42 questions · suggested time 90 minutes · calculator permitted · half of the practice weighting.
Individual questions
Question 1
A public library must decide whether to trial an earlier opening time. Which question most directly measures unmet demand among the residents it serves, rather than only satisfaction among current visitors?
- Do people entering at noon say that the library is excellent?
- Among adult residents in the service area, what proportion wanted to use the library before its current opening time at least once during the last four complete weeks?
- How many books did staff shelve yesterday?
- Which earlier opening time do the first ten current visitors prefer?
Review question 1: answer and reasoning
Answer B. The second question specifies the target population, a defined unmet-use outcome and a reference period. It can guide a collection plan that reaches nonvisitors as well as visitors. Current visitor satisfaction, shelving volume or a convenience group’s preferences do not directly estimate unmet demand among all service-area adults. A well-formulated question still requires a defensible sampling and measurement plan; it does not guarantee that residents will actually attend an earlier opening.
Review focus: Formulate a measurable investigative question before collecting data. Primary unit 1: One-variable data and collection. Practice 1: Formulate questions.
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Question 2
Each of 12 independent components has probability 0.20 of failing a specified inspection. What is the probability that at least one component fails?
- 0.0687
- 0.9313
- 0.2000
- 0.8241
Review question 2: answer and reasoning
Answer B. The complement is no failures. Its probability is (0.80)^12=0.068719476736. Therefore at least one failure has probability 1−0.80^12=0.931280523264, approximately 0.9313. Independence and a common failure probability justify the product.
Review focus: 3C: Calculate a complement probability. Primary unit 2: Probability and distributions. Practice 3: Analyze data.
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Question 3
A training team can randomly assign consenting volunteers to immediate or next-day feedback after the same practice activity. Before collecting outcomes, which question gives the clearest operational target for the effect of feedback timing?
- Do volunteers think training is useful?
- Which group has the most people who already like the instructor?
- Does next-day rather than immediate feedback change the mean number of correctly completed new tasks on a test administered seven days after the final practice?
- Which of twenty outcomes will show the largest observed difference after the study?
Review question 3: answer and reasoning
Answer C. The third question names the treatment contrast, measurable response and assessment timing before results are available. Random allocation can support a causal comparison of the assigned feedback packages under the protocol. The other questions concern a different response, a preexisting characteristic, or post-result outcome selection. A complete plan should also fix content, practice time and scoring and state the population and recruitment limits.
Review focus: Formulate a causal comparison with an operational response. Primary unit 1: One-variable data and collection. Practice 1: Formulate questions.
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Question 4
A model treats completion times as normal with mean 80 minutes and standard deviation 6 minutes. What proportion of modeled times exceed 89 minutes?
- 0.9332
- 0.1587
- 0.0668
- 0.4332
Review question 4: answer and reasoning
Answer C. Standardize the cutoff: z=(89−80)/6=1.5. The upper-tail probability is P(Z>1.5)=0.0668072, approximately 0.0668. The value 0.9332 is the probability below the cutoff.
Review focus: 3C: Calculate a normal upper-tail probability. Primary unit 2: Probability and distributions. Practice 3: Analyze data.
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Question 5
A college wants to compare unmet tutoring demand among commuter and residential students during the first month of term. Which proposed primary question best matches this aim?
- Among enrolled students in each housing group, what proportion wanted at least one tutoring session during that month but could not obtain one?
- How many visits did each tutor record, counting repeat visits as different students?
- Among students already attending tutoring, which tutor is most popular?
- After viewing the results, which week and housing category show the largest difference in attendance?
Review question 5: answer and reasoning
Answer A. The first question uses students as units, identifies both comparison groups and measures unmet rather than completed demand over the specified month. Counts of visits are not counts of distinct students and omit demand that never became a visit. Popularity among existing users answers another question. Choosing the most favorable time window after seeing results does not preserve a prespecified primary aim.
Review focus: Align the unit and comparison with a precollection investigative aim. Primary unit 1: One-variable data and collection. Practice 1: Formulate questions.
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Question 6
A normally distributed measurement has mean 80 and standard deviation 6. Which interval contains the central 90% of its modeled values?
