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AP Statistics

AP Statistics Practice Test: Practice Test 59

Practice Test 59 for AP Statistics: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A public health official wants to estimate the mean number of servings of vegetables eaten per day by adults in a county. A random sample of 100 adults was surveyed, and a 97% confidence interval for the population mean was reported as $(2.1,\ 2.8)$ servings. Which interpretation is correct?

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Question 1

About 97% of adults in the county eat between 2.1 and 2.8 servings of vegetables per day.
We are 97% confident that the true mean number of servings of vegetables eaten per day by adults in the county is between 2.1 and 2.8. (correct answer)
There is a 97% chance that the sample mean is between 2.1 and 2.8 servings.
If 97% confidence intervals are repeatedly constructed, 97% of the time the population mean will change to be inside the interval.
Because the confidence level is 97%, exactly 97 of the 100 sampled adults must have reported between 2.1 and 2.8 servings per day.

Explanation: This problem asks about interpreting a 97% CI for mean vegetable servings. The correct answer (B) properly states we are 97% confident the true mean is between 2.1 and 2.8 servings. Choice A incorrectly applies the interval to individual adults. Choice C wrongly suggests the sample mean is uncertain after calculation. Choice D nonsensically claims the population parameter changes. Choice E makes an incorrect claim about the sample data. A confidence interval provides a range estimate for the population mean based on sample data, with the confidence level indicating the procedure's reliability.

Question 2

A school district compared the mean number of minutes of homework completed per night for students using a new study app versus students not using the app. From independent random samples, a 95% confidence interval for the difference in population means $(\mu_{\text{app}}-\mu_{\text{no app}})$ was computed as $(4.2,\ 11.8)$ minutes. Which interpretation is correct?

There is a 95% chance that the true difference in means $(\mu_{\text{app}}-\mu_{\text{no app}})$ is between 4.2 and 11.8 minutes.
We are 95% confident that the mean homework time for app users is between 4.2 and 11.8 minutes.
Because 0 is not in the interval, it is plausible that there is no difference in mean homework time between the two groups.
We are 95% confident that the true difference in population means $(\mu_{\text{app}}-\mu_{\text{no app}})$ is between 4.2 and 11.8 minutes, meaning app users average that many more minutes than non-users. (correct answer)
We are 95% confident that $(\mu_{\text{no app}}-\mu_{\text{app}})$ is between 4.2 and 11.8 minutes.

Explanation: This question tests understanding of confidence intervals for the difference of two means. The interval (4.2, 11.8) represents our confidence about the true difference μ_app - μ_no app. Since both endpoints are positive, we're confident that app users average more homework time than non-users. Choice D correctly interprets this as being 95% confident about the true population mean difference, with app users doing 4.2 to 11.8 more minutes. Choice A incorrectly uses probability language instead of confidence, Choice B only mentions one group's mean, Choice C misinterprets the meaning of 0 not being in the interval (it actually suggests there IS a difference), and Choice E reverses the order of subtraction.

Question 3

A restaurant measured the wait time (in minutes) for 70 customers on a Saturday. The summary statistics were: mean $= 24.8$ min, median $= 21.0$ min, standard deviation $= 12.5$ min, five-number summary $(5, 14, 21, 30, 78)$ , and IQR $= 16$ min. Which interpretation is correct?

Because the mean is greater than the median and the maximum is far above $Q_3$ , the distribution is likely skewed right due to a few very long waits. (correct answer)
The IQR of 16 minutes means the longest wait was 16 minutes longer than the shortest wait.
The standard deviation of 12.5 minutes means most customers waited about 12.5 minutes.
Since the median is 21 minutes, 50% of customers waited exactly 21 minutes.
Because $Q_3=30$ , about 75% of customers waited at least 30 minutes.

