AP Statistics (AP Stats): Chi?Square Test for Independence (Two?Way Table, One Population)

AP Statistics – Chi?Square Test for Independence (Two?Way Table, One Population)

AP Statistics: Chi-Square Test for Independence (Two-Way Table, One Population) – Exam-Ready Study Guide

What This Is

The Chi-Square Test for Independence determines whether there is a statistically significant association between two categorical variables in a single population. For example, does gender (male/female) influence preference for a new phone model (iPhone/Android)? On the AP exam, you’ll analyze two-way tables, check conditions, compute the test statistic, and interpret results in context. This test is a staple in FRQs and multiple-choice questions, often paired with real-world scenarios like medical studies, marketing surveys, or social science research.

Key Terms & Formulas

• Chi-Square Test for Independence: Tests whether two categorical variables are associated in a single population.
• H?: The two variables are independent (no association).

• H?: The two variables are not independent (association exists).

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

• For a 2×2 table, df = 1; for a 3×4 table, df = 6.

• Conditions for Chi-Square Test:

• Random: Data comes from a random sample or randomized experiment.
• Large Counts: All expected counts-5 (check after calculating E).

• Independent: Observations are independent (10% condition if sampling without replacement).

• Calculator Command (TI-84):

• STAT-TESTS-?²-Test (enter observed counts in a matrix).

• Store expected counts: STAT-EDIT-[B] (matrix B).

• P-value Interpretation:

• If p-value <? (e.g., 0.05), reject H? (evidence of association).

• If p-value-?, fail to reject H? (no evidence of association).

• Contribution to ?²:

• Larger ((O - E)^2 / E) values indicate cells that contribute most to the test statistic.

Step-by-Step / Process Flow

For a typical FRQ (e.g., "Is there an association between gender and phone preference?"):

1. State Hypotheses:
2. H?: Gender and phone preference are independent (no association).

3. H?: Gender and phone preference are not independent (association exists).

4. Check Conditions:

5. Random: "The data comes from a random sample of 200 adults."
6. Large Counts: Calculate expected counts (all-5).

7. Independent: "Assume the sample is <10% of the population."

8. Compute Test Statistic:

9. Calculate expected counts (E) for each cell.
10. Compute ?² using the formula or ?²-Test on the TI-84.

11. Report ?² and df (e.g., ?² = 6.25, df = 1).

12. Find P-value:

13. Use ?²cdf(?², 1E99, df) on TI-84 (e.g., ?²cdf(6.25, 1E99, 1) = 0.0124).

14. Make Conclusion in Context:

15. Compare p-value to? (e.g., 0.0124 < 0.05).

16. "Since the p-value (0.0124) is less than-= 0.05, we reject H?. There is convincing evidence of an association between gender and phone preference."

17. Follow-Up (if asked):

18. Identify the largest ?² contribution to explain the association (e.g., "Males were more likely to prefer Android than expected").

Common Mistakes

• Mistake: Forgetting to check Large Counts (expected counts-5).

• Correction: Always calculate E for every cell and verify the condition. If any E < 5, the test is invalid.

• Mistake: Using row/column proportions instead of counts in the ?² formula.

• Correction: The test requires counts, not percentages or proportions.

• Mistake: Miscalculating df as (n - 1) or (rows + columns - 2).

• Correction: df = (rows - 1) × (columns - 1). For a 2×3 table, df = 2.

• Mistake: Interpreting a significant result as causation.

• Correction: The test shows association, not causation (e.g., "Gender is associated with phone preference"-"Gender causes phone preference").

• Mistake: Ignoring the 10% condition for independence.

• Correction: If sampling without replacement, ensure n-10% of the population.

AP Exam Insights

• FRQ Setup: You’ll often get a two-way table (e.g., 2×2 or 3×2) and be asked to:
• State hypotheses.
• Check conditions (especially Large Counts).
• Calculate ?² or identify the largest contribution.

• Interpret the p-value in context.

• Tricky Distinction: Chi-Square Goodness-of-Fit (one categorical variable) vs. Chi-Square Test for Independence (two variables). The exam may test both—know the difference!

• Calculator Pitfall: Entering data into the wrong matrix (use [A] for observed counts, [B] for expected).

• Context Matters: Always conclude with direction (e.g., "Females were more likely to prefer iPhones than expected").

Quick Check Questions

1. Multiple Choice: A study examines the association between exercise frequency (low/medium/high) and stress level (low/high). The two-way table has 3 rows and 2 columns. What are the degrees of freedom for the chi-square test?
2. (A) 1
3. (B) 2
4. (C) 3
5. (D) 5

6. Answer: (B) 2. df = (3 - 1)(2 - 1) = 2.

7. FRQ Part: A researcher tests whether political affiliation (Democrat/Republican) is associated with support for a policy (Yes/No). The ?² test yields p-value = 0.03. Interpret this p-value in context.

8. Answer: "Assuming political affiliation and policy support are independent, there is a 3% chance of observing a ?² statistic as extreme or more extreme than the one calculated."

9. Multiple Choice: Which condition is not required for a chi-square test for independence?

10. (A) Random sampling
11. (B) All expected counts-5
12. (C) Normal distribution of the response variable
13. (D) Independent observations
14. Answer: (C). The chi-square test does not assume normality.

Last-Minute Cram Sheet

1. Hypotheses: H?: Independent; H?: Not independent.
2. Formula: ?² = ?(O - E)²/E.
3. Expected Count: E = (row total × column total) / grand total.
4. df: (rows - 1)(columns - 1).
5. Conditions: Random, Large Counts (E-5), Independent (10% rule).
6. Calculator: ?²-Test-enter observed counts in [A].
7. P-value: ?²cdf(?², 1E99, df).
8. Conclusion: Compare p-value to ?; state association in context.
9. Never use ?² for proportions—only counts!
10. Association-causation—avoid causal language.