Biostatistics Review
Everything worth knowing before the UWorld biostatistics block, in one place: the formulas, the memory tricks, and the worked questions from the three source videos, merged with handwritten review notes. Each topic links to the exact moment in the video where it is explained.
1Diagnostic tests
Sensitivity, specificity, predictive values, cutoffs and likelihood ratios all come from one 2×2 table.
The 2×2 table and the arrow trick
SnNOut: a very Sensitive test, when Negative, rules Out.
A new UTI test is compared with the gold standard. Among women with a UTI: 40 test positive, 160 test negative. Among women without: 20 test positive, 180 test negative. What are the specificity and sensitivity?
Solution
Specificity = TN / (TN + FP) = 180 / (180 + 20) = 90%.
Sensitivity = TP / (TP + FN) = 40 / (40 + 160) = 20%.
Sensitivity is 70% and specificity is 90%. 100 patients truly have the disease and 200 truly do not. How many false negatives are there? How many false positives?
Solution
Disease column totals 100: TP = 70% of 100 = 70, so FN = 100 − 70 = 30.
No-disease column totals 200: TN = 90% of 200 = 180, so FP = 200 − 180 = 20.
Whatever they give you, write the table and work backwards.
Specificity 0.95, sensitivity 0.65. 100 people have the disease and 100 do not. Find all four cells, then PPV and NPV.
Solution
Disease column (TP + FN = 100): TP = 65, FN = 35. No-disease column (TN + FP = 100): TN = 95, FP = 5.
PPV = 65 / (65 + 5) = 65/70 ≈ 93%. NPV = 95 / (95 + 35) = 95/130 ≈ 73%.
Moving the cutoff
| Move | What changes | Goes up | Goes down |
|---|---|---|---|
| Lower the cutoff (slide left) | FN ↓, FP ↑ | Sensitivity, NPV | Specificity, PPV |
| Raise the cutoff (slide right) | FP ↓, FN ↑ | Specificity, PPV | Sensitivity, NPV |
| Curves overlap less (better test) | FN ↓ and FP ↓ | Sensitivity and specificity | Neither |
- 100% sensitivity needs FN = 0: put the cutoff at the lowest value of the diseased curve.
- 100% specificity needs FP = 0: put the cutoff at the highest value of the healthy curve.
- Reason it through with the formula. Example: FN falls, so TP / (TP + FN) rises, so sensitivity rises.
Likelihood ratios
A test has a sensitivity of 70% and a specificity of 90%. What is the positive likelihood ratio?
Solution
LR+ = 0.7 / (1 − 0.9) = 0.7 / 0.1 = 7. (LR− would be 0.3 / 0.9 ≈ 0.33.)
Accuracy vs precision
- Accuracy (validity): results are close to the true value. You need a gold standard to judge it.
- Precision (reliability): repeated results are close to each other.
The published standard for a serum level is 40. A new instrument reads 70, 68, 70, 72 and 75.
Solution
The readings are tight but far from 40: precise but not accurate.
2Study designs
Most "which study is this?" questions are really asking cohort or case-control.
| Design | Groups are split by | Time direction | Measure | Remember |
|---|---|---|---|---|
| Cross-sectional | Nothing. One snapshot of a population | A single point in time | Prevalence | Answers "what is happening?" Cannot establish causality. |
| Case-control | Disease / outcome (cases vs controls) | Looks back for the exposure | Odds ratio | The outcome is already known at the start. Prone to recall bias. |
| Cohort | Risk factor / exposure | Forward (prospective) or back (retrospective) | Relative risk | Nobody has the disease at the start; you watch who develops it. |
| Case report | No comparison group | n/a | n/a | Detailed description of one unusual patient. |
| Case series | No comparison group | n/a | n/a | Several similar unusual patients (multiple case reports lumped together). |
All five are observational: the researcher only watches, and assigns nothing.
cOhOrt has O and O: the same letters, everyone starts the same (no disease), and it pairs with Relative Risk, the same letters.
A 5-year study will assess the incidence of respiratory disease in 600 people over age 50. One group cares for a pet dog; the other has no animals at home. Cultures are taken when symptoms begin. What type of study is this?
Solution
Groups are split by exposure (dog or no dog), nobody has the disease at the start, and they are followed forward: a prospective cohort study. Its measure is relative risk.
Meta-analysis
Pools the data from several studies into one analysis. The bigger combined sample gives more statistical power, which is why it is useful when single studies are small, such as studies of rare conditions.
Clinical trial phases
| Phase | Who | Question it answers |
|---|---|---|
| I | Small number of healthy volunteers | Is it safe? Toxicity, pharmacokinetics, pharmacodynamics. |
| II | Moderate number of patients with the disease | Does it work? Efficacy, optimal dosing, common side effects. |
| III | Large number of patients, randomly assigned | Is it better? Compared against placebo or the current gold-standard treatment. |
| IV | Everyone, after approval | Post-marketing surveillance for rare or long-term adverse effects. Can lead to withdrawal from the market. |
3Measures of risk
Relative risk, odds ratio, and everything built on event rates.