- [70.13, 89.87]
- [68.24, 91.76]
- [72.31, 87.69]
- [74.00, 86.00]
Review question 6: answer and reasoning
Answer A. Central 90% leaves 5% in each tail. The standardized cutoffs are approximately ±1.644854. Convert to original units: 80±6(1.644854)=80±9.869124, giving [70.13,89.87]. Using ±1.96 instead would give central 95%.
Review focus: 3C: Calculate an interval from a normal probability. Primary unit 2: Probability and distributions. Practice 3: Analyze data.
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Question 7
A frame contains 240 distinct numbered accounts. A team wants a systematic sample of 20. Which complete procedure uses a uniform random start and the appropriate interval?
- Always begin at account 1 and select every twelfth account.
- Choose a uniform start from 1 through 20 and select every twentieth account.
- Choose any convenient first account and then select 19 nearby accounts.
- Choose a uniform integer s from 1 through 12 and select s,s+12,…,s+228.
Review question 7: answer and reasoning
Answer D. The interval is 240/20=12. A uniform start among the first 12 positions followed by 19 increments of 12 selects exactly 20 distinct accounts, with last position s+228≤240. Each account has inclusion probability 1/12. This is systematic sampling, not an SRS of 20: only 12 sample sets are possible. Periodic ordering can make a realized sample unrepresentative even with a uniform start, so the frame order also deserves inspection.
Review focus: Specify an executable systematic sampling mechanism. Primary unit 1: One-variable data and collection. Practice 2: Collect data.
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Question 8
In a population model, P(alert given a faulty item)=0.90 and P(alert given a sound item)=0.10. Only 2% of items are faulty. Which statement is justified?
- An alerted item has a 90% chance of being faulty.
- An alerted item is more likely faulty than sound.
- The 10% false-alert rate means that 10% of all items are sound.
- Among alerted items, about 15.5% are faulty.
Review question 8: answer and reasoning
Answer D. The joint probability of faulty and alert is 0.02(0.90)=0.018. The alert probability is 0.018+0.98(0.10)=0.116. Therefore P(faulty given alert)=0.018/0.116≈0.1552. Reversing the given conditional probability ignores the much larger sound-item group.
Review focus: 4D: Interpret a conditional probability with its denominator. Primary unit 2: Probability and distributions. Practice 4: Interpret results.
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Question 9
A harmless usability experiment has 32 novice and 32 experienced volunteers. Two factors are layout (compact/spacious) and instructions (text/diagram). The researchers want experience balanced across all four treatment combinations. Which allocation achieves that goal while randomizing treatments?
- Assign all novices to compact layouts and all experienced volunteers to spacious layouts.
- Within each experience block, uniformly shuffle the 32 IDs and assign eight to each of compact/text, compact/diagram, spacious/text and spacious/diagram.
- Let every volunteer choose a combination, then call experience a block.
- Randomly select one combination for all 64 volunteers, so everyone receives equal treatment.
Review question 9: answer and reasoning
Answer B. Randomizing separately within each experience block assigns eight novices and eight experienced participants to every combination, for 16 per treatment. All four combinations are represented, and experience is balanced by design. Confounding layout with experience, allowing self-selection or using only one treatment does not meet the stated comparison. This balances a named characteristic; it does not make volunteer recruitment a random population sample.
Review focus: Allocate a replicated two-factor experiment within blocks. Primary unit 1: One-variable data and collection. Practice 2: Collect data.
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Question 10
A score of 65 comes from a normal model with mean 50 and standard deviation 10. Which interpretation is correct?
- About 93.3% of modeled scores exceed 65.
- About 93.3% of modeled scores are at or below 65.
- The score is 1.5% above the mean.
- Every random sample has exactly 6.7% of scores above 65.
Review question 10: answer and reasoning
Answer B. The standardized score is (65−50)/10=1.5, and Φ(1.5)≈0.9332. Thus about 93.3% of the model distribution is below the score. Continuity makes below and at-or-below probabilities equal. A population-model proportion does not force the same fraction in every sample.
Review focus: 4C: Describe relative position within a normal distribution. Primary unit 2: Probability and distributions. Practice 4: Interpret results.
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Question 11
Researchers enroll a random sample of workers, record each worker’s existing shift, and then measure sleep for the next six months. They never change shifts. Which description correctly characterizes the collection plan?
- A randomized experiment, because future outcomes are measured.
