Chi-Square Goodness of Fit (Test) - AP Statistics

1. A school district predicts that the distribution of students' primary mode of transportation is 40% bus, 35% car, and 25% walk/bike. A random sample of 120 students is surveyed, and a chi-square goodness-of-fit test is conducted at $\alpha=0.05$. The summary is below. What conclusion is appropriate? Observed/Expected counts and p-value: - Bus: Observed 44, Expected 48 - Car: Observed 39, Expected 42 - Walk/Bike: Observed 37, Expected 30 - p-value = 0.18
   Answer: Because $p>\alpha$, fail to reject $H_0$; there is not convincing evidence that the population distribution of transportation modes differs from the district's prediction. This question focuses on the chi-square goodness-of-fit test in AP Statistics to evaluate if observed transportation modes fit the district's predicted proportions. Since the p-value of 0.18 exceeds α=0.05, we fail to reject the null hypothesis, meaning there is not convincing evidence that the population distribution differs from the prediction. The sample variations are plausible under the null model. Choice D distracts by claiming non-rejection proves the prediction true for all students, but it only indicates a lack of evidence against it, not definitive proof. A mini-lesson on chi-square conclusions: Failing to reject H0 suggests the data is consistent with the expected distribution, but other distributions might also fit; it's about insufficient evidence for deviation, not confirmation. Consider sample size and power when interpreting non-significant results.
2. A political analyst claims that in a certain county, party registration is 45% A, 40% B, and 15% C. A random sample of 200 registered voters produced the counts below. A chi-square goodness-of-fit test at $\alpha=0.05$ gives $p$-value $=0.07$. What conclusion is appropriate?
   Answer: Fail to reject $H_0$; there is not convincing evidence that the population party-registration distribution differs from 45%/40%/15%. This question involves a p-value slightly above the significance level. The p-value (0.07) is greater than α (0.05), so we fail to reject the null hypothesis. The null hypothesis states that party registration follows 45% A, 40% B, 15% C. Choice A incorrectly claims we reject H₀ and misunderstands that p > α means we fail to reject, not that differences are significant. Choice C commits the error of "proving" the analyst's percentages - we can only fail to find evidence against them. Choice D focuses on the sample rather than making a population inference. When p > α, we conclude there's not convincing evidence that the county's registration distribution differs from the claimed 45%/40%/15% split.
3. A wildlife biologist claims that in a certain habitat, the long-run distribution of frog colors is: Brown 50%, Green 30%, Gray 20%. A random sample of $n=60$ frogs is observed and recorded below. A chi-square goodness-of-fit test results in p-value = 0.07. What conclusion is appropriate at the $\alpha=0.05$ level? Observed/Expected counts (with p-value):

   Color	Observed	Expected
   Brown	25	30
   Green	24	18
   Gray	11	12

   p-value = 0.07
   Answer: Fail to reject $H_0$; there is not convincing evidence that the frog color distribution differs from the biologist's claim. This question assesses the chi-square goodness-of-fit test in AP Statistics for frog color distributions in a habitat. The p-value of 0.07 is greater than 0.05, leading us to fail to reject the null hypothesis. Therefore, there is not convincing evidence that the distribution differs from the biologist's claim. A common distractor is choice C, which erroneously claims failing to reject proves the percentages true, but it only means lack of evidence against them. In chi-square goodness-of-fit tests, df equals categories minus 1, here df=2. The p-value assesses compatibility of data with H0; values above alpha support retaining H0. Phrase conclusions to avoid overstating, focusing on evidence levels.
4. A restaurant owner claims that customers pay using 4 methods in these proportions: 0.50 credit card, 0.20 debit card, 0.20 cash, 0.10 mobile pay. A random sample of 150 transactions is recorded and a chi-square goodness-of-fit test is conducted for $H_0$: the population payment-method distribution matches the claim versus $H_a$: it does not. The p-value is 0.95. The observed and expected counts are shown. What conclusion is appropriate at the $\alpha=0.05$ level?
