Study aid only. Verify against current guidelines before clinical use.
Free USMLE Step 1 Biostatistics and Epidemiology practice questions
10 original exam-style questions from the StepStash Step 1 qbank, each with a full explanation, an educational objective and a link to the matching study note. Try each one before you open the answer. Reviewed by Veli Bakalov, MD · @HemeOncBuddy.
A phase III, randomized, double-blind trial enrolls 12,000 adults with atrial fibrillation and a low baseline risk of stroke. Participants are assigned to a new oral anticoagulant or placebo and followed for 3 years. A press release states that the new drug "reduces stroke by 25%." In the trial, stroke occurred in 4.0% of participants in the placebo group and 3.0% of participants in the drug group.
How many patients would need to be treated with the new drug for 3 years to prevent one stroke?
4
25
33
75
100
Show answer and explanation
Correct answer: E. 100
The number needed to treat (NNT) is the reciprocal of the absolute risk reduction (ARR). The ARR is the control event rate minus the treatment event rate: 4.0% minus 3.0% = 1.0%, or 0.01. The NNT is therefore 1/0.01 = 100: 100 patients must take the drug for 3 years to prevent one stroke. The press release reports the relative risk reduction (RRR), ARR divided by the control rate (0.01/0.04 = 25%). The RRR sounds impressive but hides how small the absolute benefit is when the baseline risk is low; the same 25% RRR in a high-risk population with a 20% event rate would give an ARR of 5% and an NNT of 20. The NNT always refers to a specific outcome and time period, and it should be weighed against the number needed to harm (for an anticoagulant, major bleeding).
A. 4 is the reciprocal of the RRR expressed as a fraction (1/0.25), a common error that confuses relative with absolute benefit.
B. 25 is the RRR as a percentage, not a number of patients.
C. 33 is the reciprocal of the treatment-group event rate (1/0.03), which has no meaning as an NNT.
D. 75 is the percentage of residual risk (the relative risk of 0.75 multiplied by 100), not an NNT.
Educational objective: The key takeaway is that NNT = 1/ARR, and a large relative risk reduction can correspond to a small absolute benefit when the baseline event rate is low.
Both drugs halve the event rate (RRR 50%), but when the event is rare the absolute benefit is small and the NNT is 100 instead of 20.
Investigators at four pediatric cancer centers identify 120 children who were diagnosed with hepatoblastoma, a rare liver tumor, over a 10-year period. They also select 360 children of similar age seen at the same centers for minor injuries who have never had cancer. Trained interviewers ask the mothers of all children about smoking during pregnancy, and birth weights are obtained from hospital records. The investigators calculate an odds ratio for the association between maternal smoking and hepatoblastoma.
Which of the following best describes this study design?
Prospective cohort study
Retrospective cohort study
Cross-sectional study
Randomized controlled trial
Case-control study
Show answer and explanation
Correct answer: E. Case-control study
The design is named by where the study starts and in which direction it looks. These investigators began with disease status, selecting children with hepatoblastoma (cases) and children without cancer (controls), and then looked back at prior exposures such as maternal smoking. That is a case-control study. Its measure of association is the odds ratio (ad/bc from the 2 x 2 table); because the investigators chose the numbers of cases and controls, incidence and relative risk cannot be calculated. When the disease is rare, as hepatoblastoma is, the odds ratio approximates the relative risk. This design is efficient for rare diseases, but it is prone to recall bias (mothers of affected children may report exposures differently) and to selection bias in choosing controls.
A. A prospective cohort would enroll pregnant women by smoking status and follow their children forward for new tumors, which would be impractical for so rare a disease.
B. A retrospective cohort uses old records to define groups by exposure and then traces outcomes; here groups were defined by disease.
C. A cross-sectional study measures exposure and disease at a single point in time in a sample of a population and yields prevalence.
D. In a randomized controlled trial the investigator assigns the exposure, which is impossible and unethical for maternal smoking.
Educational objective: The key takeaway is that a study that starts from people with and without a disease and looks back at exposures is a case-control study, which reports an odds ratio and suits rare diseases.
Name the design by who assigned the exposure and by where the study started: exposure groups followed forward make a cohort, even from old records; disease groups looked back on make a case-control study.