- A retrospective experiment, because shifts existed before enrollment.
- A prospective observational study; following outcomes forward does not remove possible confounding in shift selection.
- A census that proves shift type causes any later sleep difference.
Review question 11: answer and reasoning
Answer C. Exposure information is recorded before future follow-up, making this prospective observation. There is no random treatment assignment: workers already have shifts. Job demands, health or other variables could influence both shift selection and sleep. Random sampling can help population representation within the frame, but it does not transform an observational comparison into a randomized causal experiment.
Review focus: Identify what prospective observation does and does not control. Primary unit 1: One-variable data and collection. Practice 2: Collect data.
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Question 12
A box contains exactly 4 red and 8 blue tokens. Eight tokens are selected without replacement, and X counts the red tokens selected. A student models X as Binomial(8,1/3). Which objection is decisive?
- The draws are dependent and their conditional success probabilities change.
- A binomial count cannot include zero.
- A binomial model requires success probability 1/2.
- The number of draws is not fixed.
Review question 12: answer and reasoning
Answer A. After a first red draw, the next-red probability is 3/11 rather than the original 4/12=1/3. Without replacement, the remaining composition changes and the draws are dependent. Sampling 8 of only 12 tokens cannot be treated as a small sampling fraction. The draw count is fixed, and binomial models allow zero and probabilities other than 1/2; those are not the problem.
Review focus: 4B: Justify a probability-model claim using conditional calculations. Primary unit 2: Probability and distributions. Practice 4: Interpret results.
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Question 13
A school wants to compare typical commute duration by transport mode. Which collection plan best serves that aim while limiting unnecessary disclosure?
- Collect commute duration for a defined school day and transport mode through a private voluntary process, restrict record access, and report nonidentifying group summaries.
- Publish every student’s name, precise home address and family income beside the commute time so readers can check identities.
- Ask only students arriving by car, because their responses are easiest to collect.
- Require students to respond publicly in front of the teacher and treat missing answers as zero-minute commutes.
Review question 13: answer and reasoning
Answer A. Duration and mode directly address the question; a defined day improves measurement consistency. Private participation, restricted access and nonidentifying reporting reduce unnecessary risks. The plan still needs a defensible sampling/recruitment strategy and attention to nonresponse. Public identifiers and income are not required for this aim; car-only coverage changes the target, and pressure or coding missing data as zero creates additional problems.
Review focus: Collect necessary information while protecting participants. Primary unit 1: One-variable data and collection. Practice 2: Collect data.
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Question 14
Independent random samples from two large cities record whether each resident supports a proposal. The goal is a confidence interval for the difference in population support proportions. Which procedure matches the design?
- A paired t interval on resident support labels
- A one-proportion z interval after combining both cities
- An unpooled two-proportion z interval, after checking conditions
- A one-sample t interval on all pooled binary responses
Review question 14: answer and reasoning
Answer C. The response is binary, the samples are independent, and the target is p₁−p₂. A two-proportion z interval estimates the groups separately and uses unpooled standard error, with adequate observed success/failure counts and design conditions. Combining cities changes the parameter, and the observations are not paired quantitative differences.
Review focus: 2C: Identify the appropriate inference method. Primary unit 3: Proportions and categorical inference. Practice 2: Collect data.
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Question 15
Seven recorded durations are 4,6,6,8,10,14,18 minutes. The last value is corrected from 18 to 32, with all other values unchanged. What are the new mean and median?
- Mean≈9.429 minutes; median 8 minutes
- Mean≈11.429 minutes; median 10 minutes
- Mean≈11.429 minutes; median 8 minutes
- Mean≈11.429 minutes; median 15 minutes
Review question 15: answer and reasoning
Answer C. The original total is 66. Replacing 18 by 32 adds 14, so the new total is 80 and the mean is 80/7≈11.428571 minutes. With seven ordered values, the median remains the fourth value, 8. The large-end correction raises the mean by 14/7=2 minutes without changing the median; the mean and median measure different aspects of center.
Review focus: Compute and compare the sensitivity of center summaries. Primary unit 1: One-variable data and collection. Practice 3: Analyze data.
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Question 16
Let p_A and p_B be population approval proportions for two independent groups. Before collecting data, a researcher asks whether group A has a larger approval proportion. Which hypotheses express that question?
