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AP Statistics

AP Statistics Practice Test: Practice Test 45

Practice Test 45 for AP Statistics: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A sports scientist claims that a higher proportion of runners will experience muscle cramps when they drink water only compared with a sports drink during a race. In a random sample of 120 runners who drank water only, 42 reported cramps; in a separate random sample of 130 runners who drank a sports drink, 30 reported cramps. Which hypotheses are appropriate for a test of this claim?

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Question 1

$H_0: p_{\text{water}}-p_{\text{sports}}=0$ ; $H_a: p_{\text{water}}-p_{\text{sports}}>0$ (correct answer)
$H_0: p_{\text{sports}}-p_{\text{water}}=0$ ; $H_a: p_{\text{sports}}-p_{\text{water}}>0$
$H_0: p_{\text{water}}-p_{\text{sports}}=0$ ; $H_a: p_{\text{water}}-p_{\text{sports}}\ne 0$
$H_0: \hat p_{\text{water}}-\hat p_{\text{sports}}=0$ ; $H_a: \hat p_{\text{water}}-\hat p_{\text{sports}}>0$
$H_0: p_{\text{water}}=0.35$ ; $H_a: p_{\text{water}}>0.35$

Explanation: This question requires setting up hypotheses to test if water-only runners have a higher cramping rate than sports drink users. The claim specifies a direction (higher proportion), so we need a one-sided test. The correct answer A uses H_a: p_water - p_sports > 0, which tests if the water group has a higher cramping proportion. Choice D incorrectly uses sample proportions (p̂) in the hypotheses - we must use population parameters. Choice B reverses the subtraction order, which would test the opposite claim. Choice E incorrectly tests a single proportion against a fixed value rather than comparing two groups. When comparing two populations, the null hypothesis always states no difference (difference = 0), and the alternative reflects the research claim's direction.

Question 2

A runner’s time for the first half of a race is $A$ minutes and for the second half is $B$ minutes. Suppose $A$ and $B$ are independent with $\mu_A=24.0$ , $\sigma_A=1.5$ , $\mu_B=25.0$ , and $\sigma_B=1.8$ . Let $T=A+B$ be the total time. Which statement about $T$ is correct?

$\mu_T=49.0$ and $\sigma_T=1.5+1.8$
$\mu_T=49.0$ and $\sigma_T=\sqrt{1.5^2+1.8^2}$ (correct answer)
$\mu_T=-1.0$ and $\sigma_T=\sqrt{1.5^2+1.8^2}$
$\mu_T=49.0$ and $\sigma_T=1.8-1.5$
$\mu_T=1.0$ and $\sigma_T=\sqrt{1.5^2-1.8^2}$

Explanation: This question asks about combining random variables through addition to find total race time. When T = A + B, the mean of the sum equals the sum of the means: μT = μA + μB = 24.0 + 25.0 = 49.0 minutes. Since A and B are independent, the variance of the sum equals the sum of the variances: σ²T = σ²A + σ²B, which gives us σT = √(1.5² + 1.8²). A common mistake is to add the standard deviations directly (1.5 + 1.8), but remember that variances add, not standard deviations. This principle applies whenever we combine independent random variables, making the total time more variable than either individual segment.

Question 3

A restaurant chain is estimating the mean time (in minutes) customers wait for takeout orders at a particular location. On Monday, a random sample of 50 orders has a mean wait of 12.6 minutes. On Tuesday, an independent random sample of 50 orders from the same location has a mean wait of 14.1 minutes. Assume the ordering process did not change. Why might the sample results differ?

The Tuesday sample must be biased because it produced a different mean than Monday
Random sampling variability can produce different sample means even when the population and process are the same (correct answer)
The population mean wait time must have changed, because sample means always equal population means
One of the samples must have included extreme outliers, which is impossible in random samples
The difference proves that the sampling method was a voluntary response sample

Explanation: This question tests understanding of sampling variability in a real-world context where students might expect consistency. The difference in mean wait times (12.6 vs 14.1 minutes) between Monday and Tuesday samples is explained by random sampling variability, not by any change in the restaurant's process. Each sample of 50 orders represents a different random selection of customers, and by chance, one sample might include more complex orders or happen to coincide with slightly busier periods. The key insight is that sampling variability affects all statistics, including means, and occurs even when the underlying population and process remain constant. Students often incorrectly assume that differences in samples must indicate changes in the population or bias in sampling, but random variation alone can produce such differences. This understanding is crucial for interpreting real-world data where multiple samples might yield different results.