A randomized, double-blind trial of 400 adults with chronic low back pain finds that an oral herbal extract reduces pain scores more than placebo (p = 0.04, with alpha set at 0.05). The trial had adequate allocation concealment and less than 5% loss to follow-up. Over the next 5 years, three larger, well-conducted randomized trials enrolling more than 6,000 patients find no difference between the extract and placebo, and a meta-analysis concludes that the extract has no effect.
Assuming the later trials are correct, the conclusion of the first trial most likely represents which of the following?
Type I error
Type II error
Confounding
Selection bias
Effect modification
Show answer and explanation
Correct answer: A. Type I error
A type I error occurs when a study rejects a null hypothesis that is actually true, concluding that a difference exists when it does not: a false positive. Its probability is alpha, the significance level chosen before the study. With alpha set at 0.05, about 1 in 20 trials of a completely ineffective treatment will reach p below 0.05 by chance alone. Here the first trial declared the extract effective, but much larger, well-conducted trials and a meta-analysis show no true effect, so the original result was most likely a chance false positive. The quality features described (blinding, concealment, minimal attrition) argue against bias and point to random error.
B, type II error, is the opposite mistake: failing to reject a false null and missing a real effect (a false negative), usually because of an underpowered study; the first trial did not miss an effect, it found one that does not exist. C, confounding, is a distortion by a third variable related to both exposure and outcome; randomization balances known and unknown confounders, making it an unlikely explanation. D, selection bias, is a systematic error in who enters or stays in a study, and the stem describes concealed allocation and minimal loss to follow-up. E, effect modification, is a true difference in effect across subgroups, not a spurious overall result.
Educational objective: The key takeaway is that a type I error is a false positive, rejecting a true null hypothesis, and occurs with probability alpha.
Moving the cutoff trades α against β; only a larger sample, a larger effect or less variance shrinks β without adding false positives.
A study evaluates three serum biomarkers for distinguishing bacterial from viral pneumonia in hospitalized adults. For each biomarker, the investigators plot sensitivity against 1 - specificity across all possible cutoff values. The areas under the resulting curves are 0.93 for biomarker X, 0.74 for biomarker Y and 0.50 for biomarker Z.
Which of the following is the best interpretation of the result for biomarker Z?
It has 50% sensitivity at every cutoff
It discriminates no better than chance
It has a positive predictive value of 50%
It is useful only for ruling out disease
It correctly classifies half of the patients only at the best cutoff
Show answer and explanation
Correct answer: B. It discriminates no better than chance
The investigators constructed receiver operating characteristic (ROC) curves, which plot the true-positive rate (sensitivity) against the false-positive rate (1 - specificity) as the cutoff moves. The area under the curve (AUC) summarizes overall discrimination across all cutoffs and equals the probability that a randomly chosen patient with bacterial pneumonia has a more abnormal value than a randomly chosen patient with viral pneumonia. An AUC of 1.0 is perfect discrimination. An AUC of 0.50 means the curve lies on the diagonal: at every cutoff, the true-positive rate equals the false-positive rate, so a positive result is no more likely in diseased than nondiseased patients (LR = 1). Biomarker Z therefore performs no better than a coin flip. Biomarker X, with the curve bowing toward the upper-left corner, discriminates best.
A. On the diagonal, sensitivity varies from 0 to 100% as the cutoff changes; it simply always equals 1 - specificity. C. PPV depends on prevalence and cannot be read from the AUC. D. A test can rule out disease only if a negative result is much less likely in diseased patients (low LR-), which a test on the diagonal cannot achieve. E. The AUC is not the accuracy at a single cutoff; it is a summary over all cutoffs.
Educational objective: The key takeaway is that an AUC of 0.50 places the ROC curve on the diagonal, meaning the test discriminates no better than chance, whereas an AUC of 1.0 is perfect.
Sliding the cutoff down removes false negatives but adds false positives (sensitivity up, specificity down), and raising it does the reverse. Each cutoff is one point on the ROC curve, and the area under the curve measures overall discrimination.
Investigators study whether long-term use of a proton pump inhibitor is associated with community-acquired Clostridioides difficile infection. They identify 100 adults with a first episode of the infection and 100 adults without it, matched for age and recruited from the same primary care practices. Pharmacy records show that 60 of the patients with infection and 30 of the patients without infection had filled proton pump inhibitor prescriptions during the prior year.
