AP Statistics
Practice Test 24 for AP Statistics: real questions and explanations from the Varsity Tutors practice-test pool.
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A company tests two training programs for new employees and measures mean time (in minutes) to complete a standard task after training. From independent random samples, a 99% confidence interval for the difference in population means was computed as (μProgram 1−μProgram 2)∈(−8.5,−2.0). Which interpretation is correct?
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A company tests two training programs for new employees and measures mean time (in minutes) to complete a standard task after training. From independent random samples, a 99% confidence interval for the difference in population means was computed as (μProgram 1−μProgram 2)∈(−8.5,−2.0). Which interpretation is correct?
Explanation: This question involves interpreting a negative confidence interval for μ_Program1 - μ_Program2. The interval (-8.5, -2.0) being entirely negative means Program 1 has a lower mean completion time than Program 2. Choice D correctly states we're 99% confident the true difference is between -8.5 and -2.0 minutes. Choice A misinterprets the direction. Choice B incorrectly assigns probability to the parameter. Choice C gets the direction wrong. Choice E incorrectly applies the interval to individuals. Remember: negative values for μ_1 - μ_2 mean μ_1 < μ_2.
A medical screening test is used on a population where, for a randomly selected person, the probability the test returns a false positive is modeled as P(False positive)=0.02. For one test, the sample space is {False positive,Not false positive}, and the event of interest is False positive. Which statement correctly describes the probability 0.02?
Explanation: This question examines probability in a medical testing context. The sample space for each test is {False positive, Not false positive} with P(False positive) = 0.02. Option C correctly interprets this as a long-run relative frequency - over many tests under similar conditions, about 2% will be false positives. Option A incorrectly assumes exactly 1 false positive per 50 tests. Option B misunderstands odds notation (the odds are 0.02 to 0.98, or 1 to 49). Option D incorrectly suggests that low probability events cannot occur - even rare events happen. Option E falsely assumes that past results affect future independent tests. Understanding false positive rates is crucial in medical testing, as even a 2% rate can lead to many incorrect diagnoses in large-scale screening programs, highlighting why follow-up testing is often necessary.
A bank models deposits. Let X be the amount (in dollars) deposited on Monday and Y be the amount (in dollars) deposited on Tuesday. Assume X and Y are independent, with μX=1200, σX=300, μY=900, and σY=200. Define N=X−Y, the net difference (Monday minus Tuesday). Which statement about N is correct?
Explanation: This question involves the net difference N = X - Y between Monday and Tuesday deposits. The mean is μ_N = μ_X - μ_Y = 1200 - 900 = 300 dollars. For independent variables, variances add even when subtracting: σ²_N = σ²_X + σ²_Y = 300² + 200² = 90000 + 40000 = 130000, so σ_N = √130000 = √(300² + 200²). Choice A incorrectly uses 100 for the standard deviation, confusing it with the difference of standard deviations. Choice E has the wrong sign for the mean. The fundamental principle remains: when combining independent random variables, variances always add, and the operation only affects how means combine.
A health clinic asked 210 patients whether they received a flu shot this season and whether they reported having the flu in the past year. The table summarizes responses. Which comparison is appropriate for evaluating whether flu incidence is associated with receiving a flu shot using conditional distributions?
| Had flu | Did not have flu | |
|---|---|---|
| Flu shot | 24 | 96 |
| No flu shot | 36 | 54 |
Explanation: The skill here is representing two categorical variables via conditional distributions to evaluate association between flu shot receipt and flu incidence. The correct comparison conditions on flu shot status and contrasts flu proportions: 24/120 = 0.20 for shot versus 36/90 = 0.40 for no shot, revealing an association. This follows the row-based conditioning in the table. Distractor choice D reverses it by conditioning on flu status and comparing shot receipt. Choice B uses overall flu proportions without conditioning, missing the association check. Mini-lesson: Conditional distributions are obtained by normalizing within rows or columns (cell/row total), allowing comparison of how one variable's proportions change across the other's categories, with discrepancies pointing to association.
