AP Statistics Quiz: Chi Square Homogeneity Or Independence Test

What this quiz covers

This quiz focuses on Chi Square Homogeneity Or Independence Test, 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. Because the p-value is greater than 0.05, there is convincing evidence that the distribution of news source preference differs between the two age groups.
2. Because the p-value is greater than 0.05, there is not convincing evidence that the distribution of news source preference differs between the two age groups. (correct answer)
3. Because the p-value is greater than 0.05, age group causes people to choose the same news source.
4. Because the p-value is less than 0.05, there is not convincing evidence that the distributions differ.
5. Because the p-value is greater than 0.05, the sample proves the population distributions are identical.

Explanation: This chi-square test of homogeneity compares news source preferences between two age groups. With a p-value greater than 0.05, we fail to reject the null hypothesis that both age groups have the same distribution of news source preferences, meaning there is not convincing evidence that the distributions differ. Choice A incorrectly concludes there is evidence of differences. Choice C wrongly introduces causation. Choice D has the wrong p-value comparison. Choice E overstates by claiming to prove identical distributions. Important distinction: not finding evidence of differences doesn't prove the groups are identical—it means any differences aren't statistically significant.

Question 2

1. Because $p<0.05$, there is convincing evidence that class standing and club participation are associated.
2. Because $p=0.096$, there is not convincing evidence of an association between class standing and club participation at the 5% level. (correct answer)
3. Because $p=0.096$, we can conclude class standing and club participation are independent in the population.
4. Because $p=0.096$, club participation causes students to become seniors.
5. Because $p=0.096$, the sample shows that exactly 9.6% of students participate in clubs.

Explanation: This chi-square test of independence examines association between class standing and club participation. The p-value of 0.096 is greater than the significance level of 0.05, so we fail to reject the null hypothesis. Choice B correctly states there is not convincing evidence of association at the 5% level. Choice A incorrectly claims p < 0.05. Choice C overstates—failing to find evidence of association doesn't prove independence. Choice D absurdly suggests reverse causation. Choice E misinterprets the p-value as a percentage of participation. Important: p-values are probabilities of getting results at least as extreme as observed if the null hypothesis is true, not population percentages.

Question 3

1. Because $p<0.05$, there is convincing evidence that the weight-loss outcome distribution is not the same for all three diets. (correct answer)
2. Because $p<0.05$, the diets and weight loss are independent.
3. Because $p<0.05$, the sample proves Diet A is the best diet for all people.
4. Because $p>0.05$, there is not convincing evidence of differences in weight-loss outcomes among the diets.
5. Because $p<0.05$, volunteers' personal motivation caused the differences, not diet assignment.

Explanation: This is a chi-square test of homogeneity comparing weight-loss distributions across three diets. The p-value of 0.018 is less than 0.05, so we reject the null hypothesis that all diets have the same distribution of outcomes. Choice A correctly concludes there is evidence the distributions differ. Choice B incorrectly claims independence when we've found differences. Choice C overstates by claiming one diet is best. Choice D misreads p as greater than 0.05. Choice E incorrectly dismisses the randomized design. Since this was a randomized experiment (not observational), we could potentially make causal claims, though the correct answer focuses on the statistical conclusion.

Question 4

1. There is convincing evidence that program participation reduces stress.
2. There is convincing evidence of an association between program participation and stress.
3. There is not convincing evidence of an association between program participation and stress among students. (correct answer)
4. Because $p=0.94$, the null hypothesis must be false.
5. We can conclude that 94% of students are not stressed.

Explanation: This question involves the chi-square test of independence to investigate association between program participation and stress. With p=0.94 much greater than alpha=0.05, we fail to reject the null, finding no convincing evidence of association. Choice A is a distractor implying causation, that participation reduces stress, but no association was detected. Association means variables are linked statistically, whereas causation requires demonstrating one affects the other, impossible from this survey data. The nearly equal proportions in the table yield the high p-value, indicating chance variation. This example shows how balanced data can lead to non-significant results.

