Math 115: Introduction to Statistics – 14. Inference for Contingency Tables

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.

So how does the chi-square statistic turn that whole table into a single number? It measures how far the observed counts fall from those expected counts, added up across every cell. The larger the chi-square, the bigger the mismatch — the stronger the evidence against the null. A smaller one means the table is close to what we’d expect if the null holds. For the military table, chi-square came out to about eighteen point nine seven. Today’s question is the natural follow-up. Is eighteen point nine seven big enough to reject the independence null? We can’t answer that on its own. We need a reference distribution — the typical values of the chi-square we’d see across many samples if the null were really true. That benchmark is the chi-square distribution, and it’s what the rest of this lecture builds out.

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 \(H_0\).

The last slide left us needing a reference distribution. The chi-square distribution is exactly that — the shape that the chi-square statistic follows when the null is true, when the conditions we’ll see next are met. The plot shows three curves with different shape parameters: a red curve for two degrees of freedom, a blue curve for four, and a green curve for nine. Four properties to notice. First, all three curves stay on the positive side of zero — chi-square is a sum of squared terms, so it can never go negative. Second, every curve is right-skewed, with a single shape parameter called the degrees of freedom. Third, as the degrees of freedom grow, the curve slides to the right — typical values of the chi-square get bigger — and the right skew softens. Fourth, this is the same right-tail logic we just set up: the bigger the chi-square, the farther into the right tail, and the stronger the evidence against the null.

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.

Use the chi-square distribution with four degrees of freedom as the reference. The chi-square we observed, about eighteen point nine seven, sits far out in the right tail, and the p-value is the area beyond that observed value — the chance, if the null were true, that a random sample would produce a chi-square at least as large as eighteen point nine seven. One important point on tails. The chi-square alternative is two-sided in spirit — departures from independence count in any cell, in any direction. But because every gap is squared in the formula, every kind of departure ends up pushing the chi-square value up rather than down. So all of that two-sided evidence collapses into the right tail of the chi-square curve. The p-value uses only that tail — no doubling like in previous topics. Using Jamovi, the p-value is about zero point zero zero zero eight, well below the standard zero point zero five threshold.

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.

Putting the result together. Chi-square is about eighteen point nine seven, with four degrees of freedom and a p-value of about zero point zero zero zero eight. That’s well below zero point zero five, so we reject the null. In plain terms — among U.S. adults, opinion on military spending is associated with political party. From the cell-by-cell contributions in fourteen A, Republicans were much more “too little” and much less “too much” than the overall opinion mix would predict — more supportive of military spending. Independents leaned the opposite way more mildly — less supportive. Democrats sat almost exactly at the overall mix. So the association mostly comes from Republicans pulling hard in one direction and Independents pulling more gently in the other. On scope of inference. The General Social Survey uses a random sample of U.S. adults, so the conclusion generalizes to the U.S. adult population in twenty sixteen. But neither variable was assigned — they were just measured — so this is observational data and there’s no causal claim. We’re describing a real population-level association between party and military-spending opinion, not saying party “causes” the opinion or the other way around. One last caution. The chi-square test answers “is there any association at all?” — a yes-or-no question about the table as a whole. It doesn’t tell you which specific groups differ from each other, or in which direction. Looking at the cell-by-cell contributions can suggest where the mismatch is largest, but that’s only descriptive — it isn’t a formal test. Pinning down which groups really differ takes a follow-up: the pairwise comparisons in topic sixteen.

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.

Now your turn on a second example. The same one hundred forty-nine General Social Survey respondents, but a different opinion question — this time, “do we spend too little, about right, or too much on space exploration?” The new three by three table has the same parties down the rows and the same opinion levels across the columns. Pre-computed for you: chi-square is about one point three three, and degrees of freedom is four — the same as the military table since the table is still three by three. This time, use the shortcut yourself to find the smallest expected count. State the hypotheses and check the chi-square conditions before advancing. Hypotheses are the same shape as for the military example, just with a new response variable. Null: party and space-spending opinion are independent. Alternative: they are associated. Conditions both check out. Independence holds because we have the same random sample of U.S. adults. Sample size: using the shortcut, the smallest row total is Republicans at thirty-four and the smallest column total is “too little” at thirty, so the smallest expected count is thirty-four times thirty over one hundred forty-nine — about six point eight five, above five — so all nine cells clear the threshold. The chi-square distribution with four degrees of freedom is a valid reference for the new test.

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.

Time to read the result off the chi-square curve. The shaded region to the right of the chi-square we observed, one point three three, is the p-value. Remember, the total area under the whole curve is one. Use the plot to decide whether the p-value is below the zero point zero five significance level, then reach a conclusion before advancing. The shaded right-tail area is the p-value, and it’s a large share of the total area of one — nowhere near as small as zero point zero five. So the p-value is above zero point zero five; in fact it’s about zero point eight six — most random samples under the null would produce a chi-square at least this big, just by chance. We fail to reject the null. No evidence that party is associated with opinion on space spending. On scope: same random sample of U.S. adults, so the conclusion generalizes to the U.S. adult population in twenty sixteen — but still observational, so no causal claim. Notice the framing of “fail to reject” — we’re not concluding that party and space opinion are independent, only that there isn’t evidence to rule out independence in this sample.

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.

The chi-square test is the natural generalization of the two-proportion test from topic ten. In topic ten the table was always two by two and the statistic was the difference of two p hats, compared to a standard normal. Today the table can be any size, the statistic is now chi-square, and the reference is the chi-square distribution with degrees of freedom set by the table dimensions. We still test whether the two variables move together, but now phrased as independence rather than as a single difference in proportions. The biggest practical change is the tail use. In topic ten we could choose a one-sided or two-sided alternative depending on the question, and a two-sided p-value used both tails of the normal curve. With the chi-square test, the alternative is non-directional — two-sided in spirit, like a two-sided topic-ten test — but because every cell’s gap gets squared, all of that two-sided evidence collapses into the right tail of the chi-square curve. So we use the right tail only, with no doubling. Nice confirmation: for a two by two table specifically, the chi-square p-value matches the two-sided topic-ten p-value exactly. Nothing is lost by going to chi-square; we just gain the ability to handle 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.