Which of the following is the odds ratio for the association between proton pump inhibitor use and infection?
0.29
0.43
1.5
2.0
3.5
Show answer and explanation
Correct answer: E. 3.5
This is a case-control study: subjects were selected by outcome (infection vs no infection) and exposure was measured backward. Because the investigators chose how many cases and controls to enroll, incidence cannot be calculated, so the correct measure is the odds ratio.
Set up the 2 x 2 table with exposure in rows: a = exposed cases = 60, b = exposed controls = 30, c = unexposed cases = 40, d = unexposed controls = 70. OR = ad/bc = (60 x 70)/(30 x 40) = 4200/1200 = 3.5. Equivalently, the odds of exposure among cases (60/40 = 1.5) divided by the odds of exposure among controls (30/70 = 0.43) is 3.5. Proton pump inhibitor users had 3.5 times the odds of infection. Because community-acquired infection is uncommon, this OR approximates the relative risk under the rare disease assumption.
A (0.29) is the inverted ratio, bc/ad, which would wrongly suggest protection. B (0.43) is only the odds of exposure among controls (30/70). C (1.5) is only the odds of exposure among cases (60/40), without dividing by the control odds. D (2.0) divides the proportion of exposed cases (60%) by the proportion of exposed controls (30%); a ratio of exposure proportions is neither a relative risk nor the OR.
Educational objective: The key takeaway is that a case-control study reports the odds ratio, calculated as ad/bc from the 2 x 2 table.
Cohort studies and RCTs compare risks across the rows (RR); case-control studies compare odds of exposure down the columns (OR = ad/bc), which approximates the RR only when the disease is rare.
Investigators review records of 2,400 patients who underwent resection of stage II or III colon cancer. Patients who received adjuvant chemotherapy were younger, had better performance status, and more often had stage III disease than those who did not. Follow-up ranges from 4 months to 9 years, and many patients are alive at last contact. A log-rank test shows that overall survival differs between the groups (p = 0.01). The investigators now want an estimate of the effect of chemotherapy on the rate of death that accounts for age, stage, and performance status and uses all follow-up time.
Which of the following analytic methods is most appropriate?
Cox proportional hazards regression
Multiple linear regression
Multivariable logistic regression
Spearman rank correlation
Log-rank test
Show answer and explanation
Correct answer: A. Cox proportional hazards regression
The outcome is time to an event (death) with censoring: patients were followed for different lengths of time and many were alive at last contact. The investigators also need to adjust for several confounders at once. Cox proportional hazards regression handles both requirements. It models the hazard, the instantaneous death rate among patients still at risk, as a function of several covariates (chemotherapy, age, stage, performance status) and reports an adjusted hazard ratio for each. It is semiparametric: it does not require specifying the shape of the baseline hazard, but it does assume that hazard ratios stay constant over time. Like any regression, it adjusts only for measured confounders.
B. Multiple linear regression is for a continuous outcome such as blood pressure, and it cannot handle censored survival times.
C. Multivariable logistic regression models a binary outcome (dead or alive at a fixed time) and gives an adjusted odds ratio. It discards information on when deaths occurred and handles patients censored before that time point poorly.
D. Spearman correlation measures the strength of a monotonic association between two ranked variables; it does not compare survival or adjust for covariates.
E. The log-rank test has already been done. It compares whole survival curves and returns a single p-value, but it is unadjusted and gives no effect size.
Educational objective: The key takeaway is that a covariate-adjusted comparison of censored time-to-event data calls for Cox proportional hazards regression, which yields adjusted hazard ratios, whereas the log-rank test is unadjusted.
Read events as steps and censored patients as ticks, take the median where a curve crosses 0.5, and trust the late tail less because few patients remain at risk.
Case-control studies in several countries found that mothers of infants with spina bifida and anencephaly had lower periconceptional folate intake than mothers of unaffected infants. A national public health agency then required folic acid fortification of enriched cereal grain products. Surveillance of births before and after the policy showed that the prevalence of neural tube defects fell by about one third within a few years of fortification, with no comparable change in other major birth defects.