Question 4

A hospital tracks the next 30 patients who arrive at the emergency room. For each patient, a success is “patient is admitted to the hospital” and a failure is “patient is treated and released.” The hospital records the number admitted among these 30 patients. Does this meet binomial conditions?

Yes, because there are 30 trials and each patient results in admission or release, assuming the admission probability is the same for each arrival. (correct answer)
No, because arrivals occur over time, so the number of trials is not fixed.
No, because each patient could have more than two outcomes (admitted, released, transferred).
No, because the probability of admission must be $0.5$ for a binomial model.
Yes, because independence is guaranteed whenever the sample size is 30 or less.

Explanation: This situation meets all binomial conditions. There are exactly 30 patients (fixed n), each patient has two outcomes (admitted or released), and assuming similar emergency room conditions, each patient has the same probability of admission. The patients' medical conditions are independent of each other. Choice B incorrectly suggests that events occurring over time prevent a fixed trial count, C introduces a third outcome not mentioned in the problem, D incorrectly claims the probability must be 0.5 (binomial works for any constant p between 0 and 1), and E makes an irrelevant claim about sample size and independence. The key is that we're examining a fixed group of 30 patients, regardless of when they arrive, making this a valid binomial setting.

Question 5

A marketing firm wants to estimate the mean amount spent per visit by customers at a grocery chain. They randomly select 25 stores from all stores in the chain and then record the total spent by every customer who shops at those stores on a single Saturday. Which conclusion is justified based on the design?

(Assume the 25 stores are selected by simple random sampling.)

The results can be generalized to all customers of the chain on Saturdays, but not necessarily to all days of the week. (correct answer)
Because stores were randomly selected, the study proves that shopping on Saturday causes customers to spend more.
Because every customer at selected stores was included, the results can be generalized to all grocery shoppers everywhere.
Because customers were not randomly assigned to stores, no generalization to the chain is possible.
Because the sample includes all customers at selected stores, the results can be generalized to all customers of the chain on all days.

Explanation: This question examines AP Statistics inference principles in sampling designs, particularly generalization without causation. Random selection of stores enables generalization to all chain stores, but limiting data to one Saturday means results apply only to Saturdays, not all days, due to potential weekly variations. There is no random assignment or treatment, so this observational census of customers at selected stores cannot prove causation, like Saturday causing more spending. Choice E distracts by overgeneralizing to all days despite the single-day scope. In a mini-lesson on inference types, random sampling supports population estimates but not causality, which requires experimental manipulation; here, it's for estimating Saturday spending across the chain. The design justifies generalization to Saturday customers chain-wide.

Question 6

A nutrition researcher models systolic blood pressure (mmHg) as a linear function of daily sodium intake (mg) using a random sample of adults. A 90% confidence interval for the population slope is $(0.002,\ 0.006)$ . Which interpretation is correct?

There is a 90% chance that the sample slope will fall between 0.002 and 0.006 if the same people are measured again.
We are 90% confident that for each additional 1 mg of sodium intake, the mean systolic blood pressure increases by between 0.002 and 0.006 mmHg in the population of adults like those sampled. (correct answer)
Since the interval is positive, 90% of individuals’ blood pressures increase by 0.002 to 0.006 mmHg for each 1 mg increase in sodium.
We are 90% confident that the correlation between sodium intake and systolic blood pressure is between 0.002 and 0.006.
Because 0 is not included, sodium intake is proven to cause higher blood pressure.

Explanation: This question tests understanding of confidence intervals for slope in a health context. The interval (0.002, 0.006) represents the change in blood pressure per mg of sodium. Choice B correctly interprets this as being 90% confident about the mean increase in systolic blood pressure per mg of sodium for the population. Choice A incorrectly refers to repeated sampling of the same people. Choice C misapplies the interval to individual responses. Choice D confuses slope with correlation. Choice E incorrectly claims causation. A confidence interval for slope estimates the average linear relationship in the population, not causal effects or individual responses.

