- A public health group took independent random samples from three neighborhoods (North, Central, South) and recorded whether residents are up-to-date on a flu vaccine (Yes/No). The results are shown in the two-way table. A chi-square test of homogeneity produced p-value =0.0006. What conclusion is appropriate at the 5% significance level?
Answer: Because p<0.05, there is convincing evidence that vaccination status is associated with neighborhood (the distributions are not all the same).
This chi-square test of homogeneity compares vaccination rates across three neighborhoods. With p-value = 0.0006, far less than 0.05, we have very strong evidence against the null hypothesis of equal distributions. Choice A correctly concludes there is convincing evidence that vaccination status is associated with neighborhood (distributions differ). Choice B incorrectly implies causation from observational data. Choice C contradicts the finding by claiming equal proportions. Choice D misreads p as greater than 0.05. Choice E wrongly attributes the result to sampling bias. The very small p-value indicates the observed differences are unlikely due to chance alone. - A museum studies whether visitor age group is associated with preferred exhibit type. A random sample of visitors is surveyed and classified by age group (Child, Adult, Senior) and preferred exhibit (Art, History, Science). The results are shown below. A chi-square test of independence is performed and results in p=0.030. What conclusion is appropriate at the α=0.05 level?
Two-way table (counts):
| Age group | Art | History | Science |
|---|
| Child | 20 | 15 | 45 |
| Adult | 55 | 60 | 35 |
| Senior | 40 | 55 | 15 |
Answer: There is convincing evidence that age group and preferred exhibit type are associated among museum visitors.
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.
- An educator wants to know whether attendance category is related to whether a student passes a course. A random sample of students was selected, and each student was classified by attendance (High, Medium, Low) and course result (Pass, Fail). A chi-square test of independence was conducted at α=0.05. The observed counts and p-value are shown.
What conclusion is appropriate?
Answer: 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.
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.
- A school district wants to know whether preferred after-school activity is associated with grade level. A random sample of students was selected, and each student reported one preferred activity. A chi-square test of independence was performed at the α=0.05 level. The two-way table of observed counts and the computer output p-value are shown.
What conclusion is appropriate?
Answer: 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.
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.
- A technology teacher wants to know whether device type used for homework is associated with whether assignments are submitted on time. A random sample of students is taken and categorized by primary device (Phone, Tablet, Laptop) and submission status (On time, Late). The two-way table below summarizes the data. A chi-square test of independence gives p=0.006. What conclusion is appropriate at the α=0.05 level?
Two-way table (counts):
| Device | On time | Late |
|---|
| Phone | 35 | 25 |
| Tablet | 38 | 12 |
| Laptop | 70 | 20 |
Answer: There is convincing evidence that device type and on-time submission are associated among students.
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.
- A restaurant chain wants to compare customer satisfaction ratings across four locations. Independent random samples of customers were surveyed at each location and rated their experience as Satisfied or Not satisfied. A chi-square test of homogeneity was performed at α=0.05. The observed counts and p-value are shown.
What conclusion is appropriate?
Answer: Because the p-value is greater than 0.05, there is not convincing evidence that satisfaction rates differ among the four locations.
This chi-square test of homogeneity compares satisfaction rates across four restaurant locations. With a p-value greater than 0.05, we fail to reject the null hypothesis that all locations have the same satisfaction distribution, meaning there is not convincing evidence that satisfaction rates differ among locations. Choice B overstates by claiming exact equality in the population. Choice C has the wrong p-value comparison. Choice D incorrectly introduces causation. Choice E misinterprets the result as showing association. Key point: failing to find evidence of differences doesn't prove the locations are identical—it means any differences aren't statistically significant at the 0.05 level.
- A researcher wants to compare preferred news source across two age groups. Independent random samples of adults ages 18–34 and ages 35+ were surveyed, and each person selected one primary news source (TV, Online, Print). A chi-square test of homogeneity was performed at α=0.05. The observed counts and p-value are shown.
What conclusion is appropriate?
Answer: 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.
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.
- A researcher randomly assigned 180 volunteers to one of three diets (Diet A, Diet B, Diet C) and recorded whether each volunteer lost weight after 8 weeks (Yes/No). The two-way table is shown. A chi-square test of homogeneity for the distribution of weight-loss outcomes across diets produced p-value =0.018. What conclusion is appropriate at the 5% significance level?
Answer: Because p<0.05, there is convincing evidence that the weight-loss outcome distribution is not the same for all three diets.
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. - A city council wants to know whether support for a new recycling ordinance is related to neighborhood. A random sample of residents was taken, and each resident was classified by neighborhood (North, South, East) and response (Support, Oppose). A chi-square test of independence was conducted at α=0.05. The observed counts and p-value are shown.
What conclusion is appropriate?
Answer: 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.
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.
- A school wants to know whether students' preferred study location is associated with grade level. A random sample of students is taken, and each student reports their preferred location (Home, Library, or Study Hall). The results are summarized below. A chi-square test of independence is performed and yields p=0.018. What conclusion is appropriate at the α=0.05 level?
Two-way table (counts):
| Grade level | Home | Library | Study Hall |
|---|
| 9th | 30 | 18 | 12 |
| 10th | 22 | 25 | 13 |
| 11th | 18 | 28 | 14 |
| 12th | 20 | 15 | 25 |
Answer: There is convincing evidence that grade level and preferred study location are associated in the population of students at this school.
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.
