Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

AP Statistics

AP Statistics Practice Test: Practice Test 79

Practice Test 79 for AP Statistics: real questions and explanations from the Varsity Tutors practice-test pool.

0 / 25 answered

Question 1 of 25

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?

Question Navigator

All questions

Question 1

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 2

A student researches whether the number of pages read for a class ( $x$ ) predicts the time spent studying in hours ( $y$ ). The student believes that reading more pages is associated with more study time in the population. Which hypotheses are appropriate for testing this claim about the slope in the regression of $y$ on $x$ ?

$H_0: b=0$ vs. $H_a: b>0$
$H_0: \beta=0$ vs. $H_a: \beta>0$ (correct answer)
$H_0: r=0$ vs. $H_a: r>0$
$H_0: \rho=0$ vs. $H_a: \rho>0$
$H_0: \beta=0$ vs. $H_a: \beta<0$

Explanation: This question evaluates setting up slope hypotheses in AP Statistics. Student believes more pages mean more study time, positive β, so H₀: β = 0 vs Hₐ: β > 0, choice B. Distractors A sample b, C and D correlations, E wrong direction. Lesson: β quantifies relationship; null zero, alternative per claim. Population focus. Infers pages' positive association with study time.

Question 3

A biology lab expects a 1:2:1 ratio of flower colors (white:pink:red) in a large population of plants. A random sample of 80 flowers is observed, and a chi-square goodness-of-fit test is conducted at $\alpha=0.05$ . The results are below. What conclusion is appropriate?

Observed/Expected counts and p-value:

White: Observed 18, Expected 20
Pink: Observed 44, Expected 40
Red: Observed 18, Expected 20
p-value = 0.54

Because $p>\alpha$ , reject $H_0$ ; there is convincing evidence the population ratio is 1:2:1.
Because $p>\alpha$ , fail to reject $H_0$ ; the sample provides insufficient evidence that the population distribution differs from a 1:2:1 ratio. (correct answer)
Because $p<\alpha$ , fail to reject $H_0$ ; there is insufficient evidence of a difference from 1:2:1.
Because $p>\alpha$ , fail to reject $H_0$ ; this proves exactly half of all flowers are pink.
Because $p>\alpha$ , fail to reject $H_0$ ; therefore the sample must have exactly 20 white, 40 pink, and 20 red flowers.

Explanation: This question involves the chi-square goodness-of-fit test in AP Statistics to test if flower colors follow a 1:2:1 ratio in the population. With p=0.54 greater than α=0.05, we fail to reject the null hypothesis, providing insufficient evidence that the population distribution differs from the expected ratio. The sample is consistent with the lab's expectation. Choice D misleads by suggesting non-rejection proves exactly half are pink, but it doesn't; it only shows no strong evidence against the ratio. For a mini-lesson on chi-square conclusions, failing to reject means the observed data could plausibly come from the hypothesized model, but it doesn't rule out other models—use confidence intervals or further tests for more insight. Contextualize to the population, noting that 'proof' is not a statistical term here.

Question 4

A company tests two advertising designs and records the mean number of website visits per day. Design 1 had a sample mean of 510 visits and Design 2 had a sample mean of 525 visits. A 90% confidence interval for the difference in population means, defined as $\mu_1-\mu_2$ , is $(-30, -5)$ . A manager claims, “Design 1 results in fewer mean daily visits than Design 2.” Is the claim supported by the confidence interval?

Yes, because the entire interval is below 0, which supports $\mu_1<\mu_2$ . (correct answer)
No, because the interval is negative, so it shows Design 1 is better.
No, because 0 is not in the interval, so we cannot make any comparison.
Yes, because the sample mean for Design 2 is larger, so the population mean must be larger.
No, because the interval is for $\mu_2-\mu_1$ , not $\mu_1-\mu_2$ .

Explanation: This question requires interpreting a confidence interval for the difference μ_1 - μ_2 = (-30, -5). Since the entire interval contains only negative values, we can be 90% confident that μ_1 - μ_2 < 0, which means μ_1 < μ_2. This directly supports the manager's claim that Design 1 results in fewer mean daily visits than Design 2. A common error is getting confused by the negative values, but negative values for μ_1 - μ_2 indicate that the first mean is smaller than the second. When a confidence interval for a difference is entirely below zero, it provides evidence that the first population mean is less than the second population mean.