Question 7

A coach analyzes the relationship between practice sessions attended ( $x$ ) and free-throw percentage ( $y$ ) for 18 players. A 90% confidence interval for the true slope is $( -0.5,\ 2.0)$ percentage points per session. Which interpretation is correct?

Because 0 is in the interval, there is definitely no association between practice sessions and free-throw percentage.
We are 90% confident that for each additional practice session, the mean free-throw percentage changes by between $-0.5$ and $2.0$ percentage points, on average. (correct answer)
There is a 90% chance that the correlation between sessions and free-throw percentage is between $-0.5$ and $2.0$ .
We are 90% confident that 90% of players will improve by between $-0.5$ and $2.0$ percentage points for each extra session.
If we repeated the study many times, 90% of the time the true slope would vary and fall in this interval.

Explanation: This question involves interpreting a confidence interval that contains zero. The interval (-0.5, 2.0) includes 0, meaning we cannot conclude there's a significant relationship at the 10% level. Choice B correctly interprets this: we're 90% confident the true slope (change in mean free-throw percentage per session) lies between -0.5 and 2.0 percentage points. Choice A overstates the conclusion - we can't definitively say there's no association. Choice C confuses slope with correlation. Choice D wrongly applies this to individual players with a specific probability. Choice E misunderstands how confidence intervals relate to the true parameter. Key point: intervals containing zero suggest the relationship might be positive, negative, or nonexistent.

Question 8

A company wants to estimate the mean number of hours per week its 3,000 employees work from home. The population is all employees. The company divides employees into departments (Engineering, Sales, HR, Support) and then randomly selects 50 employees from each department using a random number generator, for a total of 200 employees surveyed. Which statement best describes the sample representativeness?

The sample is likely representative because it is a stratified random sample with random selection within each department. (correct answer)
The sample is not representative because stratified sampling is not random sampling.
The sample is representative only if each employee had exactly the same chance of being selected.
The sample is representative because 200 employees is a large sample regardless of how they were chosen.
The sample is not representative because employees were not randomly assigned to departments.

Explanation: This question tests understanding of stratified random sampling, a valid probability sampling method. The company divided employees into strata (departments) and then randomly selected 50 employees from each stratum. This ensures representation from all departments and can actually improve precision compared to simple random sampling when groups differ. The key is that within each stratum, selection was random using a random number generator. While this gives different selection probabilities if departments have different sizes, stratified sampling is still a representative method. The distractors incorrectly suggest that only simple random sampling is valid or confuse random sampling with random assignment.

Question 9

A grocery store tests whether a new checkout layout reduces mean customer wait time. A random sample of $n=25$ customers in the new layout had mean wait time $\bar{x}=3.9$ minutes, and a random sample of $n=27$ customers in the old layout had mean wait time $\bar{x}=4.6$ minutes. The store’s claim is that the new layout has a lower population mean wait time than the old layout. Which hypotheses are appropriate?

$H_0: \mu_{new}=3.9$ ; $H_a: \mu_{new}<3.9$
$H_0: \mu_{old}-\mu_{new}=0$ ; $H_a: \mu_{old}-\mu_{new}<0$
$H_0: \mu_{new}-\mu_{old}=0$ ; $H_a: \mu_{new}-\mu_{old}<0$ (correct answer)
$H_0: \bar{x}_{new}-\bar{x}_{old}=0$ ; $H_a: \bar{x}_{new}-\bar{x}_{old}<0$
$H_0: \mu_{new}-\mu_{old}=0$ ; $H_a: \mu_{new}-\mu_{old}\ne 0$

Explanation: This question tests hypothesis setup for the difference of two means in AP Statistics, examining wait times between new and old checkout layouts. The claim is that the new layout has a lower population mean wait time (μ_new < μ_old), so hypotheses are H0: μ_new - μ_old = 0 and Ha: μ_new - μ_old < 0. Distractor choice B reverses the subtraction order, testing if old is less than new, which opposes the claim. Choice D uses sample means, but hypotheses must involve population parameters. Mini-lesson: For two-sample mean tests, subscript parameters clearly (e.g., new vs. old); null assumes no difference, alternative direction (<0) reflects 'lower' for the first group. Sample means like 3.9 and 4.6 guide the test statistic but stay out of hypotheses. Check subtraction order to ensure it matches the claimed inequality.

