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AP Statistics

AP Statistics Practice Test: Practice Test 84

Practice Test 84 for AP Statistics: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A company has 80 employees. Of these, 30 work remotely (R), 28 are in the sales department (S), and 12 are both remote and in sales. One employee is selected at random. Let Event A be “the employee works remotely” and Event B be “the employee is in sales.” Which statement about Events A and B is correct?​

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Question 1

A company has 80 employees. Of these, 30 work remotely (R), 28 are in the sales department (S), and 12 are both remote and in sales. One employee is selected at random. Let Event A be “the employee works remotely” and Event B be “the employee is in sales.” Which statement about Events A and B is correct?​

  1. Events A and B are independent because 12 employees are in both categories.
  2. Events A and B are independent because P(A∪B)=P(A)+P(B)P(A\cup B)=P(A)+P(B)P(A∪B)=P(A)+P(B).
  3. Events A and B are not independent because P(A∩B)≠P(A)P(B)P(A\cap B)\ne P(A)P(B)P(A∩B)=P(A)P(B). (correct answer)
  4. Events A and B are independent because remote work and sales are unrelated job features.
  5. Events A and B are not independent because Events A and B are mutually exclusive.

Explanation: This question involves checking independence using given frequencies. From the data: P(A) = 30/80, P(B) = 28/80, and P(A∩B) = 12/80. To verify independence, we check if P(A∩B) = P(A)·P(B). Calculating: P(A)·P(B) = (30/80)·(28/80) = 840/6400 = 21/160, while P(A∩B) = 12/80 = 24/160. Since 21/160 ≠ 24/160, the events are not independent. The fact that 12 employees are in both categories (choice A) doesn't determine independence—we must check if the probability relationship holds. Events that overlap can still be independent if they satisfy the multiplication rule.

Question 2

A botanist measures the number of seeds produced by each of 36 plants grown under the same conditions. The dotplot shows values from 90 to 130 seeds, with the highest concentration around 108–112, and roughly similar frequencies on both sides of that center; there are no isolated points far from the rest.

Which statement best describes the distribution?

  1. The distribution is approximately symmetric and unimodal, centered near about 110 seeds, with no apparent outliers. (correct answer)
  2. The distribution is strongly right-skewed because 130 is larger than 90.
  3. The distribution is strongly left-skewed because the smallest values are below 100.
  4. The distribution is bimodal because there are values on both sides of 110.
  5. The distribution is uniform because the range is about 40 seeds.

Explanation: This question tests describing quantitative distributions like seed counts in a dotplot, focusing on shape, center, spread, and outliers. The distribution is approximately symmetric and unimodal, centered near 110 seeds with balanced frequencies on both sides from 90-130 and no outliers. Distractor B incorrectly calls it strongly right-skewed because 130 > 90, but skew requires an imbalanced tail, not just range asymmetry. Mini-lesson: Symmetry means mirror-image halves; unimodal has one peak. Bimodality shows two peaks, uniformity even spread, left skew tails low. Always verify balance around the center and check for detached points as outliers.

Question 3

A teacher claims a student is guessing on a 20-question true/false quiz. The student’s sequence of answers, in order, is: T T T T T T T T T T T T T T T T T T T T. If the student were truly guessing, you would expect a mix of T and F with occasional short runs, not an all-one-letter pattern. Is the pattern consistent with random behavior?

  1. Yes; random guessing often produces long runs, so 20 T’s is typical.
  2. Yes; randomness means any sequence is equally likely, so this looks random.
  3. No; an all-T sequence is so extreme that it suggests the student was not guessing. (correct answer)
  4. No; random guessing should produce exactly 10 T’s and 10 F’s every time.
  5. Yes; because there is no alternating pattern, it must be random.

Explanation: This question tests understanding of random versus non-random patterns in true/false guessing. When someone truly guesses randomly on a true/false quiz, we expect roughly equal numbers of T's and F's with occasional short runs (like 2-4 in a row), but getting all 20 answers the same is extremely unlikely - the probability is (1/2)^20 or about 1 in a million. Such an extreme pattern strongly suggests the student was not guessing randomly but instead had a deliberate strategy (like marking all T). Random behavior produces variation and mixing, not perfect uniformity. The key insight is that while any specific sequence is theoretically possible, extreme patterns like all one answer are so improbable under random guessing that they indicate non-random behavior.

Question 4

A principal wants to estimate the proportion of parents who are satisfied with school communication. The plan is to send a survey link by email and analyze only the first 200 responses received. The research question is: “What proportion of all parents at this school are satisfied with school communication?” Which aspect is most important to address before collecting data?

