AP Statistics Quiz: Chi Square Homogeneity Or Independence Setup

What this quiz covers

This quiz focuses on Chi Square Homogeneity Or Independence Setup, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Chi-square goodness-of-fit; $H_0$: pass and fail occur in a 50/50 split; $H_a$: not a 50/50 split
2. Chi-square test for homogeneity; $H_0$: the distribution of attendance is the same among passers and failers; $H_a$: different distributions
3. Chi-square test of independence; $H_0$: review-session attendance and exam outcome are independent; $H_a$: they are associated (correct answer)
4. Two-proportion $z$ test; $H_0$: $p_{\text{Pass,Attend}}=p_{\text{Pass,NoAttend}}$; $H_a$: not equal
5. Matched-pairs test; $H_0$: no difference before vs after review; $H_a$: a difference

Explanation: Assessing AP Statistics skill in chi-square homogeneity versus independence setups, this uses one random sample of students with two categorical variables (attendance and outcome), ideal for chi-square test of independence on association. Choice C hypothesizes null independence and alternative association correctly. Choice B is a distractor, suggesting homogeneity, but that requires separate samples, not one here. Mini-lesson: Chi-square includes goodness-of-fit for single-variable proportions, independence for two-variable ties in one group, and homogeneity for multi-group comparisons. Choice A misapplies goodness-of-fit to outcomes. C is accurate.

Question 2

1. Chi-square test of independence; $H_0$: program and outcome are independent for the employees in this study; $H_a$: they are associated.
2. Chi-square test of homogeneity; $H_0$: the distribution of outcomes (Pass/Fail/No-show) is the same for Programs A, B, and C; $H_a$: at least one program has a different outcome distribution. (correct answer)
3. Chi-square goodness-of-fit; $H_0$: each outcome occurs with probability $1/3$; $H_a$: the probabilities are not all $1/3$.
4. One-way ANOVA; $H_0$: the mean outcome is the same for all programs; $H_a$: at least one mean differs.
5. Chi-square test of homogeneity; $H_0$: the programs have equal sample sizes; $H_a$: the sample sizes differ.

Explanation: This scenario calls for a chi-square test of homogeneity. The company randomly assigns employees to three different training programs, creating three separate groups, and then compares the distribution of outcomes (Pass/Fail/No-show) across these groups. Option B correctly identifies this setup, with the null hypothesis stating that all three programs have the same outcome distribution. Option A incorrectly suggests independence, but random assignment creates distinct groups rather than measuring two variables on the same individuals. Option C (goodness-of-fit) would test against theoretical proportions, not compare groups. Remember: when subjects are randomly assigned to different treatments and we compare categorical outcomes across treatments, use the test of homogeneity.

Question 3

1. Chi-square test of homogeneity; $H_0$: the distribution of class year is the same across housing types; $H_a$: at least one housing type has a different distribution.
2. Chi-square goodness-of-fit; $H_0$: class years occur in the ratio 1:1:2; $H_a$: the ratio is different.
3. Chi-square test of independence; $H_0$: housing type and class year are independent in the undergraduate population; $H_a$: they are associated. (correct answer)
4. Matched-pairs test; $H_0$: each student's housing type matches their class year category; $H_a$: they do not match.
5. Two-proportion $z$ test; $H_0$: the proportion living on-campus is the same for First-years and Juniors/Seniors; $H_a$: the proportions differ.

Explanation: This scenario requires a chi-square test of independence. The university takes one random sample of 300 undergraduates and records two categorical variables for each student: housing type and class year. Option C correctly identifies this setup, with the null hypothesis stating that these two variables are independent in the undergraduate population. Option A incorrectly suggests homogeneity, but we don't have separate samples for each housing type - we have one sample where both variables are recorded. The distinction is crucial: one sample with two categorical variables measured on each unit leads to a test of independence, while multiple samples comparing distributions leads to a test of homogeneity.

Question 4

1. Chi-square goodness-of-fit; $H_0$: the distribution of club preference is the same for all students in the district; $H_a$: the distribution is different.
2. Chi-square test of independence; $H_0$: grade level and club preference are independent for all students in the district; $H_a$: they are associated.
3. Chi-square test of homogeneity; $H_0$: the distribution of club preference is the same for 9th graders and 12th graders; $H_a$: at least one grade has a different distribution. (correct answer)
4. Two-proportion $z$ test; $H_0$: the proportion preferring Sports is the same for 9th and 12th graders; $H_a$: the proportions differ.
5. Chi-square test of homogeneity; $H_0$: the counts in each cell match expected counts exactly; $H_a$: at least one observed count differs from its expected count.

