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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Page 1 of 1 Questions 110 of 40
  1. Q1 Past Paper · PPSC/FPSC/CSS easy

    Pearson chi-square statistic for goodness of fit is

    1. A χ² = Σ[(Oi − Ei)²/Ei]
    2. B Σ(Oi·Ei)
    3. C Σ(Oi−Ei)
    4. D Oi/Ei only
    💡 Explanation:

    Compares observed Oi to expected Ei frequencies.

  2. Q2 hard

    For testing Poisson goodness of fit, cells with small λ may need

    1. A combining categories to ensure adequate Ei
    2. B ignoring df
    3. C using z-test on means only
    4. D dropping χ²
    💡 Explanation:

    Low expected Poisson counts require pooling.

  3. Q3 Past Paper · PPSC/FPSC/CSS medium

    χ²_{0.05, 1} is approximately

    1. A 1.96
    2. B 6.635
    3. C 3.841
    4. D 0
    💡 Explanation:

    Common critical value for 2×2 at α = 0.05.

  4. Q4 Past Paper · PPSC/FPSC/CSS medium

    χ² test p-value is area to the right of observed χ² under

    1. A normal with n−1 df
    2. B t distribution always
    3. C χ² distribution with appropriate df
    4. D uniform
    💡 Explanation:

    Right-tail p-value for chi-square statistic.

  5. Q5 Past Paper · PPSC/FPSC/CSS medium

    Before chi-square, categories should be

    1. A overlapping to boost n
    2. B without totals
    3. C mutually exclusive and collectively exhaustive
    4. D always continuous
    💡 Explanation:

    Each observation falls in exactly one cell.

  6. Q6 hard

    Phi coefficient is special case of association measure for

    1. A 10×10 tables only
    2. B continuous variables
    3. C paired t tests
    4. D 2×2 tables
    💡 Explanation:

    φ = √(χ²/n) for fourfold tables.

  7. Q7 hard

    Cramér's V measures

    1. A only Type I error
    2. B only expected counts
    3. C strength of association in contingency tables
    4. D only regression R²
    💡 Explanation:

    V = √[χ²/(n·min(r−1,c−1))].

  8. Q8 hard

    Simpson's paradox can appear in

    1. A only continuous regression
    2. B contingency tables when marginal and conditional associations differ
    3. C only one cell tables
    4. D only goodness of fit
    💡 Explanation:

    Aggregated and stratified tables may show opposite associations.

  9. Q9 hard

    Likelihood-ratio chi-square G² is

    1. A an alternative statistic that also approaches χ² distribution asymptotically
    2. B always negative
    3. C equal to t² always
    4. D unrelated to contingency tables
    💡 Explanation:

    G² = 2ΣOi ln(Oi/Ei) parallels Pearson χ².

  10. Q10 Past Paper · PPSC/FPSC/CSS easy

    If all Oi equal Ei exactly, χ² equals

    1. A 1
    2. B df
    3. C undefined
    4. D 0
    💡 Explanation:

    Perfect agreement yields zero statistic.

  11. Q11 Past Paper · PPSC/FPSC/CSS medium

    For 4×3 contingency table, minimum df is

    1. A 5
    2. B 11
    3. C 6
    4. D 2
    💡 Explanation:

    (4−1)(3−1) = 6.

  12. Q12 Past Paper · PPSC/FPSC/CSS easy

    Chi-square test is classified as

    1. A a test on paired normal means only
    2. B a test on categorical/count data
    3. C only for regression slopes
    4. D only for time series trend
    💡 Explanation:

    χ² methods apply to frequencies and contingency tables.

  13. Q13 Past Paper · PPSC/FPSC/CSS easy

    Expected counts under independence sum to

    1. A zero
    2. B each Oi
    3. C df
    4. D grand total n in the full table
    💡 Explanation:

    ΣΣEij = N.

  14. Q14 Past Paper · PPSC/FPSC/CSS medium

    Rejecting independence implies

    1. A causation is proven
    2. B evidence of association between the two categorical variables
    3. C means are different
    4. D variances differ
    💡 Explanation:

    Association does not alone establish causality.