A researcher administers a short survey to randomly selected people walking through a mall until the first person agrees to participate. Each person either “agrees” or “declines,” and the researcher stops after the first agreement. Assume the probability a person agrees is approximately constant and each selection is independent. The researcher wants to model the number of people asked until the first agreement. Why is a geometric model appropriate?
Explanation: This AP Statistics item examines the geometric distribution, suitable for counting independent Bernoulli trials until the first success with unchanging p. Approaching each person is a trial with 'success' as agreement, stopping at the first success. The 'until first success' model applies since the researcher persists until the first agreement, with variable trials. A distractor like A confuses it with binomial by assuming a fixed sample. Mini-lesson: Geometric distribution gives P(X=k) = (1-p)^{k-1} p for k=1,2,...; it requires binary outcomes, independence, and constant p, as in the independent selections with approximate constant agreement probability.
A nutrition researcher uses linear regression to predict resting heart rate (y) from daily caffeine intake in mg (x). The researcher claims the slope is 0 (no linear relationship). A 95% confidence interval for the slope is (−0.03, 0.11) beats per minute per mg. Is the claim supported by the confidence interval?
Explanation: This question tests understanding of confidence intervals that contain zero. The 95% confidence interval (-0.03, 0.11) includes 0, meaning a slope of 0 (no linear relationship) is a plausible value. This supports the researcher's claim that the slope equals 0. Choices B and C incorrectly conclude the slope must be positive or negative based on the interval containing those values, but when 0 is included, we cannot rule out no relationship. When evaluating whether a slope equals 0, check if 0 falls within the confidence interval; if it does, the claim is supported.
A fitness coach wants to know whether type of workout is related to change in resting heart rate. She assigns 36 clients to one of three workout plans (cardio, strength, or mixed) and measures each client’s resting heart rate at the start and after 8 weeks; she records the change in heart rate (after − before) in beats per minute. The observational units are the clients, and the goal is to compare changes across workout types. Which classification is correct?
Explanation: This question in AP Statistics checks classification in fitness studies. Underlined 'type of workout' is explanatory for heart-rate changes. Workout type (cardio, etc.) is categorical, distinct plans. Heart-rate change in bpm is quantitative, numeric difference. Distractor A swaps types, maybe mistaking types for durations. Mini-lesson: Categorical for non-numeric groups, quantitative for calculable numbers; explanatory is the treatment-like variable, response measures the effect.
A student collects data on 10 phones: battery capacity (mAh) and battery life (hours). The scatterplot shows a positive linear association. Which statement about correlation is correct?
Explanation: AP Statistics teaches that correlation is invariant under linear transformations of variables, like unit conversions. Choice B correctly explains that converting mAh to Ah (dividing by 1000) won't change the correlation, as it's a positive linear rescaling. The positive linear association remains, unaffected by units. Distractor A wrongly suggests units alter correlation, but it's standardized. Time units like minutes vs. hours wouldn't flip the sign, countering C. Correlation indicates association strength, not causation or slope equality.
A librarian tests whether the mean time students spend studying in the library on weeknights is 2 hours. A random sample of 20 students is selected, and conditions for a one-sample t test are met. The hypotheses are H0:μ=2 and Ha:μ=2 with α=0.05. The test is completed and the p-value is 0.051. What conclusion is appropriate?
Explanation: This question assesses carrying out a one-sample t-test for a population mean, particularly with p-values near α in academic settings. Since p=0.051 slightly exceeds α=0.05, we fail to reject H0: μ=2, indicating no convincing evidence that the mean study time differs from 2 hours. Choice A, a distractor, suggests rejection by rounding or misjudging the p-value, but strict inequality governs the decision. In mean hypothesis testing, failing to reject doesn't prove H0 true or imply causation, as erred in C and E. A mini-lesson: state conclusions cautiously about evidence levels for the population, avoiding assumptions from sample means alone, like in D. This approach prevents common misinterpretations in two-tailed tests.
A company wants to estimate the proportion of all its employees who would prefer a 4-day workweek. The company surveys 200 employees chosen by a manager who selects “reliable team members who give thoughtful feedback” from across departments. Which issue most threatens the validity of using this sample to represent all employees?