Question 5

A wildlife biologist tests whether the proportion of a fish species carrying a parasite is less than 0.30. The hypotheses are $H_0: p=0.30$ versus $H_a: p<0.30$ at $\alpha=0.05$ . The test is statistically significant, so the biologist concludes the parasite rate is below 0.30. In reality, the true population proportion is $p=0.34$ . Which type of error was made?

No error was made because the alternative hypothesis was supported
Type II error (failing to reject a false null hypothesis)
Type I error (rejecting a true null hypothesis) (correct answer)
Type I error (failing to reject a false null hypothesis)
Type II error (rejecting a true null hypothesis)

Explanation: This question presents a Type I error in a one-sided test context. The biologist rejected the null hypothesis (concluded p < 0.30) when in reality the null hypothesis was true (p = 0.34, which is not less than 0.30). This is a Type I error - rejecting a true null hypothesis. The actual proportion (0.34) is even higher than the null value (0.30), making this clearly a false rejection. Students should note that with a left-tailed test (H_a: p < 0.30), any true proportion at or above 0.30 makes H₀ true. Type I errors can occur in any direction of testing and represent false alarms where we claim an effect exists when it doesn't.

Question 6

A mechanic studies whether tire pressure in psi ( $x$ ) predicts gas mileage in miles per gallon ( $y$ ). The mechanic believes that higher tire pressure (within safe limits) increases gas mileage in the population. Which hypotheses are appropriate for testing this claim about the slope of the population regression line?

$H_0: \beta=0$ vs. $H_a: \beta>0$ (correct answer)
$H_0: b=0$ vs. $H_a: b>0$
$H_0: \rho=0$ vs. $H_a: \rho>0$
$H_0: r=0$ vs. $H_a: r>0$
$H_0: \beta=0$ vs. $H_a: \beta<0$

Explanation: AP Statistics: setup for slope hypotheses. Higher tire pressure increases mileage, positive β, H₀: β = 0 vs Hₐ: β > 0, choice A. Distractors B sample b, C and D correlations (ρ and r), E negative. Mini-lesson: β > 0 for positive change. Use population parameter. Tests pressure's positive effect on mileage.

Question 7

Let $X$ be the discrete random variable representing the number of absences a randomly selected student has in a semester. At a school, $\mu_X=2.3$ absences and $\sigma_X=2.0$ absences. Which interpretation of the mean is correct?

A randomly selected student is most likely to have exactly 2.3 absences.
In the long run, students at the school average about 2.3 absences per semester. (correct answer)
Each student has between 0.3 and 4.3 absences because $2.3\pm2.0$ gives the range.
The probability a student has exactly 2 absences is 0.23.
Because the mean is 2.3, no student can have 0 absences.

Explanation: This question tests understanding of the mean for student absences. The mean μ_X = 2.3 represents the average number of absences per semester across all students at the school. Option A incorrectly suggests 2.3 is the most likely exact value for an individual. Option C wrongly assumes all students fall within one standard deviation of the mean. Option D incorrectly relates the mean to a specific probability. Option E falsely claims a positive mean eliminates the possibility of zero absences. The correct interpretation is that when averaging across all students at the school, the mean number of absences per semester is approximately 2.3.

Question 8

A technology company claims that the proportion of users who enable two-factor authentication is $p = 0.35$ . A random sample of users is selected and a one-proportion test is performed for $H_0: p = 0.35$ versus $H_a: p > 0.35$ at $\alpha = 0.05$ . The p-value is $p = 0.006$ . Which interpretation of the p-value is correct?

There is a 0.6% chance that $p$ is greater than 0.35.
If $p=0.35$ , the probability of getting a sample proportion at least as large as the observed sample proportion (in the direction of $H_a$ ) is 0.006. (correct answer)
There is a 0.6% chance that the null hypothesis is true.
A p-value of 0.006 means 0.6% of users enable two-factor authentication.
If the alternative hypothesis is true, there is a 0.6% chance of observing the sample result.

