AP Statistics
Practice Test 31 for AP Statistics: real questions and explanations from the Varsity Tutors practice-test pool.
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A student randomly draws one card from a well-shuffled standard 52-card deck and records whether it is a heart or not a heart. Let event H be “the card is a heart,” so P(H)=5213=0.25. Which statement correctly describes the probability P(H)=0.25?
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A student randomly draws one card from a well-shuffled standard 52-card deck and records whether it is a heart or not a heart. Let event H be “the card is a heart,” so P(H)=5213=0.25. Which statement correctly describes the probability P(H)=0.25?
Explanation: This question examines probability interpretation in a classic card-drawing scenario. The sample space contains 52 cards, with 13 hearts, giving P(H) = 13/52 = 0.25. Choice C correctly interprets this: over many independent single-card draws with replacement, about 25% will be hearts. Choice A incorrectly assumes probability guarantees exact outcomes in small samples. Choice D wrongly suggests that because P(H) = 0.25, three hearts in a row is impossible, when actually P(three hearts) = (0.25)³ = 0.015625. The crucial insight is that probability describes long-run behavior in repeated trials, not deterministic patterns or impossibilities based on the probability value.
A company wants to estimate the mean job satisfaction score (1–10) for all employees. The HR department selects a simple random sample of 200 employees and asks them to complete a satisfaction survey in a group meeting led by their direct managers, who can see who turns in each survey. Which issue most threatens the validity of the results?
Explanation: This question in AP Statistics examines response bias as a potential sampling issue. The company took a simple random sample of 200 employees but conducted surveys in manager-led meetings where responses weren't anonymous, potentially causing employees to inflate satisfaction scores out of fear. Validity is threatened because the lack of privacy may lead to dishonest answers, not reflecting true job satisfaction across all employees. Undercoverage (choice B) distracts by misapplying the concept, as all employees were in the frame regardless of having managers. Diagnosing sampling problems involves checking not just selection but also how data is collected; here, the non-anonymous setting introduces response bias by influencing honesty.
A library tracks the genres of books checked out by patrons. A relative frequency table shows the following: Fiction: 0.52, Non-Fiction: 0.28, Young Adult: 0.15. The rest are Children's books. If 400 books were checked out in total, what is the frequency of Children's books?
Explanation: First, find the relative frequency of Children's books. The sum of relative frequencies must be 1. 1−(0.52+0.28+0.15)=1−0.95=0.05. Then, find the frequency by multiplying this relative frequency by the total number of books: 0.05×400=20.
An environmental group claims that more than 35% of households in a county compost food scraps. A survey produced a 95% confidence interval for the true proportion p of households that compost: (0.31, 0.36). Is the claim supported by the confidence interval?
Explanation: This question asks whether a confidence interval supports a claim about exceeding a proportion. The claim states that more than 35% compost (p > 0.35), but the interval (0.31, 0.36) includes values both above and below 0.35, including 0.35 itself. Since the interval contains values less than or equal to 0.35, the claim is not supported. Choice A incorrectly thinks containing 0.35 supports p > 0.35. Choice B misinterprets confidence levels. Choice E wrongly focuses only on the upper bound. To support a strict inequality claim (p > value), the entire interval must exceed that value.
A city council wants to know whether residents from two different neighborhoods (Eastside and Westside) have the same distribution of preferred public transportation option (Bus, Train, Bike). Independent random samples of residents are taken from each neighborhood. 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.
Explanation: The problem states "Independent random samples of residents are taken from each neighborhood," which clearly indicates two separate samples (one from Eastside, one from Westside). This is a multiple groups design requiring a chi-square test of homogeneity. We're comparing whether the distribution of transportation preferences is the same between the two neighborhoods. The null hypothesis states that both neighborhoods have the same distribution of preferences, while the alternative states the distributions differ. Choice C correctly identifies this as a test of homogeneity with appropriate hypotheses. Note that choice E incorrectly focuses on row totals being equal, which is not what homogeneity tests examine—they compare distributions (proportions), not raw counts.
A school claims that students choose among 5 lunch options with equal frequency. A student randomly records the lunch choice of 150 students on one day and obtains the counts shown in the table. A chi-square goodness-of-fit test is conducted for H0: all 5 options are equally likely vs. Ha: not all options are equally likely. The p-value is 0.62. What conclusion is appropriate at the α=0.05 level?