  1. Increase the sample size beyond 200 to reduce sampling variability
  2. Ensure the responses are not biased toward parents who respond quickly by using a sampling plan that represents all parents (correct answer)
  3. Add an open-ended question so parents can explain what they like or dislike
  4. Decide whether to present results as a percent or as a fraction
  5. Plan to exclude incomplete surveys to simplify analysis

Explanation: This question highlights response bias in survey sampling. The principal's plan to analyze only the first 200 responses creates bias because parents who respond quickly may systematically differ from the general parent population - they might be more engaged, have stronger opinions, or have more time available. Option B correctly identifies that ensuring responses aren't biased toward quick responders by using a proper sampling plan is most important. Simply taking the first 200 responses is a form of convenience sampling that may not represent all parents. While the sample size of 200 (A) might be adequate, how those 200 are selected matters more. The other options about question format (C), presentation style (D), or handling incomplete surveys (E) are secondary to ensuring representative sampling. In survey research, the method of selecting respondents is crucial for valid inference about the population.

Question 5

A company tests two website designs. Among 200 randomly selected visitors shown Design A, 54 made a purchase; among 180 randomly selected visitors shown Design B, 63 made a purchase. A 90% confidence interval for (pA−pB)(p_A - p_B)(pA​−pB​) is (−0.18,−0.02)(-0.18, -0.02)(−0.18,−0.02). Which interpretation is correct?

  1. We are 90% confident that Design A’s purchase proportion is between −0.18-0.18−0.18 and −0.02-0.02−0.02.
  2. Because the interval is negative, we are 90% confident that pBp_BpB​ is between 0.02 and 0.18 greater than pAp_ApA​.
  3. There is a 90% chance that (pA−pB)(p_A - p_B)(pA​−pB​) is negative for this experiment.
  4. We are 90% confident that the true difference in purchase proportions (pA−pB)(p_A - p_B)(pA​−pB​) is between −0.18-0.18−0.18 and −0.02-0.02−0.02. (correct answer)
  5. Since 0 is not in the interval, the two sample proportions must be equal.

Explanation: This question involves interpreting a negative confidence interval for the difference of two proportions. The interval (-0.18, -0.02) estimates (pₐ - pᵦ), where negative values indicate Design A has a lower purchase proportion than Design B. Choice D correctly interprets this as being 90% confident that the true difference in purchase proportions is between -0.18 and -0.02. Choice A incorrectly refers to a single proportion rather than the difference. Choice B correctly notes that pᵦ is greater than pₐ but reverses the order of subtraction. Choice C incorrectly uses probability language about the parameter. Choice E incorrectly concludes equality when 0 is NOT in the interval. When interpreting negative intervals, pay attention to which proportion is subtracted from which.

Question 6

An environmental group tests whether the mean nitrate level in a river exceeds the legal limit of 10 mg/L. They test H0:μ=10H_0: \mu = 10H0​:μ=10 versus Ha:μ>10H_a: \mu > 10Ha​:μ>10 at α=0.05\alpha = 0.05α=0.05. After collecting water samples, they reject H0H_0H0​ and announce the mean level is above the limit. In reality, the true mean nitrate level is not above 10 mg/L (it equals 10 mg/L). Which type of error was made?

  1. Type II error: failing to reject H0H_0H0​ when H0H_0H0​ is false
  2. Type I error: rejecting H0H_0H0​ when H0H_0H0​ is true (correct answer)
  3. No error was made because the alternative is one-sided
  4. Type I error: failing to reject H0H_0H0​ when H0H_0H0​ is false
  5. Type II error: rejecting H0H_0H0​ when H0H_0H0​ is true

Explanation: The skill here is identifying hypothesis testing errors by analyzing the decision and the true parameter value. The group rejected H0 (μ = 10), but the true mean is exactly 10, making H0 true. Thus, rejecting a true H0 is a Type I error. Option E could distract by describing Type II error incorrectly as rejecting when true, which is actually Type I. In a mini-lesson, Type I errors (false positives) are limited by setting α, such as 0.05, while Type II errors (false negatives) can be minimized by increasing sample size or effect size to boost test power.

Question 7

A school district wants to estimate the proportion of all parents of students in the district who support switching to a year-round calendar. The district emailed a survey link to every parent but only counted responses submitted within 24 hours. Most responses came from parents who regularly check the school app during the workday. Which issue most threatens the validity of using these results to represent all district parents?