The change in prevalence after fortification most directly supports which of the following Bradford Hill criteria?
Analogy
Experiment
Coherence
Specificity
Temporality
Show answer and explanation
Correct answer: B. Experiment
The experiment criterion asks whether changing the exposure changes the risk of the outcome. Evidence can come from randomized trials, from people who remove an exposure themselves (lung cancer risk declines progressively after quitting smoking), or from public health interventions. Here, low folate was the suspected cause, the population's folate intake was deliberately raised by fortification, and neural tube defects fell afterward. Folate supplies one-carbon units for nucleotide synthesis, which the rapidly dividing neuroectoderm needs when the neural tube closes during week 4 of development, before many women know they are pregnant; this is why fortification of a staple food works better than relying on supplements started after pregnancy is recognized. Showing that risk falls when an exposure is altered is among the most persuasive kinds of causal evidence, because it mimics the logic of a trial.
A. Analogy means similar exposures cause similar effects, as when thalidomide made other drugs in pregnancy suspect; the data here concern folate itself. C. Coherence means the association fits known natural history and biology, such as population trends following exposure trends; the key observation is the response to a deliberate intervention. D. Specificity means one exposure causes one disease; it is the weakest criterion, and an unchanged rate of other defects does not show that folate deficiency causes only neural tube defects. E. Temporality, exposure preceding outcome, is the only essential criterion and is satisfied here, but timing alone is not what these data add; the distinctive evidence is that deliberately changing the exposure changed the risk.
Educational objective: The key takeaway is that the Bradford Hill experiment criterion is met when removing or reducing an exposure lowers disease risk, as folic acid fortification lowered neural tube defects.
Exclude chance, bias, confounding and reverse causality before calling an association causal; among the Bradford Hill criteria only temporality is essential.
Question 8 · Interpreting Drug Advertisements · medium
A randomized, double-blind trial compares a new drug with placebo in 6,000 adults with heart failure and preserved ejection fraction. The prespecified primary endpoint, cardiovascular death or heart failure hospitalization over 3 years, shows a hazard ratio of 0.94 (95% CI, 0.83 to 1.06; p = 0.31). Among 14 subgroup analyses, women have a hazard ratio of 0.72 (95% CI, 0.58 to 0.90). An advertisement later states, "Proven to reduce cardiovascular death or hospitalization by 28% in women with heart failure."
Which of the following is the most accurate interpretation of the subgroup finding?
It establishes that the drug is effective in women
It shows that the drug is harmful in men
It invalidates the result of the primary endpoint
It shows that the drug reduces the primary endpoint in the overall population
It is hypothesis-generating and requires confirmation in a new trial
Show answer and explanation
Correct answer: E. It is hypothesis-generating and requires confirmation in a new trial
The trial's sample size and alpha (usually 0.05) protect only the prespecified primary endpoint. That endpoint failed: the hazard ratio of 0.94 has a CI including 1. Once the primary analysis is negative, any positive secondary endpoint, subgroup or post hoc finding is hypothesis-generating, not confirmatory. With 14 subgroups tested at 0.05, the chance that at least one appears significant by chance alone is about 1 minus 0.95 to the 14th power, or roughly 50%. A Bonferroni threshold would be 0.05 / 14, about 0.0036. A credible subgroup effect should be prespecified, few in number, supported by a significant test of interaction (showing the effect truly differs between women and men), biologically plausible and replicated. The advertisement's "proven" headline built on a subgroup after a failed primary endpoint is a classic red flag.
A is wrong because a single subgroup result from many comparisons after a failed primary endpoint cannot establish efficacy. B is wrong because a smaller or absent benefit in men is not evidence of harm, and no harm data are given. C is wrong because the primary result is the trial's valid, prespecified answer; subgroup findings do not overturn it. D is wrong because the overall hazard ratio of 0.94 is not significant.
Educational objective: The key takeaway is that when the prespecified primary endpoint fails, a positive subgroup or secondary result is only hypothesis-generating and must be confirmed in a new trial.
Eliminate options that stretch the result past the trial: to excluded patients, unmeasured hard outcomes, other drugs in the class, a better-than-standard claim from a placebo trial, or longer use than studied.