   Answer: Fail to reject $H_0$ because the p-value is 0.95; there is not convincing evidence the population distribution differs from the claim. This restaurant payment methods question features an extremely large p-value. With p-value = 0.95 >> α = 0.05, we fail to reject H₀. This very large p-value indicates the observed data is highly consistent with the claimed payment method proportions, though it doesn't prove the claim is exactly true. Choice A incorrectly interprets a large p-value as evidence against H₀. Choice C overstates by claiming the null is proven true. Choice D incorrectly claims p < 0.05. Choice E misunderstands that sampling variability means observed counts won't exactly match expected counts. Key concept: large p-values suggest good agreement between data and H₀, but never prove H₀ is true.
5. A coffee shop manager believes customers order drinks in these proportions: 45% hot coffee, 30% iced coffee, 15% latte, 10% tea. Over a randomly selected set of 200 customers, the orders are recorded and a chi-square goodness-of-fit test is performed at $\alpha=0.05$. The summary is below. What conclusion is appropriate? Observed/Expected counts and p-value: - Hot coffee: Observed 78, Expected 90 - Iced coffee: Observed 70, Expected 60 - Latte: Observed 32, Expected 30 - Tea: Observed 20, Expected 20 - p-value = 0.029
   Answer: Because $p<\alpha$, reject $H_0$; there is convincing evidence the population ordering distribution differs from the manager's belief. This question evaluates the chi-square goodness-of-fit test in AP Statistics, comparing observed drink orders to the manager's believed proportions. Since p=0.029 is less than α=0.05, we reject the null hypothesis, providing convincing evidence that the population distribution differs from the belief. The differences are unlikely due to chance alone. Choice D is a distractor, wrongly interpreting rejection as proving an exact proportion like 35% for iced coffee, but the test assesses the overall fit, not specific values. In a mini-lesson on chi-square conclusions, rejecting H0 means strong evidence against the hypothesized distribution, pointing to an alternative, but it doesn't specify what that is—residual analysis can highlight discrepancies. Focus on evidence, not proof or bias causation.
6. A genetics lab expects offspring phenotypes in a 9:3:3:1 ratio across four categories. In a random sample of 320 offspring, the lab records counts in each category and runs a chi-square goodness-of-fit test of $H_0$: the population follows the 9:3:3:1 ratio versus $H_a$: it does not. The p-value is 0.008. The observed and expected counts are shown. What conclusion is appropriate at the $\alpha=0.01$ level?
   Answer: Reject $H_0$ because the p-value is less than 0.01; there is evidence the population ratio is not 9:3:3:1. This genetics problem tests chi-square conclusions at a different significance level. The p-value of 0.008 is less than α = 0.01, so we reject H₀. Rejecting the null hypothesis means we have evidence that the population ratio differs from the expected 9:3:3:1 ratio. Choice B incorrectly states we fail to reject when p < 0.01. Choices C and D incorrectly claim p > 0.01 when it's actually 0.008. Choice E incorrectly suggests chi-square can't be used with ratios. Important principle: chi-square goodness-of-fit tests can compare observed data to any specified distribution, including ratios that convert to proportions.
7. A biology lab expects a 1:2:1 ratio of flower colors (white:pink:red) in a large population of plants. A random sample of 80 flowers is observed, and a chi-square goodness-of-fit test is conducted at $\alpha=0.05$. The results are below. What conclusion is appropriate? Observed/Expected counts and p-value: - White: Observed 18, Expected 20 - Pink: Observed 44, Expected 40 - Red: Observed 18, Expected 20 - p-value = 0.54
   Answer: Because $p>\alpha$, fail to reject $H_0$; the sample provides insufficient evidence that the population distribution differs from a 1:2:1 ratio. This question involves the chi-square goodness-of-fit test in AP Statistics to test if flower colors follow a 1:2:1 ratio in the population. With p=0.54 greater than α=0.05, we fail to reject the null hypothesis, providing insufficient evidence that the population distribution differs from the expected ratio. The sample is consistent with the lab's expectation. Choice D misleads by suggesting non-rejection proves exactly half are pink, but it doesn't; it only shows no strong evidence against the ratio. For a mini-lesson on chi-square conclusions, failing to reject means the observed data could plausibly come from the hypothesized model, but it doesn't rule out other models—use confidence intervals or further tests for more insight. Contextualize to the population, noting that 'proof' is not a statistical term here.