Question 10

A linear regression model is used to predict monthly sales over a period of three years. The residuals are plotted against the month number (1, 2, ..., 36). The plot shows that residuals for consecutive months are often close in value, with strings of positive residuals followed by strings of negative residuals.

Which assumption for regression inference is most likely violated according to this plot?

The assumption that the residuals are independent of each other. (correct answer)
The assumption that the relationship between sales and time is linear.
The assumption that the variance of the residuals is constant.
The assumption that the residuals are approximately normally distributed.

Explanation: When residuals are plotted against a time variable, a non-random pattern such as strings of positive or negative values suggests that the error term for one observation is related to the error term of a previous observation. This pattern indicates a lack of independence among the residuals (autocorrelation).

Question 11

A commuter records $X$ , the number of red lights encountered on the drive to work. The distribution of $X$ has mean $\mu_X=5.8$ and standard deviation $\sigma_X=1.9$ . Which interpretation of the mean is correct?

The commuter will encounter exactly 5.8 red lights on most drives.
In the long run, the commuter’s average number of red lights per drive is about 5.8. (correct answer)
The commuter will always encounter between 3.9 and 7.7 red lights.
The standard deviation 1.9 means the maximum number of red lights is 1.9.
Because the mean is 5.8, the commuter cannot ever encounter 6 red lights.

Explanation: This question assesses understanding of the mean for red lights encountered while commuting. The mean μ_X = 5.8 represents the long-run average number of red lights per drive. Choice B correctly interprets this as "in the long run, the commuter's average number of red lights per drive is about 5.8." This properly treats the mean as an expected value over many drives. Choice A wrongly suggests exactly 5.8 red lights on most drives, choice C incorrectly creates rigid boundaries (3.9 to 7.7), choice D misinterprets the standard deviation as a maximum, and choice E incorrectly claims 6 red lights are impossible. Since red lights are discrete, individual drives will have whole numbers of red lights, but the average over many drives can be 5.8.

Question 12

A city reports that 6% of parking meters are broken at any given time. Many random samples of $n=300$ meters are inspected and $\hat p$ is recorded each time. Which statement about the sampling distribution is correct?

The sampling distribution of $\hat p$ has standard deviation $\sqrt{0.06(0.94)/300}$ . (correct answer)
The sampling distribution of $\hat p$ has standard deviation $\sqrt{0.06(0.94)\cdot 300}$ .
The sampling distribution of $\hat p$ has standard deviation $0.06/300$ .
The sampling distribution of $\hat p$ has standard deviation 0 because $p$ is fixed.
The sampling distribution of $\hat p$ has standard deviation 300.

Explanation: This question tests the standard deviation of sampling distributions for sample proportions in AP Statistics. Sd $\sqrt{0.06 \times 0.94 / 300}$ , spread around $p=0.06$ . Choice B multiplies by n. A distractor is choice D, sd=0. In a mini-lesson, formula for variability; here $n p = 18 \geq 10$ , $n(1-p)=282 \geq 10$ , normal ok. Essential for understanding sampling error.

Question 13

Two brands of batteries are tested for how long they last (hours) in the same device. The results are summarized below.

Which comparison is supported?

Statistic	Brand X	Brand Y
Mean	9.8	10.1
Median	10.2	9.9
Shape	left-skewed	right-skewed
IQR	1.4	1.4

Brand Y has a higher mean and a higher median than Brand X.
Brand X and Brand Y have the same center because their IQRs are equal.
Brand Y has a slightly higher mean, but Brand X has a higher median; the IQRs are equal. (correct answer)
Brand X is right-skewed and Brand Y is left-skewed, so Brand X must have a lower mean.
Brand X has a larger IQR, so it must have the larger mean.

Explanation: This question evaluates comparing battery life distributions using means, medians, shapes, and IQRs in AP Statistics. Brand Y has a slightly higher mean (10.1 vs. 9.8 hours) but lower median (9.9 vs. 10.2), with equal IQRs of 1.4, reflecting shape influences—Y is right-skewed, X left-skewed. Skewness explains why Y's mean exceeds its median, unlike X. Choice A distracts by claiming Y has both higher mean and median, ignoring the reversal. In comparisons, note how shape affects center measures; equal IQRs indicate similar variabilities. Mini-lesson: In skewed distributions, medians resist tail pulls unlike means—left-skew pulls mean below median, right-skew above; compare both for insights.