Relative risk and odds ratio
| Disease + | Disease − | |
|---|---|---|
| Exposed | a | b |
| Not exposed | c | d |
Risk: one number over two (the cell over its row total).
Read the question top to bottom, left to right: the group named first goes on top.
< 1 → exposure is protective
= 1 → no association EXTRA
Older students: 60 had low scores, 200 had normal scores. Younger students: 40 low, 160 normal. What is the odds ratio of older students having a low score, relative to younger students?
Solution
Older students are named first, so they go on top. One number over one number:
OR = (60 / 200) ÷ (40 / 160) = 0.30 ÷ 0.25 = 1.2.
If the question had asked for relative risk: (60 / 260) ÷ (40 / 200).
Men with no children: 80 had prostate cancer, 920 did not. Men with children: 220 had prostate cancer, 1,280 did not. What is the relative risk of prostate cancer in men with no children compared with men who had children?
Solution
One number over two: RR = (80 / 1,000) ÷ (220 / 1,500) = 0.080 ÷ 0.147 ≈ 0.55.
The odds ratio from the same table would be (80 / 920) ÷ (220 / 1,280).
Event rates: ARR, RRR, NNT, attributable risk
ER = event rate, the fraction of a group that had the event. The control group is the placebo or standard therapy.
By physician assessment, the outcome rate was 60% in one arm and 35% in the other (placebo vs drug X). How many patients must be treated with drug X for one to benefit?
Solution
ARR = 0.60 − 0.35 = 0.25. NNT = 1 / 0.25 = 4.
The table also listed patient-assessed data. Use only the row the question asks about.
New drug: 15 developed psychosis, 985 did not. Placebo: 30 developed psychosis, 970 did not. What is the number needed to treat?
Solution
ARR = 30/1,000 − 15/1,000 = 15/1,000 = 0.015. NNT = 1 / 0.015 = 66.7, rounded up to 67.
A prospective study finds a 20-year relative risk of 4.0 for prostate cancer in men with elevated PSA compared with normal PSA (p = 0.03). What percent of prostate cancer in men with elevated PSA can be attributed to the elevated PSA?
Solution
"What percent" → ARP = (RR − 1) / RR = (4 − 1) / 4 = 75%.
The absolute risk reduction for drug ABC versus standard therapy is 6%. The incidence of UTI with standard therapy is 8%. There are 29 UTI patients in the drug ABC group. How many subjects are in the drug ABC group?
Solution
ARR = ERcontrol − ERtreatment, so 0.06 = 0.08 − ERtreatment, and ERtreatment = 0.02.
29 / x = 0.02, so x = 29 / 0.02 = 1,450.
When a question looks impossible, ask "what formula do I know here?", write it down, and plug in.
Incidence, prevalence and case fatality
- Incidence = new cases over a period of time.
- Prevalence = all existing cases (old and new) at one point in time.
- Prevalence ≈ incidence × disease duration EXTRA, so when the disease is short-lived, prevalence ≈ incidence. Chronic disease: prevalence > incidence.
A table of spinal cord injuries lists 20 falls, 4 of them fatal. Case fatality for falls?
Solution
4 / 20 = 20%.
4Hypothesis testing
The null hypothesis table, p-values, power, and which test to pick.
The null hypothesis table
H0, the null hypothesis: there is no association. H1 is the alternative: there is one. Draw it exactly like the diagnostic 2×2, with reality on top and the study on the side.
| Reality: association exists (H1) | Reality: no association (H0) | |
|---|---|---|
| Study finds an association (H1) | Power = 1 − βcorrectly finds a real association | α error · Type Ifinds an association that is not there (false positive) |
| Study finds no association (H0) | β error · Type IImisses an association that is there (false negative) | Correctcorrectly finds nothing |
Mean HDL was 42 in non-diabetic patients and 35 in diabetic patients. The probability this was due to chance was 0.05. There is also a 15% probability of concluding there is no difference in HDL when in reality there is one. What is the p-value? What is the power?
Solution
p-value = probability due to chance = 0.05.
"Concluding no difference when there really is one" is the β box, so β = 0.15 and power = 1 − 0.15 = 0.85.
Which statistical test?
| Test | Compares | Use when |
|---|---|---|
| t-test | Means | 2 groups |
| ANOVA | Means | 3 or more groups |
| Chi-square (χ²) | Outcomes as percentages or categories | Large sample |
| Fisher exact | Outcomes as percentages or categories | Small sample |
Correlation
- The correlation coefficient r runs from −1 to +1.
- Sign = direction. Positive: both rise together (more study, higher score). Negative: one rises as the other falls (more alcohol, lower score).
- Size = strength. Near ±1 the dots hug the line; near 0 they are a cloud.
- Correlation does not prove causation. EXTRA
5Distributions
The 68, 95, 99.7 rule, and the tail trap.