8. A music streaming service claims that listening time is distributed across 4 genres as follows: Pop 40%, Hip-hop 30%, Rock 20%, Classical 10%. A random sample of 300 listening sessions produced the counts below. A chi-square goodness-of-fit test at $\alpha=0.10$ gives $p$-value $=0.62$. What conclusion is appropriate?
   Answer: Fail to reject $H_0$; there is not convincing evidence that the population distribution of genres differs from the claimed percentages. This question involves a chi-square goodness-of-fit test with a large p-value. The p-value (0.62) is much greater than α (0.10), so we fail to reject the null hypothesis. The null hypothesis states that the population follows the claimed genre distribution (40% Pop, 30% Hip-hop, 20% Rock, 10% Classical). Choice A incorrectly interprets a large p-value as evidence against H₀ - actually, large p-values support H₀. Choice C commits the error of claiming we've proven the exact percentages - failing to reject H₀ doesn't prove it's true. Choice E misinterprets the p-value as relating to expected counts. When p > α, we conclude there's not convincing evidence that the actual distribution differs from the claimed percentages.
9. A political scientist claims that support for three candidates in a district is 35% for Candidate A, 40% for Candidate B, and 25% for Candidate C. A random sample of 240 likely voters is surveyed. A chi-square goodness-of-fit test is performed for $H_0$: the population proportions match 0.35/0.40/0.25 versus $H_a$: they do not. The p-value is 0.057. The observed and expected counts are shown. What conclusion is appropriate at the $\alpha=0.05$ level?
   Answer: Fail to reject $H_0$ because the p-value is 0.057; there is not sufficient evidence at $\alpha=0.05$ that population support differs from the claimed proportions. This political science question involves a p-value very close to the significance level. With p-value = 0.057 > α = 0.05, we fail to reject H₀. Even though 0.057 is close to 0.05, we must follow the strict decision rule: we don't have sufficient evidence at the 0.05 level to conclude population support differs from the claimed proportions. Choice A incorrectly rejects H₀. Choices C and D confuse the decision rules. Choice E incorrectly suggests we should reject just because p is close to α. Important principle: hypothesis testing uses strict cutoffs; "close" doesn't count, and we don't round p-values to make different decisions.
10. A city's transit authority claims riders use 5 payment methods with the following population proportions: 40% Card, 25% Mobile, 20% Cash, 10% Pass, 5% Other. In a random sample of 400 rides, the observed counts are: Card 150, Mobile 120, Cash 70, Pass 45, Other 15. A chi-square goodness-of-fit test of $H_0$: the population proportions match the claimed values gives $p$-value = 0.003. Expected counts under $H_0$ are: Card 160, Mobile 100, Cash 80, Pass 40, Other 20. What conclusion is appropriate at $\alpha=0.01$?​
    Answer: Reject $H_0$; there is convincing evidence that the population distribution of payment methods differs from the claimed proportions. This question tests understanding of significance levels in chi-square tests. The null hypothesis states the population proportions match the transit authority's claimed values. With p-value = 0.003 < α = 0.01, we reject H₀. This provides convincing evidence that the population distribution of payment methods differs from the claimed proportions. Choice A incorrectly compares 0.003 to 0.05 when α = 0.01 is specified. Choice C fails to reject despite the small p-value. Choice D misunderstands the scope of inference. Choice E irrelevantly mentions the expected count condition. When p-value < α, we always reject H₀ regardless of how small the p-value is.
