Unit 8: Inference for Categorical Data: Chi-Square

Chi-Square Test for Goodness-of-Fit

• Goodness-of-Fit: Evaluate the fit of a theoretical distribution to observed data.
• Null Hypothesis: The given theoretical distribution correctly describes the situation.
• Chi-Square Statistic (𝜒²): Measures the sum of weighted differences between observed and expected frequencies.
  • The smaller the 𝜒²-value, the better the fit.
  • P-value: Probability of obtaining a 𝜒²-value as extreme as the one observed if the null hypothesis is true.
  • A large 𝜒²-value indicates a poor fit and leads to rejecting the null hypothesis.
• Chi-Square Distribution:
  • Nonnegative values, not symmetric, skewed to the right.
  • Different distributions for different degrees of freedom (df).
  • Larger df results in a distribution closer to normal.

Example 8.1

• Scenario: Distribution of liquor stores across four city regions.
• Hypotheses:
  • H₀: Liquor stores are distributed according to region area percentages.
  • Hₐ: Distribution differs from region area percentages.
• Procedure: Chi-square test for Goodness-of-fit.
• Checks:
  • Random sample and expected values > 5.
  • Sample size < 10% of population.
• Calculation: 𝜒² = 6.264, P = 0.099.
• Conclusion: P > 0.05, insufficient evidence to reject H₀.

Chi-Square Test for Independence

• Test of Independence: Determine if there is a significant association between two categorical variables.
• Null Hypothesis: The two classifications are independent.
• Degrees of Freedom (df): Calculated as (number of rows - 1) * (number of columns - 1).
• Limitations: Even if independence is rejected, causality cannot be assumed.*

Example 8.2

• Scenario: Relationship between party affiliation and support for marijuana legalization.
• Hypotheses:
  • H₀: Party affiliation and support are independent.
  • Hₐ: Party affiliation and support are not independent.
• Procedure: Chi-square test for independence.
• Checks:
  • Random sample, n < 10% of the population, expected cells > 5.
• Calculation: 𝜒² = 94.5, P = 0.000.
• Conclusion: P < 0.05, sufficient evidence to reject H₀.

Chi-Square Test for Homogeneity

• Test for Homogeneity: Compare samples from two or more populations regarding a single categorical variable.
• Conditions:
  • Samples must be simple random samples.
  • Samples should be independent.
  • Large populations relative to sample sizes.
  • Expected values > 5.

Example 8.3

• Scenario: Job satisfaction among school employees.
• Hypotheses:
  • H₀: Job satisfaction is the same across job categories.
  • Hₐ: At least two categories differ in job satisfaction.
• Procedure: Chi-square test for homogeneity.
• Checks:
  • Independent random samples, expected cells > 5, samples < 10% of populations.
• Calculation: 𝜒² = 8.707, P = 0.0335.
• Conclusion: P < 0.05, sufficient evidence to reject H₀; job satisfaction differs across categories.