Explanation: This question tests recognition of sampling bias from non-random, judgment-based selection. The company wants to estimate preferences of ALL employees, but the manager hand-picks "reliable team members who give thoughtful feedback." This judgment-based selection creates systematic bias because the manager's criteria for selection (reliability, thoughtfulness) might correlate with work preferences - perhaps these selected employees are more invested in the company or have different work-life balance needs than average employees. This is neither simple random sampling nor any form of probability sampling; it's convenience sampling based on subjective criteria. The manager might unconsciously select employees who share similar views or exclude those seen as complainers or troublemakers. To identify this bias, look for subjective selection criteria that could correlate with the variable being measured. True random sampling gives every employee an equal chance regardless of perceived qualities.
A store tracks the next 45 customers and records whether each customer uses a coupon. A success is “uses a coupon,” and exactly 45 customers are observed. A big coupon promotion starts halfway through, increasing coupon use for later customers. Does this meet binomial conditions?
Explanation: This question in AP Statistics checks constant p in customer behavior. Fixed n=45, binary (uses coupon or not), but promotion changes p midway, so no. Violates constant p. Distractor A ignores change, focusing on binary. Choice C denies promotion effect, which it has. Mini-lesson: Interventions mid-process break constant p; analyze pre- and post-separately.
An agriculture researcher wants to study the effects of fertilizer type (Type 1 vs. Type 2) and watering schedule (daily vs. every other day) on tomato yield. The researcher has 40 similar tomato plants and can randomly assign each plant to one fertilizer and one watering schedule for the entire growing season. The researcher is also interested in whether the effect of fertilizer depends on watering schedule. Which experimental design is most appropriate?
Explanation: Selecting an experimental design in AP Statistics here involves studying multiple factors and their interactions efficiently. The completely randomized design with four treatment groups (combinations of fertilizer and watering) is appropriate, as it randomly assigns plants to all combinations, allowing assessment of main effects and interactions without blocking needs. This fits the 40 similar plants and single-assignment constraint, supporting the goal of exploring if fertilizer effects depend on watering. Distractors: B is mismatched for time-based pairing; C blocks unnecessarily; D ignores one factor; E is not experimental. Mini-lesson: For factorial experiments with independent factors, use completely randomized assignment to treatment combinations; this enables interaction analysis, but ensure units are homogeneous to avoid needing blocks.
A university dining hall records the amounts (in grams) of pasta served in a random sample of n=58 lunch plates to model portion sizes and estimate the fraction below a target serving amount. Why is a normal model reasonable in this situation?
Explanation: This question addresses normal models for food service portions. The correct answer identifies that portion sizes vary due to many small factors (server differences, scoop variations, settling), which can combine to produce approximately normal distributions. With n=58 plates sampled, the dining hall can check if portions cluster symmetrically around a typical serving size. The main distractor (E) wrongly claims that n≥58 guarantees population normality, confusing sample size thresholds with distribution requirements. Normal models are reasonable when measurements result from multiple small, additive effects, not because they eliminate inconsistency or require specific mean values.
An engineer wants to compare the mean battery life (hours) of two brands of rechargeable batteries. The engineer randomly selects 30 batteries of Brand X and 30 batteries of Brand Y from large shipments and tests each battery once under identical conditions. The engineer wants to test whether the mean battery life differs between the two brands. Which inference procedure is most appropriate?
Explanation: This scenario compares means from two independent groups (Brand X and Brand Y batteries). With 30 batteries randomly selected from each brand and tested separately, we have independent samples of quantitative data (battery life in hours). The two-sample t-test compares the population means of the two brands. Choice B is incorrect because the batteries aren't paired - each battery is tested only once. Choice C tests proportions, not means. Choice D estimates a single mean, not comparing two. Choice E tests homogeneity of categorical distributions across populations. When comparing means from two independent groups, the two-sample t-test is the appropriate choice.