Explanation: This question evaluates recognition of right-skewed distributions in wait time data. The mean (24.8 minutes) exceeding the median (21.0 minutes) indicates right skew, with the distribution having a tail extending toward longer wait times. The maximum value (78 minutes) being far above Q3 (30 minutes) confirms this interpretation - a few customers experienced unusually long waits that pull the mean upward. The incorrect options demonstrate common errors: IQR is the difference between Q3 and Q1, not between max and min; standard deviation measures typical variation from the mean, not actual wait time; the median divides the data in half, not indicating that 50% waited exactly 21 minutes; and Q3=30 means 75% waited 30 minutes or less, not at least 30 minutes.

Question 4

A school nurse measures the heights (in inches) of 52 ninth-grade students. The histogram is described as having two distinct peaks: one around 61–63 inches and another around 67–69 inches, with a noticeable dip in frequency around 64–66 inches.

Which statement best describes the distribution?

The distribution is bimodal, suggesting two clusters of heights. (correct answer)
The distribution is approximately symmetric because heights are measured on a continuous scale.
The distribution is right-skewed because the tallest students are far from the shortest.
The distribution is left-skewed because the histogram has a dip in the middle.
The distribution is uniform because there are values across many height intervals.

Explanation: AP Statistics skills include describing distributions of variables like heights via histograms, noting shape, center, spread, and outliers. The distribution is bimodal with two distinct peaks at 61-63 and 67-69 inches, suggesting subgroups like genders, and a dip in between. Distractor C mislabels it as right-skewed by noting the range from short to tall, but bimodality is about multiple peaks, not just spread. Mini-lesson: Bimodal distributions often indicate mixed populations; symmetry lacks peaks and tails imbalance. Left skew would tail toward lower values, uniformity means flat frequencies, and continuous scales don't inherently imply symmetry. Look for clusters to identify modality.

Question 5

A movie theater manager wants a point estimate of the population proportion $p$ of all customers who buy popcorn. Over one weekend, the manager records whether each customer in the first 300 tickets sold on Saturday purchased popcorn, computes the sample proportion $\hat p$ , and reports $\hat p$ as the point estimate for $p$ for all customers over the entire weekend. Which statement describes whether the estimate is biased?

The estimate is biased because the first customers on Saturday may differ from other customers (for example, by showtime), so the sampling method may systematically misrepresent $p$ . (correct answer)
The estimate is unbiased because $\hat p$ is always unbiased for $p$ even when the sample is a convenience sample.
The estimate is unbiased because 300 is a large sample size, which removes any bias.
The estimate is biased because $\hat p$ will vary from sample to sample.
The estimate is unbiased only if the manager also records soda purchases and adjusts $\hat p$ .

Explanation: This question tests recognizing bias from convenience sampling. The manager records only the first 300 Saturday customers, which is not a random sample of all weekend customers. Early Saturday customers might differ systematically from later customers or Sunday customers (different showtimes, different demographics), making this a biased sampling method. Choice A correctly identifies this bias. Choice B is wrong because $\hat{p}$ is only unbiased for random samples, not convenience samples. Choice C incorrectly suggests large samples eliminate bias—but 300 biased observations are still biased. The key insight: bias comes from how you select your sample, and convenience samples almost always introduce bias because they systematically over- or underrepresent certain groups.

Question 6

A game app records the discrete random variable $X$ , the number of attempts a player needs to pass a level (up to 4 attempts). The probability distribution for $X$ is shown in the table. Which statement correctly interprets the random variable?

$X$ is the probability that a player passes the level on the first attempt.
$X$ is the event that a player needs at most 2 attempts to pass the level.
$X$ is the number of attempts a randomly selected player needs to pass the level. (correct answer)
$X$ is the set of possible outcomes for whether the player passes or fails the level.
$X$ is the probability distribution of attempts for a specific player over many levels.

Explanation: This question evaluates understanding of random variables in gaming contexts. The random variable X represents the number of attempts needed to pass a level, which can take values 1, 2, 3, or 4. Choice A incorrectly describes X as a probability rather than a count of attempts. Choice B mistakes X for an event (needing at most 2 attempts) rather than a variable. Choice D confuses X with a set of pass/fail outcomes rather than a numerical count. Choice E misinterprets X as a probability distribution across multiple levels rather than the variable for a single level. When interpreting random variables, focus on what numerical quantity is being measured or counted in the random experiment - here it's the number of attempts, not probabilities or events.