  1. Measurement bias from poorly worded questions that lead parents to support the change
  2. Undercoverage/voluntary response bias because parents who respond quickly may differ from those who do not respond in time (correct answer)
  3. A placebo effect because parents may change opinions after taking the survey
  4. Random assignment bias because parents were not randomly assigned to receive the email
  5. Confounding because the calendar type is mixed up with parents’ grade levels

Explanation: This question tests understanding of undercoverage and voluntary response bias in sampling. The school district wants to estimate opinions of ALL parents but only counts responses submitted within 24 hours, creating a systematic exclusion of certain groups. Parents who respond quickly are likely those who regularly check school communications and have flexible schedules during the workday - this group may differ systematically from parents who work in jobs without email access, have multiple jobs, or check messages less frequently. This is undercoverage because entire segments of the parent population have little chance of being included, combined with voluntary response bias since only motivated parents self-select to respond quickly. When diagnosing sampling problems, look for whether the sampling method gives all population members a fair chance of selection - here, the 24-hour deadline effectively excludes many parents regardless of their opinions.

Question 8

A student tosses a thumbtack 20 times and records whether it lands point-up. A success is “point-up,” and exactly 20 tosses are made under similar conditions. Does this meet binomial conditions?

  1. Yes, if each toss can be treated as independent and the probability of point-up stays constant across tosses. (correct answer)
  2. No, because thumbtacks have more than two outcomes.
  3. No, because the probability of point-up must be 0.50.50.5 for binomial.
  4. No, because 20 tosses is not a fixed number.
  5. Yes, but only if the thumbtack is perfectly symmetric.

Explanation: This AP Statistics question assesses binomial fit for physical experiments. Fixed n=20 tosses, binary (point-up or not), assuming independent with constant p, yes. Meets conditions. Distractor B claims more outcomes, but thumbtacks typically have two stable positions. Choice C requires p=0.5, but binomial allows any p. Mini-lesson: Physical randomness can be binomial if conditions hold; estimate p from data.

Question 9

A school principal believes that fewer than 40% of students eat breakfast at home. In a random sample of 150 students, 52 reported eating breakfast at home. A one-sample zzz test for a population proportion was conducted with H0:p=0.40H_0: p=0.40H0​:p=0.40 versus Ha:p<0.40H_a: p<0.40Ha​:p<0.40 at significance level α=0.01\alpha=0.01α=0.01. The test produced a ppp-value of 0.084, so the null hypothesis was not rejected. Which conclusion is appropriate?

  1. There is not enough evidence at the 1% level to conclude that the population proportion of all students who eat breakfast at home is less than 0.40. (correct answer)
  2. Because H0H_0H0​ was not rejected, we have proven that exactly 40% of students eat breakfast at home.
  3. There is convincing evidence that fewer than 40% of the sampled students eat breakfast at home.
  4. Since the ppp-value is 0.084, there is an 8.4% chance that the alternative hypothesis is true.
  5. Failing to reject H0H_0H0​ shows that eating breakfast at home causes better academic performance.

Explanation: This question tests understanding of concluding a one-sample z-test when we fail to reject the null hypothesis. Since the p-value (0.084) is greater than α (0.01), we fail to reject H₀: p = 0.40. Choice A correctly states that there is not enough evidence at the 1% level to support the alternative hypothesis (p < 0.40). Choice B incorrectly claims we've proven H₀ is true - failing to reject never proves the null. Choice C focuses on the sample rather than the population. Choice D misinterprets the p-value as the probability of Hₐ being true. Choice E introduces causation that wasn't tested. When we fail to reject H₀, we conclude there's insufficient evidence for the alternative hypothesis.

Question 10

A gym tracked members by membership type (Monthly or Annual) and whether they attended at least 8 times in the last month (Yes/No). To interpret the relationship using conditional distributions, choose the correct comparison.

Which comparison is appropriate?

  1. Compare the percent who attended at least 8 times (Yes) among Monthly members to the percent who attended at least 8 times (Yes) among Annual members. (correct answer)
  2. Compare the overall percent who are Monthly members to the overall percent who are Annual members.
  3. Compare the overall percent who attended at least 8 times (Yes) to the overall percent who did not (No).
  4. Compare the percent who are Annual members among those who attended at least 8 times (Yes) to the percent who are Annual members among those who did not (No).
  5. Compare the number of Monthly members who attended at least 8 times (Yes) to the number of Annual members who attended at least 8 times (Yes).