- H₀: p_A−p_B=0; Hₐ: p_A−p_B>0
- H₀: p̂_A−p̂_B=0; Hₐ: p̂_A−p̂_B>0
- H₀: p_A−p_B>0; Hₐ: p_A−p_B=0
- H₀: p_A−p_B=0; Hₐ: p_A−p_B≠0
Review question 16: answer and reasoning
Answer A. Hypotheses concern population proportions, not the sample estimates. Equality is the null and the prespecified directional question gives a positive difference alternative. A two-sided alternative would ask a different question.
Review focus: 2E: Specify population hypotheses. Primary unit 3: Proportions and categorical inference. Practice 2: Collect data.
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Question 17
A dataset has the five-number summary shown. Using the 1.5×IQR rule, which statement follows?
- IQR=7; the stated minimum and maximum are both inside the outlier fences.
- IQR=20; the maximum is an outlier.
- IQR=7;22 exceeds the upper fence of 21.
- IQR=11; the minimum is an outlier.
Review question 17: answer and reasoning
Answer A. IQR=Q 3−Q 1=15−8=7. The fences are 8−1.5·7=−2.5 and 15+1.5·7=25.5. The minimum 2 and maximum 22 are inside, so no observed value falls beyond these fences. This rule does not say the data are normal or that no observation could be substantively unusual; it supplies a specific numerical flag.
Review focus: Apply an interquartile-range outlier rule to supplied summaries. Primary unit 1: One-variable data and collection. Practice 3: Analyze data.
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Question 18
A test uses H₀: p=0.08 versus Hₐ: p>0.08, where p is a factory’s population defect proportion. What is a Type I error?
- Failing to find evidence of an increase when p is 0.12
- Obtaining a sample defect proportion different from 0.08
- Concluding that the null has probability 0.08
- Concluding that the defect proportion exceeds 0.08 when it is actually 0.08
Review question 18: answer and reasoning
Answer D. A Type I error rejects a true null. Here it means concluding there is evidence of a defect proportion above 0.08 even though the population value is 0.08. Failing to detect a real increase is a Type II error.
Review focus: 2D: Identify a Type I error in context. Primary unit 3: Proportions and categorical inference. Practice 2: Collect data.
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Question 19
A workplace invites all 300 employees to a confidential survey, and 180 respond. Of the respondents, 72% report satisfaction. Which conclusion is best supported without further information?
- Exactly 72% of all 300 employees are satisfied because everyone was invited.
- The result is unbiased because 180 is more than half the workforce.
- The other 120 employees must be dissatisfied.
- About 72% of respondents reported satisfaction; nonresponse could make this differ from the workforce proportion.
Review question 19: answer and reasoning
Answer D. The percentage describes respondents. Inviting everyone attempts a census, but it does not obtain complete measurement when 120 responses are missing. Satisfaction may relate to response propensity, so the direction and size of nonresponse error are unknown. A high response count does not by itself establish unbiasedness, and missing answers cannot be classified as dissatisfied without evidence.
Review focus: Interpret nonresponse in an attempted census. Primary unit 1: One-variable data and collection. Practice 4: Interpret results.
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Question 20
For planning a one-proportion interval, a researcher fixes the confidence level and the planning value of p. How must the sample size change to halve the planned margin of error?
- Double it
- Quadruple it
- Halve it
- Multiply it by the confidence level
Review question 20: answer and reasoning
Answer B. The planned margin is z*√(p(1−p)/n), proportional to 1/√n when the other quantities are held fixed. Replacing n by 4n divides the margin by 2. This planning relation does not repair sampling bias or guarantee a particular realized p̂.
Review focus: 2D: Relate sample size to interval precision. Primary unit 3: Proportions and categorical inference. Practice 2: Collect data.
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Question 21
The graph and table show monthly service requests. Which statement correctly describes the change from month 2 to month 6 without claiming a cause?
Graph description: Connected monthly counts at(1,12),(2,18),(3,15),(4,24),(5,21),(6,30); lines show chronological order, not observations between months. Observed requests (solid): straight segments through (1, 12), (2, 18), (3, 15), (4, 24), (5, 21), (6, 30). Axes use the labeled scales; no values outside the specified domain are supplied.
- Requests rose by 40%, proving that the service improved.