Observed/Expected counts and p-value:
Explanation: This question assesses interpreting chi-square test results when the p-value is large. With p-value = 0.62, which is much greater than α = 0.05, we fail to reject the null hypothesis. Failing to reject H₀ means we don't have convincing evidence against the claim that lunch choices are equally likely. Choice B correctly states this conclusion. Choice A incorrectly rejects H₀ when p > α, while choice E makes the common error of claiming a large p-value proves H₀ is true. Remember that failing to reject H₀ doesn't prove it's true; it simply means we lack evidence to conclude it's false. Large p-values indicate the observed data is consistent with the null hypothesis.
A cafeteria manager randomly selects one customer and defines a discrete random variable X as the number of pieces of fruit the customer buys (0, 1, 2, or 3). The probability distribution of X is shown in the table. Which statement correctly interprets the random variable?
Explanation: This question tests understanding of what a random variable represents in probability. The random variable X is defined as "the number of pieces of fruit the customer buys," which means X takes on numerical values (0, 1, 2, or 3) representing a count. Choice A incorrectly interprets X as a probability value itself, when X is actually the variable whose probabilities are given in the distribution table. Choice C mistakenly treats X as a specific event rather than a variable that can take multiple values. Choice D confuses the individual customer's purchase with the total for all customers. Choice E misunderstands X as the set of probabilities rather than the variable itself. A random variable is a function that assigns numerical values to the outcomes of a random phenomenon—in this case, the number of fruit pieces purchased by one randomly selected customer.
A school district recorded the number of minutes 120 high school students spent on homework last night. The summaries were: mean = 62.4 min, median = 55 min, SD = 28.1 min, five-number summary = (min 10, Q1 40, median 55, Q3 75, max 160), and IQR = 35 min. Which interpretation is correct?
Explanation: This question tests understanding of how summary statistics reveal distribution shape and typical values. The mean (62.4) exceeding the median (55) indicates right skewness, as the mean is pulled toward higher values by extreme observations like the maximum of 160 minutes. The IQR of 35 minutes tells us the middle 50% of students did homework between Q1=40 and Q3=75 minutes, not that the total range is 35 minutes. The standard deviation of 28.1 minutes measures typical deviation from the mean, not a typical homework time. When interpreting medians, remember that 50% fall below and 50% fall above, not that half fall between the median and maximum.
In a clinical trial, 200 patients were randomly assigned to either a new treatment or a placebo. After one month, their condition was assessed as 'Improved' or 'Not Improved'. In the treatment group of 100 patients, 70 showed improvement. In the placebo group of 100 patients, 50 showed improvement. Overall, 120 patients showed improvement.
To conduct a chi-square test for homogeneity, what is the expected number of patients in the treatment group who would show improvement if the treatment had no effect compared to the placebo?
Explanation: The expected count is (row total × column total) / grand total. The row total (treatment group) is 100. The column total (Improved) is 120. The grand total is 200. The expected count for the treatment/improved cell is (100×120)/200=12000/200=60.
In a region, the distribution of monthly household electricity use (in kWh) is approximately Normal for the population of all households. The mean is μ=900 kWh with standard deviation σ=150 kWh. The value 1050 kWh is marked on the normal curve. Which statement about the marked value is correct?
Explanation: This problem requires locating an electricity usage value on a normal distribution. Given μ = 900 kWh and σ = 150 kWh, we need to find where 1050 kWh falls. Calculate the number of standard deviations from the mean: (1050 - 900)/150 = 150/150 = 1. So 1050 kWh is exactly 1 standard deviation above the mean. In a normal distribution, values within 1 standard deviation of the mean are fairly typical, as approximately 68% of all values fall within μ ± 1σ. Being 1σ above the mean indicates this household uses somewhat more electricity than average, but this usage level is not unusual or rare. About 16% of households would use even more electricity than this. The key is recognizing that values within 1σ of μ are common, while those beyond 2σ are uncommon.
A public health researcher compared the proportion of adults who received a flu shot in two counties. In a random sample of 250 adults from County 1, 155 reported getting a flu shot; in a random sample of 220 adults from County 2, 110 reported getting a flu shot. A two-proportion z test was conducted for H0:p1=p2 versus Ha:p1>p2, and the p-value was 0.018. Using α=0.05, what conclusion is appropriate?
