Chi-Square Goodness of Fit (Test)
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AP Statistics › Chi-Square Goodness of Fit (Test)
A wildlife biologist expects, based on a long-term model, that sightings of a bird species across 5 habitats occur in proportions: 0.15 Wetland, 0.25 Forest, 0.20 Grassland, 0.30 Shrubland, 0.10 Urban. In a random sample of 500 sightings, the observed counts are: Wetland 92, Forest 118, Grassland 93, Shrubland 146, Urban 51. A chi-square goodness-of-fit test gives $p$-value = 0.049.
Expected counts under $H_0$ are: Wetland 75, Forest 125, Grassland 100, Shrubland 150, Urban 50.
What conclusion is appropriate at $\alpha=0.05$?
Reject $H_0$; there is convincing evidence that the sample habitat distribution differs from the model, so the population must match the model.
Fail to reject $H_0$; there is convincing evidence that the population habitat distribution differs from the model proportions.
Fail to reject $H_0$; this proves the model proportions are correct for the population.
Reject $H_0$; there is convincing evidence that the population habitat distribution of sightings differs from the model proportions.
Fail to reject $H_0$; because $p=0.049>0.05$, there is not convincing evidence of a difference.
Explanation
This question examines a borderline p-value that leads to rejection. The null hypothesis claims the population habitat distribution matches the model proportions. With p-value = 0.049 < α = 0.05, we reject H₀. This provides convincing evidence that the population habitat distribution of bird sightings differs from the model proportions. Choice A incorrectly states 0.049 > 0.05. Choice C incorrectly claims this "proves" the model. Choice D misinterprets the relationship between sample and population. Choice E contradicts itself. When p-value is just below α, we still reject H₀ - there's no "almost significant" in hypothesis testing.
A university expects the distribution of students’ primary commute methods to be: 50% Car, 20% Bus, 15% Bike, 10% Walk, 5% Other. A random sample of 300 students reports: Car 141, Bus 72, Bike 39, Walk 33, Other 15. A chi-square goodness-of-fit test of $H_0$: the population distribution matches the expected proportions yields $p$-value = 0.27.
Expected counts under $H_0$ are: Car 150, Bus 60, Bike 45, Walk 30, Other 15.
What conclusion is appropriate at $\alpha=0.05$?
Reject $H_0$; there is convincing evidence that the population distribution matches the university’s expected proportions.
Fail to reject $H_0$; because the sample size is 300, the expected distribution must be correct.
Reject $H_0$; there is convincing evidence that the population distribution differs from the university’s expected proportions.
Fail to reject $H_0$; there is not convincing evidence that the population distribution differs from the university’s expected proportions.
Fail to reject $H_0$; there is convincing evidence that the population distribution differs from the university’s expected proportions.
Explanation
This question examines interpretation of a large p-value. The null hypothesis claims the population distribution matches the university's expected proportions for commute methods. With p-value = 0.27 > α = 0.05, we fail to reject H₀. This means there is not convincing evidence that the population distribution differs from the university's expected proportions. Choice A incorrectly interprets rejecting H₀. Choice B incorrectly rejects H₀. Choice C contradicts itself. Choice E irrelevantly mentions sample size. A large p-value (0.27) indicates the observed data is quite consistent with the null hypothesis.
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?
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.
Reject $H_0$ because the p-value is greater than 0.10; there is evidence the population distribution is uniform.
Because the expected counts are all 20, the correct conclusion is to reject $H_0$.
Reject $H_0$ because the p-value is 0.20, which is less than 0.10; there is evidence topic choices are not equally likely.
Fail to reject $H_0$ because the p-value is less than 0.10; there is evidence the population distribution is not uniform.
Explanation
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.
A genetics model predicts offspring phenotypes in a 3-category ratio of 1:2:1 (Type A, Type B, Type C). A researcher records 160 offspring: A=35, B=92, C=33. A chi-square goodness-of-fit test is conducted with $H_0$: the population follows the 1:2:1 distribution. The test produces $p$-value = 0.18.
Observed vs. Expected counts (under $H_0$):
- Type A: Observed 35, Expected 40
- Type B: Observed 92, Expected 80
- Type C: Observed 33, Expected 40
What conclusion is appropriate at $\alpha=0.05$?
Fail to reject $H_0$; this proves the population distribution is exactly 1:2:1.
Reject $H_0$ because some observed counts are not equal to expected counts.
Fail to reject $H_0$; there is not convincing evidence that the population distribution differs from 1:2:1.
Reject $H_0$; there is convincing evidence that the population distribution differs from 1:2:1.
Reject $H_0$; there is convincing evidence the sample distribution is 1:2:1.
Explanation
This question examines chi-square goodness-of-fit test interpretation for a genetics model. The null hypothesis claims the population follows a 1:2:1 ratio for the three phenotypes. With p-value = 0.18 > α = 0.05, we fail to reject H₀. This means there is not convincing evidence that the population distribution differs from the expected 1:2:1 ratio. Choice C incorrectly claims this "proves" the null hypothesis. Choice D misunderstands that observed counts will rarely equal expected counts exactly due to sampling variability. Choice E misinterprets what rejecting H₀ would mean. Remember: failing to reject H₀ never proves it true; it only indicates insufficient evidence against it.
A streaming service claims that, among its users, time-of-day for starting a movie is distributed as: 20% Morning, 30% Afternoon, 35% Evening, 15% Night. A random sample of 200 movie starts yields: Morning 33, Afternoon 52, Evening 80, Night 35. A chi-square goodness-of-fit test of $H_0$: the population distribution matches the claim gives $p$-value = 0.006.
Expected counts under $H_0$ are: Morning 40, Afternoon 60, Evening 70, Night 30.
What conclusion is appropriate at $\alpha=0.05$?