Explanation: This question tests your ability to identify when to use a chi-square test of homogeneity. The key indicator is that we have two separate random samples (80 ninth graders and 90 twelfth graders) and we want to compare the distribution of a categorical variable (club preference) across these groups. Option C correctly identifies this as a test of homogeneity with the proper null hypothesis stating that the distribution of club preference is the same for both grade levels. Option B incorrectly suggests independence, but that's for when we have one sample with two variables measured on each individual. Option A (goodness-of-fit) is for comparing one sample to a theoretical distribution, and option D only compares one category rather than the entire distribution. Remember: multiple separate samples comparing distributions = homogeneity; one sample with two variables = independence.

Question 5

1. Chi-square goodness-of-fit; $H_0$: payment methods are equally preferred overall; $H_a$: not equally preferred
2. Chi-square test of independence; $H_0$: location and payment preference are independent; $H_a$: they are not independent
3. Chi-square test for homogeneity; $H_0$: the distribution of payment preference is the same at both locations; $H_a$: the distributions differ (correct answer)
4. Two-proportion $z$ test; $H_0$: $p_{\text{Credit,Mall}}=p_{\text{Credit,Downtown}}$; $H_a$: not equal
5. Chi-square goodness-of-fit; $H_0$: the distribution of locations is 50/50; $H_a$: not 50/50

Explanation: This tests the AP Statistics skill of choosing chi-square for homogeneity or independence via design. Separate samples per location, with one categorical response (payment method), warrant chi-square homogeneity to compare distributions. Choice C properly nulls same distributions and alternatives with difference. Choice B distracts as independence, needing one sample with two variables, not multiple samples. Mini-lesson: Goodness-of-fit tests one variable's fit; independence assesses association in one sample; homogeneity checks uniformity across groups. Choice A wrongly uses goodness-of-fit on methods overall. C is correct.

Question 6

1. Chi-square test of independence; $H_0$: species and feeding time are independent; $H_a$: they are associated
2. Chi-square test for homogeneity; $H_0$: the distribution of feeding time is the same for all species; $H_a$: at least one species differs (correct answer)
3. Chi-square goodness-of-fit; $H_0$: feeding times are equally likely; $H_a$: not equally likely
4. One-way ANOVA; $H_0$: mean feeding time is the same for all species; $H_a$: at least one mean differs
5. Two-proportion $z$ test; $H_0$: $p_{\text{Morning,Meerkat}}=p_{\text{Morning,Otter}}$; $H_a$: not equal

Explanation: This AP Statistics question targets the skill of chi-square test setup for homogeneity or independence. With separate random samples of feeding events per species and one categorical variable (feeding time), chi-square for homogeneity tests if distributions are identical across species. Choice B's hypotheses null same distributions and alternative at least one difference. Choice A distracts with independence, incorrect without two variables per observation in one sample. Mini-lesson: Goodness-of-fit matches one variable to expected; independence links two variables in a sample; homogeneity equates distributions from multiple groups. Choice C wrongly applies goodness-of-fit to times overall. B fits the design.

Question 7

1. Chi-square test of independence; $H_0$: reminder method and appointment outcome are independent; $H_a$: they are associated
2. Chi-square goodness-of-fit; $H_0$: outcomes are equally likely; $H_a$: outcomes are not equally likely
3. Two-proportion $z$ test; $H_0$: the no-show rates are equal for text and phone; $H_a$: they differ
4. Chi-square test of homogeneity; $H_0$: the distribution of outcomes is the same for text and phone reminders; $H_a$: the distributions differ (correct answer)
5. Chi-square test of independence; $H_0$: the distribution of outcomes is the same for both reminder methods; $H_a$: the distribution differs

Explanation: This question tests the skill of selecting the chi-square test and hypotheses for comparing categorical outcomes across experimentally assigned groups in AP Statistics. The design randomly assigns patients to reminder methods (like treatments), effectively creating independent groups, and records one categorical outcome per group, akin to multiple populations. Hence, the chi-square test of homogeneity is appropriate to see if outcome distributions differ by reminder method. Mini-lesson: Goodness-of-fit checks one sample against a model, homogeneity tests for same distributions in independent samples or treatment groups, and independence examines two variables' link in one sample. Distractor A calls it independence with associated hypotheses, but that ignores the assignment-based grouping better suited to homogeneity. Option D correctly identifies homogeneity with fitting hypotheses for this experimental context.