Explanation: This question tests understanding of conditional distributions when analyzing the relationship between membership type and gym attendance. To assess whether membership type is associated with attendance frequency, we should condition on membership type and compare attendance rates. Option A correctly compares the percent who attended at least 8 times among Monthly members to the percent among Annual members - this conditional comparison directly shows if attendance patterns differ by membership type. Options B and C examine marginal distributions that don't reveal the relationship, while option D reverses the conditioning by looking at membership type given attendance. Option E uses counts instead of percentages, which doesn't properly control for different group sizes. Remember: to study associations between categorical variables, condition on one variable and compare the distribution of the other across those conditions.

Question 11

A delivery service claims the mean delivery time for a certain route is μ=45\mu=45μ=45 minutes. A random sample of deliveries produces a 95% confidence interval for μ\muμ of (41, 44)(41,\ 44)(41, 44) minutes. Is the claim supported by the confidence interval?

  1. Yes, because the interval is close to 45 minutes, so the claim is supported.
  2. No, because 45 is not in the interval, so the claim is not supported. (correct answer)
  3. Yes, because 95% confidence means the true mean is definitely between 41 and 44 minutes.
  4. Yes, because 95% of delivery times are between 41 and 44 minutes.
  5. No, because the confidence interval is based on a sample, so it can never be used to assess a claim.

Explanation: This question asks whether a claim that μ = 45 minutes is supported by a 95% confidence interval of (41, 44) minutes. The claimed value of 45 is NOT in the interval - it exceeds the upper bound of 44. This means we have evidence against the claim at the 95% confidence level. Choice A vaguely suggests being "close" is sufficient, which is incorrect. Choice C misinterprets confidence intervals as providing certainty about the parameter's location. Choice D incorrectly describes individual delivery times rather than the mean. The fundamental rule: a claim is supported only if the claimed value falls within the confidence interval boundaries.

Question 12

A campus dining hall observes whether a student chooses a vegetarian entree (event EEE) and whether the student also selects a salad (event SSS). The menu layout places salads next to vegetarian entrees, so these choices may be associated. How does knowing that a student chose a vegetarian entree (EEE occurred) affect the likelihood that the student selected a salad (SSS)?

  1. It increases the likelihood of SSS if P(S∣E)>P(S)P(S\mid E) > P(S)P(S∣E)>P(S). (correct answer)
  2. It decreases the likelihood of SSS if P(S∣E)>P(S)P(S\mid E) > P(S)P(S∣E)>P(S).
  3. It has no effect because P(S∣E)P(S\mid E)P(S∣E) always equals P(S)P(S)P(S).
  4. It has no effect because P(S∣E)=P(E∣S)P(S\mid E)=P(E\mid S)P(S∣E)=P(E∣S).
  5. It increases the likelihood of SSS only if P(E∣S)<P(E)P(E\mid S) < P(E)P(E∣S)<P(E).

Explanation: This question tests how choosing a vegetarian entree affects salad selection probability. When we know E occurred (student chose vegetarian entree), we're evaluating P(S|E). The menu layout suggests association between these choices, so P(S|E) > P(S). Option A correctly states that knowing E occurred increases the likelihood of S if P(S|E) > P(S). Option B reverses the effect, C and D make false claims about probability relationships, and E introduces an irrelevant condition. The key is recognizing that physical proximity in the menu layout can create positive association between food choices.

Question 13

A principal wants to estimate the mean number of times per week students buy lunch at school. Two independent simple random samples of 75 students are selected from the same school roster in the same month. Sample 1 reports a mean of 2.1 days per week; Sample 2 reports a mean of 2.6 days per week. Why might the sample results differ?

  1. The difference indicates one sample must have been chosen incorrectly, since random sampling should give the same mean.
  2. The difference is due to random sampling variability: different random samples can produce different sample means. (correct answer)
  3. The difference means both samples are biased because they do not match exactly.
  4. Sampling variability only occurs when the population is small; in a school it does not occur.
  5. The difference shows the population mean must be exactly halfway between 2.1 and 2.6.

Explanation: This question tests understanding of sampling variability in educational settings. The difference between 2.1 and 2.6 days per week for lunch purchases is exactly what we expect from random sampling - each sample of 75 students includes a different mix of frequent and infrequent lunch buyers by chance. Choice A wrongly assumes incorrect sampling, choice C mistakenly labels both samples as biased, choice D incorrectly limits variability to large populations, and choice E misinterprets what sample differences tell us about the population. The fundamental principle: whenever we use samples to learn about populations, we must expect and account for sampling variability in our interpretations.

Question 14

A tech company claims that at most 10% of its users experience a certain app crash each month. From a random sample of 600 users, a 90% confidence interval for the true proportion who experience the crash is (0.09,0.12)(0.09, 0.12)(0.09,0.12). Is the claim supported by the confidence interval?

