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The Chi-Square Test for Homogeneity determines whether the distribution of a categorical variable is the same across multiple populations (e.g., "Do high school students, college students, and adults have the same preferences for social media platforms?"). Unlike the Chi-Square Test for Independence (which analyzes one sample with two categorical variables), homogeneity compares two or more independent samples on one categorical variable. This test is essential for the AP exam because it appears in FRQs (often with two-way tables) and requires careful interpretation of hypotheses, conditions, and conclusions.
Real-world example: A researcher wants to know if voter support for a new policy (Support/Oppose/Neutral) differs across three regions (North, South, West). The homogeneity test answers: "Is the distribution of support the same in all three regions?"
H?: The distribution of the categorical variable is not the same for at least one population.
Expected Count Formula: [ E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}} ]
E = expected count for a cell in the two-way table.
Chi-Square Test Statistic (?²): [ \chi^2 = \sum \frac{(O - E)^2}{E} ]
O = observed count, E = expected count.
Degrees of Freedom (df) for Homogeneity: [ df = (\text{number of rows} - 1) \times (\text{number of columns} - 1) ]
Columns = populations being compared (e.g., North/South/West).
Conditions for Chi-Square Tests (RICE):
Expected Counts-5: All expected counts must be at least 5 (check with ?²-Test output).
?²-Test
Calculator Command (TI-84):
STAT-TESTS-?²-Test
2nd-x?¹-EDIT
Output: ?² test statistic, p-value, and expected counts (store in a matrix).
P-value Interpretation:
If p-value > ?, fail to reject H?-no evidence that distributions differ.
Follow-Up Analysis (if H? is rejected):
Example: H?: The distribution of voter support (Support/Oppose/Neutral) is the same in the North, South, and West.
Check Conditions (RICE)
Expected Counts-5: Use calculator output to verify (or show expected counts table).
Compute Test Statistic & P-value
Record ?² statistic and p-value.
Make a Conclusion in Context
Example: "Since the p-value (0.02) < 0.05, we reject H?. There is convincing evidence that the distribution of voter support differs across the three regions."
Follow-Up (If Required)
Why? The hypotheses and interpretations differ (e.g., "same distribution" vs. "no association").
Mistake: Forgetting to check expected counts-5.
Why? The test is invalid if expected counts are too small.
Mistake: Miscalculating degrees of freedom.
(rows – 1) × (columns – 1)
Why? Incorrect df leads to wrong p-values and conclusions.
Mistake: Writing a generic conclusion (e.g., "Reject H?").
Why? AP graders deduct points for lack of context.
Mistake: Using proportions instead of counts in the matrix.
Follow up (e.g., "Which group differs most?").
Tricky Distinction: Homogeneity vs. Independence
Independence: "Is there an association between two variables in one population?" (e.g., "Is there an association between gender and brand preference in adults?")
Calculator Pitfall: Forgetting to store expected counts in a matrix.
Always check 2nd-x?¹-[B] (or another matrix) to verify expected counts-5.
2nd-x?¹-[B]
Common Follow-Up: If H? is rejected, AP may ask:
(O – E)²/E
A study compares the distribution of favorite ice cream flavors (Vanilla, Chocolate, Strawberry) across three age groups (Kids, Teens, Adults). The ?² test for homogeneity yields a p-value of 0.03. Which conclusion is correct? (A) There is no association between age group and ice cream preference. (B) The distribution of ice cream preferences is the same for all age groups. (C) There is convincing evidence that the distribution of ice cream preferences differs across age groups. (D) The sample size was too small to draw a conclusion.
Answer: (C) Explanation: A p-value of 0.03 < 0.05 means we reject H?, concluding the distributions differ.
A researcher surveys 500 people from three cities (A, B, C) about their primary mode of transportation (Car, Public Transit, Bike). The observed counts are:
Part (a): State the hypotheses for a ?² test for homogeneity. Part (b): Are the conditions for inference met? Justify your answer.
Answer (a): - H?: The distribution of transportation modes is the same for all three cities. - H?: The distribution of transportation modes is not the same for at least one city.
Answer (b): - Random: Assume the samples are random (if not stated, note the assumption). - Independent: The cities are independent populations. - Categorical: Transportation mode is categorical. - Expected Counts-5: All expected counts (e.g., for City A, Car: (250×310)/750-103.3) are-5.
(250×310)/750-103.3
(row total × column total) / grand total
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