Question 8

1. Chi-square test of independence; $H_0$: neighborhood and recycling status are independent; $H_a$: they are associated
2. Two-proportion $z$ test; $H_0$: $p_A=p_B$; $H_a$: $p_A\ne p_B$
3. Chi-square goodness-of-fit; $H_0$: the overall proportion recycling is the same as a known value; $H_a$: it differs
4. Chi-square test of homogeneity; $H_0$: the distribution of recycling status is the same across neighborhoods; $H_a$: at least one neighborhood has a different distribution (correct answer)
5. Chi-square test of homogeneity; $H_0$: neighborhood and recycling status are independent; $H_a$: not independent

Explanation: This problem focuses on the skill of identifying the chi-square test and hypotheses for comparing a binary categorical variable across independent samples from neighborhoods in AP Statistics. The design takes independent random samples from each neighborhood (populations) and measures recycling status (yes/no) in each, fitting a multi-population comparison. The chi-square test of homogeneity is thus appropriate to check if recycling distributions are the same across neighborhoods. Mini-lesson: Use goodness-of-fit for single-sample distribution testing, homogeneity for equivalence across independent groups, and independence for two-variable association in one sample. Distractor E correctly names homogeneity but wrongly applies independence hypotheses, which doesn't align with the setup. Option D is accurate in test and hypotheses.

Question 9

1. Chi-square goodness-of-fit test; $H_0$: the three schools have the same number of students sampled in each activity category; $H_a$: at least one school has a different sample size pattern
2. Chi-square test of homogeneity; $H_0$: the distribution of preferred activity is the same across the three schools; $H_a$: at least one school has a different distribution (correct answer)
3. Chi-square test of independence; $H_0$: school and preferred activity are independent for this one random sample; $H_a$: school and preferred activity are not independent
4. Two-proportion $z$ test; $H_0$: the proportion preferring Sports is equal across the three schools; $H_a$: at least one proportion differs
5. Chi-square goodness-of-fit test; $H_0$: preferred activities are equally likely overall; $H_a$: preferred activities are not equally likely overall

Explanation: This question asks about comparing the distribution of preferred after-school activity across three schools, where separate random samples are taken from each school. Since we have separate samples from distinct populations (the three schools) and want to compare if the categorical distribution is the same across these populations, we use a chi-square test of homogeneity. The null hypothesis states that the distribution of preferred activity is the same across all three schools, while the alternative states that at least one school has a different distribution. Option A incorrectly focuses on sample sizes rather than distributions, Option C wrongly suggests independence (which requires one sample), Option D only examines one category (Sports) rather than the full distribution, and Option E examines overall preferences rather than comparing across schools.

Question 10

1. Chi-square goodness-of-fit; $H_0$: overall brand preferences follow a specified distribution; $H_a$: they do not
2. Chi-square test of independence; $H_0$: age group and brand preference are independent; $H_a$: they are associated
3. Chi-square test of homogeneity; $H_0$: the distribution of brand preference is the same across the four age groups; $H_a$: at least one age group has a different distribution (correct answer)
4. One-way ANOVA; $H_0$: mean age is the same for each brand; $H_a$: at least one mean differs
5. Chi-square test of homogeneity; $H_0$: age group and brand preference are independent; $H_a$: they are not independent

Explanation: This question evaluates the skill of choosing the chi-square test and hypotheses for comparing brand preference distributions across independent age group samples in AP Statistics. The design involves independent random samples from each age group (populations), with one categorical variable (brand preference) per sample. This setup warrants the chi-square test of homogeneity to test for identical distributions across groups. Quick mini-lesson: Chi-square goodness-of-fit is for one sample's match to expected, homogeneity for same distributions in independent multi-group samples, and independence for two variables in one sample. Distractor E names homogeneity but uses incorrect independence hypotheses, blending concepts erroneously. Option C is correct with proper test and hypotheses.