Question 14

A bottling plant targets an average fill volume of $\mu=2.00$ liters. A random sample of 50 bottles is measured, and a one-sample $t$ test is carried out with $H_0:\mu=2.00$ and $H_a:\mu\neq 2.00$ at $\alpha=0.05$ . The p-value is 0.62. What conclusion is appropriate?

Reject $H_0$ ; there is convincing evidence the population mean fill volume differs from 2.00 liters.
Fail to reject $H_0$ ; there is not convincing evidence that the population mean fill volume differs from 2.00 liters. (correct answer)
Fail to reject $H_0$ ; therefore the population mean is exactly 2.00 liters.
Because p-value = 0.62, there is a 62% chance that $H_0$ is true.
Since the sample mean was close to 2.00, we conclude the sample mean fill volume equals 2.00 liters.

Explanation: This problem tests understanding of a two-tailed test with a large p-value. With p-value = 0.62 and α = 0.05, we fail to reject H₀ because 0.62 > 0.05, indicating no convincing evidence that the population mean fill volume differs from 2.00 liters. Choice C incorrectly concludes that failing to reject H₀ proves the mean equals exactly 2.00 liters. Choice D misinterprets the p-value as the probability H₀ is true. Choice E discusses only the sample mean rather than making an inference about the population. A large p-value suggests our sample data is consistent with H₀, but we cannot conclude H₀ is definitely true.

Question 15

A researcher compares mean systolic blood pressure for adults who follow Diet A and Diet B. The claim is that Diet A leads to a lower mean systolic blood pressure than Diet B. A 95% confidence interval for $(\mu_{A}-\mu_{B})$ is $(-6.2,\ 1.5)$ mmHg. Is the claim supported by the confidence interval?

Yes, because the interval includes negative values, so Diet A must be lower.
Yes, because the interval includes 0, meaning the diets are the same and Diet A is lower.
No, because the interval includes 0, so there is not convincing evidence that $\mu_A < \mu_B$ . (correct answer)
No, because the interval crosses 0, which proves Diet A is higher.
Yes, because 95% confidence guarantees Diet A is lower for all adults.

Explanation: In AP Statistics, this question evaluates justifying claims about the difference of two means via confidence intervals. The 95% interval for μ_A - μ_B is (-6.2, 1.5) mmHg, which includes zero, so it does not provide convincing evidence that Diet A leads to lower mean blood pressure than Diet B. Choice B is a common distractor, mistakenly thinking inclusion of zero means the diets are the same and thus A is lower, but it actually indicates possible equality or either direction. A mini-lesson: For a claim that μ1 < μ2, the interval for μ1 - μ2 must be entirely negative; crossing zero suggests the difference could be zero or positive. Here, the interval spans negative and positive values, failing to support the claim. Always check the order of subtraction to match the claim's direction.

Question 16

A bookstore owner claims that the distribution of customers’ preferred format is 60% print, 25% e-book, and 15% audiobook. A random sample of 160 customers is surveyed; results are shown in the table. Which hypotheses are appropriate for a chi-square goodness-of-fit test of the owner’s claim?

$H_0$ : The observed counts are exactly the expected counts. $H_a$ : The observed counts are not exactly the expected counts.
$H_0$ : The population distribution of preferred format is 60% print, 25% e-book, 15% audiobook. $H_a$ : The population distribution of preferred format differs from 60%, 25%, 15%. (correct answer)
$H_0$ : Preferred format is independent of customer gender. $H_a$ : Preferred format is not independent of customer gender.
$H_0$ : The sample proportions are 0.60, 0.25, 0.15. $H_a$ : The sample proportions are not 0.60, 0.25, 0.15.
$H_0$ : $p_{ebook}=0.25$ . $H_a$ : $p_{ebook}\ne 0.25$ .