11. A city transit agency claims that riders use 4 ticket types in the following proportions: Regular 50%, Student 20%, Senior 20%, and Day Pass 10%. A random sample of 200 ticket purchases is recorded, and a chi-square goodness-of-fit test is performed at $\alpha=0.05$. The results are summarized below (including the p-value). What conclusion is appropriate? Observed/Expected counts and p-value: - Regular: Observed 92, Expected 100 - Student: Observed 48, Expected 40 - Senior: Observed 36, Expected 40 - Day Pass: Observed 24, Expected 20 - p-value = 0.041
    Answer: Because $p<\alpha$, reject $H_0$; there is convincing evidence that the distribution of ticket types in the population differs from the claimed proportions. This question assesses the chi-square goodness-of-fit test in AP Statistics, which determines if observed categorical data fits an expected distribution. Here, the p-value of 0.041 is less than the significance level α=0.05, so we reject the null hypothesis that the population proportions match the claimed ones. This provides convincing evidence that the actual distribution of ticket types differs from the agency's claims. A common distractor is choice D, which incorrectly states that the test proves exact proportions like 50% for Regular tickets, but statistical tests do not prove exact values; they only assess evidence against the null. In a mini-lesson on chi-square conclusions, remember that rejecting H0 suggests the data does not fit the expected model, but it doesn't specify how or why it differs—further analysis like residuals can help identify which categories contribute most to the discrepancy. Always contextualize the conclusion to the population, not just the sample.
12. A school claims that students choose among 5 lunch options with equal frequency. A student randomly records the lunch choice of 150 students on one day and obtains the counts shown in a table. A chi-square goodness-of-fit test is conducted for $H_0$: all 5 options are equally likely vs. $H_a$: not all options are equally likely. The p-value is 0.62. What conclusion is appropriate at the $\alpha=0.05$ level?
    Answer: Fail to reject $H_0$ because the p-value is greater than $\alpha$; there is not convincing evidence the lunch choices differ from equal likelihood. This question assesses interpreting chi-square test results when the p-value is large. With p-value = 0.62, which is much greater than α = 0.05, we fail to reject the null hypothesis. Failing to reject H₀ means we don't have convincing evidence against the claim that lunch choices are equally likely. Choice B correctly states this conclusion. Choice A incorrectly rejects H₀ when p > α, while choice E makes the common error of claiming a large p-value proves H₀ is true. Remember that failing to reject H₀ doesn't prove it's true; it simply means we lack evidence to conclude it's false. Large p-values indicate the observed data is consistent with the null hypothesis.
13. A manufacturer claims defects in its products fall into 4 categories with proportions 0.40 cosmetic, 0.30 packaging, 0.20 functional, 0.10 labeling. A quality-control team inspects a random sample of 100 defective products and performs a chi-square goodness-of-fit test of $H_0$: the population defect-category proportions match the claim versus $H_a$: they do not. The p-value is 0.001. The observed and expected counts are shown. What conclusion is appropriate at the $\alpha=0.05$ level?
    Answer: Reject $H_0$ because the p-value is 0.001; there is evidence the population defect-category distribution differs from the claimed proportions. This quality control problem involves a very small p-value. With p-value = 0.001 << α = 0.05, we strongly reject H₀. This provides convincing evidence that the population defect-category distribution differs from the manufacturer's claimed proportions. Choice A incorrectly fails to reject when p is extremely small. Choice C overstates the conclusion by claiming we've found the "true" distribution. Choice D has both the wrong decision and interpretation. Choice E incorrectly suggests the test is invalid for defective products. Key insight: very small p-values provide strong evidence against H₀, but rejecting H₀ doesn't tell us what the true distribution is.
14. A genetics model predicts that offspring phenotypes should occur in a 9:3:3:1 ratio across four categories. A researcher observes 160 offspring and records the counts in the table. A chi-square goodness-of-fit test is run for $H_0$: the phenotype distribution follows the 9:3:3:1 ratio vs. $H_a$: it does not. The p-value is 0.041. What conclusion is appropriate at the $\alpha=0.01$ level? Observed/Expected counts and p-value: - p-value = 0.041
    Answer: Fail to reject $H_0$ because the p-value is greater than $\alpha$; the data do not provide convincing evidence the distribution differs from 9:3:3:1. This problem tests understanding of significance levels in hypothesis testing. The p-value is 0.041, and we're testing at α = 0.01 level. Since 0.041 > 0.01, we fail to reject H₀. Choice B correctly identifies this: we fail to reject because p-value > α, meaning we don't have convincing evidence the distribution differs from 9:3:3:1. Choice A would be correct if α = 0.05, but not for α = 0.01. Choice E incorrectly suggests we always reject when p < 0.05 regardless of the chosen α level. The significance level α is our threshold for evidence; we only reject H₀ when p-value < α, not when p-value < 0.05.