- A university randomly sampled students and recorded their class standing (First-year, Sophomore, Junior, Senior) and whether they participate in a campus club (Yes/No). The two-way table is shown. A chi-square test of independence gave p-value =0.096. What conclusion is appropriate at the 5% significance level?
Answer: Because p=0.096, there is not convincing evidence of an association between class standing and club participation at the 5% level.
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. - A school district surveyed a random sample of students to see whether preferred learning format (In-person, Hybrid, Online) is associated with grade level (9th, 10th, 11th, 12th). The results are shown in the two-way table. A chi-square test of independence produced p-value =0.032. What conclusion is appropriate at the 5% significance level?
Answer: 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.
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. - A company randomly sampled employees and recorded department (Sales, Engineering, HR) and whether the employee prefers remote work (Yes/No). The two-way table is shown. A chi-square test of independence resulted in p-value =0.67. What conclusion is appropriate at the 5% significance level?
Answer: Because p=0.67, there is not convincing evidence of an association between department and remote-work preference.
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. - A counselor examines whether participation in an after-school program is associated with whether students report feeling stressed. A random sample of students is surveyed and classified as Program (Yes/No) and Stressed (Yes/No). The results are shown below. A chi-square test of independence yields p=0.94. What conclusion is appropriate at the α=0.05 level?
Two-way table (counts):
| Program participation | Stressed: Yes | Stressed: No |
|---|
| Yes | 48 | 52 |
| No | 47 | 53 |
Answer: There is not convincing evidence of an association between program participation and stress among students.
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.
- A researcher wants to determine whether political affiliation is associated with preferred news source among adults in a state. A random sample of adults is surveyed and classified by affiliation (Democrat, Republican, Independent) and preferred news source (TV, Online, Print). The results are shown below. A chi-square test of independence yields p=0.072. What conclusion is appropriate at the α=0.05 level?
Two-way table (counts):
| Affiliation | TV | Online | Print |
|---|
| Democrat | 60 | 85 | 15 |
| Republican | 75 | 70 | 20 |
| Independent | 55 | 80 | 20 |
Answer: There is not convincing evidence of an association between preferred news source and political affiliation in the state.
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.
- A political scientist wants to compare party affiliation across three colleges (College X, Y, Z). Independent random samples of students were selected from each college and classified as Democrat, Republican, or Independent. A chi-square test of homogeneity was performed at α=0.05. The observed counts and p-value are shown.
What conclusion is appropriate?
Answer: Because the p-value is less than 0.05, there is convincing evidence that party affiliation distributions differ among the three colleges.
This chi-square test of homogeneity compares party affiliation distributions across three colleges. With a p-value less than 0.05, we reject the null hypothesis that all colleges have the same party affiliation distribution and conclude there is convincing evidence that the distributions differ among the colleges. Choice A has the wrong p-value comparison. Choice C incorrectly claims causation—attending a college doesn't necessarily cause party affiliation. Choice D contradicts the finding. Choice E overstates by claiming to prove exact proportions. The test shows differences exist but doesn't explain why—colleges may attract different student populations or have different regional influences.
- A community center randomly sampled adults and recorded whether they have a gym membership (Yes/No) and their transportation type to work (Car, Public transit, Walk/Bike). The two-way table is shown. A chi-square test of independence gave p-value =0.12. What conclusion is appropriate at the 5% significance level?
Answer: Because p=0.12, there is not convincing evidence of an association between gym membership and transportation type.
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. - A public health researcher wants to compare smoking status across three regions of a state. Independent random samples of adults were taken from each region, and each adult was classified as a current smoker, former smoker, or never smoker. A chi-square test of homogeneity was conducted at α=0.05. The observed counts and p-value are shown.
What conclusion is appropriate?
Answer: Because the p-value is less than 0.05, there is convincing evidence that at least one region's smoking-status distribution differs from the others.
This question involves a chi-square test of homogeneity, which compares distributions across multiple populations. With a p-value less than 0.05, we reject the null hypothesis that all regions have the same smoking-status distribution and conclude at least one region differs from the others. Choice A incorrectly states the p-value is greater than 0.05. Choice C wrongly claims causation, which statistical tests cannot establish. Choice D overstates the conclusion by claiming to prove exact percentages. Choice E misinterprets the result as showing independence. The key distinction: homogeneity tests compare distributions across groups, while independence tests examine relationships between variables within one population.
- A marketing team wants to know whether device type is related to whether a customer completes an online purchase. From a random sample of site visits, each visit was classified by device (Phone, Tablet, Computer) and outcome (Purchase, No purchase). A chi-square test of independence was performed at the α=0.05 level. The observed counts and p-value are shown.
What conclusion is appropriate?
Answer: 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.
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.
- A city randomly sampled residents and asked two questions: whether they own a dog (Yes/No) and whether they live in an apartment, townhouse, or single-family home. The results are shown in the two-way table. A chi-square test of independence resulted in p-value =0.004. What conclusion is appropriate at the 1% significance level?
Answer: Because p<0.01, there is convincing evidence of an association between dog ownership and housing type in the city.
This chi-square test of independence examines whether dog ownership and housing type are associated. With p-value = 0.004, which is less than the significance level of 0.01, we reject the null hypothesis and conclude there is convincing evidence of an association. Choice B correctly states this conclusion. Choice A incorrectly states p > 0.01. Choice C wrongly implies causation from an observational study. Choice D contradicts the finding of association. Choice E overstates the conclusion to every resident rather than the population. Remember: association means the variables are related, but doesn't indicate which causes which.