Explanation: For testing the bookstore owner's claim about format preferences, we need hypotheses about the population distribution. The null hypothesis should state that the population prefers formats in the proportions 60% print, 25% e-book, and 15% audiobook, while the alternative states that the population distribution differs from these percentages. Option B correctly expresses both hypotheses in terms of population parameters. Option A incorrectly focuses on exact count matching rather than testing distributions. Option C describes a test of independence between two variables, which isn't appropriate here. Option D incorrectly refers to sample proportions rather than population parameters. Option E only tests one format rather than the entire distribution.

Question 17

A lab claims that 11% of test results are inconclusive. Many random samples of $n=80$ tests are taken and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?

The sampling distribution of $\hat p$ is centered at 0.11. (correct answer)
The sampling distribution of $\hat p$ is centered at 80.
The sampling distribution of $\hat p$ is centered at 0.89.
The sampling distribution of $\hat p$ has no variability because 11% is fixed.
The sampling distribution of $\hat p$ is centered at $\sqrt{0.11(0.89)/80}$ .

Explanation: This question assesses the center of sampling distributions for sample proportions in AP Statistics. Centered at $p=0.11$ , with variability $sqrt(p(1-p)/n)$ . Choice B centers at n=80. A distractor is choice D, claiming no variability. In a mini-lesson, mean always p; here n=80, $n p=8.8<10$ , so normality questionable despite $n(1-p)=71.2 \geq 10$ . Center fixed regardless.

Question 18

A manager wants to estimate the mean time (in minutes) customers spend waiting in line at a coffee shop during the morning rush. A random sample of 25 customers is observed, and a 90% confidence interval for the population mean waiting time is $(3.2, 4.6)$ minutes. Which interpretation is correct?

We are 90% confident that the population mean waiting time is between 3.2 and 4.6 minutes. (correct answer)
There is a 90% probability that the true mean waiting time is between 3.2 and 4.6 minutes.
About 90% of customers wait between 3.2 and 4.6 minutes.
If we repeated the study many times, 90% of the sample means would fall between 3.2 and 4.6 minutes.
Exactly 90% of all possible 90% confidence intervals from samples of size 25 will have endpoints 3.2 and 4.6.

Explanation: This question assesses proper interpretation of a confidence interval for mean waiting time. The correct answer (A) properly states that we are 90% confident the population mean waiting time is between 3.2 and 4.6 minutes. Option B incorrectly assigns probability to the parameter after data collection. Option C wrongly interprets the interval as describing individual customer wait times rather than the mean. Option D confuses the sampling distribution of the mean with the confidence interval. Option E makes an incorrect claim about specific interval endpoints. Key concept: confidence intervals estimate population parameters, not individual values, and confidence refers to the method's long-run success rate.

Question 19

A school compares the proportion of students who pass an Algebra end-of-course exam after using two different review programs. In a random sample, 78 of 120 students using Program A passed, and 64 of 120 students using Program B passed. A two-proportion $z$ test was performed for $H_0: p_A-p_B=0$ versus $H_a: p_A-p_B>0$ , and the p-value was $0.018$ . Using $\alpha=0.05$ , what conclusion is appropriate?

Fail to reject $H_0$ ; there is not convincing evidence that $p_A>p_B$ in the populations.
Reject $H_0$ ; there is convincing evidence that the population pass rate is higher with Program A than with Program B. (correct answer)
Reject $H_0$ ; there is convincing evidence that the population pass rate is higher with Program B than with Program A.
Because the p-value is small, Program A caused more students in the sample to pass than Program B.
Reject $H_0$ ; we can conclude only that 78/120 is greater than 64/120 for these samples, not for the populations.

Explanation: This question tests understanding of hypothesis test conclusions for comparing two population proportions. Since the p-value (0.018) is less than α = 0.05, we reject the null hypothesis. The alternative hypothesis states p_A - p_B > 0, which means p_A > p_B, so we have convincing evidence that Program A has a higher population pass rate than Program B. Choice D incorrectly claims causation and refers only to the sample rather than making an inference about populations. Choice E incorrectly states we can only conclude about samples, not populations, but the entire purpose of hypothesis testing is to make inferences about populations. When we reject H₀ in a one-sided test with Ha: p_A - p_B > 0, we conclude there is convincing evidence that the first population proportion exceeds the second.

Question 20

A political analyst claims that in a certain county, party registration is 45% A, 40% B, and 15% C. A random sample of 200 registered voters produced the counts below. A chi-square goodness-of-fit test at $\alpha=0.05$ gives $p$ -value $=0.07$ .