Question 5

1. Because $p<0.01$, there is convincing evidence that the distribution of tree condition differs between Park A and Park B. (correct answer)
2. Because $p<0.01$, the park location causes individual trees to become diseased.
3. Because $p>0.01$, there is not convincing evidence that the distribution of tree condition differs between the parks.
4. Because $p=0.007$, the probability a tree is diseased is 0.007 in each park.
5. Because $p<0.01$, the trees in the two parks must have been sampled without randomness.

Explanation: This chi-square test of homogeneity compares tree condition distributions between two parks. The p-value of 0.007 is less than the significance level of 0.01, so we reject the null hypothesis and conclude the distributions differ. Choice A correctly states there is convincing evidence of different distributions. Choice B incorrectly implies the park location causes disease in individual trees. Choice C misreads 0.007 as greater than 0.01. Choice D misinterprets the p-value as a disease probability. Choice E wrongly questions the sampling method. The small p-value indicates the observed differences in tree conditions between parks are unlikely due to chance alone.

Question 6

1. Because the p-value is less than 0.05, there is convincing evidence that support for the ordinance is associated with neighborhood in the city. (correct answer)
2. Because the p-value is less than 0.05, living in a particular neighborhood causes a resident to support the ordinance.
3. Because the p-value is less than 0.05, the results show no association between neighborhood and support.
4. Because the p-value is greater than 0.05, there is convincing evidence of an association between neighborhood and support.
5. Because the p-value is less than 0.05, the sample proves the exact proportion supporting in each neighborhood.

Explanation: This chi-square test of independence examines whether support for a recycling ordinance is related to neighborhood. With a p-value less than 0.05, we reject the null hypothesis of independence and conclude there is convincing evidence of an association between neighborhood and support for the ordinance. Choice B incorrectly claims causation—living in a neighborhood doesn't necessarily cause support; other factors may be involved. Choice C contradicts the finding by claiming no association. Choice D has the wrong p-value comparison. Choice E overstates by claiming to prove exact proportions. Association indicates a relationship exists but doesn't explain why or establish cause-and-effect.

Question 7

1. Because the p-value is greater than 0.05, there is not convincing evidence of an association between device type and purchase outcome for site visits. (correct answer)
2. Because the p-value is greater than 0.05, device type and purchase outcome are proven to be independent in all circumstances.
3. Because the p-value is greater than 0.05, device type causes purchase outcome to be the same across devices.
4. Because the p-value is less than 0.05, there is not convincing evidence of an association between device type and purchase outcome.
5. Because the p-value is greater than 0.05, the sample shows that exactly the same percentage purchase on each device in the population.

Explanation: This chi-square test of independence examines whether device type and purchase outcome are associated. With a p-value greater than 0.05, we fail to reject the null hypothesis of independence, meaning there is not convincing evidence of an association between the variables. Choice B overstates by claiming to prove independence in all circumstances. Choice C incorrectly introduces causation. Choice D has the wrong p-value comparison. Choice E misinterprets the result as proving exact percentages. Important concept: failing to reject the null hypothesis doesn't prove independence—it simply means we lack sufficient evidence to claim association.

Question 8

1. Because the p-value is greater than 0.05, there is convincing evidence that preference and department are associated.
2. Because the p-value is greater than 0.05, there is not convincing evidence of an association between department and preferred work arrangement among employees. (correct answer)
3. Because the p-value is greater than 0.05, department causes employees to have the same preferences.
4. Because the p-value is less than 0.05, there is not convincing evidence of an association.
5. Because the p-value is greater than 0.05, the sample proves the population preferences are identical across departments.

Explanation: This chi-square test of independence examines whether work arrangement preference is associated with department. With a p-value greater than 0.05, we fail to reject the null hypothesis of independence, meaning there is not convincing evidence of an association between department and preferred work arrangement. Choice A incorrectly concludes there is evidence of association. Choice C wrongly introduces causation. Choice D has the wrong p-value comparison. Choice E overstates by claiming to prove identical preferences. Key concept: failing to find evidence of association doesn't mean the variables are definitely independent—it means we lack sufficient evidence to claim they're related at the 0.05 significance level.