Question 7

A company recorded whether employees work remotely (Yes/No) and whether they report being satisfied with work-life balance (Satisfied/Not). The purpose is to compare the satisfaction rates between remote and non-remote employees using conditional proportions. Which comparison is supported by the data?

Counts: Remote: Satisfied 84, Not 36 (Total 120). Not remote: Satisfied 150, Not 50 (Total 200).

Because more not-remote employees are satisfied (150) than remote employees are satisfied (84), not-remote employees are more likely to be satisfied.
The proportion satisfied among remote employees is higher than the proportion satisfied among not-remote employees.
The proportion satisfied among remote employees is lower than the proportion satisfied among not-remote employees. (correct answer)
The proportion remote is higher among satisfied employees than among not satisfied employees.
Remote and not-remote employees have the same satisfaction rate.

Explanation: This question asks us to compare satisfaction rates between remote and non-remote employees using conditional proportions. For remote employees, the satisfaction rate is 84/120 = 0.70 or 70%. For non-remote employees, the satisfaction rate is 150/200 = 0.75 or 75%. Since 70% < 75%, the proportion satisfied among remote employees is lower than among non-remote employees. Choice A incorrectly compares raw counts (150 vs 84) instead of proportions, which doesn't account for the different group sizes. When analyzing two-way tables, calculate the proportion within each category of the explanatory variable to make valid comparisons.

Question 8

A principal wants to compare the average time it takes students to finish a standardized reading test when taken on paper versus on a computer. She randomly assigns 120 students to take the paper version and 120 students to take the computer version. The paper group’s mean time is 54 minutes; the computer group’s mean time is 49 minutes. Which conclusion is supported by the data?

Students who take the test on a computer tend to finish faster than students who take it on paper, in this randomized experiment. (correct answer)
Students at all schools will finish 5 minutes faster on a computer than on paper.
Students who prefer computers were more likely to be placed in the computer group.
Taking the test on a computer causes every student to finish faster than they would on paper.
Because the computer group had a smaller mean time, the distribution of times must have been less variable for the computer group.

Explanation: This randomized experiment compared test completion times between paper and computer formats. With random assignment of 120 students to each group, we can make causal conclusions. Option A correctly states that computer-takers "tend to finish faster" in this experiment - appropriate language for experimental results. Options B and D overgeneralize (to all schools or every student). Option C incorrectly suggests assignment bias in a randomized study. Option E makes an unsupported claim about variability. Experimental conclusions should describe the observed effect without overgeneralizing beyond the experimental context.

Question 9

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?

Events A and B are independent because $P(A\cap B)=P(A)P(B)$ . (correct answer)
Events A and B are not independent because they overlap.
Events A and B are independent because $P(A\cup B)=P(A)+P(B)$ .
Events A and B are not independent because $P(A\cap B)=0$ .
Events A and B are independent because “even” and “greater than 4” describe different properties.

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.

Question 10

A bookstore tracked the number of books purchased per transaction for 52 customers on a Saturday, counting the total items on each receipt. The dotplot shows the distribution.

Which feature of the distribution is most evident?

The distribution is skewed left, with a tail toward larger numbers of books
The distribution is skewed right, with a tail toward larger numbers of books (correct answer)
The distribution is symmetric because the smallest value is 0
The distribution is bimodal because the maximum value is far from the minimum
The distribution is uniform across all counts from 0 to 10

Explanation: This question tests your understanding of skewness in count data. Looking at the dotplot showing books purchased per transaction, you can observe that most customers bought a small number of books (many dots at 1, 2, or 3 books), with progressively fewer customers buying larger numbers of books extending toward the right. This creates a distribution with a long tail toward the larger values, which defines a right-skewed distribution. Choice B correctly identifies this pattern - the distribution is skewed right with a tail toward larger numbers of books. This shape is typical for purchase data because while most customers buy just a few items, occasional customers make large purchases. Count data often exhibits right skewness because there's a natural lower bound (0 or 1) but no strict upper limit for enthusiastic customers.