A runner recorded her heart rate (in beats per minute) at the end of each of 45 training runs over a season (population: all 45 recorded heart rates). The five-number summary is: min =118, Q1=132, median =145, Q3=156, max =170. A boxplot is made. Which feature is consistent with the summary statistics?
Explanation: This question assesses understanding of boxplot construction and feature identification. The box spans from Q₁ (132) to Q₃ (156), with the median line at 145 positioned inside this box. The whiskers extend from the box to the minimum (118) and maximum (170) values. Choice B correctly states that the median is at 145 and the box spans from 132 to 156. Common mistakes in the distractors include: calculating IQR incorrectly (it's 156-132=24, not 52), confusing whisker direction, mixing up box and whisker extents, and assuming large values must be outliers. A key principle: the five-number summary provides all necessary values for a standard boxplot without outliers.
A school counselor models the relationship between hours of sleep the night before an exam (x) and exam score (y) for a group of students using a least-squares regression line. A scatterplot with the fitted regression line is shown. Which statement about the regression line is correct?
Explanation: This question tests understanding of least-squares regression properties. The least-squares regression line has a special property: it always passes through the point (x̄, ȳ), where x̄ is the mean of all x-values and ȳ is the mean of all y-values. This point is called the centroid or center of mass of the data. Option A is incorrect because least-squares minimizes the sum of squared residuals, not absolute values. Option C is wrong because residuals are vertical distances (y - ŷ), not horizontal. Option D is false because extrapolation far outside the observed range is unreliable. Option E is incorrect because it's the sum of residuals that equals zero, not the sum of y-values.
A city measured the amount of water used per household in one month (in hundreds of gallons) for a random sample of 90 households. The summaries were: mean =62.5, median =58, SD =21.0, five-number summary (18, 45, 58, 74, 130), and IQR =29. Which interpretation is correct?
Explanation: This question assesses understanding of the interquartile range. The IQR represents the range of the middle 50% of data, from Q1 (45) to Q3 (74). This means about 50% of households used between 45 and 74 hundred gallons. Choice A incorrectly equates IQR with the full range. Choice B misinterprets SD as a typical usage amount. Choice D confuses the median value with sample size. Choice E reverses the skewness direction - when mean > median, the distribution is right-skewed, not left-skewed.
A call center records the number of calls an operator handles in an hour. Let X be the number of calls for a randomly selected operator-hour; X is discrete. The distribution of X has mean μX=22 calls and standard deviation σX=5 calls. Which interpretation of the mean is correct?
Explanation: This question assesses mean interpretation for a discrete random variable in AP Statistics. The mean μ_X = 22 is the long-run average calls handled per operator-hour, correctly in choice C. Over numerous hours, the average would be about 22. Distractor A confuses mean with mode, suggesting 22 is most common, but mean is an average, not necessarily the most frequent. Choice D misuses SD to set hard limits of 17 to 27, whereas SD measures typical deviation. In a mini-lesson, the mean is the expected value E(X) = Σ x P(x) for discrete variables, representing the center of mass and long-term average outcome.
A school counselor fits a least-squares regression model predicting students’ first-semester GPA (y) from the number of hours per week spent in a tutoring program (x). The counselor claims that each additional tutoring hour increases mean GPA by at least 0.05 points. A 95% confidence interval for the slope is (0.02, 0.08) GPA points per hour. Is the claim supported by the confidence interval?
Explanation: This question tests understanding of how confidence intervals support claims about regression slopes. The 95% confidence interval (0.02, 0.08) contains all plausible values for the true slope. The counselor claims the slope is at least 0.05, meaning slope ≥ 0.05. However, the interval includes values less than 0.05 (like 0.02, 0.03, 0.04), so we cannot be confident the true slope is at least 0.05. Choice A incorrectly focuses on the interval being above 0 rather than above 0.05. When evaluating a claim that a parameter is at least some value, the entire confidence interval must be at or above that value to support the claim.
A streaming service believes that the proportion of users who watch content on a mobile device is different from 35%. The company randomly samples 160 users and finds that 44 primarily watch on mobile (p^=0.275). Which hypotheses are appropriate for testing the company’s belief?