Explanation: This problem involves a one-tailed test for comparing two population proportions. The alternative hypothesis H_a: p₁ > p₂ indicates we're testing if County 1's proportion is greater than County 2's. With p-value = 0.018 < α = 0.05, we reject H₀. The correct conclusion is that there's sufficient evidence that the population flu-shot proportion in County 1 is greater than in County 2 (choice A). Choice C reverses the direction of the inequality. Choice D incorrectly implies causation from an observational study. Choice E refers to sample proportions instead of population proportions. In one-tailed tests, the conclusion must match the direction specified in the alternative hypothesis, and we always make inferences about population parameters, not sample statistics.
A coffee shop owner wants to estimate the mean amount (in ounces) of coffee dispensed by a machine. She randomly selects 50 cups and computes a 99% confidence interval for the population mean amount dispensed: (11.6,12.4) ounces. Which interpretation is correct?
Explanation: This question evaluates understanding of confidence interval interpretation for a population mean. The correct interpretation (C) states we are 99% confident the true mean amount dispensed is between 11.6 and 12.4 ounces. Choice A incorrectly applies the interval to individual cups rather than the mean. Choice B wrongly treats confidence as probability after the interval is computed. Choice D misinterprets the interval as describing the sampling distribution of sample means. Choice E absurdly suggests the population parameter changes. Remember: confidence intervals estimate the population mean, not individual values, and confidence describes the reliability of the interval construction method over repeated sampling.
A school district wants to know whether a new text-message reminder system changes the proportion of parents who attend parent-teacher conferences. Two similar middle schools are chosen: one uses reminders (n = 180 parents invited) and the other does not (n = 200 parents invited). Attendance is recorded as Yes/No, and the counts are 108 Yes in the reminder school and 90 Yes in the no-reminder school. Which inference procedure is most appropriate to determine whether the reminder system is associated with a different attendance rate?
Explanation: This question assesses the skill of selecting appropriate inference procedures for categorical data in AP Statistics, specifically comparing proportions from two independent samples. The scenario involves two similar middle schools, one with a reminder system and one without, recording Yes/No attendance, making it a comparison of two proportions to see if the reminder is associated with a different rate. The two-sample z-test for a difference in proportions is most appropriate because it handles independent samples and tests for a difference in population proportions based on the given counts. Other options like the one-sample z-interval are incorrect as they apply to single proportions, while the chi-square goodness of fit tests distributions against expected values, not comparisons between groups; the matched-pairs t-test is for quantitative data, and the two-sample t-interval is also for means. A common distractor is choosing chi-square for association, but since there are only two groups and binary outcomes, the two-sample z-test is equivalent and often preferred for direct proportion comparison. To align methods, remember that for two independent samples with categorical responses, check conditions like large sample sizes (np and n(1-p) ≥10 for each) before proceeding with z-procedures. This mini-lesson highlights that procedure selection depends on the number of samples, variable type, and whether you're estimating or testing.
A fair six-sided die is rolled once. Let Event A be “the result is even” and Event B be “the result is greater than 4.” Which statement about Events A and B is correct?
Explanation: This problem tests calculating independence for events with a fair die. Event A (even) includes {2, 4, 6}, so P(A) = 3/6 = 1/2. Event B (greater than 4) includes {5, 6}, so P(B) = 2/6 = 1/3. The intersection A∩B = {6}, so P(A∩B) = 1/6. Checking independence: P(A)·P(B) = (1/2)·(1/3) = 1/6 = P(A∩B). Since the equation holds, events A and B are independent. Choice C incorrectly applies the addition rule for mutually exclusive events, but these events overlap at outcome 6. Independence means knowing one event occurred doesn't change the probability of the other, which is true here.
A city health department collected data on neighborhood type (Urban or Suburban) and whether residents get a flu shot (Yes or No). The results are shown in the table. Which comparison is appropriate for assessing the relationship between neighborhood type and flu-shot status using conditional distributions?
Explanation: This question asks about using conditional distributions to assess the relationship between neighborhood type and flu-shot status. To determine if these variables are associated, we should compare the proportion of residents who get flu shots within each neighborhood type - that is, the percent who get flu shots among Urban residents versus the percent who get flu shots among Suburban residents. Choice C correctly identifies this comparison of conditional distributions. Options A and E examine marginal distributions, B compares raw counts instead of proportions, and D conditions on flu-shot status rather than neighborhood type. The principle here is to condition on the explanatory variable (neighborhood type) and examine how the response variable (flu-shot status) varies across those conditions.