Question 11

A city compares mean wait times (minutes) at two DMV locations. An independent random sample of $n_1=12$ customers from Location 1 and $n_2=12$ customers from Location 2 is selected, and $\bar{x}_1-\bar{x}_2$ is computed. Which statement is correct about using a normal model for the sampling distribution of $\bar{x}_1-\bar{x}_2$ ?

A normal model is automatically appropriate whenever the two sample sizes are equal.
A normal model is appropriate only if the two population distributions are approximately normal (since the sample sizes are small). (correct answer)
A normal model is never appropriate for differences of sample means.
A normal model is appropriate only if $\sigma_1=\sigma_2$ .
A normal model is appropriate because independence alone guarantees normality.

Explanation: This question examines when a normal model is appropriate for the sampling distribution of a difference in means with small samples. With sample sizes of only 12 each, we cannot rely on the Central Limit Theorem to ensure approximate normality. Instead, we need the population distributions themselves to be approximately normal for the sampling distribution to be normal. Choice A incorrectly suggests equal sample sizes guarantee normality. Choice C is too extreme, as normal models can be appropriate under certain conditions. Choice D incorrectly requires equal population standard deviations. The critical lesson is that sample size matters: large samples (typically n ≥ 30) allow normal approximations regardless of population shape, but small samples require the populations themselves to be approximately normal.

Question 12

A chemistry teacher wants to test whether three different lab instruction formats (video demo, written instructions, and live teacher demo) affect lab report scores. There are 3 lab sections meeting on different days (Monday, Wednesday, Friday), and the teacher suspects day-of-week differences (fatigue, scheduling) could affect scores. Each section must use only one format to avoid confusion during the lab. Which experimental design is most appropriate?

Completely randomized design assigning individual students within each section to different formats during the same lab period
Randomized block design blocking by student GPA, then assign formats to students across sections
Cluster randomized design: randomly assign each entire lab section (day) to one of the three formats (correct answer)
Matched-pairs design: match students across sections by prior chemistry grade, then assign formats within pairs
Latin square design using day of week and instruction format as blocking variables, then assign students

Explanation: This scenario requires cluster randomized design due to the practical constraint that each lab section must use only one instruction format to avoid confusion. The teacher cannot mix formats within a section, making each section (day) an indivisible cluster. Option C correctly identifies this by randomly assigning each entire lab section to one of the three formats. While the teacher suspects day-of-week effects, this becomes a secondary consideration to the primary constraint of needing uniform instruction within each section. Options A and D impossibly suggest mixing formats within sections, while B blocks by an irrelevant variable (GPA) and ignores the section constraint. Option E misapplies Latin squares. When practical constraints require treating groups as units, cluster randomization is necessary even if it means accepting some potential confounding with group characteristics.

Question 13

A nutritionist compares mean sodium intake (mg/day) for adults following Diet A versus Diet B. Using independent random samples, a 90% confidence interval for the difference in population means was found to be $(\mu_A-\mu_B)\in(120,\,340)$ . Which interpretation is correct?

We are 90% confident that the true difference in mean sodium intake, $\mu_A-\mu_B$ , is between 120 and 340 mg/day. (correct answer)
There is a 90% chance that Diet A adults have sodium intake 120 to 340 mg/day higher than Diet B adults.
We are 90% confident that $\mu_B-\mu_A$ is between 120 and 340 mg/day.
Because the interval is positive, 90% of adults on Diet A consume between 120 and 340 mg/day more sodium than adults on Diet B.
Since 0 is not in the interval, there is a 90% probability that the computed sample difference will be between 120 and 340 mg/day in future samples.

