AP Statistics Quiz: Chi Square Goodness Of Fit Test

20 questions · exam conditions

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Chi Square Goodness Of Fit TestQuestion 1 of 20

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 $\alpha=0.05$ ?

Fail to reject

H_0

; because

p=0.049>0.05

, there is not convincing evidence of a difference.

Reject

H_0

; there is convincing evidence that the population habitat distribution of sightings 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 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.

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AP Statistics Quiz

AP Statistics Quiz: Chi Square Goodness Of Fit Test

Practice Chi Square Goodness Of Fit Test in AP Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Chi Square Goodness Of Fit 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.

All questions

Question 1

Expected counts under $H_0$ are: Wetland 75, Forest 125, Grassland 100, Shrubland 150, Urban 50.

What conclusion is appropriate at $\alpha=0.05$ ?

Fail to reject $H_0$ ; because $p=0.049>0.05$ , there is not convincing evidence of a difference.
Reject $H_0$ ; there is convincing evidence that the population habitat distribution of sightings differs from the model proportions. (correct answer)
Fail to reject $H_0$ ; this proves the model proportions are correct for the population.
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.

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.

Question 2

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.
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 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. (correct answer)
Fail to reject $H_0$ ; because the sample size is 300, the expected distribution must be correct.

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.

Question 3

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?

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 0.20, which is greater than 0.10; there is not sufficient evidence that topic choices are not equally likely in the population. (correct answer)
Reject $H_0$ because the p-value is greater than 0.10; there is evidence the population distribution is uniform.
Fail to reject $H_0$ because the p-value is less than 0.10; there is evidence the population distribution is not uniform.
Because the expected counts are all 20, the correct conclusion is to reject $H_0$ .

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.

Question 4

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 $\alpha=0.05$ ?

Reject $H_0$ ; there is convincing evidence that the population distribution differs from 1:2:1.
Fail to reject $H_0$ ; there is not convincing evidence that the population distribution differs from 1:2:1. (correct answer)
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.
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.

Question 5

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 $\alpha=0.05$ ?

Fail to reject $H_0$ ; there is not convincing evidence that the population distribution differs from the claimed distribution.
Reject $H_0$ ; there is convincing evidence that the population distribution of start times differs from the claimed distribution. (correct answer)
Fail to reject $H_0$ ; since $p=0.006<0.05$ , the claim is supported.
Reject $H_0$ ; this proves the service's claim is false for the sample but could still be true for the population.
Reject $H_0$ ; because the expected counts are all at least 5, the null hypothesis must be rejected.

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.

Question 6

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 $\alpha=0.05$ ?

Reject $H_0$ ; there is convincing evidence the population defect-type distribution differs from the stated proportions.
Fail to 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. (correct answer)
Reject $H_0$ ; since $p=0.061>0.05$ , the results are statistically significant.
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.

Question 7

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 $p<\alpha$ , reject $H_0$ ; there is convincing evidence that the distribution of ticket types in the population differs from the claimed proportions. (correct answer)
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 sample distribution matches 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.

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.

Question 8

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

Reject $H_0$ ; there is convincing evidence the shipping-speed distribution differs from the claim.
Fail to reject $H_0$ ; the data do not provide convincing evidence that the shipping-speed distribution differs from the retailer's claim. (correct answer)
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.
Conclude that the probability an order is Overnight is 0.11.

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.

Question 9

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?

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 greater than 0.05; there is not convincing evidence the population distribution differs from the stated proportions. (correct answer)
Fail to reject $H_0$ because the p-value is less than 0.05; there is evidence the population distribution matches 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.

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.

Question 10

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?

Fail to reject $H_0$ because the p-value is 0.001; there is insufficient evidence of a difference.
Reject $H_0$ because the p-value is 0.001; there is evidence the population defect-category distribution differs from the claimed proportions. (correct answer)
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.
Fail to reject $H_0$ because the p-value is less than 0.05; there is evidence the claimed proportions are correct.
Because the sample includes only defective products, a goodness-of-fit test cannot be used.

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.

Question 11

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?