Question 11

1. Chi-square goodness-of-fit; $H_0$: study locations are equally preferred; $H_a$: not equally preferred
2. Chi-square test of homogeneity; $H_0$: the distribution of majors is the same across study locations; $H_a$: at least one location differs
3. Chi-square test of independence; $H_0$: major and preferred study location are independent; $H_a$: major and preferred study location are associated (correct answer)
4. Two-sample $t$ test; $H_0$: mean study time is the same for STEM and Non-STEM; $H_a$: means differ
5. Chi-square test of homogeneity; $H_0$: major and preferred study location are independent; $H_a$: they are associated

Explanation: The skill assessed is choosing the correct chi-square test and hypotheses for investigating the relationship between two categorical variables in a single student sample for AP Statistics. The sampling design is one random sample where each student reports both major and preferred study location, indicating a single population with two variables. This requires the chi-square test of independence to test for an association between them. Brief mini-lesson: Chi-square goodness-of-fit is for one variable's fit to expectations, homogeneity compares distributions across separate samples, and independence tests variable independence in one sample. Option E is a distractor as it names homogeneity but uses independence hypotheses, confusing test types with sampling. Choice C properly states the independence test and hypotheses.

Question 12

1. Chi-square test of independence; $H_0$: exercise level and sleep quality are independent; $H_a$: they are associated (correct answer)
2. Chi-square test of homogeneity; $H_0$: the distribution of sleep quality is the same across exercise levels; $H_a$: at least one exercise level has a different distribution
3. Chi-square goodness-of-fit; $H_0$: sleep quality is equally likely to be Poor, Fair, or Good; $H_a$: not all categories are equally likely
4. Two-sample $t$ test; $H_0$: mean sleep quality is the same for all exercise levels; $H_a$: at least one mean differs
5. Chi-square test of independence; $H_0$: the distribution of sleep quality is the same across exercise levels; $H_a$: the distribution differs

Explanation: The skill being tested is identifying the correct chi-square test and hypotheses for examining the relationship between two categorical variables in a single sample for AP Statistics. The sampling design features one random sample of adults where both exercise level and sleep quality are recorded for each individual, indicating a single population with two variables. Therefore, the chi-square test of independence is suitable to check if these variables are associated. Mini-lesson: Chi-square goodness-of-fit tests a single distribution against expectations, homogeneity compares distributions across groups from independent samples, and independence assesses if two variables are related in one sample. Option E is a distractor because it names the test as independence but uses hypotheses worded like homogeneity, which mismatches the sampling design. The marked answer A correctly specifies the test and hypotheses.

Question 13

1. Chi-square goodness-of-fit; $H_0$: the distribution of lunch preferences is the same across grades; $H_a$: the distribution differs across grades
2. Chi-square test of homogeneity; $H_0$: the distribution of lunch preferences is the same for 6th, 7th, and 8th graders; $H_a$: at least one grade has a different distribution (correct answer)
3. Chi-square test of independence; $H_0$: lunch preference and grade level are independent in the population; $H_a$: lunch preference and grade level are associated
4. Two-proportion $z$ test; $H_0$: $p_{6}=p_{7}=p_{8}$ for choosing Pizza; $H_a$: not all proportions are equal
5. Chi-square test of homogeneity; $H_0$: grade level and lunch preference are independent; $H_a$: grade level and lunch preference are not independent

Explanation: This question assesses the skill of selecting the appropriate chi-square test and hypotheses for comparing distributions of a categorical variable across multiple groups in AP Statistics. The sampling design involves taking independent random samples from each of three distinct grade levels, which are treated as separate populations, and measuring one categorical variable (lunch preference) in each sample. Given this design, the chi-square test of homogeneity is appropriate to determine if the distribution of lunch preferences is the same across the grades. In a mini-lesson on chi-square tests: the goodness-of-fit test is for one sample matching a specified distribution, homogeneity is for comparing distributions across multiple populations with independent samples, and independence is for testing association between two variables in one sample. A common distractor here is option E, which correctly names the test as homogeneity but incorrectly states the hypotheses in terms of independence, confusing the conceptual framing. The correct choice is B, which properly identifies the test and hypotheses for this setup.