Explanation: This question asks about interpreting a confidence interval for μ_A - μ_B. The interval (120, 340) means we're 90% confident that Diet A's mean sodium intake is between 120 and 340 mg/day higher than Diet B's. Choice A correctly states this interpretation. Choice B incorrectly assigns probability to individual adults. Choice C reverses the order of subtraction. Choice D misinterprets the interval as describing individual differences. Choice E incorrectly relates the interval to future sample statistics. Key concept: confidence intervals estimate population parameters, not individual values or sample statistics.

Question 14

A school counselor wants to study whether the amount of time students spend on social media is associated with sleep. For each student in a random sample of 80 juniors, the counselor records daily social media time (in minutes) and hours of sleep last night. The research goal is to use social media time to help predict sleep. Observational units are the individual juniors. Which classification is correct for the two variables?

Social media time: categorical explanatory; Sleep hours: quantitative response
Social media time: quantitative explanatory; Sleep hours: quantitative response (correct answer)
Social media time: quantitative response; Sleep hours: quantitative explanatory
Social media time: quantitative response; Sleep hours: categorical explanatory
Social media time: categorical response; Sleep hours: categorical explanatory

Explanation: This question tests your ability to classify variables as quantitative or categorical and identify explanatory versus response variables. The key phrase "use social media time to help predict sleep" tells us that social media time is the explanatory variable (predictor) and sleep hours is the response variable (outcome). Both "daily social media time (in minutes)" and "hours of sleep" are measured numerically, making them both quantitative variables. Choice B correctly identifies social media time as quantitative explanatory and sleep hours as quantitative response. Choice A incorrectly calls social media time categorical, while choices C and D reverse the explanatory/response roles. Remember: quantitative variables are measured with numbers that have mathematical meaning, while categorical variables place individuals into groups or categories.

Question 15

In a region, daily high temperatures in April are approximately Normal for the population of April days, with mean $\mu=70^\circ\!F$ and standard deviation $\sigma=5^\circ\!F$ . A forecasted high of 80 $^\circ\!F$ is noted. Which statement about the marked value is correct?

80°F is about 1 standard deviation above the mean, so it is very typical for April.
80°F is about 2 standard deviations above the mean, so it is warmer than most April days and somewhat unusual. (correct answer)
80°F is about 2 standard deviations below the mean, so it is cooler than most April days.
80°F is about 2 standard errors above the mean, so it would occur about 50% of the time.
80°F is about 3 standard deviations above the mean, so it is essentially impossible in April.

Explanation: This question tests interpreting temperatures above the mean using standard deviations. Given μ = 70°F and σ = 5°F, we calculate (80 - 70)/5 = 10/5 = 2, meaning 80°F is 2 standard deviations above the mean. The distractor about "standard errors" incorrectly applies sampling concepts to individual daily temperatures. In a normal distribution, temperatures more than 2 standard deviations above the mean occur in only about 2.5% of days. An 80°F day in April would be warmer than about 97.5% of all April days, making it somewhat unusual but not impossible for the region.

Question 16

A psychologist compares mean reaction time (ms) for participants after drinking caffeine vs drinking no caffeine. From two independent random samples, a 92% confidence interval for $(\mu_{caff}-\mu_{no\ caff})$ is $(-25, -5)$ . Which interpretation is correct?

We are 92% confident that caffeine reduces the population mean reaction time by between 5 and 25 ms compared with no caffeine. (correct answer)
There is a 92% chance that caffeine reduces reaction time for each individual by between 5 and 25 ms.
We are 92% confident that $(\mu_{no\ caff}-\mu_{caff})$ is between $-25$ and $-5$ ms.
Because the interval does not include 0, there is no evidence of a difference in population means.
There is a 92% probability that $(\mu_{caff}-\mu_{no\ caff})$ is exactly $-15$ ms.