Question 9

1. There is convincing evidence that the distribution of growth categories differs among the fertilizers. (correct answer)
2. There is not convincing evidence of differences because $p<0.001$ is too small to be reliable.
3. There is convincing evidence that fertilizer choice and growth category are independent.
4. We can conclude Fertilizer B causes high growth for all plants.
5. We should fail to reject $H_0$ because the sample size is the same in each fertilizer group.

Explanation: This question tests the chi-square test of homogeneity, comparing distributions of growth categories across fertilizer groups. The p-value less than 0.001 is well below a typical alpha=0.05, so we reject the null hypothesis and conclude there is convincing evidence of differences in distributions. Choice D is a distractor that overstates causation, claiming Fertilizer B causes high growth for all plants, but the test only shows differing distributions. Association in this context means the growth outcomes vary by fertilizer, but causation would need confirmation that the fertilizer directly affects growth, perhaps through controlled trials. The table reveals Fertilizer B has more high growth, contributing to the significant result. Equal sample sizes per group support the test's assumptions, but the conclusion hinges on the p-value.

Question 10

1. There is convincing evidence that preferred news source and political affiliation are associated in the state.
2. There is not convincing evidence of an association between preferred news source and political affiliation in the state. (correct answer)
3. Because $p=0.072$, we should reject $H_0$ at $\alpha=0.05$.
4. The results show that political affiliation causes people to choose different news sources.
5. We can conclude that 7.2% of adults prefer TV news.

Explanation: This question involves the chi-square test of independence to check for an association between political affiliation and preferred news source. Since the p-value of 0.072 exceeds alpha=0.05, we fail to reject the null hypothesis, indicating no convincing evidence of an association. Choice D serves as a distractor by incorrectly suggesting causation, that affiliation causes news source preferences, even though no association was found. Association means the variables covary statistically, whereas causation implies a directional influence, which this observational study cannot prove. The table's counts show similar patterns across affiliations, explaining the higher p-value. This outcome suggests any observed differences could be due to random sampling variation.

Question 11

1. There is convincing evidence that the distribution of entree preference differs by region.
2. There is not convincing evidence that entree preference differs by region. (correct answer)
3. Because $p=0.20$, we should reject $H_0$ at $\alpha=0.05$.
4. The results prove that region has no effect on entree preference.
5. We can conclude 20% of all customers prefer beef.

Explanation: This question assesses the chi-square test of homogeneity to see if entree preferences differ by region. The p-value of 0.20 is greater than alpha=0.05, so we fail to reject the null hypothesis, concluding no convincing evidence of differences. Choice D distracts by claiming the results 'prove' no effect, but failing to reject only means insufficient evidence, not proof of no difference. While association would indicate varying preferences, causation might suggest region influences choice, but neither is supported here. The similar counts across regions explain the high p-value, suggesting uniformity. This test underscores that large p-values reflect patterns consistent with the null hypothesis.

Question 12

1. Because the p-value is less than 0.05, the data provide convincing evidence that attendance category and course result are associated for students in this school. (correct answer)
2. Because the p-value is less than 0.05, higher attendance causes passing the course.
3. Because the p-value is less than 0.05, there is not convincing evidence of an association.
4. Because the p-value is greater than 0.05, there is convincing evidence of an association.
5. Because the p-value is less than 0.05, the sample proves the exact passing rate for each attendance category in the population.

Explanation: This chi-square test of independence examines the relationship between attendance category and course result. With a p-value less than 0.05, we reject the null hypothesis of independence and conclude there is convincing evidence that attendance and course result are associated for students in this school. Choice B incorrectly claims causation—while the variables are associated, we cannot conclude that attendance causes passing without a controlled experiment. Choice C contradicts the finding. Choice D has the wrong p-value comparison. Choice E overstates by claiming to prove exact rates. Association suggests a relationship but doesn't establish whether attendance affects grades or if both are influenced by other factors.