Explanation: This question evaluates interpreting p-values in a one-sided right-tailed test for a proportion in AP Statistics. The p-value of 0.006 is the conditional probability of getting a sample proportion at least as large as observed, given H0: p = 0.35 is true, in the direction of Ha: p > 0.35. A frequent distractor is choice C, which mistakes the p-value for the probability that H0 is true. As a mini-lesson, p-values assess evidence by computing data extremity under H0: a small 0.006 provides strong grounds to reject H0 at α = 0.05, suggesting more than 35% enable authentication. Choice B accurately specifies 'at least as large... (in the direction of Ha)'. Misinterpretations in A, D, and E include reversing conditioning or applying to population subsets.

Question 9

A teacher studies whether students’ quiz average (percent) can be predicted from the number of class absences during a grading period. From a random sample of 40 students, the teacher fits a least-squares regression with quiz average as the response and absences as the explanatory variable. The teacher’s claim is that more absences are associated with lower average quiz scores in the population, implying a negative population slope. Which hypotheses are appropriate for testing this claim about the slope?

$H_0: \beta_1=0$ vs. $H_a: \beta_1<0$ (correct answer)
$H_0: r=0$ vs. $H_a: r<0$
$H_0: b=0$ vs. $H_a: b<0$
$H_0: \beta_0=0$ vs. $H_a: \beta_0<0$
$H_0: \beta_1<0$ vs. $H_a: \beta_1\ge 0$

Explanation: This question involves testing for a negative relationship between class absences and quiz average. The teacher claims that more absences are associated with lower quiz scores, indicating a negative population slope. For regression slope hypothesis tests, we use β₁ as the population slope parameter, with the null hypothesis H₀: β₁ = 0 (no relationship). Since the claim is about a negative association, the alternative hypothesis is Hₐ: β₁ < 0. Choice A correctly presents these hypotheses. Students should distinguish between the population parameter β₁ and sample statistics like b or r, and ensure the alternative hypothesis matches the research claim's direction.

Question 10

A student observes 20 cars passing an intersection and records whether each car runs a red light. A success is “runs red light,” and exactly 20 cars are observed during the same signal cycle length. Does this meet binomial conditions?

Yes, if each car’s behavior can be treated as independent and the probability of running the light is approximately constant during the observation period. (correct answer)
No, because cars have more than two outcomes (stop, slow, go).
No, because the number of trials is not fixed.
Yes, but only if exactly one car runs the red light.
No, because binomial requires sampling without replacement.

Explanation: This AP Statistics question evaluates binomial in observations. Fixed n=20 cars, binary (runs light or not), assuming independent with constant p, yes. Meets if no interactions. Distractor B posits more outcomes, but it's binary. Choice C says n not fixed, but it is. Mini-lesson: Observational data can be binomial if conditions hold; watch for dependencies.

Question 11

A school district wants to know whether the proportion of students who prefer block scheduling differs between two grade levels. A random sample of 120 ninth-graders and 150 twelfth-graders was surveyed. Among ninth-graders, 54 preferred block scheduling; among twelfth-graders, 93 preferred block scheduling. Which inference procedure is most appropriate to answer the research question?

One-sample $z$ interval for a population proportion
Two-sample $z$ test for a difference in proportions (correct answer)
Chi-square test for goodness of fit
Two-sample $t$ test for a difference in means
Matched-pairs $t$ interval for a mean difference

Explanation: This question asks about comparing proportions between two independent groups (ninth-graders and twelfth-graders), making the two-sample z test for a difference in proportions the appropriate choice. We have two distinct populations with binary outcomes (prefer block scheduling or not), and we want to test if the population proportions differ. The chi-square test for goodness of fit (option C) tests against a specific distribution, not comparing two groups. The t-tests (options D and E) are for quantitative data, not categorical. When comparing proportions between two independent samples, always use the two-sample z test.

Question 12

A bookstore tracks the number of books purchased in a single transaction by a randomly selected customer. Let $X$ be the number of books, a discrete random variable with $\mu_X=1.8$ books and $\sigma_X=1.3$ books. Which interpretation of the standard deviation is correct?

About 1.3 books is the typical distance between a customer’s purchase count and 1.8 books. (correct answer)
About 1.3 books means the average number of books purchased is 1.3.
About 1.3 books is the difference between the largest and smallest purchase counts.
About 1.3 books means most customers buy exactly 1.3 books.
About 1.3 books means 1.8 books occurs about 13% of the time.