15. A school cafeteria claims students choose among 4 lunch options (Pizza, Salad, Sandwich, Pasta) in equal proportions. A random sample of 200 student purchases is recorded. A chi-square goodness-of-fit test is performed with $H_0$: the distribution is uniform across the 4 options and $H_a$: the distribution is not uniform. Results are shown below. Observed vs. Expected counts (under $H_0$) and test result: - Pizza: Observed 62, Expected 50 - Salad: Observed 41, Expected 50 - Sandwich: Observed 55, Expected 50 - Pasta: Observed 42, Expected 50 Test output: $p$-value = 0.041 What conclusion is appropriate at the $\alpha=0.05$ level?​
    Answer: Reject $H_0$; there is convincing evidence that the population distribution of lunch choices is not uniform. This question tests understanding of chi-square goodness-of-fit test conclusions. The null hypothesis states the population distribution is uniform (equal proportions across all 4 lunch options), while the alternative hypothesis states it is not uniform. With p-value = 0.041 < α = 0.05, we reject H₀. This means we have convincing evidence that the population distribution of lunch choices is not uniform. Choice A incorrectly fails to reject and claims to "prove" H₀. Choice C incorrectly interprets rejecting H₀. Choice D contradicts itself by failing to reject while claiming evidence against H₀. Choice E correctly fails to reject but applies the wrong p-value comparison.
16. A political scientist claims that, in a certain county, party affiliation among registered voters is 45% Party X, 35% Party Y, and 20% Independent. A random sample of 300 registered voters yields: Party X 120, Party Y 118, Independent 62. A chi-square goodness-of-fit test is conducted and reports $p$-value = 0.014. Expected counts under $H_0$ are: Party X 135, Party Y 105, Independent 60. What conclusion is appropriate at the $\alpha=0.05$ level?​
    Answer: Reject $H_0$; there is convincing evidence that the county's population distribution of affiliations differs from the claimed proportions. This question examines chi-square goodness-of-fit test conclusions for political data. The null hypothesis claims the population proportions are 45% Party X, 35% Party Y, 20% Independent. With p-value = 0.014 < α = 0.05, we reject H₀. This provides convincing evidence that the county's population distribution of party affiliations differs from the claimed proportions. Choice B contradicts itself by failing to reject while claiming evidence against H₀. Choice C incorrectly fails to reject. Choice D misinterprets what the test proves. Choice E irrelevantly mentions sample size. When p-value < α, we have statistically significant evidence against H₀.
17. A candy company advertises that its assorted bag contains 30% red, 25% blue, 25% green, and 20% yellow candies. A student randomly samples 160 candies from many bags and performs a chi-square goodness-of-fit test at $\alpha=0.10$. The results are below. What conclusion is appropriate? Observed/Expected counts and p-value: - Red: Observed 52, Expected 48 - Blue: Observed 36, Expected 40 - Green: Observed 44, Expected 40 - Yellow: Observed 28, Expected 32 - p-value = 0.62
    Answer: Because $p>\alpha$, fail to reject $H_0$; there is not convincing evidence that the population candy-color distribution differs from the advertised distribution. This question evaluates understanding of the chi-square goodness-of-fit test in AP Statistics, used to check if sample data aligns with a hypothesized distribution of categories. With a p-value of 0.62 greater than α=0.10, we fail to reject the null hypothesis, indicating insufficient evidence that the population candy-color distribution differs from the advertised one. This means the observed differences could reasonably occur by chance if the advertisement is true. Choice D is a frequent distractor, as it wrongly claims failing to reject proves the exact advertised proportions for the population, but failing to reject simply means we lack evidence to disprove the null—it doesn't confirm it's true. For a mini-lesson on chi-square conclusions, note that a high p-value supports retaining the null but doesn't prove it; alternative explanations like insufficient sample size could also lead to non-rejection. Always interpret in terms of evidence, not proof.