Reject $H_0$ because the p-value is less than 0.01; there is evidence the population ratio is not 9:3:3:1. (correct answer)
Fail to reject $H_0$ because the p-value is less than 0.01; there is evidence the population ratio is 9:3:3:1.
Fail to reject $H_0$ because the p-value is greater than 0.01; there is evidence the population ratio differs from 9:3:3:1.
Reject $H_0$ because the p-value is greater than 0.01; there is evidence the sample ratio differs from 9:3:3:1.
Because expected counts are based on a ratio, a chi-square test cannot be used here.

Explanation: 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.

Question 12

A game developer claims players choose among 3 character classes in the population with proportions 0.50 Warrior, 0.30 Mage, 0.20 Rogue. In a random sample of 250 new players, the observed counts are Warrior 132, Mage 63, Rogue 55. A chi-square goodness-of-fit test is performed and yields $p$ -value = 0.072.

Expected counts under $H_0$ are Warrior 125, Mage 75, Rogue 50.

What conclusion is appropriate at $\alpha=0.10$ ?

Fail to reject $H_0$ ; there is not convincing evidence that the population proportions differ from 0.50/0.30/0.20.
Reject $H_0$ ; there is convincing evidence that the population distribution of character classes differs from the claimed proportions. (correct answer)
Fail to reject $H_0$ ; since $p=0.072<0.10$ , the population distribution must match the claimed proportions.
Reject $H_0$ ; since $p=0.072>0.10$ , the results are statistically significant at $\alpha=0.10$ .
Reject $H_0$ ; there is convincing evidence that the sample distribution is different from the expected counts, so the claim is false.

Explanation: This question examines chi-square test conclusions with α = 0.10. The null hypothesis claims the population proportions are 0.50 Warrior, 0.30 Mage, 0.20 Rogue. With p-value = 0.072 < α = 0.10, we reject H₀. This means there is convincing evidence that the population distribution of character classes differs from the claimed proportions. Choice A incorrectly fails to reject. Choice C misinterprets the p-value comparison. Choice D has the comparison backwards (0.072 < 0.10, not >). Choice E confuses sample and population inference. Remember to always compare p-value directly to the stated significance level.

Question 13

A jar of candies is advertised to contain colors in these population proportions: 30% Red, 25% Blue, 20% Green, 15% Yellow, 10% Orange. A student randomly samples 100 candies and records counts: Red 34, Blue 22, Green 19, Yellow 16, Orange 9. A chi-square goodness-of-fit test of $H_0$ : the population proportions match the advertisement gives $p$ -value = 0.86.

Expected counts under $H_0$ are: Red 30, Blue 25, Green 20, Yellow 15, Orange 10.

What conclusion is appropriate at $\alpha=0.05$ ?

Reject $H_0$ ; there is convincing evidence that the population color distribution matches the advertisement.
Fail to reject $H_0$ ; there is not convincing evidence that the population color distribution differs from the advertised proportions. (correct answer)
Reject $H_0$ ; because the observed counts are not exactly equal to expected counts, the advertisement must be false.
Fail to reject $H_0$ ; this proves the jar's population proportions are exactly the advertised values.
Reject $H_0$ ; because $p=0.86>0.05$ , the result is statistically significant.

Explanation: This question tests interpretation of a large p-value in chi-square testing. The null hypothesis states the population proportions match the advertised values. With p-value = 0.86 > α = 0.05, we fail to reject H₀. This means there is not convincing evidence that the population color distribution differs from the advertised proportions. Choice A incorrectly rejects and misinterprets the conclusion. Choice C misunderstands sampling variability. Choice D incorrectly claims this "proves" H₀. Choice E completely misunderstands p-value interpretation (large p-values lead to failing to reject, not rejecting). A large p-value indicates the observed data is consistent with H₀.

Question 14

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 $\alpha=0.05$ ?

Reject $H_0$ ; there is convincing evidence that the population distribution differs from 1:2:1.
Fail to reject $H_0$ ; there is not convincing evidence that the population distribution differs from 1:2:1. (correct answer)
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.
Reject $H_0$ ; there is convincing evidence the sample distribution is 1:2:1.

Question 15

Expected counts under $H_0$ are: Wetland 75, Forest 125, Grassland 100, Shrubland 150, Urban 50.

What conclusion is appropriate at $\alpha=0.05$ ?