A health system reviews its annual low-dose CT lung cancer screening program. Tumors detected on scheduled screens are mostly small, low-grade adenocarcinomas with low proliferation indices. Tumors diagnosed in the intervals between screens, after patients developed symptoms, are mostly high-grade, rapidly growing carcinomas. Even after the investigators statistically correct for the earlier date of diagnosis, 5-year survival is much better for screen-detected cancers than for interval cancers.
Which of the following best explains the survival difference between screen-detected and interval cancers?
Lead-time bias
Hawthorne effect
Recall bias
Berkson bias
Length-time bias
Show answer and explanation
Correct answer: E. Length-time bias
Length-time bias arises because slowly progressive tumors have a long detectable preclinical phase, so a periodic screening test is much more likely to catch them. Aggressive tumors have a short preclinical window; they grow from undetectable to symptomatic between scheduled screens and appear as interval cancers. The screen-detected group is therefore enriched for indolent, biologically favorable disease, and its better survival reflects tumor biology rather than any benefit of early detection. The extreme form is overdiagnosis, finding cancers that would never have caused symptoms in the patient's lifetime. The proper way to judge a screening program is by disease-specific mortality in a randomized comparison of screened and unscreened populations.
A, lead-time bias, makes survival look longer because the clock starts earlier at diagnosis even though death is not delayed; the stem states that survival remained better after correcting for the earlier diagnosis date, and the clue here is the difference in tumor grade and growth rate. B, the Hawthorne effect, is a change in subjects' behavior because they know they are being observed. C, recall bias, is differential memory of past exposures in a case-control study. D, Berkson bias, is a selection bias from using hospitalized controls in a case-control study.
Educational objective: The key takeaway is that length-time bias occurs because screening preferentially detects slow-growing disease with a long preclinical phase, inflating the apparent survival of screen-detected cases.
Lead time: same death date, earlier diagnosis, so judge screening by mortality. Length time: the screen finds the slow, forgiving tumors; overdiagnosis is its extreme form.
A randomized, double-blind trial compares a proposed biosimilar anti-TNF monoclonal antibody with its reference product in 600 adults with rheumatoid arthritis. The prespecified equivalence margin for the difference in 24-week response rates (biosimilar minus reference) is plus or minus 10 percentage points. Response occurs in 68% of the biosimilar arm and 64% of the reference arm, a difference of 4 percentage points (95% CI, minus 3 to plus 11 percentage points). Intention-to-treat and per-protocol analyses agree.
Which of the following is the most accurate conclusion?
Equivalence is shown because the point estimate lies within the margin
The biosimilar is superior to the reference product
The biosimilar is inferior to the reference product
Noninferiority is shown, but equivalence is not
Neither noninferiority nor equivalence can be concluded
Show answer and explanation
Correct answer: D. Noninferiority is shown, but equivalence is not
Each design has its own success criterion, judged from the confidence interval against the prespecified margin (delta), not from the point estimate or the p-value.
Equivalence requires the entire CI to lie between minus delta and plus delta, showing the new product is neither meaningfully worse nor meaningfully better. Here the lower limit (minus 3) is inside minus 10, but the upper limit (plus 11) extends beyond plus 10, so a clinically meaningful difference in favor of the biosimilar cannot be excluded and equivalence is not established.
Noninferiority requires only that the limit on the unfavorable side stay within the margin. The unfavorable limit is minus 3, well inside minus 10, so the biosimilar is not worse than the reference by more than the acceptable margin. Agreement of intention-to-treat and per-protocol results supports this, because nonadherence pulls ITT toward the null and is anti-conservative in noninferiority and equivalence trials.
Superiority would require the CI to exclude 0, the no-difference value for a difference; this CI includes 0, so superiority is not shown.
A is wrong because equivalence depends on the whole interval, not the point estimate. B is wrong because the CI crosses 0. C is wrong because the entire interval, including the unfavorable limit, lies above minus 10. E is wrong because the noninferiority criterion is clearly met.
Educational objective: The key takeaway is that equivalence requires the whole confidence interval within plus or minus the margin, whereas noninferiority requires only the unfavorable limit within the margin.
Superiority needs the CI to exclude no effect; noninferiority needs the unfavorable limit inside the margin; equivalence needs the whole CI within plus or minus the margin.