18. A website designer claims that visitors click on 5 menu tabs with equal probability (20% each). A random sample of 250 first-clicks is collected and a chi-square goodness-of-fit test is run at $\alpha=0.05$. The output gives p-value = 0.003, and the observed/expected counts are shown below. What conclusion is appropriate? Observed/Expected counts and p-value: - Tab 1: Observed 70, Expected 50 - Tab 2: Observed 45, Expected 50 - Tab 3: Observed 40, Expected 50 - Tab 4: Observed 55, Expected 50 - Tab 5: Observed 40, Expected 50 - p-value = 0.003
    Answer: Because $p<\alpha$, reject $H_0$; there is convincing evidence that first-click probabilities are not all equal in the population. This question tests the chi-square goodness-of-fit test in AP Statistics, specifically for assessing if data fits a uniform distribution across categories. The p-value of 0.003 is less than α=0.05, leading us to reject the null hypothesis of equal probabilities (20% each) for first-clicks on the menu tabs. Thus, there is convincing evidence that the population probabilities are not all equal. A typical distractor is choice D, which misinterprets rejection as proving an exact probability like 28% for Tab 1, but the test only indicates deviation from the null, not precise alternative values. In a mini-lesson on chi-square conclusions, rejecting H0 means the observed counts deviate significantly from expected, suggesting the model doesn't fit, but it doesn't identify the true distribution—examine observed vs. expected to see patterns. Conclusions apply to the population based on the sample evidence.
19. A city's transportation department believes that cars arrive at a toll booth in the following proportions: 25% compact, 45% midsize, 30% SUV. A random sample of 160 cars is classified. A chi-square goodness-of-fit test is conducted for $H_0$: the population distribution matches the stated proportions versus $H_a$: it does not. The p-value from the test is 0.41. The observed and expected counts are shown. What conclusion is appropriate at the $\alpha=0.05$ level?
    Answer: Fail to reject $H_0$ because the p-value is greater than 0.05; there is not convincing evidence the population distribution differs from the stated proportions. This question assesses interpretation of chi-square test results when we fail to reject the null hypothesis. The p-value of 0.41 is greater than α = 0.05, so we fail to reject H₀. This means we don't have convincing evidence that the population distribution differs from the stated proportions (25% compact, 45% midsize, 30% SUV). Choice A incorrectly states we reject when p > 0.05. Choices C and D confuse the decision rule by mixing up when p < 0.05 versus p > 0.05. Choice E makes an incorrect absolute claim about the population. Key concept: failing to reject H₀ doesn't prove H₀ is true; it means we lack evidence to conclude it's false.
20. A teacher believes students choose among 4 project topics equally often. In a random sample of 80 students, the topic choices are recorded and a chi-square goodness-of-fit test is run for $H_0$: all 4 topics are equally likely versus $H_a$: not all are equally likely. The p-value is 0.20. The observed and expected counts are shown. What conclusion is appropriate at the $\alpha=0.10$ level?
    Answer: Fail to reject $H_0$ because the p-value is 0.20, which is greater than 0.10; there is not sufficient evidence that topic choices are not equally likely in the population. This final question tests understanding of chi-square conclusions with a different significance level. With p-value = 0.20 > α = 0.10, we fail to reject H₀. This means we don't have sufficient evidence to conclude that topic choices are not equally likely in the population. Choice A incorrectly claims 0.20 < 0.10. Choice C incorrectly rejects when p > α. Choice D has the decision correct but misinterprets what it means. Choice E makes an irrelevant claim about expected counts. Remember: the significance level α determines our threshold for evidence; with α = 0.10, we need p < 0.10 to reject, and 0.20 is clearly above this threshold.

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