- Requests rose by about 66.7%; the graph alone does not establish why.
- Requests rose by 12%, because 30−18=12.
- Requests doubled, and every intervening month had more requests than the previous month.
Review question 21: answer and reasoning
Answer B. The increase is 30−18=12 requests relative to the starting value 18, giving 12/18≈0.6667 or 66.7%. Dividing by the final 30 would answer a different relative-change question. The series also includes declines, and the graph does not identify a cause for the overall increase. Chronological association alone is not a randomized comparison.
Review focus: Interpret a time graph with the correct percentage-change baseline. Primary unit 1: One-variable data and collection. Practice 4: Interpret results.
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Question 22
In a random sample of 120 people from a much larger population, 72 have a specified preference. Assuming conditions hold, which is the approximate 95% one-proportion z interval?
- [0.424, 0.776]
- [0.550, 0.650]
- [0.512, 0.688]
- [0.438, 0.562]
Review question 22: answer and reasoning
Answer C. The estimate is p̂=72/120=0.60. Its estimated SE is √(0.6·0.4/120)=0.04472136. The 95% margin is 1.959964 times this, about 0.0876523, giving [0.51235,0.68765]. The interval uses observed counts, which are 72 and 48.
Review focus: 3C: Calculate a one-proportion confidence interval. Primary unit 3: Proportions and categorical inference. Practice 3: Analyze data.
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Question 23
A study predicts weekly spending in dollars from hours worked, using ŷ=12+4 x with correlation r=0.60. The same spending values are converted to cents, while hours remain unchanged. Which statement is correct?
- The new slope is 4 cents per hour and r=0.006.
- The new slope is 400 cents per hour and r=0.60.
- The new slope is 400 cents per hour and r=60.
- The new slope is 0.04 cents per hour and r=0.60.
Review question 23: answer and reasoning
Answer B. Multiplying every response by 100 multiplies the intercept and slope by 100: predicted cents=1200+400 x. The slope is 400 cents per hour. Correlation is unitless and unchanged by a positive linear rescaling of either variable, so r stays 0.60. A negative rescaling would reverse its sign, but that is not the conversion here.
Review focus: Interpret positive unit changes in regression and correlation. Primary unit 5: Regression. Practice 4: Interpret results.
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Question 24
Independent samples have 65 successes out of 100 in group 1 and 45 out of 100 in group 2. For H₀: p₁=p₂, what is the pooled-null z statistic for group 1 minus group 2?
- 2.843
- 2.000
- 2.907
- 0.200
Review question 24: answer and reasoning
Answer A. The pooled estimate is (65+45)/200=0.55. Under equality, SE=√[0.55·0.45·(1/100+1/100)]=√0.00495≈0.0703562. Thus z=(0.65−0.45)/0.0703562≈2.84268. The interval’s unpooled SE is not the specified null-test SE.
Review focus: 3E: Calculate a pooled-null two-proportion statistic. Primary unit 3: Proportions and categorical inference. Practice 3: Analyze data.
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Question 25
For paired data, x has mean 12 and sample SD 4 hours, y has mean 70 and sample SD 15 points, and r=0.60. What does the least-squares line predict about a 4-hour increase in x?
- Every individual’s score must rise by 0.60 points.
- The predicted score rises by 15 points because that is the response SD.
- The predicted score rises by 9 points; this describes the fitted change, not a guaranteed individual or causal effect.
- The predicted score rises by 2.25 points, regardless of the size of the increase in x.
Review question 25: answer and reasoning
Answer C. The slope is r·sy/sx=0.60·15/4=2.25 points per hour. A 4-hour increase therefore changes the fitted response by 4·2.25=9 points. Equivalently, one predictor SD corresponds to r response SDs, 0.60·15=9. The fitted comparison does not guarantee an individual change or establish a causal effect; those require additional evidence.
Review focus: Interpret the regression slope obtained from standardized association. Primary unit 5: Regression. Practice 4: Interpret results.
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Question 26
Two independently sampled populations produce the response counts in the table. For a chi-square test of homogeneity, which pair gives the Pearson statistic and degrees of freedom?