Question 13

1. Because the p-value is less than 0.05, there is convincing evidence that preferred activity and grade level are associated in the population of students in the district. (correct answer)
2. Because the p-value is less than 0.05, grade level causes students to prefer different activities.
3. Because the p-value is less than 0.05, the two variables are independent in the population.
4. Because the p-value is greater than 0.05, there is convincing evidence of an association between preferred activity and grade level.
5. Because the p-value is less than 0.05, the sample proves that every grade level has the same distribution of preferences.

Explanation: This question tests understanding of chi-square tests of independence, which examine whether two categorical variables are associated. With a p-value less than 0.05, we reject the null hypothesis of independence and conclude there is convincing evidence of an association between preferred activity and grade level. Choice B incorrectly claims causation, which chi-square tests cannot establish. Choice C contradicts the p-value interpretation by claiming independence. Choice D has the wrong p-value comparison. Choice E misinterprets the test as proving identical distributions. Remember: association does not imply causation—the test only shows the variables are related, not that one causes the other.

Question 14

1. There is not convincing evidence of an association because $p=0.030$ is greater than $0.05$.
2. There is convincing evidence that age group and preferred exhibit type are associated among museum visitors. (correct answer)
3. We can conclude that visiting the museum causes seniors to prefer history exhibits.
4. We can conclude that the distributions of preferred exhibit type are the same for all age groups.
5. Because $p=0.030$, we should fail to reject $H_0$ at $\alpha=0.05$.

Explanation: This question evaluates the chi-square test of independence to determine if age group and preferred exhibit are associated. With p=0.030 less than alpha=0.05, we reject the null, finding convincing evidence of association. A common distractor is choice C, which wrongly infers causation, suggesting museum visits cause seniors' preferences, but association does not imply causation. Association means the preferences differ by age, while causation would require evidence that age directly drives preferences. The table shows distinct patterns, like children favoring science, leading to the significant p-value. This result demonstrates how demographic factors can correlate with interests without causal links.

Question 15

1. There is convincing evidence that grade level and preferred study location are associated in the population of students at this school. (correct answer)
2. There is convincing evidence that grade level causes students to choose different study locations.
3. There is not convincing evidence of an association because the sample sizes in some cells are not equal.
4. There is not convincing evidence of an association because $p=0.018$ is greater than $0.05$.
5. We can conclude that the proportion who prefer the library is the same for all grade levels.

Explanation: This question assesses the chi-square test of independence, which examines whether two categorical variables are associated in a population. The p-value of 0.018 is less than the significance level alpha=0.05, so we reject the null hypothesis of no association and conclude there is convincing evidence of an association between grade level and preferred study location. A common distractor is choice B, which incorrectly infers causation from the association, suggesting grade level causes study location preferences. Remember, association means the variables are statistically related, but it does not imply causation; causation would require evidence from a controlled experiment showing one variable directly influences the other. In this case, the test only shows that the distribution of study locations differs by grade level, not that grade level causes the difference. The table provides observed counts, and the low p-value indicates the observed differences are unlikely under the null hypothesis.

Question 16

1. There is not convincing evidence of an association because $p=0.006$ is less than $0.05$.
2. There is convincing evidence that device type and on-time submission are associated among students. (correct answer)
3. The data show that using a laptop causes students to submit on time.
4. We can conclude that exactly 0.6% of submissions are late.
5. We should fail to reject $H_0$ because there are three device categories.

Explanation: This question tests the chi-square test of independence for association between device type and submission timeliness. The p-value of 0.006 is less than alpha=0.05, so we reject the null and conclude convincing evidence of association. Choice C distracts by claiming causation, that laptops cause on-time submissions, but the test only identifies association. Association implies statistical dependence, like higher on-time rates for laptops, but causation needs experimental proof of influence. The table's varying late rates by device contribute to the low p-value. This highlights how categorical data can reveal practical associations in educational settings.

