14. Inference for Contingency Tables

b. Chi-Square Inference
Math 115

Yurk

Recap: Observed vs Expected

14a — GSS 2016, party × military-spending opinion, a 3×3 table of 149 U.S. adults. Each cell: observed count on top, expected count (if H_0 / independence held) below.

Party	Too Little	About Right	Too Much	Total
Dem	17 (17.6)	14 (14.4)	12 (11)	43
Ind	20 (29.5)	28 (24.2)	24 (18.4)	72
Rep	24 (13.9)	8 (11.4)	2 (8.7)	34
Total	61	50	38	149

H_0: party and military-spending opinion are independent · H_A: they are associated.

From Statistic to Decision

Recall the logic (14a). X^2 = \sum_{\text{all cells}} \frac{(O - E)^2}{E} adds up how far each observed count falls from its expected count, across the whole table:

Larger X^2 = bigger mismatch → stronger evidence against H_0.
Smaller X^2 = table close to what H_0 predicts.

For the military table, X^2 \approx 18.97.

Today’s question: is 18.97 “big enough” to reject H_0? We need a reference distribution — the X^2 values we’d expect if H_0 were true.

The Chi-Square Distribution

When conditions are met (next slide), X^2 under H_0 follows a chi-square distribution with degrees of freedom set by the table size.

Always non-negative — X^2 is a sum of squared standardized gaps.
Right-skewed with a single shape parameter, df.
As df grows, the curve slides right (typical values get larger) and grows less skewed.
Bigger X^2 → farther into the right tail → stronger evidence against H_0.

Degrees of Freedom for r×c Tables

For an r \times c contingency table:

df = (r - 1) \times (c - 1)

Intuition. Once the row and column totals are fixed, only (r-1)(c-1) cells can vary freely — the rest are pinned down by the totals.

Try it: what’s df for the 3×3 military table (3 parties × 3 opinion levels)?

df = (3 - 1) \times (3 - 1) = 2 \times 2 = 4. So we compare X^2 = 18.97 to a chi-square distribution with df = 4.

Conditions for the Chi-Square Test

Two conditions before we trust the chi-square distribution as the reference:

Condition	Rule
Independence	Each individual independent of the others — a random sample or random assignment, otherwise justified from study context.
Sample size	At least 5 expected counts in every cell. Check E (expected), not the observed counts.

Military table. Independence holds — GSS draws a random sample of U.S. adults ✓. The sample-size condition needs the expected counts; rather than compute all nine, we’ll find just the smallest one next.

Checking Sample Size: a Shortcut

Every cell needs expected ≥ 5. Since \text{Expected} = \dfrac{\text{row} \times \text{col}}{\text{grand}}, the smallest expected count sits where the row and column totals are both smallest — check one cell, not nine.

	Too Little	About Right	Too Much	Total
Dem	17	14	12	43
Ind	20	28	24	72
Rep	24	8	2	34
Total	61	50	38	149

Smallest row total: Rep = 34 (orange).

Smallest column total: Too Much = 38 (blue).

Rep × Too Much — observed count here is just 2.

Expected: E_{\min} = \dfrac{34 \times 38}{149} \approx 8.67 \ge 5 ✓ — every cell clears.

The P-Value for Military

P-value = area to the right of X^2 = 18.97 under the chi-square curve with df = 4.
The alternative is two-sided in spirit — departures from H_0 count in any cell, in any direction — but squaring sends every gap to a positive contribution, so all of that evidence lands in the right tail. Use the right tail only — no doubling.
Using Jamovi: p-value \approx 8\times 10^{-4} — far below 0.05.

Conclusion for Military Spending

Results: X^2 = 18.97, df = 4, p-value \approx 8\times 10^{-4} — well below 0.05 → reject H_0.

In context: Among U.S. adults, opinion on military spending is associated with political party. Republicans were much more “Too Little” and much less “Too Much” than expected (more supportive of military spending); Independents leaned the opposite way more mildly (less supportive).

Scope of inference.

Generalization: random sample of U.S. adults → generalizes to U.S. adults (2016).
Causation: observational (variables measured, not assigned) → no causal claim, just a population-level association.

Heads up: the chi-square test answers “is there any association?” — not which groups differ or which direction. The cell contributions only suggest where the mismatch is largest; formally pinning down which groups differ waits for the pairwise follow-up tests in topic 16.

Space Exploration — Setup

Same 149 GSS respondents, same parties — does opinion on space exploration depend on party? New 3×3 table:

Party	Too Little	About Right	Too Much	Total
Dem	8	22	13	43
Ind	13	37	22	72
Rep	9	17	8	34
Total	30	76	43	149

Computed for you: X^2 \approx 1.33, df = 4.

State H_0 and H_A, then use the shortcut to find the smallest expected count and check both conditions. H_0: party and space-spending opinion are independent. H_A: they are associated. Independence: GSS random sample of U.S. adults ✓. Sample size (shortcut): smallest row total is Rep (34), smallest column total is Too Little (30), so E_{\text{min}} = \frac{34 \times 30}{149} \approx 6.85 \geq 5 ✓. Both conditions met.

Space Exploration — P-Value & Conclusion

Recall: X^2 \approx 1.33, df = 4. The shaded region to the right of X^2 is the p-value.

The total area under the curve is 1. Using the figure, is the p-value less than the \alpha = 0.05 significance level? Conclude (include scope). X^2 \approx 1.33 sits in the bulk of the df = 4 curve, so the shaded area is a large share of the total area of 1 — nowhere near as small as 0.05. So the p-value > 0.05 (in fact \approx 0.86). Fail to reject H_0. No evidence that party is associated with opinion on space spending. Scope: GSS random sample → conclusion generalizes to U.S. adults (2016); observational, so no causation.

Same Sample, Different Stories

Question	X^2	df	p-value	Decision (α = 0.05)
Military spending vs party	18.97	4	\approx 8\times 10^{-4}	Reject H_0 — clear association
Space spending vs party	1.33	4	\approx 0.86	Fail to reject H_0 — no evidence of association

Same 149 people, same three parties, same three opinion levels — but the topic changes everything. Party views split on military spending; party views look indistinguishable on space spending.

The chi-square test answers each question on its own — for these variables, is there an association at all?

Connecting to Topic 10

The chi-square test generalizes the two-proportion test from topic 10:

	Two Proportions (Topic 10)	Chi-Square Test (Topic 14)
Table size	2 × 2 only	Any r \times c
Statistic	\hat{p}_1 - \hat{p}_2	X^2
Reference distribution	Standard normal	Chi-square, df = (r-1)(c-1)
What we test	p_1 - p_2 = 0	Independence
Tail(s) used for p-value	One- or two-sided, depending on H_A	Always right tail

For a 2×2 table the chi-square test is equivalent to a two-sided topic-10 test — they give the same p-value and same decision. Chi-square is the right tool once the table grows beyond 2×2, because no single “difference in proportions” can summarize bigger tables.

Looking Ahead

We’ve added the chi-square test to the toolkit — for two categorical variables of any size, asking whether the two are associated.

Next: multiple means. Topic 15 tackles the matching question for numerical data — given three or more groups, are any of their mean responses different? — using the F-statistic and ANOVA.

References

Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin
Section 18.2
General Social Survey, 2016 — National Opinion Research Center.