14. Inference for Contingency Tables

Extending Our Framework

Previously, we compared two groups with a binary outcome (2×2 table).

• Used difference in proportions: \(\hat{p}_1 - \hat{p}_2\)
• Used Z-statistic for hypothesis testing

But what if we have:

• More than 2 groups? (e.g., Dem, Ind, Rep)
• Non-binary outcomes? (e.g., Spending: Too Little, About Right, Too Much)

We need a new approach!

Two Ways to Generalize

Generalization 1: More groups

• Explanatory variable has 3+ categories
• Example: Democrat, Independent, Republican

Generalization 2: Non-binary response

• Response variable has 3+ categories
• Example (Government Spending): “Too Little”, “About Right”, “Too Much”

With either generalization, we can’t use a single difference in proportions!

The Hypothesis Framework

Same concept as before, different statistic:

• \(H_0\): No association between the variables (independence)
• \(H_A\): There is an association (variables are not independent)

New statistic: Chi-square, \(X^2\), measures deviation from independence.

What Would We Expect Under \(H_0\)?

If \(H_0\) is true (no association), each party should have the same distribution of opinions.

Expected count for any cell:

\[\text{Expected} = \frac{\text{row total} \times \text{column total}}{\text{grand total}}\]

This distributes the overall opinion proportions equally across all parties.

The Chi-Square Statistic

Measures total deviation of observed from expected:

\[X^2 = \sum_{\text{all cells}} \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}}\]

Why this formula?

• Squaring: Makes all contributions positive
• Dividing by Expected: Standardizes (a difference of 5 matters more when Expected is 10 than when Expected is 100)

What Does \(X^2\) Tell Us?

• \(X^2 = 0\) means observed = expected exactly (perfect independence)
• Larger \(X^2\) means greater deviation from independence
• Larger \(X^2\) → stronger evidence against \(H_0\)

But how large is “large enough” to reject \(H_0\)?

We need a reference distribution → chi-square distribution

Degrees of Freedom

For contingency tables:

\[df = (\text{rows} - 1) \times (\text{columns} - 1)\]

For our 3×3 table:

\[df = (3-1) \times (3-1) = 2 \times 2 = 4\]

Intuition: After fixing row and column totals, only \(df\) cells can vary freely.

The Chi-Square Distribution

• Always right-skewed, starts at 0
• As df increases, distribution shifts right

The Null Distribution

When conditions are met, we can use the chi-square distribution as our reference:

• Use the chi-square distribution with \(df = (r-1) \times (c-1)\)
• This distribution describes what \(X^2\) values we’d expect if \(H_0\) is true

Computing the p-value:

• P-value = area to the right of observed \(X^2\) under the chi-square curve
• Larger \(X^2\) values are more extreme → stronger evidence against \(H_0\)

Conditions for Chi-Square Test

Independence:

• Random sample ✓
• Observations independent of each other ✓

Expected Counts (under \(H_0\)):

• At least 5 expected in each cell ✓

Check expected counts, not observed counts!

Randomization Test for Independence

• We can randomly permute the response (opinion) between the three explanatory groups (party)
  • To simulate the null hypothesis being true
• For each permuted sample, we calculate value of the \(X^2\) statistic
• In this course we won’t cover randomization for contingency tables
  • But you should know that you can use randomization for these problems

Summary

Chi-square test extends two-proportion test to larger tables:

Conditions: Independence + Expected counts ≥ 5 in each cell

Connection to Topic 10

Chi-square generalizes the two-proportion test to larger tables!

(For 2×2 tables, both methods give equivalent results)

References

• Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin
• Section 18.2