- χ²=1.2, df=2
- χ²=2.4, df=5
- χ²=4.8, df=2
- χ²=2.4, df=2
Review question 26: answer and reasoning
Answer D. Each row totals 60; column totals are 30,30,60, with grand total 120. Each expected row is therefore 15,15,30. The four nonzero contributions are each 9/15=0.6, totaling 2.4. Degrees of freedom are (2−1)(3−1)=2. All expected counts exceed 5.
Review focus: 3E: Calculate a contingency-table test result. Primary unit 3: Proportions and categorical inference. Practice 3: Analyze data.
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Question 27
Across 20 towns, researchers find a strong positive correlation between the town’s average income and its average annual library visits per resident. Which conclusion is justified by this result alone?
- In every town, higher-income individuals visit the library more often than lower-income individuals.
- Raising an individual’s income will cause more library visits.
- The correlation must remain the same if individual residents replace towns as observational units.
- Among these towns, larger average incomes tend to occur with larger average visits; individual relationships and causal effects require other evidence.
Review question 27: answer and reasoning
Answer D. The observational units are towns, so the reported association concerns town averages. Relationships among aggregates need not equal relationships among individuals within towns. Resources, access, age composition and other variables could also affect both averages. Neither an individual-level association nor causality follows from the town correlation alone.
Review focus: Limit a correlation conclusion to the observed units and design. Primary unit 5: Regression. Practice 4: Interpret results.
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Question 28
A valid 95% confidence interval for p₁−p₂ is [−0.02,0.14]. Which conclusion is supported by this interval?
- There is a 95% probability that the fixed population difference is zero.
- Zero is compatible with the interval, but equality has not been established.
- The two population proportions are equal.
- Exactly 95% of individual responses lie between −0.02 and 0.14.
Review question 28: answer and reasoning
Answer B. The interval contains zero and both negative and positive differences. It does not establish a nonzero difference using the corresponding two-sided interval criterion, but it also does not prove equality or equivalence. Confidence describes the long-run coverage of the procedure, not a probability that the fixed parameter equals zero.
Review focus: 4F: Interpret an interval for a difference. Primary unit 3: Proportions and categorical inference. Practice 4: Interpret results.
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Question 29
A random sample of 80 independent binary responses contains 8 successes. A student considers a Wald confidence interval and a z test of H₀: p=0.20. Using the rule that both relevant expected or observed counts should be at least 10, which assessment is correct?
- Both count checks pass because n is greater than 30.
- Both fail because the observed success count is 8.
- The interval count check fails, while the null-test count check passes.
- The interval count check passes, while the null-test count check fails.
Review question 29: answer and reasoning
Answer C. The interval uses observed counts 8 and 72, so its success count fails the stated rule. The test uses null expected counts 80(0.20)=16 and 80(0.80)=64, so that count check passes. Passing these numerical checks would not by itself establish random sampling or independence.
Review focus: 4E: Distinguish interval and test count conditions. Primary unit 3: Proportions and categorical inference. Practice 4: Interpret results.
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Question 30
Twenty participants each complete a task before and after a training session. The target is the population mean within-person change. Which analysis uses the design appropriately?
- Treat the 40 measurements as one independent sample
- Use the 20 after-minus-before differences in a one-sample t procedure
- Use an independent two-sample t procedure on the two time points
- Subtract the before and after sample SDs to obtain the SD of change
Review question 30: answer and reasoning
Answer B. Each person supplies a matched pair. Form 20 after-minus-before differences and use their mean and sample SD for paired t inference, checking the distribution of differences and independence across people. The two measurements within one person are not independent samples, and the SD of differences is not the difference of marginal SDs.
Review focus: 2C: Select paired-mean inference from the design. Primary unit 4: Means and quantitative inference. Practice 2: Collect data.
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Question 31
Separate random samples from two large populations record a quantitative response. The sample sizes and sample SDs differ, and equal population variances are not assumed. The target is μ₁−μ₂. Which method is appropriate if shape and design conditions are reasonable?
- A paired t procedure after sorting each sample
- A two-proportion z procedure
- A one-sample t procedure on all pooled measurements
- A two-sample Welch t procedure
Review question 31: answer and reasoning
Answer D. Welch t inference uses independent quantitative samples and allows unequal population variances. Its SE is √(s₁²/n₁+s₂²/n₂), with approximate, potentially fractional degrees of freedom. Sorting does not create genuine pairs, and pooling all measurements changes the target parameter.