Question 14

1. Chi-square goodness-of-fit; $H_0$: device types occur in equal proportions; $H_a$: not all device proportions are equal
2. Chi-square test for homogeneity; $H_0$: the distribution of device type is the same for purchases and no purchases; $H_a$: different distributions
3. Chi-square test of independence; $H_0$: device type and purchase outcome are independent; $H_a$: device type and purchase outcome are associated (correct answer)
4. Two-proportion $z$ test; $H_0$: $p_{\text{Purchase,Mobile}}=p_{\text{Purchase,Desktop}}$; $H_a$: not equal
5. Chi-square goodness-of-fit; $H_0$: purchase and no purchase occur equally often; $H_a$: they do not

Explanation: This AP Statistics question focuses on the skill of distinguishing chi-square test setups for homogeneity versus independence. With a single random sample of site visits where each visit records two categorical variables (device type and purchase outcome), the chi-square test of independence is correct to test for association. Choice C properly states the null as independence between variables and the alternative as association. Choice B is a distractor because it suggests homogeneity, which would require separate samples from groups like 'purchases' and 'no purchases,' but here it's one sample. Mini-lesson: Chi-square types include goodness-of-fit for one variable matching a model, independence for two variables in one sample, and homogeneity for one variable across multiple samples. Choice A incorrectly applies goodness-of-fit to device types alone. Overall, C fits the design.

Question 15

1. Chi-square test of independence; $H_0$: class year and preferred study location are independent; $H_a$: they are associated (correct answer)
2. Chi-square test for homogeneity; $H_0$: the distribution of class year is the same across study locations; $H_a$: different distributions
3. Chi-square goodness-of-fit; $H_0$: study locations are equally preferred; $H_a$: not equally preferred
4. Two-proportion $z$ test; $H_0$: $p_{\text{Library,First}}=p_{\text{Library,Senior}}$; $H_a$: not equal
5. Chi-square test of independence; $H_0$: the distribution of study locations is the same for each class year; $H_a$: at least one class year differs

Explanation: This question evaluates the AP Statistics skill of chi-square test selection for homogeneity or independence. A single random sample of students, each reporting two categorical variables (class year and study location), calls for a chi-square test of independence to assess if they are related. Choice A accurately hypothesizes independence in the null and association in the alternative. Choice B distracts by suggesting homogeneity, but that's for multiple samples, not one sample like this. Mini-lesson: Distinguish chi-square as goodness-of-fit for one variable's fit to expectations, independence for two variables' linkage in one group, and homogeneity for comparing one variable across populations. Choice E resembles A but uses homogeneity phrasing, which isn't precise for independence. A is the correct setup.

Question 16

1. Chi-square goodness-of-fit; $H_0$: support and oppose are equally common; $H_a$: they are not equally common
2. Chi-square test for homogeneity; $H_0$: the distribution of housing types is the same for supporters and opposers; $H_a$: different distributions
3. Chi-square test of independence; $H_0$: housing type and support for the proposal are independent; $H_a$: they are associated (correct answer)
4. Two-proportion $z$ test; $H_0$: $p_{\text{Support,Apartment}}=p_{\text{Support,Single}}$; $H_a$: not equal
5. Chi-square goodness-of-fit; $H_0$: housing types occur in equal proportions; $H_a$: not equal proportions

Explanation: The AP Statistics skill here is differentiating chi-square tests for homogeneity and independence. One random sample of residents, each giving two categorical variables (housing type and support), makes chi-square test of independence appropriate for association testing. Choice C correctly nulls independence and alternatives with association. A distractor, choice B, proposes homogeneity, but that needs separate samples from groups like 'supporters' versus 'opposers,' not one sample. Mini-lesson: Chi-square goodness-of-fit checks one category's proportions; independence explores two categories' relationship in one sample; homogeneity tests equality across multiple samples. Choice E misuses goodness-of-fit on housing alone. C is the proper choice.

Question 17

1. Chi-square test of independence; $H_0$: campaign and reaction are independent; $H_a$: they are not independent
2. Chi-square goodness-of-fit; $H_0$: reactions occur equally often overall; $H_a$: not equally often
3. Chi-square test for homogeneity; $H_0$: the distribution of reactions is the same for all campaigns; $H_a$: at least one campaign differs (correct answer)
4. One-way ANOVA; $H_0$: mean reaction score is the same for all campaigns; $H_a$: at least one mean differs
5. Two-proportion $z$ test; $H_0$: $p_{\text{Positive,A}}=p_{\text{Positive,B}}$; $H_a$: not equal

Explanation: In AP Statistics, this tests the skill of chi-square setup for homogeneity or independence based on sampling. Separate samples per campaign, each with one categorical response (reaction), suit a chi-square test for homogeneity to see if distributions match across campaigns. Choice C's hypotheses state the null as same distributions and alternative as at least one difference. Choice A is a distractor as independence requires one sample with dual variables, not separate groups here. Mini-lesson: Goodness-of-fit evaluates one variable against theory; independence tests two variables' connection in a single sample; homogeneity compares one variable's distribution over multiple samples. Choice E is too narrow, focusing on proportions for one reaction type. C aligns perfectly.