Explanation: This question tests understanding of standard deviation interpretation. The standard deviation σ = 1.3 books measures the typical distance between individual purchase counts and the mean of 1.8 books. Choice A correctly identifies this - it's the typical distance between a customer's actual purchase count and the mean of 1.8. Choice B incorrectly defines standard deviation as the average. Choice C confuses it with the range. Choice D misinterprets it as an exact value. Choice E makes an incorrect probability claim. Standard deviation tells us about the spread or variability in the data, specifically how far values typically fall from the mean.

Question 13

A random sample of 80 packaged salads from a factory found that 12 were underweight. A 99% confidence interval for the true proportion $p$ of all packaged salads from that factory that are underweight is $(0.06,\ 0.24)$ . Which interpretation is correct?

There is a 99% chance that the sample proportion of underweight salads is between 0.06 and 0.24.
We are 99% confident that the true proportion $p$ of all packaged salads from the factory that are underweight is between 0.06 and 0.24. (correct answer)
About 99% of all packaged salads from the factory are underweight.
The probability that $p$ is between 0.06 and 0.24 is 0.99.
If we repeatedly sample 80 salads, 99% of the time the true proportion $p$ will fall between 0.06 and 0.24.

Explanation: Interpreting a confidence interval for a population proportion is the skill being tested here in AP Statistics. The accurate interpretation is that we are 99% confident the true proportion p of underweight salads is between 0.06 and 0.24, reflecting that the interval construction method succeeds in capturing p 99% of the time over many samples. Choice D distracts by claiming the probability that p is between 0.06 and 0.24 is 0.99, which is misleading because once the interval is calculated, it either contains p or not; the probability refers to the method, not this fixed interval. In a mini-lesson, a confidence interval means that if we repeatedly sample and compute intervals, 99% will include the true parameter, providing a measure of reliability for our estimation process. This concept prevents errors like treating the CI as a direct probability statement about p. It also highlights why we use 'confidence' to describe our trust in the interval based on statistical theory.

Question 14

A dataset of the last digit of 1,000 telephone numbers is collected. Which of the following best describes the likely shape of the distribution of these digits?

Skewed to the left, because smaller digits are typically less common in phone numbers.
Skewed to the right, because larger digits are typically more common in phone numbers.
Approximately uniform, because each digit from 0 to 9 is expected to occur with roughly equal frequency. (correct answer)
Unimodal and symmetric, because the middle digits such as 4, 5, and 6 are most likely to occur.

Explanation: The correct answer is C. In a large, random set of telephone numbers, there is no reason to believe any one digit (0-9) would appear more frequently than another as the last digit. Therefore, the distribution of these digits is expected to be approximately uniform, with each digit having a roughly equal count.

Question 15

A number is selected at random from the integers $1$ through $20$ , inclusive. Define event $A$ as “the number is even” and event $B$ as “the number is a multiple of 5.” Are the two events mutually exclusive?

Yes; no number from 1 to 20 can be both even and a multiple of 5.
Yes; the events are mutually exclusive because they involve different properties.
No; the events are not mutually exclusive because some numbers are both even and multiples of 5. (correct answer)
No; the events are independent, so they cannot be mutually exclusive.
Cannot be determined without counting how many evens and multiples of 5 there are.

Explanation: This problem asks about the mutual exclusivity of selecting an even number versus a multiple of 5 from 1-20. To determine this, we need to check if any numbers satisfy both conditions. The numbers 10 and 20 are both even AND multiples of 5, meaning both events can occur with the same outcome. Since there exist outcomes where both events happen simultaneously, the events are not mutually exclusive. Choice A incorrectly claims no overlap exists, while choice D confuses independence with mutual exclusivity. Remember: mutually exclusive events have no outcomes in common.

Question 16

A nonprofit claims that 50% of donors prefer to be contacted by email rather than phone. A random sample of 100 donors finds that 61 prefer email. A one-sample proportion test is performed at $\alpha=0.05$ with hypotheses $H_0:p=0.50$ and $H_a:p\ne0.50$ . The test yields a $p$ -value of 0.04, so the decision is to reject $H_0$ . Which conclusion is appropriate?

Since 61 of the sampled donors prefer email, we can conclude only that 61% of the sample prefers email, not the population.
There is sufficient evidence at the 0.05 level that the true proportion of all donors who prefer email is different from 0.50. (correct answer)
Rejecting $H_0$ proves that more than half of all donors prefer email.
The $p$ -value of 0.04 means there is a 4% chance that $H_0$ is true.
Preferring email is caused by the nonprofit’s decision to contact donors.