Review focus: 2C: Select Welch inference for independent quantitative groups. Primary unit 4: Means and quantitative inference. Practice 2: Collect data.
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Question 32
For paired measurements, let d be after minus before, and let μ_d be the population mean difference. The prespecified question asks whether there is any average change. Which hypotheses are correct?
- H₀: μ_d=0; Hₐ: μ_d≠0
- H₀: d̄=0; Hₐ: d̄≠0
- H₀: μ_d=0; Hₐ: μ_d>0
- H₀: μ_d≠0; Hₐ: μ_d=0
Review question 32: answer and reasoning
Answer A. The parameter is the population mean of the defined differences. Any change makes the alternative two-sided, while the null is zero mean change. The observed sample mean is evidence used to evaluate the hypotheses, not the parameter named in them.
Review focus: 2E: State hypotheses for a paired population mean. Primary unit 4: Means and quantitative inference. Practice 2: Collect data.
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Question 33
Group 1 has n₁=10, x̄₁=14 and s₁=4. Independent group 2 has n₂=15, x̄₂=10 and s₂=3. For a Welch test of H₀: μ₁−μ₂=0, what is the test statistic?
- 1.818
- 4.000
- 2.697
- 0.741
Review question 33: answer and reasoning
Answer C. The estimated difference is 14−10=4. The Welch SE is √(4²/10+3²/15)=√(1.6+0.6)=√2.2≈1.4832397. Thus t=4/√2.2≈2.69680. Dividing by the variance 2.2 instead of its square root gives the distractor 1.818.
Review focus: 3E: Calculate an independent-means test statistic. Primary unit 4: Means and quantitative inference. Practice 3: Analyze data.
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Question 34
A valid 90% t interval for a population mean service time is [11.2,13.8] minutes. Which is the appropriate confidence interpretation?
- 90% of individual service times lie in this interval.
- There is a 90% probability that the next service takes 11.2–13.8 minutes.
- Exactly 90% of all possible sample means lie in this particular interval.
- The procedure used would cover the fixed population mean in about 90% of repeated samples under its conditions.
Review question 34: answer and reasoning
Answer D. Confidence is a repeated-sampling coverage property of the interval procedure. It concerns the population mean, not individual times or a next-observation prediction. The particular computed interval either covers the fixed mean or does not; the frequentist procedure supplies the stated coverage rate.
Review focus: 4F: Interpret a mean confidence interval. Primary unit 4: Means and quantitative inference. Practice 4: Interpret results.
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Question 35
Independent random samples compare mean commute times for residents who choose two different travel modes. A valid two-sided Welch test gives p=0.012. At a prespecified α=0.05, which conclusion is justified?
- Choosing the first mode causes a different mean commute time.
- There is evidence of different population means, but mode choice was not randomized, so causation is not established.
- The probability that the two population means are equal is 0.012.
- The practical size of the difference must be important because p is below 0.05.
Review question 35: answer and reasoning
Answer B. Since 0.012<0.05, reject the equal-means null under the model and report evidence of a population mean difference. Random sampling can support population scope, but observational travel-mode choice leaves possible confounding and does not establish a causal effect. A p-value is neither a posterior probability of equality nor a measure of practical importance.
Review focus: 4G: Justify a contextual inference without overclaiming. Primary unit 4: Means and quantitative inference. Practice 4: Interpret results.
Item stat-q-24
Question 36
A random sample of five independent repair times is 4,5,5,6 and 48 minutes. A student says that using t instead of z automatically makes a mean interval reliable. Which response is best?
- The small sample has a severe outlier and skew; using t does not automatically justify the normal-population approximation.
- The interval is justified because the sample SD can always be calculated.
- The interval is justified because five observations are independent.
- Delete 48 without investigation, then use the remaining observations as an unchanged random sample.
Review question 36: answer and reasoning
Answer A. Random sampling and independence do not remove distribution-shape concerns. With only five observations and a strong high outlier, a t interval lacks the usual normal-population justification and small-sample robustness. A computable SD is not sufficient, and deleting an inconvenient observation without investigating its validity changes the evidence and can bias inference.
Review focus: 4E: Evaluate small-sample t conditions. Primary unit 4: Means and quantitative inference. Practice 4: Interpret results.
Item stat-q-25
