Two-Way Tables: Homogeneity and Independence — Solutions and Rubrics

Use a chi-square test for homogeneity: independent samples were selected from three town populations and one categorical variable, primary commute choice, is compared. H₀: the walk/bus/drive distributions are the same across towns A, B and C. Hₐ: the distributions are not all the same across these town populations. The alternative need not mean every town or every category differs.

For each cell E=(row total)(column total)/(table total)=90(90)/270=30, so the expected table has 30 in all nine cells. The independent SRSs establish randomization. Each sample 90 is at most 10% of its town population (90≤500 even at the stated minimum). Every expected count 30 is strictly greater than 5, satisfying the official expected-count condition. The condition uses the null expected counts, not the observed 15, 30 or 45.

Cells with O=45 or 15 contribute (15)²/30=7.5 each; there are four. Five O=30 cells contribute zero. Thus χ²=4(7.5)=30. df=(3−1)(3−1)=4. Under H₀ the statistic is approximately chi-square with df=4; p=P(χ²₄≥30)=0.00000489444. The sum measures observed-versus-expected squared discrepancy relative to expected counts. Values are nonnegative and large values are the evidence against H₀; do not double the upper tail. Chi-square curves are right-skewed; skew diminishes as df increases.

If all three town populations had the same commute distribution, the probability of obtaining a chi-square discrepancy at least 30 from these independent samples is approximately 0.00000489444. Because this is ≤0.05, reject H₀. There is convincing statistical evidence that the primary commute distributions are not all the same across working adults in towns A, B and C. The omnibus test alone does not say which specific town/category differences are responsible or establish that town residence caused them.

Use chi-square independence: one SRS from one city measures dwelling type and park-visit frequency on every person. H₀: no association between the variables among city residents; equivalently, the variables are independent in that population. Hₐ: an association exists; equivalently, they are not independent. Separate row labels do not turn this single sample into two independently sampled populations.

Each row’s expected counts are 100(80)/200=40, 100(70)/200=35 and 100(50)/200=25. SRS gives randomization; 200≤2,500 meets 10%; all expected counts (40,35,25 in both rows) are >5. Contributions per row: (50−40)²/40 or (30−40)²/40=2.5; squared 5 difference over 35=0.7142857; squared 5 difference over 25=1. Thus χ²=2(2.5+0.7142857+1)=8.428571. df=(2−1)(3−1)=2; p≈0.01478288, upper tail.

Assuming dwelling type and visit frequency are independent among city residents, the chance of a sample table with χ² at least 8.428571 is about 0.01478. Since 0.01478>0.01, fail to reject H₀: there is not convincing evidence of an association at the 1% level. This does not establish independence or rule out a real association. At a prespecified α=0.05 the same p≤0.05 would reject H₀ and provide evidence of association. The threshold changes the decision, not the observed data or p-value.

A supports a homogeneity analysis across randomized treatments; random assignment supplies randomization, the sampling-without-replacement condition is unnecessary for the experiment, and each expected count is >5. Its volunteers do not establish population-wide representativeness; random assignment can support a treatment effect for these participants under the design. B is an independence design if it records two categorical variables per household, but 80>0.10(400)=40 violates the sampling condition despite adequate counts. C is a homogeneity design with randomization and 10% satisfied, but the expected count exactly 5 fails the required strictly-greater-than-5 condition. D’s numerical travel-time mean is not a categorical count distribution; a mean method may fit, subject to its own design/conditions.

χ² adds (O−E)²/E over all cells; positive expected counts make contributions nonnegative. Its distribution is positive/nonnegative and skewed right. At df=10 the skew is less pronounced than at df=2. Large upper-tail values represent more discrepancy, so the p-value is an upper-tail probability. A negative sum or a lower-tail p-value contradicts the construction and direction of evidence.