Reject $H_0$; there is convincing evidence that the population distribution of start times differs from the claimed distribution.
Reject $H_0$; this proves the service’s claim is false for the sample but could still be true for the population.
Fail to reject $H_0$; there is not convincing evidence that the population distribution differs from the claimed distribution.
Reject $H_0$; because the expected counts are all at least 5, the null hypothesis must be rejected.
Fail to reject $H_0$; since $p=0.006<0.05$, the claim is supported.
Explanation
This question tests chi-square goodness-of-fit conclusions with a small p-value. The null hypothesis states the population distribution matches the streaming service's claim. With p-value = 0.006 < α = 0.05, we reject H₀. This provides convincing evidence that the population distribution of movie start times differs from the claimed distribution. Choice A incorrectly fails to reject. Choice C misinterprets what a small p-value means. Choice D incorrectly limits the conclusion to the sample. Choice E irrelevantly mentions the expected count condition. A small p-value indicates strong evidence against H₀, leading to rejection.
A manufacturer states that defects in a production line fall into 4 categories with population proportions: 0.10 Scratch, 0.20 Dent, 0.30 Paint, 0.40 Other. In a random sample of 200 defects, the observed counts are Scratch 28, Dent 36, Paint 59, Other 77. A chi-square goodness-of-fit test of $H_0$: the defect-type distribution matches the stated proportions yields $p$-value = 0.061.
Expected counts under $H_0$ are Scratch 20, Dent 40, Paint 60, Other 80.
What conclusion is appropriate at $\alpha=0.05$?
Fail to reject $H_0$; there is convincing evidence the population defect-type distribution differs from the stated proportions.
Reject $H_0$; since $p=0.061>0.05$, the results are statistically significant.
Reject $H_0$; there is convincing evidence the population defect-type distribution differs from the stated proportions.
Fail to reject $H_0$; there is not convincing evidence that the population defect-type distribution differs from the stated proportions.
Fail to reject $H_0$; this proves the manufacturer’s stated proportions are exactly correct.
Explanation
This question tests understanding of borderline p-values in chi-square testing. The null hypothesis states the defect-type distribution matches the manufacturer's stated proportions. With p-value = 0.061 > α = 0.05, we fail to reject H₀. This means there is not convincing evidence that the population defect-type distribution differs from the stated proportions. Choice A incorrectly rejects H₀. Choice B contradicts itself. Choice D misunderstands p-value comparison (0.061 > 0.05 leads to failing to reject, not rejecting). Choice E incorrectly claims this "proves" H₀. Even when p-values are close to α, we must strictly follow the decision rule.
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
Because $p<\alpha$, fail to reject $H_0$; there is not convincing evidence that the sample distribution differs from the claimed proportions.
Because $p<\alpha$, reject $H_0$; there is convincing evidence that the distribution of ticket types in the population differs from the claimed proportions.
Because $p<\alpha$, reject $H_0$; the sample proves that exactly 50% of all riders use Regular tickets.
Because $p<\alpha$, reject $H_0$; there is convincing evidence that the sample proportions differ from the claimed proportions, so the population proportions must differ by the same amounts.
Because $p>\alpha$, reject $H_0$; there is convincing evidence that the sample distribution matches the claimed proportions.
Explanation
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.
An online retailer claims that customers choose shipping speed in these proportions: Standard 0.55, Expedited 0.30, Overnight 0.15. A random sample of $n=200$ orders yields the counts below. A chi-square goodness-of-fit test returns p-value = 0.11.
What conclusion is appropriate at the $\alpha=0.05$ level?
Observed/Expected counts (with p-value):
| Shipping speed | Observed | Expected |
|---|---|---|
| Standard | 120 | 110 |
| Expedited | 54 | 60 |
| Overnight | 26 | 30 |
p-value = 0.11
Fail to reject $H_0$; the data do not provide convincing evidence that the shipping-speed distribution differs from the retailer’s claim.
Reject $H_0$; there is convincing evidence the shipping-speed distribution differs from the claim.
Conclude that the probability an order is Overnight is 0.11.
Fail to reject $H_0$; therefore exactly 55% of all customers choose Standard shipping.
Reject $H_0$ because the p-value is greater than 0.05.
Explanation
This question evaluates the chi-square goodness-of-fit test in AP Statistics for shipping speed choices. The p-value of 0.11 exceeds 0.05, so we fail to reject the null hypothesis. This means the data do not provide convincing evidence of a difference from the retailer's claim. A key distractor is choice C, which misstates failing to reject as proving exact proportions, like 55% for Standard, but it doesn't. In chi-square tests, we use claimed proportions to find expected counts and compute the statistic. A moderate p-value like this suggests the sample is plausible under H0. Always remember, failing to reject H0 is about insufficient evidence, not confirmation.
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?
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.
Reject $H_0$ because the p-value is less than 0.05; there is evidence the sample distribution must match the stated proportions.
Conclude the stated proportions are exactly correct in the population because the sample size is large.
Reject $H_0$ because the p-value is greater than 0.05; there is convincing evidence the distribution differs from the stated proportions.
Fail to reject $H_0$ because the p-value is less than 0.05; there is evidence the population distribution matches the stated proportions.
Explanation
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.
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?
Reject $H_0$ because the p-value is 0.001; there is evidence the population defect-category distribution differs from the claimed proportions.
Fail to reject $H_0$ because the p-value is less than 0.05; there is evidence the claimed proportions are correct.
Reject $H_0$ because the p-value is 0.001; therefore 0.40/0.30/0.20/0.10 must be the true distribution in the population.
Because the sample includes only defective products, a goodness-of-fit test cannot be used.
Fail to reject $H_0$ because the p-value is 0.001; there is insufficient evidence of a difference.
Explanation
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.