Explanation: This question asks about interpreting a negative confidence interval for reaction time differences. The interval (-25, -5) for (μ_caff - μ_no caff) is entirely negative, meaning μ_caff < μ_no caff. Since lower reaction times are better, this indicates caffeine improves (reduces) reaction time. The correct answer states we are 92% confident that caffeine reduces the population mean reaction time by between 5 and 25 ms compared with no caffeine. Choice D incorrectly states that excluding 0 means no evidence of a difference—actually, it means strong evidence of a difference. When interpreting confidence intervals for differences, pay attention to what direction indicates improvement: for reaction time, negative differences mean the first group is faster.

Question 17

An auto manufacturer compares average stopping distance (in feet) for cars equipped with new brake pads versus standard brake pads, and wants to know if the new pads result in a different mean stopping distance. A random sample of 25 cars with new pads had a mean stopping distance of 128 feet, and a random sample of 27 cars with standard pads had a mean of 133 feet. Let $\mu_{new}$ and $\mu_{std}$ be the true mean stopping distances for the two pad types. Which hypotheses are appropriate?

$H_0: \mu_{new}-\mu_{std}=0$ ; $H_a: \mu_{new}-\mu_{std}\ne 0$ (correct answer)
$H_0: \mu_{new}-\mu_{std}=0$ ; $H_a: \mu_{new}-\mu_{std}<0$
$H_0: \mu_{std}-\mu_{new}=0$ ; $H_a: \mu_{std}-\mu_{new}>0$
$H_0: \bar{x}_{new}-\bar{x}_{std}=0$ ; $H_a: \bar{x}_{new}-\bar{x}_{std}\ne 0$
$H_0: \mu_{new}=128$ ; $H_a: \mu_{new}\ne 128$

Explanation: The manufacturer wants to know if new brake pads result in a DIFFERENT mean stopping distance (not specifically shorter or longer). This calls for a two-sided test with alternative hypothesis μ_new - μ_std ≠ 0. Choice B would only be appropriate if testing specifically for shorter stopping distances with new pads. Choice D incorrectly uses sample means (x̄) instead of population means (μ) - remember that hypotheses are always about population parameters. Choice E tests only the new pads against a specific value rather than comparing two groups. For comparing two groups with no directional claim, always use a two-tailed test with ≠ in the alternative hypothesis.

Question 18

A streaming service wants to estimate the proportion of all its subscribers who are satisfied with a new interface. The company emails a survey link to every subscriber and analyzes the first 10,000 responses received within 24 hours. No random selection is used among respondents. Which statement best describes the sample representativeness?

The sample is representative because it includes 10,000 subscribers, which is large.
The sample is likely biased because it is a voluntary response sample; those with strong opinions may be overrepresented. (correct answer)
The sample is representative because everyone had an equal chance to respond to the email.
The sample is biased because the company did not randomly assign subscribers to use the old or new interface.
The sample is representative because the first 10,000 responses are essentially a simple random sample.

Explanation: This question tests recognition of voluntary response bias in sampling. When a survey is sent to everyone but only those who choose to respond are included, this creates a voluntary response sample. Such samples are notoriously biased because people with strong opinions (either very satisfied or very dissatisfied) are more motivated to respond than those with neutral views. The fact that 10,000 responses were collected doesn't eliminate this bias - it's the self-selection mechanism that creates the problem. Random sampling requires the researcher to select participants, not participants to select themselves. A better approach would be to randomly select subscribers and follow up to achieve a high response rate from the selected sample.

Question 19

A grocery chain wants to estimate the mean amount (in dollars) customers spend per visit. Two different analysts each take an independent simple random sample of 60 receipts from the same month. Analyst 1 finds a mean of $42.10; Analyst 2 finds a mean of$ 45.80. Why might the sample results differ?