What conclusion is appropriate?

Reject $H_0$ ; since $p=0.07$ is greater than 0.05, the observed differences are statistically significant.
Fail to reject $H_0$ ; there is not convincing evidence that the population party-registration distribution differs from 45%/40%/15%. (correct answer)
Fail to reject $H_0$ ; this proves the analyst’s percentages are correct for the entire county.
Reject $H_0$ ; there is convincing evidence that the sample distribution differs from 45%/40%/15%.
Because $p=0.07$ , about 7% of the county belongs to party C.

Explanation: This question involves a p-value slightly above the significance level. The p-value (0.07) is greater than α (0.05), so we fail to reject the null hypothesis. The null hypothesis states that party registration follows 45% A, 40% B, 15% C. Choice A incorrectly claims we reject H₀ and misunderstands that p > α means we fail to reject, not that differences are significant. Choice C commits the error of "proving" the analyst's percentages - we can only fail to find evidence against them. Choice D focuses on the sample rather than making a population inference. When p > α, we conclude there's not convincing evidence that the county's registration distribution differs from the claimed 45%/40%/15% split.

Question 21

A real estate agent summarizes the sale prices (in thousands of dollars) for 55 homes sold in one neighborhood last year. The five-number summary is: min $=190$ , $Q_1=240$ , median $=310$ , $Q_3=400$ , max $=650$ . A modified boxplot (1.5·IQR rule) is created. Which feature is consistent with the summary statistics?

The IQR is $650-190=460$ (thousand dollars).
The median line is at 310 and lies inside a box from 240 to 400 (thousand dollars). (correct answer)
The right whisker must extend to 650 because it is the maximum value.
The range is $400-240=160$ (thousand dollars).
Since $Q_3-Q_1=160$ , the median must be exactly 320.

Explanation: This question tests understanding of modified boxplot features with the 1.5·IQR rule. The box in any boxplot extends from Q₁ to Q₃ with the median line inside. Here, Q₁ = 240, Q₃ = 400, and median = 310 (all in thousands of dollars), so the box extends from 240 to 400 with the median line at 310, making choice B correct. Choice A incorrectly calculates IQR using max - min; the actual IQR = 400 - 240 = 160. Choice C misunderstands the modified rule; with IQR = 160, the upper fence is Q₃ + 1.5·IQR = 400 + 240 = 640, so 650 would be plotted as an outlier, not connected by the whisker. Choice D incorrectly calculates range using quartiles; the actual range = 650 - 190 = 460. Choice E incorrectly assumes the median must be at the arithmetic mean of Q₁ and Q₃; median position depends on data distribution.

Question 22

A school cafeteria manager wants to compare three different menu layouts (A, B, C) to see which leads to the highest average number of fruit servings selected per student at lunch. Fruit selection differs a lot by grade level (9th, 10th, 11th, 12th), and the manager can implement only one layout per lunch period. Which experimental design is most appropriate to estimate the effect of layout while accounting for grade level?

Completely randomized design: randomly assign each lunch period to layout A, B, or C
Matched-pairs design: for each grade, compare layout A vs B only, ignoring layout C
Randomized block design: block by grade level, then randomly assign lunch periods within each grade to layouts A, B, or C (correct answer)
Stratified sample: take a random sample of students from each grade and ask which layout they prefer
Randomized block design: block by day of the week and assign one layout to each day

Explanation: This question tests understanding of randomized block design when experimental units naturally group by a confounding variable. Since fruit selection varies significantly by grade level, we need to control for this variation to isolate the effect of menu layout. A randomized block design (choice C) blocks by grade level, then randomly assigns lunch periods within each grade to layouts A, B, or C, ensuring each layout is tested across all grades while controlling for grade-level differences. Choice A (completely randomized) ignores the grade effect, potentially confounding results. Choice B incorrectly limits comparison to only two layouts. Choice D is a sampling method, not an experimental design. Choice E blocks by day rather than grade, failing to address the stated source of variation.

Question 23

A hospital compares mean length of stay (days) for patients receiving a new discharge protocol vs the old protocol. From two independent random samples, a 97% confidence interval for $(\mu_{new}-\mu_{old})$ is $(0.2, 1.4)$ . Which interpretation is correct?