  1. Yes, because 0.10 is in the interval, so the claim is definitely true.
  2. No, because the interval includes values greater than 0.10, so the data do not support a claim of at most 10%. (correct answer)
  3. Yes, because 90% confidence means 90% of users crash, which is at most 10%.
  4. No, because the interval is too narrow to be believable, so it cannot be used.
  5. Yes, because the lower bound is below 0.10, proving the true proportion is at most 0.10.

Explanation: This question tests understanding of "at most" claims using confidence intervals. The company claims that AT MOST 10% of users experience the crash, meaning the true proportion is 0.10 or less. The confidence interval (0.09, 0.12) includes values above 0.10, such as 0.11 and 0.12, which means proportions greater than 0.10 are plausible. Since we cannot rule out proportions exceeding 0.10, we cannot support the claim that the proportion is at most 0.10. Choice A incorrectly suggests that containing 0.10 supports the claim, but an "at most" claim requires all plausible values to be at or below the threshold. For "at most" claims to be supported, the entire confidence interval must be at or below the claimed value.

Question 15

A fitness coach claims that clients who follow a new stretching routine can increase their sit-and-reach flexibility. A random sample of 32 clients measures sit-and-reach distance (cm) before starting the routine and again after 4 weeks on the routine. The research question is whether the routine increases the mean sit-and-reach distance. Which inference procedure is most appropriate?

  1. Matched-pairs ttt test for a mean difference (correct answer)
  2. Two-sample ttt interval for a difference in means
  3. One-sample zzz test for a proportion
  4. Chi-square goodness-of-fit test
  5. One-sample ttt interval for a mean

Explanation: This scenario involves measuring the same clients twice (before and after the stretching routine), creating paired data where each client serves as their own control. The matched-pairs t-test for a mean difference is the appropriate procedure because we're analyzing the mean of the differences between paired measurements (after minus before for each client). The two-sample t-interval (option B) would be incorrect because it assumes independent groups, not paired measurements. The one-sample procedures (options C and E) don't account for the paired nature of the data. When the same subjects are measured under two conditions, the matched-pairs design controls for individual variation and requires a matched-pairs t-test to analyze the mean difference.

Question 16

A basketball player makes 50 free throws in practice. A teammate claims the sequence of makes (M) and misses (X) is “too streaky” to be random. The record shows several clusters like MMMM and XXX, but overall there are 36 makes and 14 misses. Is the pattern consistent with random behavior? Under randomness (with a fixed make probability), you can still see clusters and streaks; randomness does not guarantee alternation.

  1. No; random sequences should alternate M and X frequently, so clusters indicate nonrandomness.
  2. Yes; streaks and clusters can occur naturally in random sequences. (correct answer)
  3. No; randomness requires exactly equal numbers of makes and misses.
  4. Yes; because the player made more than missed, the pattern must be random.
  5. No; independent events cannot produce runs like MMMM.

Explanation: This question addresses the common misconception that random sequences should alternate frequently. In reality, random processes with fixed probabilities (like a 72% free throw shooter) will naturally produce streaks and clusters - seeing runs like MMMM or XXX is expected, not suspicious. The overall proportion (36/50 = 72% makes) could reflect the player's true ability, and the presence of streaks doesn't contradict randomness. Students often think randomness means "no patterns," but random sequences actually contain many local patterns and runs. The key is that these patterns aren't predictable in advance. Hot and cold streaks in sports are often just natural random variation, not evidence of momentum or clutch performance.

Question 17

A cereal company states that the mean amount of cereal in its boxes is μ=18\mu=18μ=18 ounces, with individual box weights varying around this mean. Suppose many random samples of n=36n=36n=36 boxes are taken, and the sample mean xˉ\bar{x}xˉ is computed each time. The sampling distribution of xˉ\bar{x}xˉ is approximately normal (by the CLT). Which statement about the sampling distribution is correct?

  1. The mean of xˉ\bar{x}xˉ is 181818, and the standard deviation of xˉ\bar{x}xˉ is the same as the standard deviation of individual box weights.
  2. The mean of xˉ\bar{x}xˉ is 181818, and the standard deviation of xˉ\bar{x}xˉ is smaller than the standard deviation of individual box weights. (correct answer)
  3. The mean of xˉ\bar{x}xˉ is 18/3618/3618/36, and the standard deviation of xˉ\bar{x}xˉ is smaller than the standard deviation of individual box weights.
  4. The mean of xˉ\bar{x}xˉ is the same as the population standard deviation, and the standard deviation of xˉ\bar{x}xˉ is 181818.
  5. The sampling distribution of xˉ\bar{x}xˉ has the same center and the same spread as the distribution of individual box weights.