In 200 hospital patients, serum cholesterol is normally distributed with a mean of 210 and a standard deviation of 15. How many patients would you expect to have a cholesterol above 240?
Solution
240 = 210 + 2 × 15, which is +2 SD. 95% of patients (190) lie within ±2 SD, leaving 10 outside. Half of those are on the low side, so 5 patients are above 240. The tempting wrong answer is 10.
Bimodal distribution: a curve with two peaks. Classic example: age of onset of ulcerative colitis.
6Bias
Lead-time and length-time are the pair the exam plays against each other.
| Bias | What happens | Keywords in the stem |
|---|---|---|
| Lead-time | The disease is found earlier, so survival looks longer, but the patient dies at the same time as before. | Early detection, screening test, "survival increased" |
| Length-time | Screening picks up a disproportionate number of slow, harmless cases, so outcomes look better than they are. | Benign, slowly progressive |
| Hawthorne effect | People change their behaviour because they know they are being studied. | Participants stop smoking while being followed in a cohort |
| Recall bias | Data depend on memory, which is unreliable and differs between sick and healthy people. | Retrospective study, case-control, "asked to remember" |
| Sampling bias | The people studied do not represent everyone, so the result may not apply to other populations. | Studied in one specific population only |
Researchers test a new biomarker in asymptomatic patients over 70. Most cases of the disease progress slowly, and they worry the test's benefit will be overestimated because it detects a disproportionate number of slowly progressive, benign cases. Which bias is this?
Solution
"Benign, slowly progressive" → length-time bias. Lead-time bias is the distractor: that one is only about detecting the disease earlier (finding prostate cancer at 50 instead of 60 makes survival look 10 years longer).
7Levels of prevention
| Level | Goal | Examples |
|---|---|---|
| Primary | Stop the disease before it happens | Vaccines, education |
| Secondary | Catch the disease early to reduce its impact | Screening: blood pressure check at a health fair, Pap smear, mammogram |
| Tertiary EXTRA | Treat established disease to limit complications | Cardiac rehabilitation after MI, chemotherapy |
8Hardy-Weinberg
Two equations and one move: find q², take the square root, work backwards.
EXTRA For a rare disease p ≈ 1, so the carrier frequency 2pq ≈ 2q.
16% of a population cannot smell a certain chemical. Non-smellers are homozygous recessive for the smelling gene. (a) What percent are smellers? (b) What are the dominant and recessive allele frequencies? (c) What percent are heterozygous?
Solution
(a) Everyone else: 100 − 16 = 84%.
(b) q² = 0.16, so q = 0.4 and p = 1 − 0.4 = 0.6.
(c) 2pq = 2 × 0.6 × 0.4 = 0.48 = 48%.
A healthy woman has a child with a rare autosomal recessive disease that affects 1 in 90,000 people. She plans a second child using a new healthy sperm donor. What is the probability that this child is affected?
Solution
q² = 1 / 90,000, so q = 1 / 300. That is the chance a random donor's sperm carries the allele.
The mother is a known carrier (she already had an affected child), so she passes the allele 1 / 2 of the time.
1/2 × 1/300 = 1 in 600. The trap is stopping at 1 in 300.
9Formula sheet
Write these on the scratch sheet before the first question. Practise until it takes two minutes.
| Disease + | Disease − | |
|---|---|---|
| Test + | TP | FP |
| Test − | FN | TN |
| Reality H1 | Reality H0 | |
|---|---|---|
| Study H1 | Power = 1 − β | α (Type I) |
| Study H0 | β (Type II) | Correct |
| Measure | Formula | Cue |
|---|---|---|
| Sensitivity | TP / (TP + FN) | TP, down. SnNOut. |
| Specificity | TN / (TN + FP) | TN, up. SpPIn. |
| PPV | TP / (TP + FP) | ↑ with prevalence ↑ |
| NPV | TN / (TN + FN) | ↑ with prevalence ↓ |
| LR+ | sens / (1 − spec) | "1 −" on the bottom |
| LR− | (1 − sens) / spec | "1 −" on top |
| Relative risk | [a / (a + b)] ÷ [c / (c + d)] | Cohort. One number over two. |
| Odds ratio | (a / b) ÷ (c / d) = ad / bc | Case-control. One number over one. |
| ARR | ERcontrol − ERtreatment | Difference in risk |
| RRR | ARR / ERcontrol | = 1 − RR |
| NNT | 1 / ARR | Round up |
| Attributable risk | ERexposed − ERunexposed | NNH = 1 / AR |
| ARP | (RR − 1) / RR | "What percent…" |
| Case fatality | deaths / cases of that condition | Stay in the one category |
| Power | 1 − β | Significant if p < 0.05 |
| Normal curve | ±1 SD 68% · ±2 SD 95% · ±3 SD 99.7% | Halve the leftover for one tail |
| Hardy-Weinberg | p + q = 1 · p² + 2pq + q² = 1 | Stem gives q². Take the root. |