  15. Q15 hard

    Goodness-of-fit to binomial with unknown p estimates p from data, reducing df by

    1. A 0
    2. B 1
    3. C 2
    4. D k
    💡 Explanation:

    Estimating p uses one degree of freedom.

  16. Q16 medium

    When r = 1 or c = 1, independence test is

    1. A df = 0 still valid
    2. B χ² = 0 always
    3. C use paired t
    4. D not applicable — need at least 2×2
    💡 Explanation:

    Both dimensions need at least two categories.

  17. Q17 Past Paper · PPSC/FPSC/CSS medium

    Contribution of one cell to χ² is

    1. A (Oi−Ei)²/Ei
    2. B Oi·Ei
    3. C (Oi−Ei)
    4. D √Ei only
    💡 Explanation:

    Each cell adds a squared standardized component.

  18. Q18 Past Paper · PPSC/FPSC/CSS medium

    Pearson chi-square requires

    1. A random sampling and mutually exclusive categories
    2. B normal distribution of each cell
    3. C equal variances of means
    4. D paired differences normal
    💡 Explanation:

    Multinomial/count model assumptions apply.

  19. Q19 Past Paper · PPSC/FPSC/CSS medium

    In 5×2 table, df for independence equals

    1. A 9
    2. B 5
    3. C 4
    4. D 1
    💡 Explanation:

    (5−1)(2−1) = 4.

  20. Q20 Past Paper · PPSC/FPSC/CSS easy

    Larger discrepancy between O and E produces

    1. A smaller χ²
    2. B df = 0
    3. C automatic acceptance
    4. D larger χ² value
    💡 Explanation:

    Big relative errors inflate the statistic.

  21. Q21 Past Paper · PPSC/FPSC/CSS easy

    χ² statistic cannot be negative because

    1. A Oi are always negative
    2. B it sums squared standardized differences
    3. C Ei are zero always
    4. D df is negative
    💡 Explanation:

    Squared terms make χ² ≥ 0.

  22. Q22 medium

    Pooling adjacent categories may be done when

    1. A Ei are all large
    2. B expected counts are too small to meet chi-square rules
    3. C df is already maximum
    4. D H0 is certainly true
    💡 Explanation:

    Combining categories raises Ei at cost of detail.

  23. Q23 Past Paper · PPSC/FPSC/CSS medium

    Chi-square homogeneity test compares

    1. A distribution of a categorical variable across several populations
    2. B means of normal variables only
    3. C variances only
    4. D correlation only
    💡 Explanation:

    Homogeneity: same categories, different groups — similar mechanics to independence.

  24. Q24 hard

    McNemar's test applies to

    1. A paired nominal data in 2×2 table (before–after)
    2. B independent two samples proportion z-test
    3. C r×c independence with large n
    4. D goodness of fit to normal
    💡 Explanation:

    McNemar uses discordant pairs on diagonal-off cells.

  25. Q25 Past Paper · PPSC/FPSC/CSS hard

    Fisher's exact test is used for

    1. A large r×c tables always
    2. B continuous normal data
    3. C 2×2 tables with small expected counts
    4. D paired t situations
    💡 Explanation:

    Exact test avoids chi-square approximation issues.

  26. Q26 hard

    Standardized residual for cell (i,j) is approximately

    1. A Oi−Eij without square root
    2. B (Oi − Eij)/√Eij
    3. C Eij/Oi
    4. D χ²/df
    💡 Explanation:

    Large standardized residuals flag cells contributing to χ².

  27. Q27 Past Paper · PPSC/FPSC/CSS easy

    Test of independence H0 states

    1. A equal means across groups
    2. B variance ratio equals 1
    3. C π = 0.5
    4. D no association between row and column variables
    💡 Explanation:

    Independence: P(row i, col j) = P(row i)·P(col j).