Explanation: This question assesses understanding of sampling distributions for sample means in AP Statistics. The sampling distribution of the sample mean ar{x} has a center (mean) equal to the population mean μ = 18 ounces, so it remains 18 regardless of sample size. The spread, measured by the standard deviation of ar{x}, is the population standard deviation σ divided by √n, which for n=36 is σ/6, making it smaller than the standard deviation of individual box weights. A common distractor is choice C, which incorrectly divides the mean by 36, confusing the mean of the sampling distribution with something like a proportion or average per unit. In a mini-lesson on sampling distributions of means: when you take many samples of size n from a population with mean μ and standard deviation σ, the sample means form a distribution that is approximately normal (by the Central Limit Theorem for large n), centered at μ, with standard error σ/√n, which decreases as n increases, meaning larger samples give more precise estimates of μ. Choice B correctly captures this by stating the mean is 18 and the standard deviation is smaller.

Question 18

An economist investigates whether the unemployment rate in a region is related to the average monthly rent. For a random sample of 50 regions, the economist records unemployment rate (explanatory variable xxx) and average monthly rent (response variable yyy) and fits a linear regression predicting rent from unemployment. The claim is that there is no linear association between unemployment rate and rent, meaning the population slope is 0. Which hypotheses are appropriate for testing this claim about the slope?

  1. H0:β1=0H_0: \beta_1 = 0H0​:β1​=0 vs. Ha:β1≠0H_a: \beta_1 \ne 0Ha​:β1​=0 (correct answer)
  2. H0:β1=0H_0: \beta_1 = 0H0​:β1​=0 vs. Ha:β1>0H_a: \beta_1 > 0Ha​:β1​>0
  3. H0:r=0H_0: r = 0H0​:r=0 vs. Ha:r≠0H_a: r \ne 0Ha​:r=0
  4. H0:b1=0H_0: b_1 = 0H0​:b1​=0 vs. Ha:b1≠0H_a: b_1 \ne 0Ha​:b1​=0
  5. H0:β0=0H_0: \beta_0 = 0H0​:β0​=0 vs. Ha:β0≠0H_a: \beta_0 \ne 0Ha​:β0​=0

Explanation: This question examines the skill of hypothesis testing for regression slopes in AP Statistics. The economist's claim is no linear association (slope=0), so to test this, use H₀: β₁ = 0 versus Hₐ: β₁ ≠ 0, seeking evidence against the null of no association. This two-sided setup is appropriate for non-directional claims about existence. Distractors include choice B, which assumes a positive direction not stated, and choice C using r instead of β₁. Choice E focuses on intercept β₀, irrelevant here. Mini-lesson: null is β₁=0; use ≠0 to test for any relationship when no direction is claimed; avoid one-sided unless specified, and use β₁, not sample b₁ (choice D) or r.

Question 19

A wildlife biologist estimates that p=0.15p=0.15p=0.15 of fish in a lake are a particular species. The biologist repeatedly takes many random samples of n=40n=40n=40 fish (with each sample taken independently from the lake) and records p^\hat pp^​, the sample proportion that are that species. Which statement about the sampling distribution of p^\hat pp^​ is correct?

  1. The sampling distribution of p^\hat pp^​ is centered at 0.150.150.15, and an approximately Normal model may not be justified because npnpnp is small. (correct answer)
  2. The sampling distribution of p^\hat pp^​ is centered at 0.850.850.85 because that is the proportion of other species.
  3. The sampling distribution of p^\hat pp^​ is centered at p^\hat pp^​ because p^\hat pp^​ is the parameter.
  4. The sampling distribution of p^\hat pp^​ has no variability because the lake proportion is fixed at 0.150.150.15.
  5. The sampling distribution of p^\hat pp^​ is exactly Normal because n=40n=40n=40 is greater than 30.

Explanation: This question tests sampling distributions for sample proportions in AP Statistics. The distribution is centered at p = 0.15, but with n = 40, np = 6 < 10, an approximately normal model may not be justified due to skewness, as in choice A. Spread remains (\sqrt{\frac{p(1-p)}{n}}), but shape concerns arise. Distractor choice E asserts exact normality for n > 30, but proportions require the np/n(1-p) rule, not just n > 30. Choice B centers at 0.85, confusing the complement. Mini-lesson: sampling distributions show how (\hat{p}) scatters around p across samples; unbiased center at p, variability inversely proportional to sqrt(n), but normality is only approximate if np ≥ 10 and n(1-p) ≥ 10—small np can result in asymmetric distributions.