Question 18

1. Chi-square test of independence; $H_0$: clinic and vaccination status are independent; $H_a$: they are not independent
2. Chi-square goodness-of-fit; $H_0$: vaccination status follows a specified distribution; $H_a$: it does not
3. Chi-square test for homogeneity; $H_0$: the distribution of vaccination status is the same for all clinics; $H_a$: at least one clinic differs (correct answer)
4. Matched-pairs test; $H_0$: no change in vaccination status; $H_a$: change
5. Two-proportion $z$ test; $H_0$: $p_A=p_B$; $H_a$: $p_A\ne p_B$

Explanation: The skill tested is setting up chi-square tests for homogeneity or independence in AP Statistics contexts. Separate random samples from each clinic, with one categorical response (vaccination status), indicate a chi-square test for homogeneity to compare distributions across clinics. Choice C's hypotheses correctly nullify that distributions are the same, with the alternative that at least one differs. A distractor is choice A, independence, which is unsuitable because it needs one sample with two variables per unit, not multiple samples. Mini-lesson on chi-square: Goodness-of-fit tests against a predefined distribution; independence checks two variables' relationship in one sample; homogeneity verifies uniform distributions across groups. Choice E limits to two proportions, ignoring the full categorical nature. Thus, C is right.

Question 19

1. Chi-square test for homogeneity; $H_0$: the distribution of commuting method is the same in all neighborhoods; $H_a$: at least one neighborhood has a different distribution (correct answer)
2. Chi-square test of independence; $H_0$: neighborhood and commuting method are independent in the population; $H_a$: they are not independent
3. Chi-square goodness-of-fit; $H_0$: commuting methods follow a specified distribution; $H_a$: they do not
4. Two-proportion $z$ test; $H_0$: $p_{\text{Car,North}}=p_{\text{Car,South}}$; $H_a$: not equal
5. Chi-square goodness-of-fit; $H_0$: each neighborhood has equal counts in each commuting category; $H_a$: not all counts are equal

Explanation: The skill here is identifying the correct chi-square test setup for homogeneity or independence based on study design in AP Statistics. This scenario involves separate random samples from each neighborhood, with one categorical response (commuting method) compared across groups, so a chi-square test for homogeneity is suitable to check if distributions are the same. The hypotheses in choice A correctly state that the distribution is the same for all neighborhoods under the null, with the alternative that at least one differs. A distractor like choice B, the test of independence, is wrong because independence requires one sample with two variables per unit, not multiple samples as here. For a mini-lesson: goodness-of-fit compares one variable to expected proportions; independence examines association in one sample with two variables; homogeneity tests if multiple populations have the same distribution for one variable. Choice E misapplies goodness-of-fit by assuming equal counts rather than proportions. Therefore, A is appropriate.

Question 20

1. Chi-square goodness-of-fit; $H_0$: the satisfaction distribution is the same across the three locations; $H_a$: at least one location differs
2. Chi-square test of independence; $H_0$: satisfaction rating and location are independent; $H_a$: satisfaction rating and location are associated
3. Chi-square test of homogeneity; $H_0$: the distribution of satisfaction ratings is the same for all locations; $H_a$: at least one location has a different distribution (correct answer)
4. Matched-pairs $t$ test; $H_0$: customers rate the three locations the same on average; $H_a$: averages differ
5. Chi-square test of homogeneity; $H_0$: satisfaction rating and location are independent; $H_a$: they are not independent

Explanation: This problem evaluates the ability to choose the right chi-square test and hypotheses when comparing categorical distributions across independent samples from different locations in AP Statistics. The design uses independent random samples from each store location, treating them as separate populations, with one categorical variable (satisfaction rating) measured per sample. Thus, the chi-square test of homogeneity is the proper choice to test if satisfaction distributions are the same across locations. Quick mini-lesson: Goodness-of-fit is for one sample versus a hypothesized distribution, homogeneity for identical distributions in multiple groups via independent sampling, and independence for two variables' association in a single sample. Distractor E names the test correctly as homogeneity but errs by using independence-style hypotheses, which don't fit the multi-population context. Option C is correct in both test name and hypotheses.