Q1. Sensitivity is calculated as: A. True positives / (true positives + false positives) B. True positives / (true positives + false negatives) C. True negatives / (true negatives + false positives) D. True negatives / (true negatives + false negatives) ANSWER: B — TP / (TP + FN). Why: sensitivity asks, of everyone who HAS the disease, what proportion the test detected. That is the diseased column of the 2×2 table. A: positive predictive value. C: specificity. LEARNING POINT: read DOWN the columns for sensitivity and specificity.

Q2. Which study design starts with the OUTCOME and looks backward at exposure? A. Cohort B. Case-control C. Cross-sectional D. Randomised trial ANSWER: B — case-control. Why: cases (with the outcome) and controls (without) are selected first, then past exposure is compared. It yields an odds ratio and suits rare diseases, but is prone to recall bias. LEARNING POINT: cohort looks FORWARD from exposure; case-control looks BACKWARD from outcome.

Q3. A test with 99% sensitivity and 95% specificity is used to screen a population where the disease affects 1 in 2000. A patient tests positive. The most accurate statement is: A. He almost certainly has the disease B. Most positives in this setting will be FALSE positives; confirmation is needed C. The test must be faulty D. Specificity should be recalculated ANSWER: B. Why: a 5% false-positive rate applied to a large healthy population generates far more false positives than there are true cases at this prevalence. PPV is therefore low despite good test characteristics. LEARNING POINT: predictive value depends on prevalence — the test did not change, the population did.

Q4. A drug "reduces stroke risk by 40%", with event rates falling from 0.5% to 0.3%. The number needed to treat is approximately: A. 40 B. 100 C. 500 D. 2 ANSWER: C — about 500. Why: absolute risk reduction is 0.5% − 0.3% = 0.2%, so NNT = 1 / 0.002 = 500. The 40% figure is the RELATIVE reduction and is accurate but far less useful for a treatment decision. LEARNING POINT: always ask for the absolute figure alongside the relative one.

Q5. After introducing screening, five-year survival for a cancer improves markedly but annual mortality is unchanged. The most likely explanation is: A. The screening programme is highly effective B. Lead time and length time bias C. The cancer has become less aggressive D. Data error ANSWER: B — lead time and length time bias. Why: earlier diagnosis lengthens the measured interval between diagnosis and death without postponing death, and screening preferentially detects slower-growing tumours with a better prognosis. LEARNING POINT: judge screening programmes on mortality, not survival.

Q6. Prevalence of a chronic disease has doubled while incidence is unchanged. This most likely indicates: A. Increased transmission B. Patients are living longer with the disease C. A diagnostic error D. Falling population size ANSWER: B — longer survival. Why: prevalence ≈ incidence × duration. With incidence flat, a rising prevalence reflects increased duration — usually better treatment or earlier diagnosis. LEARNING POINT: rising prevalence can be evidence of success, and implies a burden-of-care response rather than outbreak control.

Q7. Which is the essential requirement before a screening programme is introduced? A. A test with perfect sensitivity B. An accepted treatment for the condition detected C. A very common disease D. A cheap test ANSWER: B — an accepted treatment must exist. Why: detecting disease you cannot treat causes anxiety, investigation and harm without benefit. A latent stage, an acceptable test and facilities for diagnosis and treatment are all required alongside it. LEARNING POINT: a good test is not a good programme.