A school district wanted to study whether a new online homework platform improves Algebra 1 test scores. From all 12 Algebra 1 teachers in the district, the district randomly selected 6 teachers to use the new platform for one semester; the other 6 teachers continued with the usual platform. All students in each selected teacher’s classes used that teacher’s assigned platform. At the end of the semester, the mean test score was higher in classes using the new platform. Which conclusion is justified based on the design of this study?
Explanation: This question tests understanding of causal inference from experiments with cluster randomization. The key skill is recognizing that random assignment of teachers (not students) to treatments allows causal conclusions, but only for the specific population studied. The district randomly assigned teachers to platforms, creating an experiment where the treatment was applied at the teacher level, with all students of each teacher receiving the same treatment. This random assignment eliminates selection bias and allows us to conclude that the new platform likely caused the higher scores. However, the causal conclusion is limited to students of Algebra 1 teachers in this specific district, not all Algebra 1 students everywhere. In experiments, random assignment permits causal inference for the experimental units, while random selection permits generalization to a larger population.
A company recorded the number of emails received per day by 40 employees. The summary statistics were: mean =58, median =57, standard deviation =11, five-number summary (32,50,57,65,83), and IQR =15. Which interpretation is correct?
Explanation: This question tests understanding of symmetric distributions and their characteristics. The mean (58) and median (57) being nearly equal is a key indicator of a roughly symmetric distribution. Additionally, the five-number summary shows reasonable balance - the distances from the median to Q1 and Q3 are similar (7 and 8 respectively), and the extremes are roughly equidistant from the quartiles. Common misconceptions in other options include: IQR measures the spread of the middle 50%, not the total range; standard deviation doesn't define exact boundaries for data; Q1=50 means 25% (not 75%) received fewer than 50 emails; and the median indicates the middle value when data is ordered, not that all employees received that exact number.
A school cafeteria manager claims that students choose among four lunch options in the following distribution: Pizza 40%, Salad 25%, Sandwich 20%, and Pasta 15%. On a randomly selected day, the manager records the lunch choices of n=200 students with the observed counts shown in the table. Which hypotheses are appropriate for a chi-square goodness-of-fit test of the manager’s claim?
Explanation: This question tests understanding of setting up hypotheses for a chi-square goodness-of-fit test. The null hypothesis should state the claimed population distribution (40% pizza, 25% salad, 20% sandwich, 15% pasta), while the alternative states that the true distribution differs from these claimed proportions. Option A incorrectly uses observed counts instead of proportions, Option C describes a test of independence rather than goodness-of-fit, Option D only tests one category, and Option E incorrectly uses sample proportions in the null hypothesis. The chi-square goodness-of-fit test compares observed frequencies to expected frequencies based on a hypothesized distribution. The correct setup always states population proportions in H₀, not observed counts or sample proportions.
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 H0 are: Wetland 75, Forest 125, Grassland 100, Shrubland 150, Urban 50.
What conclusion is appropriate at α=0.05?
Explanation: This question examines a borderline p-value that leads to rejection. The null hypothesis claims the population habitat distribution matches the model proportions. With p-value = 0.049 < α = 0.05, we reject H₀. This provides convincing evidence that the population habitat distribution of bird sightings differs from the model proportions. Choice A incorrectly states 0.049 > 0.05. Choice C incorrectly claims this "proves" the model. Choice D misinterprets the relationship between sample and population. Choice E contradicts itself. When p-value is just below α, we still reject H₀ - there's no "almost significant" in hypothesis testing.
A call center tracks the length of individual customer calls (in minutes). The population mean is μ=8.0 minutes with population standard deviation σ=5.0 minutes, and the population distribution is strongly right-skewed. Each day, a supervisor takes a simple random sample of n=25 calls and computes the sample mean xˉ. Over many days, which statement about the sampling distribution of xˉ is correct?
Explanation: This problem involves the sampling distribution of mean call times. The population has μ=8.0 minutes and σ=5.0 minutes with a right-skewed distribution. For samples of size n=25, the sampling distribution of x̄ has mean 8.0 (same as population) and standard deviation σ/√n = 5.0/√25 = 5.0/5 = 1.0. Despite the population being strongly right-skewed, with n=25 (close to 30), the Central Limit Theorem indicates the sampling distribution will be approximately normal. Choice B incorrectly claims it remains skewed, while C divides by n instead of √n.