Because both samples are from the same month, the means must match; otherwise, one analyst must have fabricated data
Independent random samples can produce different sample means due to random sampling variability (correct answer)
The difference shows the sampling was biased, since unbiased sampling always produces the same mean
The population mean must be exactly halfway between the two sample means
Random sampling error means the larger mean is automatically closer to the population mean

Explanation: This question assesses understanding of sampling variability when two analysts independently sample from the same population. The correct answer recognizes that independent random samples naturally produce different results due to the randomness of which receipts each analyst selected. The $3.70 difference between means ($ 42.10 vs $45.80) reflects normal sample-to-sample variation, not any problem with the sampling or analysis. The distractors represent common errors: believing identical populations must yield identical sample means (A), thinking unbiased sampling eliminates variability (C), assuming the population mean is the average of two samples (D), or believing larger values are automatically more accurate (E). This scenario emphasizes that sampling variability is expected even when multiple researchers use proper methods on the same population - the variation comes from the random selection process itself, not from errors or bias.

Question 20

A wildlife biologist claims that between 30% and 40% of a certain bird species in a region carry a particular parasite. A random sample of 250 birds is tested, and a 95% confidence interval for the true proportion carrying the parasite is $(0.33, 0.39)$ . Is the claim supported by the confidence interval?

No, because a 95% confidence interval means 95% of birds in the region were tested.
Yes, because the entire interval lies within 0.30 to 0.40, supporting the claim. (correct answer)
No, because 0.30 and 0.40 are not endpoints of the interval, so the claim is false.
Yes, because there is a 95% probability the true proportion is between 0.30 and 0.40.
No, because the interval does not include 0.35 exactly, so the claim cannot be supported.

Explanation: This question evaluates whether a confidence interval supports a range claim. The biologist claims the proportion is BETWEEN 30% AND 40%, and the confidence interval (0.33, 0.39) falls entirely within this range. Since every plausible value for the true proportion (according to our interval) satisfies the claim, the interval supports it. Choice C incorrectly assumes the claimed values must be endpoints of the interval, and Choice D misinterprets the confidence level as a probability statement about the parameter. When a confidence interval is entirely contained within a claimed range, it provides strong support for that claim. This is different from directional claims - here we need the interval to be a subset of the claimed range.

Question 21

A researcher surveys a single random sample of 300 adults and records two categorical variables for each person: exercise level (Low, Moderate, High) and sleep quality (Poor, Fair, Good). The two-way table summarizes the counts.

Which test is appropriate, and what are the correct hypotheses?

Chi-square test of independence; $H_0$ : exercise level and sleep quality are independent; $H_a$ : they are associated (correct answer)
Chi-square test of homogeneity; $H_0$ : the distribution of sleep quality is the same across exercise levels; $H_a$ : at least one exercise level has a different distribution
Chi-square goodness-of-fit; $H_0$ : sleep quality is equally likely to be Poor, Fair, or Good; $H_a$ : not all categories are equally likely
Two-sample $t$ test; $H_0$ : mean sleep quality is the same for all exercise levels; $H_a$ : at least one mean differs
Chi-square test of independence; $H_0$ : the distribution of sleep quality is the same across exercise levels; $H_a$ : the distribution differs

Explanation: The skill being tested is identifying the correct chi-square test and hypotheses for examining the relationship between two categorical variables in a single sample for AP Statistics. The sampling design features one random sample of adults where both exercise level and sleep quality are recorded for each individual, indicating a single population with two variables. Therefore, the chi-square test of independence is suitable to check if these variables are associated. Mini-lesson: Chi-square goodness-of-fit tests a single distribution against expectations, homogeneity compares distributions across groups from independent samples, and independence assesses if two variables are related in one sample. Option E is a distractor because it names the test as independence but uses hypotheses worded like homogeneity, which mismatches the sampling design. The marked answer A correctly specifies the test and hypotheses.

Question 22

A study is conducted to explore the relationship between two variables. For each of 100 participants, their daily screen time (in hours) and their score on a happiness scale (from 1 to 10) are recorded. How is this data best classified?

A set of two univariate datasets, one for screen time and one for happiness score.
A bivariate dataset with one quantitative variable and one categorical variable.
A bivariate quantitative dataset. (correct answer)
An experimental dataset designed to show a causal relationship between the variables.

