/FPSC/CSSeasy
Test of independence H0 states
💡 Explanation:Independence: P(row i, col j) = P(row i)·P(col j).
- Q23Past Paper · PPSC/FPSC/CSSmedium
For goodness of fit to equally likely k categories, each Ei equals
💡 Explanation:Uniform H0: Ei = n/k.
- Q24Past Paper · PPSC/FPSC/CSSmedium
Chi-square distribution is
💡 Explanation:χ² with ν df has mean ν and variance 2ν.
- Q25Past Paper · PPSC/FPSC/CSSeasy
Marginal totals in contingency table are
💡 Explanation:Margins Ri and Cj define Ei under independence.
- Q26Past Paper · PPSC/FPSC/CSSeasy
Observed frequencies in contingency table are
💡 Explanation:Oi are data; Ei computed under H0.
- Q27Past Paper · PPSC/FPSC/CSSeasy
If χ² calculated exceeds χ²_{α, df}, decision at level α is
💡 Explanation:Large χ² relative to critical value contradicts H0.
- Q28Past Paper · PPSC/FPSC/CSShard
Yates continuity correction for 2×2 tables
💡 Explanation:Yates improves approximation for small 2×2 samples.
- Q29Past Paper · PPSC/FPSC/CSSmedium
Chi-square approximation is unreliable when
💡 Explanation:Small Ei violate asymptotic chi-square assumptions.
- Q30Past Paper · PPSC/FPSC/CSSmedium
A 3×4 contingency table has df for independence test equal
💡 Explanation:(3−1)(4−1) = 6.
- Q31Past Paper · PPSC/FPSC/CSSeasy
A 2×2 contingency table for independence has
💡 Explanation:(2−1)(2−1) = 1.
- Q32Past Paper · PPSC/FPSC/CSSeasy
Expected frequency under independence in cell (i,j) is
💡 Explanation:Eij = Ri·Cj/N under H0 of independence.
- Q33Past Paper · PPSC/FPSC/CSSeasy
Chi-square test of independence in an r×c contingency table has df
💡 Explanation:Independence df is product of (rows−1) and (cols−1).
- Q34Past Paper · PPSC/FPSC/CSSmedium
In a goodness-of-fit test with k categories, df equal
💡 Explanation:If all probabilities specified, df = k − 1.
- Q35Past Paper · PPSC/FPSC/CSSeasy
Goodness-of-fit test evaluates whether
💡 Explanation:GoF tests multinomial counts against hypothesized probabilities.
- Q36Past Paper · PPSC/FPSC/CSSeasy
Pearson chi-square statistic for goodness of fit is
💡 Explanation:Compares observed Oi to expected Ei frequencies.
- Q37Past Paper · PPSC/FPSC/CSSmedium
χ²_{0.05, 1} is approximately
💡 Explanation:Common critical value for 2×2 at α = 0.05.
- Q38hard
For testing Poisson goodness of fit, cells with small λ may need
💡 Explanation:Low expected Poisson counts require pooling.
- Q39Past Paper · PPSC/FPSC/CSSmedium
χ² test p-value is area to the right of observed χ² under
💡 Explanation:Right-tail p-value for chi-square statistic.
- Q40Past Paper · PPSC/FPSC/CSSmedium
Before chi-square, categories should be
💡 Explanation:Each observation falls in exactly one cell.