A gym manager wants to study whether class format is associated with how many days per week members attend. For each of 120 randomly selected members, the manager records the member’s primary class format (coded 1 = cycling, 2 = yoga, 3 = strength) and the number of days the member attended the gym last week. The observational units are the individual gym members. Which classification is correct for the two recorded variables?
Explanation: This question tests your ability to classify variables as categorical or quantitative and identify explanatory versus response variables. The gym manager wants to study whether class format (cycling, yoga, strength) is associated with days attended per week. Class format is categorical because it represents distinct categories or groups, not numerical measurements. Days attended is quantitative because it's a numerical count that can be measured and averaged. Since the manager wants to see if class format affects attendance, class format is the explanatory variable (what might cause a difference) and days attended is the response variable (what might be affected). The correct answer is B: class format is categorical and explanatory, while days attended is quantitative and response.
A public health official claims that more than 10% of adults in a county have received a flu shot this month. A random sample is used to compute an 80% confidence interval for the true proportion p: (0.09, 0.13). Is the claim supported by the confidence interval?
Explanation: This question tests whether a claim that more than 10% received flu shots (p > 0.10) is supported by an 80% CI of (0.09, 0.13). The confidence interval includes values less than or equal to 0.10 (from 0.09 to 0.10), meaning some plausible values for p do not satisfy p > 0.10. Choice B correctly identifies this - when the CI contains values that don't satisfy the inequality, the claim is not supported. Choice A misinterprets confidence as probability about p, Choice C incorrectly thinks containment proves the claim, Choice D wrongly focuses only on the upper bound, and Choice E dismisses the validity of 80% confidence. The key lesson: for strict inequality claims like p > value, the entire CI must exceed that value.
Two medications are compared for the proportion of patients who report a certain side effect. A 95% confidence interval for the difference in proportions, defined as pMedX−pMedY, is (−0.02, 0.09). A nurse claims, “MedX causes side effects in a higher proportion of patients than MedY.” Is the claim supported by the confidence interval?
Explanation: This question examines whether a confidence interval (-0.02, 0.09) supports a claim about MedX having a higher proportion of side effects than MedY. The interval for p_MedX - p_MedY includes negative values, zero, and positive values. Since 0 is within the interval, "no difference" between the medications is a plausible scenario. Additionally, the negative values in the interval suggest that MedX could actually have a lower proportion of side effects than MedY. Because the interval includes values where p_MedX ≤ p_MedY, the claim that MedX causes side effects in a higher proportion is not clearly supported. When making comparative claims, the entire interval must be on one side of 0 to provide clear support.
A cereal box label states the mean net weight is μ=18 ounces. A quality-control analyst suspects the mean is less than stated. A random sample of 12 boxes is selected and a one-sample t test is performed with H0:μ=18 versus Ha:μ<18 at α=0.05. The p-value is 0.26. What conclusion is appropriate?
Explanation: In this one-tailed test for a population mean, the p-value (0.26) is much greater than the significance level (0.05), so we fail to reject the null hypothesis. This means there is not convincing evidence that the population mean weight is less than 18 ounces. Choice B incorrectly rejects H₀ when p > α. Choice C misinterprets the p-value as P(H₀ is true) rather than P(data | H₀). Choice D confuses the sample mean with the population mean claim. When we fail to reject H₀, we're saying the data doesn't provide strong enough evidence against the null hypothesis claim about the population parameter. A large p-value suggests the observed sample result is reasonably likely under the null hypothesis.
An environmental group wants to compare recycling behavior across three neighborhoods. Independent random samples of households are taken from Neighborhood A, B, and C. Each household is classified as Recycles regularly (Yes/No). Results are shown.
Which test is appropriate, and what are the correct hypotheses?
Explanation: This problem focuses on the skill of identifying the chi-square test and hypotheses for comparing a binary categorical variable across independent samples from neighborhoods in AP Statistics. The design takes independent random samples from each neighborhood (populations) and measures recycling status (yes/no) in each, fitting a multi-population comparison. The chi-square test of homogeneity is thus appropriate to check if recycling distributions are the same across neighborhoods. Mini-lesson: Use goodness-of-fit for single-sample distribution testing, homogeneity for equivalence across independent groups, and independence for two-variable association in one sample. Distractor E correctly names homogeneity but wrongly applies independence hypotheses, which doesn't align with the setup. Option D is accurate in test and hypotheses.