Explanation: This question examines conclusions from rejecting H₀ in a two-tailed test. With p-value (0.04) less than α (0.05), we reject H₀: p = 0.50 in favor of Hₐ: p ≠ 0.50. Choice B correctly states there is sufficient evidence that the true proportion differs from 0.50. Choice C incorrectly specifies a direction (more than half) when the two-tailed test only establishes that p ≠ 0.50—we'd need to examine the sample proportion to determine direction. Choice D misinterprets the p-value as the probability H₀ is true. In two-tailed tests, rejecting H₀ means the parameter significantly differs from the hypothesized value, but the conclusion must reflect the non-directional nature of the alternative hypothesis.

Question 17

A political scientist wants to compare whether voters from three political parties (Party A, Party B, Party C) have the same distribution of opinion on a policy (Support, Oppose). Independent random samples of registered voters are taken from each party list. The results are shown in the table. Which chi-square test is appropriate, and what are the correct hypotheses?

State clearly whether this is one sample or multiple groups.

Chi-square goodness-of-fit; $H_0$ : Support and Oppose are each 50%; $H_a$ : not 50/50
Chi-square test of homogeneity; $H_0$ : the distribution of opinion (Support/Oppose) is the same across Party A, Party B, and Party C; $H_a$ : at least one party has a different distribution (correct answer)
Chi-square test of independence; $H_0$ : party and opinion are independent in a single random sample of voters; $H_a$ : they are associated
Two-proportion $z$ test; $H_0$ : the proportion who support is the same in Party A and Party B; $H_a$ : it differs
Chi-square test of homogeneity; $H_0$ : each party has the same number of sampled voters; $H_a$ : at least one differs

Explanation: The problem clearly states "Independent random samples of registered voters are taken from each party list," indicating three separate samples (one from each party). This is a multiple groups design comparing distributions across independent samples, requiring a chi-square test of homogeneity. We're testing whether the distribution of opinion (Support/Oppose) is the same across all three parties. The null hypothesis states that all parties have the same distribution of opinions, while the alternative states that at least one party differs. Choice B correctly identifies this as a test of homogeneity with proper hypotheses. The key indicator is the separate sampling from each party's voter list, creating independent groups for comparison.

Question 18

A museum studies whether visitor age group is associated with preferred exhibit type. A random sample of visitors is surveyed and classified by age group (Child, Adult, Senior) and preferred exhibit (Art, History, Science). The results are shown below. A chi-square test of independence is performed and results in $p=0.030$ . What conclusion is appropriate at the $\alpha=0.05$ level?

Two-way table (counts):

Age group	Art	History	Science
Child	20	15	45
Adult	55	60	35
Senior	40	55	15

There is not convincing evidence of an association because $p=0.030$ is greater than $0.05$ .
There is convincing evidence that age group and preferred exhibit type are associated among museum visitors. (correct answer)
We can conclude that visiting the museum causes seniors to prefer history exhibits.
We can conclude that the distributions of preferred exhibit type are the same for all age groups.
Because $p=0.030$ , we should fail to reject $H_0$ at $\alpha=0.05$ .

Explanation: This question evaluates the chi-square test of independence to determine if age group and preferred exhibit are associated. With p=0.030 less than alpha=0.05, we reject the null, finding convincing evidence of association. A common distractor is choice C, which wrongly infers causation, suggesting museum visits cause seniors' preferences, but association does not imply causation. Association means the preferences differ by age, while causation would require evidence that age directly drives preferences. The table shows distinct patterns, like children favoring science, leading to the significant p-value. This result demonstrates how demographic factors can correlate with interests without causal links.

Question 19

A teacher wants to know whether participation in an optional review session is related to exam outcome. From a single random sample of 180 students enrolled in a course, the teacher records whether each student attended the review session (Yes/No) and whether they passed the exam (Pass/Fail). The results are shown in the two-way table. Which chi-square test is appropriate, and what are the correct hypotheses?

Note: One sample; two categorical variables.