  28. Q28 Past Paper · PPSC/FPSC/CSS medium

    For goodness of fit to equally likely k categories, each Ei equals

    1. A n·pi with unknown pi
    2. B Oi
    3. C n/k
    4. D k/n
    💡 Explanation:

    Uniform H0: Ei = n/k.

  29. Q29 Past Paper · PPSC/FPSC/CSS medium

    Chi-square distribution is

    1. A symmetric like normal always
    2. B negative for large df
    3. C right-skewed and defined for positive values only
    4. D identical to t
    💡 Explanation:

    χ² with ν df has mean ν and variance 2ν.

  30. Q30 Past Paper · PPSC/FPSC/CSS easy

    Marginal totals in contingency table are

    1. A only diagonal cells
    2. B only χ² statistic
    3. C only expected counts
    4. D row sums and column sums
    💡 Explanation:

    Margins Ri and Cj define Ei under independence.

  31. Q31 Past Paper · PPSC/FPSC/CSS easy

    Observed frequencies in contingency table are

    1. A always equal to expected under H0
    2. B the p-value
    3. C the actual sample counts in each cell
    4. D the degrees of freedom
    💡 Explanation:

    Oi are data; Ei computed under H0.

  32. Q32 Past Paper · PPSC/FPSC/CSS easy

    If χ² calculated exceeds χ²_{α, df}, decision at level α is

    1. A fail to reject always
    2. B accept with certainty
    3. C use t-test instead
    4. D reject H0 of independence (or goodness of fit)
    💡 Explanation:

    Large χ² relative to critical value contradicts H0.

  33. Q33 Past Paper · PPSC/FPSC/CSS hard

    Yates continuity correction for 2×2 tables

    1. A adds 1 to every cell
    2. B doubles all df
    3. C replaces χ² with F
    4. D subtracts 0.5 from |Oi−Ei| before squaring (in standard formula)
    💡 Explanation:

    Yates improves approximation for small 2×2 samples.

  34. Q34 Past Paper · PPSC/FPSC/CSS medium

    Chi-square approximation is unreliable when

    1. A all Ei exceed 100
    2. B many expected cell counts are less than 5
    3. C n is large with adequate Ei
    4. D only r = 2
    💡 Explanation:

    Small Ei violate asymptotic chi-square assumptions.

  35. Q35 Past Paper · PPSC/FPSC/CSS medium

    A 3×4 contingency table has df for independence test equal

    1. A 6
    2. B 11
    3. C 7
    4. D 3
    💡 Explanation:

    (3−1)(4−1) = 6.

  36. Q36 Past Paper · PPSC/FPSC/CSS easy

    A 2×2 contingency table for independence has

    1. A 1 degree of freedom
    2. B 0
    3. C 2
    4. D 3
    💡 Explanation:

    (2−1)(2−1) = 1.

  37. Q37 Past Paper · PPSC/FPSC/CSS easy

    Expected frequency under independence in cell (i,j) is

    1. A Oi only
    2. B (row i total × column j total) / grand total
    3. C row total only
    4. D column total − row total
    💡 Explanation:

    Eij = Ri·Cj/N under H0 of independence.

  38. Q38 Past Paper · PPSC/FPSC/CSS easy

    Chi-square test of independence in an r×c contingency table has df

    1. A r + c − 1
    2. B rc − 1
    3. C n − 1
    4. D (r − 1)(c − 1)
    💡 Explanation:

    Independence df is product of (rows−1) and (cols−1).

  39. Q39 Past Paper · PPSC/FPSC/CSS medium

    In a goodness-of-fit test with k categories, df equal

    1. A k − 1 − (number of parameters estimated from data)
    2. B k
    3. C k + 1
    4. D n − 1 always
    💡 Explanation:

    If all probabilities specified, df = k − 1.

  40. Q40 Past Paper · PPSC/FPSC/CSS easy

    Goodness-of-fit test evaluates whether

    1. A two categorical variables are independent only
    2. B sample data conform to a specified theoretical distribution
    3. C means of two groups are equal
    4. D σ² is zero
    💡 Explanation:

    GoF tests multinomial counts against hypothesized probabilities.