There is a 97% probability that $(\mu_{new}-\mu_{old})$ is between 0.2 and 1.4 days.
We are 97% confident that the new protocol increases the population mean length of stay by between 0.2 and 1.4 days compared with the old protocol. (correct answer)
Because 0 is not in the interval, the new protocol decreases the population mean length of stay.
97% of all patients under the new protocol stay between 0.2 and 1.4 days longer than patients under the old protocol.
We are 97% confident that $(\mu_{old}-\mu_{new})$ is between 0.2 and 1.4 days.

Explanation: This question involves interpreting a positive confidence interval for hospital length of stay. The interval (0.2, 1.4) for (μ_new - μ_old) is entirely positive, indicating μ_new > μ_old. This means the new protocol results in longer average hospital stays. The correct answer states we are 97% confident that the new protocol increases the population mean length of stay by between 0.2 and 1.4 days compared with the old protocol. Choice C incorrectly concludes that because 0 is not included, the new protocol decreases length of stay—the positive interval actually indicates an increase. When interpreting confidence intervals, remember that the sign of the interval indicates the direction of the difference: positive means the first group has a larger mean.

Question 24

A medical test returns a false positive 3% of the time for healthy patients. A clinic wants to estimate the probability that, among 5 healthy patients tested independently, at least one receives a false positive. They simulate 25,000 groups of 5 by generating random integers 1–100 for each patient and letting 1–3 represent “false positive.” The simulation found 3,675 groups with at least one false positive. Which interpretation of the simulation results is correct?

The probability is exactly $\frac{3675}{25000}$ because the simulation used 25,000 groups.
The simulation shows that exactly 3,675 of the next 25,000 real groups of 5 healthy patients will have at least one false positive.
The estimated probability of at least one false positive is about $\frac{3675}{25000}$ under the model, and more trials would typically reduce the random variation in the estimate. (correct answer)
The probability is $\frac{25000}{3675}$ because the simulation counted 3,675 successes.
The simulation is only valid if exactly 3% of all generated integers are 1, 2, or 3.

Explanation: This question examines medical testing probability through simulation. The clinic simulated 25,000 groups of 5 patients where integers 1-3 represent false positives (3% probability), finding 3,675 groups with at least one false positive. This estimates the probability as 3675/25000 = 0.147. Choice C correctly identifies this as an estimate subject to random variation that would stabilize with more trials. Choice A incorrectly claims exactness, B confuses simulation with prediction, D inverts the fraction, and E misunderstands that exact proportions aren't required in random samples. In probability simulation, we use random number generation to model complex scenarios, and the law of large numbers ensures our estimates improve with increased trial count.

Question 25

A fitness coach records resting heart rate ( $y$ , beats per minute) and weekly aerobic exercise time ( $x$ , minutes) for 14 clients. A least-squares regression line is fit to predict resting heart rate from exercise time: $\hat{y}=78.4-0.06x$ . The purpose of this linear model is to describe the association and predict typical resting heart rate for clients with exercise times similar to those observed. Which interpretation of the model is correct?

For each additional minute of weekly aerobic exercise, the predicted resting heart rate decreases by about 0.06 beats per minute, on average, for clients like those in the data. (correct answer)
If a client does 0 minutes of exercise per week, the client will have a resting heart rate of exactly 78.4 bpm.
Increasing weekly exercise by 10 minutes causes resting heart rate to drop by 0.6 bpm.
Because the slope is negative, the relationship must be nonlinear.
The model says 78.4% of resting heart rate is explained by exercise time.

Explanation: This question tests interpretation of a regression model with a negative slope in a health context. The equation $\hat{y}=78.4-0.06x$ indicates that for each additional minute of weekly exercise, the predicted resting heart rate decreases by 0.06 beats per minute on average. Choice A correctly interprets this relationship using appropriate statistical language and acknowledges the model applies to "clients like those in the data." Choice B incorrectly treats the y-intercept as an exact value, Choice C implies causation, Choice D incorrectly links negative slope to nonlinearity (linear models can have negative slopes), and Choice E confuses the y-intercept value with R-squared percentage. Remember that regression models describe average associations, not individual outcomes or causal effects, and their reliability depends on staying within the observed data range.