Explanation: The data consists of two variables measured for each participant. Screen time (in hours) is quantitative. A numerical scale like the happiness score is also treated as quantitative. Therefore, the data are bivariate and quantitative.

Question 23

Which of the following statements provides the best definition of a discrete random variable?

A variable whose values form a countable set of numbers, where each value is associated with a probability. (correct answer)
A variable in an experiment whose value is measured to determine the effect of a treatment.
A numerical characteristic of a sample that is calculated from data, such as the sample mean.
A variable that can take on any value within a given range or interval, measured with a certain precision.

Explanation: A discrete random variable is a variable that can take on a finite or countably infinite number of distinct values. Its probability distribution assigns a probability to each possible value. Choice B describes a response variable. Choice C describes a statistic. Choice D describes a continuous random variable.

Question 24

In an AP Statistics class, let $X$ be the number of minutes a student spends on math homework in an evening with mean $\mu_X=50$ and standard deviation $\sigma_X=10$ . Let $Y$ be the number of minutes the same student spends on reading with mean $\mu_Y=30$ and standard deviation $\sigma_Y=6$ . Assume $X$ and $Y$ are independent. Define the combined random variable $T=X+Y$ (total homework time). Which statement about $T$ is correct?

$\mu_T=80$ and $\sigma_T=16$
$\mu_T=20$ and $\sigma_T=\sqrt{10^2+6^2}$
$\mu_T=80$ and $\sigma_T=\sqrt{10^2+6^2}$ (correct answer)
$\mu_T=80$ and $\sigma_T=10-6$
$\mu_T=50+30$ and $\sigma_T=\sqrt{10-6}$

Explanation: This question assesses the AP Statistics skill of combining independent random variables, specifically for the sum T = X + Y. The mean of the sum is always the sum of the means, so μ_T = 50 + 30 = 80, regardless of independence. For the standard deviation, since X and Y are independent, the variance of T is the sum of the variances: Var_T = 10² + 6² = 100 + 36 = 136, so σ_T = √136 = √(10² + 6²). A common distractor, like choice D, subtracts the standard deviations, but that's incorrect because standard deviations don't subtract directly; variances add for both sums and differences. In a mini-lesson on combining random variables: always add means for sums and subtract for differences, but add variances in both cases when variables are independent, then take the square root for the standard deviation. This rule ensures we account for the total variability without assuming correlation.

Question 25

A call center knows that the true proportion of calls resolved on the first call is $p=0.64$ . A manager repeatedly takes many random samples of size $n=16$ from all calls and computes $\hat p$ each time. Which statement about the sampling distribution of $\hat p$ is correct?

The sampling distribution of $\hat p$ is centered at $0.64$ , and it is approximately normal because $p$ is greater than $0.50$ .
The sampling distribution of $\hat p$ is centered at $0.64$ , but it may not be approximately normal because the sample size is small. (correct answer)
The sampling distribution of $\hat p$ is centered at $\hat p=0.64$ for every sample, so it has standard deviation $0$ .
The sampling distribution of $\hat p$ is centered at $0.36$ , and it is approximately normal because $n(1-p)=0.36$ .
The sampling distribution of $\hat p$ is centered at $0.64$ , and its standard deviation is $\dfrac{0.64(0.36)}{16}$ .

Explanation: This question examines normality with a small sample size. With n = 16 and p = 0.64, we check the success-failure conditions: np = 16(0.64) = 10.24 ≥ 10 (barely satisfied) and n(1-p) = 16(0.36) = 5.76 < 10 (not satisfied). Since n(1-p) < 10, the sampling distribution may not be approximately normal despite p > 0.50. Choice A incorrectly assumes p > 0.50 ensures normality. Choice C misunderstands sampling variability. Choice D incorrectly centers at 1-p and misinterprets n(1-p). Choice E gives the wrong formula for standard deviation (missing the square root). Small samples require both np ≥ 10 and n(1-p) ≥ 10 for approximate normality.