Question 17

1. Because $p<0.05$, there is convincing evidence that preferred learning format and grade level are associated in the population of students in the district. (correct answer)
2. Because $p<0.05$, grade level causes students to prefer different learning formats.
3. Because $p<0.05$, the sample results prove that preferred learning format and grade level are independent in the population.
4. Because $p>0.05$, there is not enough evidence of an association between preferred learning format and grade level.
5. Because $p<0.05$, the distribution of grade levels is the same for all learning-format groups.

Explanation: This question tests understanding of chi-square tests of independence, which examine whether two categorical variables are associated. With a p-value of 0.032, which is less than the significance level of 0.05, we reject the null hypothesis of independence and conclude there is convincing evidence of an association between preferred learning format and grade level. Choice A correctly states this conclusion. Choice B incorrectly implies causation when the test only shows association. Choice C contradicts the p-value by claiming independence. Choice D misreads the p-value as greater than 0.05. Choice E misinterprets what association means. Remember: chi-square tests can show association but never prove causation.

Question 18

1. Because $p=0.67$, there is convincing evidence that remote-work preference depends on department.
2. Because $p=0.67$, there is not convincing evidence of an association between department and remote-work preference. (correct answer)
3. Because $p=0.67$, department causes employees to have the same remote-work preference.
4. Because $p=0.67$, we can conclude the two variables are independent for every possible employee population.
5. Because $p=0.67$, 67% of employees prefer remote work.

Explanation: This chi-square test of independence examines whether department and remote-work preference are associated. The p-value of 0.67 is much greater than 0.05, so we fail to reject the null hypothesis of independence. Choice B correctly states there is not convincing evidence of association. Choice A incorrectly claims evidence of dependence with such a large p-value. Choice C wrongly mentions causation. Choice D overstates the conclusion to every possible population. Choice E misinterprets the p-value as a percentage preferring remote work. Remember: large p-values mean the data are consistent with the null hypothesis of no association.

Question 19

1. Because $p=0.12$, there is convincing evidence that gym membership and transportation type are associated.
2. Because $p=0.12$, there is not convincing evidence of an association between gym membership and transportation type. (correct answer)
3. Because $p=0.12$, transportation type causes gym membership status.
4. Because $p=0.12$, the probability of gym membership is 0.12.
5. Because $p<0.05$, we should reject the null hypothesis of independence.

Explanation: This chi-square test of independence examines association between gym membership and transportation type. With p-value = 0.12, which is greater than 0.05, we fail to reject the null hypothesis of independence. Choice B correctly states there is not convincing evidence of association. Choice A incorrectly claims evidence of association with p > 0.05. Choice C wrongly introduces causation. Choice D misinterprets the p-value as a probability of membership. Choice E incorrectly states p < 0.05. The p-value of 0.12 means that if the variables were truly independent, we'd see results at least this extreme about 12% of the time.

Question 20

1. Because $p<0.05$, there is convincing evidence that brand preference differs between the two age groups.
2. Because $p=0.41$, there is not convincing evidence that the distribution of brand preference differs between the two age groups. (correct answer)
3. Because $p=0.41$, the two age groups have exactly the same brand-preference proportions in the population.
4. Because $p=0.41$, age group causes brand preference to be the same for everyone.
5. Because $p=0.41$, the brands are independent of each other.

Explanation: This question involves a chi-square test of homogeneity, which compares distributions across different groups. The p-value of 0.41 is much greater than the significance level of 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence that brand preference distributions differ between age groups. Choice B correctly states this conclusion. Choice A incorrectly compares p to 0.05. Choice C overstates the conclusion—failing to reject doesn't prove exact equality. Choice D incorrectly mentions causation. Choice E misunderstands what's being tested. Key lesson: when p > α, we lack evidence of a difference, but this doesn't prove the distributions are identical.