Fail to reject $H_0$ ; because $p=0.049>0.05$ , there is not convincing evidence of a difference.
Reject $H_0$ ; there is convincing evidence that the population habitat distribution of sightings differs from the model proportions. (correct answer)
Fail to reject $H_0$ ; this proves the model proportions are correct for the population.
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.

Question 16

Expected counts under $H_0$ are Warrior 125, Mage 75, Rogue 50.

What conclusion is appropriate at $\alpha=0.10$ ?

Fail to reject $H_0$ ; there is not convincing evidence that the population proportions differ from 0.50/0.30/0.20.
Reject $H_0$ ; there is convincing evidence that the population distribution of character classes differs from the claimed proportions. (correct answer)
Fail to reject $H_0$ ; since $p=0.072<0.10$ , the population distribution must match the claimed proportions.
Reject $H_0$ ; since $p=0.072>0.10$ , the results are statistically significant at $\alpha=0.10$ .
Reject $H_0$ ; there is convincing evidence that the sample distribution is different from the expected counts, so the claim is false.

Question 17

A university states that among all enrolled students, class year proportions are 27% first-year, 26% sophomore, 24% junior, and 23% senior. A random sample of 200 students is selected from the enrollment list and surveyed. A chi-square goodness-of-fit test is run at $\alpha=0.05$ with the results below. What conclusion is appropriate?

Observed/Expected counts and p-value:

First-year: Observed 60, Expected 54
Sophomore: Observed 45, Expected 52
Junior: Observed 42, Expected 48
Senior: Observed 53, Expected 46
p-value = 0.072

Because $p<\alpha$ , reject $H_0$ ; there is convincing evidence the population class-year distribution differs from the stated proportions.
Because $p>\alpha$ , fail to reject $H_0$ ; there is not convincing evidence that the population class-year distribution differs from the stated proportions. (correct answer)
Because $p>\alpha$ , reject $H_0$ ; there is convincing evidence the sample matches the stated proportions.
Because $p>\alpha$ , fail to reject $H_0$ ; therefore the sample results generalize to all universities.
Because $p>\alpha$ , fail to reject $H_0$ ; this proves the stated proportions are exactly correct for the population.

Explanation: This question tests the chi-square goodness-of-fit test in AP Statistics to see if student class years fit the university's stated proportions. With p=0.072 greater than α=0.05, we fail to reject the null hypothesis, meaning there is not convincing evidence of a difference in the population distribution. The sample aligns reasonably with the statement. Choice E distracts by claiming non-rejection proves exact correctness, but it doesn't; it indicates insufficient evidence against the null. A mini-lesson on chi-square conclusions: Non-rejection suggests consistency with the model, but absence of evidence isn't evidence of absence—larger samples might detect small differences. Always tie back to the population and avoid generalizing beyond the tested hypothesis.

Question 18

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?

Reject $H_0$ ; there is convincing evidence that the county's population distribution of affiliations differs from the claimed proportions. (correct answer)
Fail to reject $H_0$ ; there is convincing evidence that the county's population distribution differs from the claimed proportions.
Fail to reject $H_0$ ; there is not convincing evidence that the county's population distribution differs from the claimed proportions.
Reject $H_0$ ; this proves that exactly 45% of the sample is Party X in the population.
Fail to reject $H_0$ because the sample size is large, so $H_0$ must be true.

Explanation: 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.

Question 19

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?

Reject $H_0$ because the p-value is very large; there is strong evidence the distribution differs from the claim.
Fail to reject $H_0$ because the p-value is 0.95; there is not convincing evidence the population distribution differs from the claim. (correct answer)
Fail to reject $H_0$ because the p-value is 0.95; therefore the claim is proven true for every customer.
Reject $H_0$ because the p-value is less than 0.05; there is evidence the claim is correct.
Conclude $H_0$ is false because the observed counts are not exactly the same as the expected counts.

Explanation: 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.

Question 20

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?

Reject $H_0$ because the p-value is 0.057; there is evidence the population support differs from the claimed proportions.
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. (correct answer)
Reject $H_0$ because the p-value is greater than 0.05; there is evidence the sample proportions match the claim.
Fail to reject $H_0$ because the p-value is less than 0.05; there is evidence the population support differs from the claim.
Because the p-value is close to 0.05, the correct decision is to reject $H_0$ .

Explanation: 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.