Comparing Hypotheses of Chi-Square Independence and Homogeneity Tests

📚 Quick Study Guide

• 🔍 Chi-Square Test of Independence: This test determines if there's a statistically significant association between two categorical variables within a single population. Think of it as checking if two things are related to each other.
• 📊 Null Hypothesis (Independence): The two categorical variables are independent (not related).
• 📈 Alternative Hypothesis (Independence): The two categorical variables are dependent (related).
• 🧪 Chi-Square Test Statistic (Independence): $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$, where $O_i$ is the observed frequency and $E_i$ is the expected frequency.
• 🧑‍🏫 Degrees of Freedom (Independence): $(r - 1)(c - 1)$, where $r$ is the number of rows and $c$ is the number of columns in the contingency table.
• 🌍 Chi-Square Test of Homogeneity: This test determines if the distribution of a categorical variable is the same across multiple populations. Think of it as checking if different groups have similar characteristics.
• 🏘️ Null Hypothesis (Homogeneity): The distribution of the categorical variable is the same for all populations.
• 💯 Alternative Hypothesis (Homogeneity): The distribution of the categorical variable is not the same for all populations.
• ➗ Chi-Square Test Statistic (Homogeneity): $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$, where $O_i$ is the observed frequency and $E_i$ is the expected frequency.
• 📐 Degrees of Freedom (Homogeneity): $(r - 1)(c - 1)$, where $r$ is the number of rows and $c$ is the number of columns in the contingency table.
• 💡 Key Difference: Independence tests analyze relationships within one population, while homogeneity tests compare distributions across multiple populations.

Practice Quiz

1. Question 1: What is the primary purpose of the Chi-Square test of independence?
   1. A. To compare the means of two groups.
   2. B. To determine if there is an association between two categorical variables in a single population.
   3. C. To determine if the variances of two groups are equal.
   4. D. To compare the distributions of a continuous variable across multiple populations.
2. Question 2: What is the null hypothesis for a Chi-Square test of homogeneity?
   1. A. The means of the populations are equal.
   2. B. The variances of the populations are equal.
   3. C. The distribution of the categorical variable is the same for all populations.
   4. D. The categorical variables are dependent.
3. Question 3: In a Chi-Square test of independence, what do the observed frequencies represent?
   1. A. The expected counts under the null hypothesis.
   2. B. The actual counts observed in the sample.
   3. C. The mean of the sample.
   4. D. The standard deviation of the sample.
4. Question 4: What is the key difference between the Chi-Square test of independence and the Chi-Square test of homogeneity?
   1. A. The test of independence involves multiple populations, while the test of homogeneity involves only one.
   2. B. The test of independence assesses the distribution of a continuous variable, while the test of homogeneity assesses categorical variables.
   3. C. The test of independence examines relationships within one population, while the test of homogeneity compares distributions across multiple populations.
   4. D. There is no difference between the two tests.
5. Question 5: The degrees of freedom for both tests are calculated the same way. If you have a contingency table with 4 rows and 3 columns, what are the degrees of freedom?
   1. A. 12
   2. B. 7
   3. C. 6
   4. D. 5
6. Question 6: Which test is appropriate to use to see if the proportion of people who prefer Coke vs. Pepsi is the same in New York, LA, and Chicago?
   1. A. Chi-Square test of independence
   2. B. Chi-Square test of homogeneity
   3. C. T-test
   4. D. ANOVA
7. Question 7: Which test is appropriate to use to see if gender (male/female) is related to political affiliation (Democrat/Republican) in a sample of registered voters?
   1. A. Chi-Square test of homogeneity
   2. B. T-test
   3. C. Chi-Square test of independence
   4. D. ANOVA