Explanation: The skill here is crafting hypotheses for a one-proportion z-test in AP Statistics to examine if p differs from 0.35. The correct setup is H0: p = 0.35 and Ha: p ≠ 0.35, capturing the two-sided nature of 'different from,' as in choice A. A distractor is choice C, which uses ˆp incorrectly. Choice D uses a one-sided alternative, but 'different' requires checking both directions. For a mini-lesson, two-sided alternatives use ≠ for claims of difference or inaccuracy. Null is always equality, focused on p. Sample data tests the hypotheses but doesn't appear in them.
A city council uses a random sample survey to estimate support for a proposed property-tax increase to fund road repairs. If they proceed when true support is below 50%, the measure may fail at the ballot and reduce public trust. Why is it important to consider potential error when using the sample estimate to decide whether to place the measure on the ballot?
Explanation: This question examines understanding of sampling variability in political polling contexts. The city council must decide whether to place a tax measure on the ballot based on a sample survey estimate. The correct answer (B) properly identifies that the margin of error reflects sampling variability - different random samples would yield different estimates of support. If the true support is below 50% but the sample estimate suggests it's above, the council might proceed with a doomed measure, wasting resources and damaging public trust. Choice A incorrectly claims sample and population proportions are always equal, while D wrongly suggests sample size alone guarantees accuracy. Statistical inference always involves uncertainty because we're using partial information (the sample) to make conclusions about the whole (all voters).
A city planner records distance from downtown (miles) x and average rent (dollars) y for a sample of apartments, then fits a least-squares regression line to model rent from distance. Which statement about the regression line is correct?
Explanation: This AP Statistics item checks the optimization in least-squares for rent prediction from distance. Choice B is accurate: the line minimizes ∑(y - ŷ)^2, the sum of squared vertical errors. This defines the method. Choice A distracts by focusing on minimizing the maximum residual, which is minimax regression, not least-squares. Mini-lesson: Least-squares prioritizes overall squared error reduction, not ensuring all positive residuals (Choice C) or using perpendicular distances (Choice E). Predictions are most reliable within observed x, not extrapolated far beyond (Choice D).
A marketing firm surveyed 500 shoppers about their primary reason for choosing a particular grocery store. The reasons were categorized as Price, Convenience, Quality, and Service. Which of the following graphical displays is most appropriate for visualizing the distribution of these reasons?
Explanation: A bar chart is the appropriate graphical display for a single categorical variable. The variable 'reason' has distinct categories (Price, Convenience, etc.), and a bar chart effectively shows the count or proportion in each category. Histograms, dotplots, and stem-and-leaf plots are used for quantitative data, not categorical data.
A shipment contains many boxes of cereal. A consumer group randomly selects 20 boxes to check a promotional code inside each box. A success is finding a valid code, and the probability a box contains a valid code is 0.95 (assume independence). Which values of n and p correctly model the number of successes with a binomial distribution?
Explanation: This tests binomial parameter identification for valid codes in 20 boxes. n=20 and p=0.95 fit, as in choice D. Distractors: A uses complement 0.05, B swaps, C scales wrongly, E reciprocal. Binomial needs fixed trials n, success p per trial, independence. This counts valid codes. High p suggests many successes expected.
A fitness app sampled the number of steps taken yesterday by 85 users. The summaries were: mean = 7,900 steps, median = 8,150 steps, SD = 1,600 steps, five-number summary = (min 3,900; Q1 7,200; median 8,150; Q3 9,000; max 10,200), and IQR = 1,800 steps. Which interpretation is correct?
Explanation: This question assesses interpretation of summary statistics for a slightly left-skewed distribution. When the mean (7,900) is less than the median (8,150), this suggests left skewness, where a few low values pull the mean down. The five-number summary confirms this with a longer lower tail (minimum at 3,900 is 4,250 steps below the median, while maximum at 10,200 is only 2,050 steps above). The IQR of 1,800 steps represents the spread of the middle 50% of data, not the total range. The standard deviation of 1,600 steps measures variability around the mean, not a typical step count. Understanding these distinctions helps correctly identify distribution shape from summary statistics.