Chi-Square Tests and Contingency Tables MCQs 2026
40 questions with detailed answers · 30 from past papers · 4 quiz batches available
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- Q1 Past Paper · PPSC/FPSC/CSS easy
Pearson chi-square statistic for goodness of fit is
💡 Explanation:Compares observed Oi to expected Ei frequencies.
- Q2 hard
For testing Poisson goodness of fit, cells with small λ may need
💡 Explanation:Low expected Poisson counts require pooling.
- Q3 Past Paper · PPSC/FPSC/CSS medium
χ²_{0.05, 1} is approximately
💡 Explanation:Common critical value for 2×2 at α = 0.05.
- Q4 Past Paper · PPSC/FPSC/CSS medium
χ² test p-value is area to the right of observed χ² under
💡 Explanation:Right-tail p-value for chi-square statistic.
- Q5 Past Paper · PPSC/FPSC/CSS medium
Before chi-square, categories should be
💡 Explanation:Each observation falls in exactly one cell.
- Q6 hard
Phi coefficient is special case of association measure for
💡 Explanation:φ = √(χ²/n) for fourfold tables.
- Q7 hard
Cramér's V measures
💡 Explanation:V = √[χ²/(n·min(r−1,c−1))].
- Q8 hard
Simpson's paradox can appear in
💡 Explanation:Aggregated and stratified tables may show opposite associations.
- Q9 hard
Likelihood-ratio chi-square G² is
💡 Explanation:G² = 2ΣOi ln(Oi/Ei) parallels Pearson χ².
- Q10 Past Paper · PPSC/FPSC/CSS easy
If all Oi equal Ei exactly, χ² equals
💡 Explanation:Perfect agreement yields zero statistic.
- Q11 Past Paper · PPSC/FPSC/CSS medium
For 4×3 contingency table, minimum df is
💡 Explanation:(4−1)(3−1) = 6.
- Q12 Past Paper · PPSC/FPSC/CSS easy
Chi-square test is classified as
💡 Explanation:χ² methods apply to frequencies and contingency tables.
- Q13 Past Paper · PPSC/FPSC/CSS easy
Expected counts under independence sum to
💡 Explanation:ΣΣEij = N.
- Q14 Past Paper · PPSC/FPSC/CSS medium
Rejecting independence implies
💡 Explanation:Association does not alone establish causality.
- Q15 hard
Goodness-of-fit to binomial with unknown p estimates p from data, reducing df by
💡 Explanation:Estimating p uses one degree of freedom.
- Q16 medium
When r = 1 or c = 1, independence test is
💡 Explanation:Both dimensions need at least two categories.
- Q17 Past Paper · PPSC/FPSC/CSS medium
Contribution of one cell to χ² is
💡 Explanation:Each cell adds a squared standardized component.
- Q18 Past Paper · PPSC/FPSC/CSS medium
Pearson chi-square requires
💡 Explanation:Multinomial/count model assumptions apply.
- Q19 Past Paper · PPSC/FPSC/CSS medium
In 5×2 table, df for independence equals
💡 Explanation:(5−1)(2−1) = 4.
- Q20 Past Paper · PPSC/FPSC/CSS easy
Larger discrepancy between O and E produces
💡 Explanation:Big relative errors inflate the statistic.
- Q21 Past Paper · PPSC/FPSC/CSS easy
χ² statistic cannot be negative because
💡 Explanation:Squared terms make χ² ≥ 0.
- Q22 medium
Pooling adjacent categories may be done when
💡 Explanation:Combining categories raises Ei at cost of detail.
- Q23 Past Paper · PPSC/FPSC/CSS medium
Chi-square homogeneity test compares
💡 Explanation:Homogeneity: same categories, different groups — similar mechanics to independence.
- Q24 hard
McNemar's test applies to
💡 Explanation:McNemar uses discordant pairs on diagonal-off cells.
- Q25 Past Paper · PPSC/FPSC/CSS hard
Fisher's exact test is used for
💡 Explanation:Exact test avoids chi-square approximation issues.
- Q26 hard
Standardized residual for cell (i,j) is approximately
💡 Explanation:Large standardized residuals flag cells contributing to χ².
- Q27 Past Paper · PPSC/FPSC/CSS easy
Test of independence H0 states
💡 Explanation:Independence: P(row i, col j) = P(row i)·P(col j).
- Q28 Past Paper · PPSC/FPSC/CSS medium
For goodness of fit to equally likely k categories, each Ei equals
💡 Explanation:Uniform H0: Ei = n/k.
- Q29 Past Paper · PPSC/FPSC/CSS medium
Chi-square distribution is
💡 Explanation:χ² with ν df has mean ν and variance 2ν.
- Q30 Past Paper · PPSC/FPSC/CSS easy
Marginal totals in contingency table are
💡 Explanation:Margins Ri and Cj define Ei under independence.
- Q31 Past Paper · PPSC/FPSC/CSS easy
Observed frequencies in contingency table are
💡 Explanation:Oi are data; Ei computed under H0.
- Q32 Past Paper · PPSC/FPSC/CSS easy
If χ² calculated exceeds χ²_{α, df}, decision at level α is
💡 Explanation:Large χ² relative to critical value contradicts H0.
- Q33 Past Paper · PPSC/FPSC/CSS hard
Yates continuity correction for 2×2 tables
💡 Explanation:Yates improves approximation for small 2×2 samples.
- Q34 Past Paper · PPSC/FPSC/CSS medium
Chi-square approximation is unreliable when
💡 Explanation:Small Ei violate asymptotic chi-square assumptions.
- Q35 Past Paper · PPSC/FPSC/CSS medium
A 3×4 contingency table has df for independence test equal
💡 Explanation:(3−1)(4−1) = 6.
- Q36 Past Paper · PPSC/FPSC/CSS easy
A 2×2 contingency table for independence has
💡 Explanation:(2−1)(2−1) = 1.
- Q37 Past Paper · PPSC/FPSC/CSS easy
Expected frequency under independence in cell (i,j) is
💡 Explanation:Eij = Ri·Cj/N under H0 of independence.
- Q38 Past Paper · PPSC/FPSC/CSS easy
Chi-square test of independence in an r×c contingency table has df
💡 Explanation:Independence df is product of (rows−1) and (cols−1).
- Q39 Past Paper · PPSC/FPSC/CSS medium
In a goodness-of-fit test with k categories, df equal
💡 Explanation:If all probabilities specified, df = k − 1.
- Q40 Past Paper · PPSC/FPSC/CSS easy
Goodness-of-fit test evaluates whether
💡 Explanation:GoF tests multinomial counts against hypothesized probabilities.