Chi-square goodness-of-fit; $H_0$ : pass and fail occur in a 50/50 split; $H_a$ : not a 50/50 split
Chi-square test for homogeneity; $H_0$ : the distribution of attendance is the same among passers and failers; $H_a$ : different distributions
Chi-square test of independence; $H_0$ : review-session attendance and exam outcome are independent; $H_a$ : they are associated (correct answer)
Two-proportion $z$ test; $H_0$ : $p_{\text{Pass,Attend}}=p_{\text{Pass,NoAttend}}$ ; $H_a$ : not equal
Matched-pairs test; $H_0$ : no difference before vs after review; $H_a$ : a difference

Explanation: Assessing AP Statistics skill in chi-square homogeneity versus independence setups, this uses one random sample of students with two categorical variables (attendance and outcome), ideal for chi-square test of independence on association. Choice C hypothesizes null independence and alternative association correctly. Choice B is a distractor, suggesting homogeneity, but that requires separate samples, not one here. Mini-lesson: Chi-square includes goodness-of-fit for single-variable proportions, independence for two-variable ties in one group, and homogeneity for multi-group comparisons. Choice A misapplies goodness-of-fit to outcomes. C is accurate.

Question 20

A marine biologist studies whether ocean depth in meters ( $x$ ) predicts water temperature in °C ( $y$ ) at sampling locations. The biologist expects that greater depth corresponds to lower temperature in the population. Which hypotheses are appropriate for testing this expectation about the slope in the regression of $y$ on $x$ ?

$H_0: \beta=0$ vs. $H_a: \beta<0$ (correct answer)
$H_0: b=0$ vs. $H_a: b<0$
$H_0: r=0$ vs. $H_a: r<0$
$H_0: \rho=0$ vs. $H_a: \rho<0$
$H_0: \beta=0$ vs. $H_a: \beta>0$

Explanation: This AP Statistics question focuses on hypothesizing about the regression slope. The biologist expects greater depth to lower temperature, implying negative β, so H₀: β = 0 versus Hₐ: β < 0 fits, per choice A. Distractors B, C, D use sample b or correlations r and ρ, and E has the opposite direction. Mini-lesson: β represents population slope; test H₀: β = 0 against directional Hₐ based on expectation. Greek symbols denote parameters. This evaluates if depth negatively affects temperature in oceanic populations.

Question 21

In a survey of 200 students, 90 reported that they play a sport (S), 120 reported that they have a part-time job (J), and 50 reported both playing a sport and having a part-time job. One student is selected at random. Let Event A be “the student plays a sport” and Event B be “the student has a part-time job.” Which statement about Events A and B is correct?

Events A and B are independent because $P(A\cup B)=P(A)+P(B)$ .
Events A and B are independent because $P(A\cap B)=P(A)P(B)$ .
Events A and B are not independent because the events overlap.
Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$ . (correct answer)
Events A and B are independent because playing a sport implies having a job.

Explanation: This problem tests independence with survey data. Given: P(A) = 90/200, P(B) = 120/200, and P(A∩B) = 50/200. For independence, we need P(A∩B) = P(A)·P(B). Calculating: P(A)·P(B) = (90/200)·(120/200) = 10800/40000 = 54/200. However, P(A∩B) = 50/200. Since 54/200 ≠ 50/200, events A and B are not independent. Choice C incorrectly states that overlapping events cannot be independent—events can overlap and still be independent if they satisfy the multiplication rule. The key is checking the probability relationship, not just whether events can occur together.

Question 22

A school district studies the height (in inches) of individual 10th-grade students. The population mean is $\mu=66$ inches and the population standard deviation is $\sigma=3.5$ inches. A researcher repeatedly selects simple random samples of size $n=100$ and computes the sample mean height $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?

The sampling distribution of $\bar{x}$ has mean $66$ and standard deviation $3.5/\sqrt{100}$ . (correct answer)
The sampling distribution of $\bar{x}$ has mean $66$ and standard deviation $3.5$ because the sample size does not affect variability.
The sampling distribution of $\bar{x}$ has mean $66/100$ and standard deviation $3.5/100$ .
The sampling distribution of $\bar{x}$ has mean $100$ and standard deviation $66/\sqrt{3.5}$ .
The sampling distribution of $\bar{x}$ is approximately normal only if the population distribution is exactly normal.