Question 20

A teacher compares two study methods to see if Method A leads to a higher proportion of students earning an A on the next quiz. Students were randomly assigned to methods. In Method A, 33 of 80 earned an A; in Method B, 25 of 80 earned an A. A two-proportion zzz test for H0:pA−pB=0H_0: p_A-p_B=0H0​:pA​−pB​=0 versus Ha:pA−pB>0H_a: p_A-p_B>0Ha​:pA​−pB​>0 gave a p-value of 0.090.090.09. At α=0.05\alpha=0.05α=0.05, what conclusion is appropriate?

  1. Reject H0H_0H0​; there is convincing evidence that Method A results in a higher population proportion of A grades than Method B.
  2. Fail to reject H0H_0H0​; there is not convincing evidence that Method A has a higher population A-rate than Method B. (correct answer)
  3. Reject H0H_0H0​; there is convincing evidence that Method B results in a higher population proportion of A grades than Method A.
  4. Because students were randomly assigned, the p-value of 0.09 proves Method A causes a higher A-rate.
  5. Fail to reject H0H_0H0​; therefore the two population proportions of A grades are exactly equal.

Explanation: This problem involves testing whether Method A leads to a higher proportion of A grades than Method B. The p-value (0.09) is greater than α = 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence that Method A has a higher population A-rate than Method B, despite the sample showing 33/80 versus 25/80. Choice D incorrectly claims the p-value proves causation. Choice E incorrectly states that failing to reject H₀ proves the proportions are exactly equal. When the p-value exceeds the significance level, we lack sufficient evidence to support the alternative hypothesis, but this doesn't prove the null hypothesis is true.

Question 21

An online retailer wants to know whether device type is associated with whether a customer completes a purchase. From a single random sample of 500 site visits, the company records device type (Mobile, Tablet, Desktop) and outcome (Purchase, No purchase). The results appear in the two-way table below. Which test is appropriate, and what are the correct hypotheses?

Note: One sample; two categorical variables measured on each visit.

  1. Chi-square goodness-of-fit; H0H_0H0​: device types occur in equal proportions; HaH_aHa​: not all device proportions are equal
  2. Chi-square test for homogeneity; H0H_0H0​: the distribution of device type is the same for purchases and no purchases; HaH_aHa​: different distributions
  3. Chi-square test of independence; H0H_0H0​: device type and purchase outcome are independent; HaH_aHa​: device type and purchase outcome are associated (correct answer)
  4. Two-proportion zzz test; H0H_0H0​: pPurchase,Mobile=pPurchase,Desktopp_{\text{Purchase,Mobile}}=p_{\text{Purchase,Desktop}}pPurchase,Mobile​=pPurchase,Desktop​; HaH_aHa​: not equal
  5. Chi-square goodness-of-fit; H0H_0H0​: purchase and no purchase occur equally often; HaH_aHa​: they do not

Explanation: This AP Statistics question focuses on the skill of distinguishing chi-square test setups for homogeneity versus independence. With a single random sample of site visits where each visit records two categorical variables (device type and purchase outcome), the chi-square test of independence is correct to test for association. Choice C properly states the null as independence between variables and the alternative as association. Choice B is a distractor because it suggests homogeneity, which would require separate samples from groups like 'purchases' and 'no purchases,' but here it's one sample. Mini-lesson: Chi-square types include goodness-of-fit for one variable matching a model, independence for two variables in one sample, and homogeneity for one variable across multiple samples. Choice A incorrectly applies goodness-of-fit to device types alone. Overall, C fits the design.

Question 22

A public health team studies whether average systolic blood pressure differs between adults who exercise at least 150 minutes per week and adults who exercise less than 150 minutes per week. A random sample of 38 high-exercise adults had a mean systolic pressure of 124 mmHg, and a random sample of 41 low-exercise adults had a mean of 129 mmHg. Let μH\mu_HμH​ and μL\mu_LμL​ be the true mean systolic pressures for the two groups. Which hypotheses are appropriate?

