Chi-Square Tests — AP Statistics Unit 8 practice.
This unit covers goodness of fit, test for independence and test for homogeneity — essential concepts for AP Statistics. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers goodness of fit, test for independence and test for homogeneity — essential concepts for AP Statistics. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Chi-Square Goodness Of Fit Test
This test determines whether observed sample counts match a claimed or expected distribution for a single categorical variable. Students must be able to state hypotheses, calculate expected counts, compute the test statistic, and draw conclusions in context. The null hypothesis always claims the population follows a specific stated distribution.
Key Points
- H₀: The population follows the specified distribution; Hₐ: The population does NOT follow the specified distribution
- Expected count for each category = n × (claimed proportion); all expected counts must be ≥ 5
- χ² = Σ (Observed − Expected)² / Expected; degrees of freedom = number of categories − 1
- Always use a one-tailed (right-tail) p-value; a large χ² means more deviation from H₀
A six-sided die is rolled 120 times. Results: 1→18, 2→22, 3→17, 4→24, 5→15, 6→24. Test at α = 0.05 whether the die is fair.
Under H₀ (fair die), each face has probability 1/6, so expected count = 120/6 = 20 for every face. Compute χ² = (18−20)²/20 + (22−20)²/20 + (17−20)²/20 + (24−20)²/20 + (15−20)²/20 + (24−20)²/20 = 0.2 + 0.2 + 0.45 + 0.8 + 1.25 + 0.8 = 3.70. With df = 5, the p-value ≈ 0.593 > 0.05, so we fail to reject H₀ — the data do not provide convincing evidence the die is unfair.
2 Chi-Square Test For Independence
This test determines whether two categorical variables are associated (not independent) within a single population, using a two-way table of observed counts. Students must recognize this design — one sample, two variables measured on each individual — and distinguish it from homogeneity. The null hypothesis always states the two variables are independent.
Key Points
- H₀: [Variable A] and [Variable B] are independent in the population; Hₐ: they are NOT independent (there is an association)
- Expected count for each cell = (row total × column total) / table total; all must be ≥ 5
- df = (rows − 1)(columns − 1); χ² statistic formula is the same as goodness of fit
- A significant result means association exists — it does NOT establish causation or direction
A random sample of 200 adults is asked their age group (Under 40 / 40 or Over) and preferred news source (TV / Online / Print). Test at α = 0.05 whether age group and news preference are independent.
Because one sample was drawn and two categorical variables were recorded on each person, this is a test for independence. Compute expected counts using (row total × column total)/200 for each of the 6 cells, verify all are ≥ 5, then calculate χ² = Σ(O − E)²/E with df = (2−1)(3−1) = 2. If the resulting p-value < 0.05, reject H₀ and conclude there is convincing evidence of an association between age group and news source preference in the population.
3 Chi-Square Test For Homogeneity
This test determines whether two or more populations (or treatment groups) share the same distribution of a single categorical variable. The critical design feature is that separate samples are drawn from each population or group — this is what distinguishes it from the test for independence. The null hypothesis states that all populations have the same distribution of the categorical variable.
Key Points
- H₀: The distribution of [categorical variable] is the same across all populations/groups; Hₐ: The distributions are NOT all the same
- Separate random samples are taken from each group — this is the key design distinction from independence
- Mechanics are identical to the test for independence: same expected-count formula, same χ² statistic, df = (rows−1)(cols−1)
- Conditions: independent random samples from each group, all expected counts ≥ 5
Researchers randomly sample 150 teenagers, 150 adults (25–54), and 150 seniors (55+) and ask each whether they exercise Daily, Weekly, or Rarely. Test whether the distribution of exercise frequency is the same across the three age groups.
Because three separate samples were drawn (one per age group) and one categorical variable (exercise frequency) was recorded, this is a test for homogeneity with df = (3−1)(3−1) = 4. Compute expected counts as (row total × column total)/450 for each cell, check that all are ≥ 5, then find χ² = Σ(O−E)²/E. If the p-value is below the chosen α, reject H₀ and state there is convincing evidence that the distribution of exercise frequency is not the same across all three age groups.
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What is Chi-Square Tests?
Chi-Square Tests is Unit 8 of AP Statistics, covering goodness of fit, test for independence and test for homogeneity.
How to study for AP Statistics Unit 8?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 27+ review questions across 5 different game modes.