Explanation: For student heights with μ=66 inches and σ=3.5 inches, samples of size n=100 yield a sampling distribution of x̄ with mean 66 and standard deviation σ/√n = 3.5/√100 = 3.5/10 = 0.35. The large sample size (n=100) ensures approximate normality through the Central Limit Theorem. Choice B incorrectly claims sample size doesn't affect variability - a fundamental misunderstanding since larger samples produce more precise estimates. Choice E incorrectly states normality requires a normal population; with n≥30, the CLT provides approximate normality.

Question 23

A school district completed a study to evaluate whether a new online homework platform improves Algebra I quiz scores. From all 12 high schools in the district, administrators randomly selected 240 Algebra I students (20 per school). Within each school, the selected students were randomly assigned to either use the new platform for 6 weeks or continue with the usual homework system. At the end of 6 weeks, all selected students took the same district quiz, and the platform group had a higher average score. Which conclusion is justified based on the design of this study?

Because students were randomly assigned, the higher scores prove the platform works for all Algebra I students in the state.
Because students were randomly sampled from the district and randomly assigned, it is reasonable to generalize to Algebra I students in the district and to conclude the platform caused higher quiz scores. (correct answer)
Because students were randomly sampled, any difference in scores must have been caused by the platform.
Because students were randomly assigned within schools, it is reasonable to conclude the platform caused higher scores for the 240 students but not to generalize beyond them.
Because the district used multiple schools, it is reasonable to generalize to all U.S. high school students, but causation cannot be concluded.

Explanation: This question tests understanding of when both causation and generalization are justified in experimental design. The study uses random sampling (240 students randomly selected from all district Algebra I students) which allows generalization to the population sampled from - the district's Algebra I students. It also uses random assignment (students randomly assigned to platform or usual homework within schools) which allows causal conclusions. The key distractor is choice A, which overextends generalization to all state students when only district students were sampled. In statistical inference, random sampling permits generalization to the sampled population, while random assignment permits causal conclusions. Since this study has both features, we can both generalize to district Algebra I students AND conclude the platform caused the improvement.

Question 24

A sports scientist examines whether the number of hours of sleep the night before a race ( $x$ ) predicts a runner’s 5K time ( $y$ , in minutes). Using data from 35 runners, a least-squares regression of time on sleep is fit. The scientist’s claim is that more sleep leads to faster times (smaller minutes), implying a negative population slope. Which hypotheses are appropriate for testing the population regression slope consistent with this claim?

$H_0: \beta_1=0$ ; $H_a: \beta_1>0$
$H_0: r=0$ ; $H_a: r\ne 0$
$H_0: b_1=0$ ; $H_a: b_1<0$
$H_0: \beta_0=0$ ; $H_a: \beta_0<0$
$H_0: \beta_1=0$ ; $H_a: \beta_1<0$ (correct answer)

Explanation: This question assesses setting up slope hypotheses in AP Statistics, where more sleep is claimed to predict faster race times (lower minutes), implying a negative slope for sleep hours versus time. The appropriate hypotheses are H₀: β₁ = 0 versus H_a: β₁ < 0, matching the one-sided claim of negative association. A common distractor is choice A, which uses >0, reversing the expected direction. Choice C uses sample b₁ incorrectly instead of β₁. Mini-lesson: Hypotheses for slopes target β₁, with null at 0 and alternative reflecting the claim's direction—here <0 since increased x (sleep) decreases y (time). This one-sided test is suitable for directional research questions to maximize detection power.

Question 25

A researcher tests whether a new fertilizer changes the mean height of a certain plant. They test $H_0: \mu=18$ cm vs. $H_a: \mu\neq18$ cm at $\alpha=0.05$ . The p-value is 0.64, so they fail to reject $H_0$ . In reality, the true mean height with the fertilizer is $\mu=18$ cm. Which type of error was made?

Type I error: failing to reject a false null hypothesis
Type II error: failing to reject a false null hypothesis
No error was made because the null hypothesis was true and they failed to reject it (correct answer)
Type I error: rejecting a true null hypothesis
Type II error: rejecting a true null hypothesis

Explanation: This question tests understanding of correct decisions in hypothesis testing. The researcher failed to reject the null hypothesis (concluded no evidence of height change) and in reality the null hypothesis was true (mean height is indeed 18 cm). This represents a correct decision: failing to reject a true null hypothesis. The p-value of 0.64 being much greater than α = 0.05 led to failing to reject, which was the correct conclusion. No error was made because the test correctly concluded there was no effect when indeed no effect existed, demonstrating proper control of Type I error.