  1. H0:μH−μL=0H_0: \mu_H-\mu_L=0H0​:μH​−μL​=0; Ha:μH−μL≠0H_a: \mu_H-\mu_L\ne 0Ha​:μH​−μL​=0 (correct answer)
  2. H0:μH−μL=0H_0: \mu_H-\mu_L=0H0​:μH​−μL​=0; Ha:μH−μL<0H_a: \mu_H-\mu_L<0Ha​:μH​−μL​<0
  3. H0:μL−μH=0H_0: \mu_L-\mu_H=0H0​:μL​−μH​=0; Ha:μL−μH<0H_a: \mu_L-\mu_H<0Ha​:μL​−μH​<0
  4. H0:xˉH−xˉL=0H_0: \bar{x}_H-\bar{x}_L=0H0​:xˉH​−xˉL​=0; Ha:xˉH−xˉL≠0H_a: \bar{x}_H-\bar{x}_L\ne 0Ha​:xˉH​−xˉL​=0
  5. H0:μH=124H_0: \mu_H=124H0​:μH​=124; Ha:μH≠124H_a: \mu_H\ne 124Ha​:μH​=124

Explanation: The public health team wants to test whether blood pressure DIFFERS between high and low exercise groups, with no specified direction. This requires a two-sided test with H_a: μ_H - μ_L ≠ 0. Choice B would test specifically if high-exercise adults have lower blood pressure, but the problem asks only about a difference. Choice D incorrectly uses sample means (x̄) rather than population means (μ) in the hypotheses. Choice E tests only one group against a fixed value rather than comparing two groups. When investigating whether two groups differ without specifying direction, always use a two-tailed alternative hypothesis with ≠.

Question 23

A quality-control engineer inspects the next 25 light bulbs coming off an assembly line. Each bulb is classified as a success if it is defective and a failure if it is not defective. The engineer plans to model the number of defective bulbs in the 25 inspected. Does this situation meet the conditions for using a binomial model?

  1. No, because the number of trials is not fixed.
  2. Yes, because there are a fixed number of trials, two outcomes per trial, and the probability of defect is constant for each bulb. (correct answer)
  3. No, because there are more than two outcomes for each bulb (defective, not defective, or unclear).
  4. No, because the probability of defect changes from bulb to bulb as the inspection continues.
  5. No, because the trials are not independent since the same engineer inspects all bulbs.

Explanation: This question tests understanding of binomial distribution conditions. A binomial setting requires: (1) fixed number of trials, (2) exactly two outcomes per trial, (3) constant probability of success, and (4) independent trials. Here we have 25 fixed trials (bulbs inspected), two outcomes per bulb (defective or not defective), and assuming consistent manufacturing, the probability of defect should be constant for each bulb. The trials are independent since one bulb being defective doesn't affect whether another is defective. The fact that the same engineer inspects all bulbs doesn't violate independence - what matters is whether the outcome of one bulb affects another bulb's outcome, which it doesn't.

Question 24

A scientist plots the concentration of a chemical reactant over time during a reaction. The points on the scatterplot clearly follow a curve that decreases rapidly at first and then flattens out as time progresses. Which of the following is the best description of the form of this association?

  1. Linear, because the relationship can be approximated with a straight line.
  2. Non-linear, because the points clearly follow a curved pattern. (correct answer)
  3. Weak, because the rate of change is not constant throughout the reaction.
  4. Positive, because both time and concentration are positive measurements.

Explanation: The form of an association is its general shape. Since the points are described as following a curve rather than a straight line, the form is non-linear.

Question 25

A researcher wants to estimate the mean number of minutes per day that students at a large high school spend on homework. She takes a random sample of 40 students and constructs a 95% confidence interval for the population mean homework time: (52.1,68.7)(52.1, 68.7)(52.1,68.7) minutes. Which interpretation is correct?

  1. There is a 95% chance that the population mean homework time is between 52.1 and 68.7 minutes.
  2. We are 95% confident that the interval from 52.1 to 68.7 minutes captures the true population mean homework time. (correct answer)
  3. About 95% of all students at the school spend between 52.1 and 68.7 minutes per day on homework.
  4. If many random samples of 40 students were taken, 95% of the sample means would fall between 52.1 and 68.7 minutes.
  5. If the study were repeated many times, 95% of the time the population mean would fall between 52.1 and 68.7 minutes.

Explanation: This question tests understanding of confidence interval interpretation for a population mean. The correct interpretation (B) states that we are 95% confident the interval captures the true population mean homework time. Choice A incorrectly treats the confidence level as a probability about the parameter itself. Choice C misinterprets the interval as describing individual student values rather than the mean. Choice D incorrectly describes a sampling distribution of sample means. Choice E wrongly suggests the population mean changes between studies. Remember: a confidence interval provides a range of plausible values for the population parameter based on our sample data, and the confidence level describes the long-run success rate of the method.