USMLE Step 2 CK

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.

Basics 0:31 style links open the video at that timestamp EXTRA not in the videos or the notes; added because Step 2 tests it

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

Reality (gold standard) Disease + Disease − Test positive Test negative TP true positive FP false positive FN false negative TN true negative Sens Spec PPV NPV
Start at the circle (that cell goes on top). The bottom is both cells the arrow touches.
How to draw it every time Reality goes on top, the test goes on the side, positive first. Every cell is named after the test result: test negative but disease present = false negative.
Rule in / rule out SpPIn: a very Specific test, when Positive, rules In.
SnNOut: a very Sensitive test, when Negative, rules Out.
Prevalence Sensitivity and specificity belong to the test and do not change with prevalence. Predictive values do: PPV ↑ when prevalence ↑, NPV ↑ when prevalence ↓.
Sensitivity
TPTP + FN
Probability the test is positive when the disease is present. Circle TP, go down.
Specificity
TNTN + FP
Probability the test is negative when the disease is absent. Circle TN, go up.
Positive predictive value
TPTP + FP
Probability the person has the disease when the test is positive. From TP, go across.
Negative predictive value
TNTN + FN
Probability the person is disease-free when the test is negative. From TN, go across.
Worked example · read it off a table

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%.

Worked example · no table given, so build it

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.

Worked example · from the handwritten notes

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

cutoff ← test negative test positive → Healthy Diseased TN TP FN FP biomarker value →
Only the two middle slivers change when the cutoff moves. TN and TP stay put for exam purposes.
Labelling trick "Everybody keeps their last name." The sliver on the true negative side is the false negative; the sliver on the true positive side is the false positive. Sliding the cutoff squashes one sliver and grows the other.
MoveWhat changesGoes upGoes down
Lower the cutoff (slide left)FN ↓, FP ↑Sensitivity, NPVSpecificity, PPV
Raise the cutoff (slide right)FP ↓, FN ↑Specificity, PPVSensitivity, NPV
Curves overlap less (better test)FN ↓ and FP ↓Sensitivity and specificityNeither

Likelihood ratios

Positive likelihood ratio
LR+ = sensitivity1 − specificity
The "1 −" goes on the bottom.
Negative likelihood ratio
LR− = 1 − sensitivityspecificity
The "1 −" goes on top.
Memory hook Both are "sensitivity over specificity". The only difference is where the "1 −" goes. Use decimals, not percents.
Worked example

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

Accurate close to the gold standard Precise, not accurate close to each other
  • 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.
Worked example

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.

DesignGroups are split byTime directionMeasureRemember
Cross-sectionalNothing. One snapshot of a populationA single point in timePrevalenceAnswers "what is happening?" Cannot establish causality.
Case-controlDisease / outcome (cases vs controls)Looks back for the exposureOdds ratioThe outcome is already known at the start. Prone to recall bias.
CohortRisk factor / exposureForward (prospective) or back (retrospective)Relative riskNobody has the disease at the start; you watch who develops it.
Case reportNo comparison groupn/an/aDetailed description of one unusual patient.
Case seriesNo comparison groupn/an/aSeveral similar unusual patients (multiple case reports lumped together).

All five are observational: the researcher only watches, and assigns nothing.

Memory hook · the vowels cAse-cOntrol has an A and an O: two different letters, two different groups at the start (one has the disease, one does not), and it pairs with Odds Ratio, also two different letters.
cOhOrt has O and O: the same letters, everyone starts the same (no disease), and it pairs with Relative Risk, the same letters.
PAST NOW FUTURE Cross-sectional Exposure + disease, once Case-control Cases vs controls exposed? Retrospective cohort Exposed vs unexposed disease? Prospective cohort Exposed vs unexposed disease?
Where each study starts and which way it looks. Gold groups by outcome; indigo groups by exposure.
Retrospective = look back. Faster and cheaper, but you can only trust what was documented or what people remember, which invites recall bias.
Prospective = look forward. Follow participants over time and wait for the outcome.
Worked example

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

PhaseWhoQuestion it answers
ISmall number of healthy volunteersIs it safe? Toxicity, pharmacokinetics, pharmacodynamics.
IIModerate number of patients with the diseaseDoes it work? Efficacy, optimal dosing, common side effects.
IIILarge number of patients, randomly assignedIs it better? Compared against placebo or the current gold-standard treatment.
IVEveryone, after approvalPost-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 −
Exposedab
Not exposedcd
Memory hook Odds = Vegas odds, "3 to 1": one number over one number.
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.
Relative risk · cohort
RR = a / (a + b)c / (c + d)
Risk in the exposed divided by risk in the unexposed.
Odds ratio · case-control
OR = a / bc / d = adbc
Odds in one group divided by odds in the other.
Reading the result
> 1 → exposure is risky
< 1 → exposure is protective
= 1 → no association EXTRA
Same reading for RR and OR.
Worked example · odds ratio

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).

Worked example · relative risk

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.

Absolute risk reduction
ARR = ERcontrol − ERtreatment
The plain difference in risk (X% − Y%).
Number needed to treat
NNT = 1ARR
Use the decimal. Always round up to a whole person.
Relative risk reduction
RRR = ERcontrol − ERtreatmentERcontrol
ARR as a share of the control risk. Equals 1 − RR.
Attributable risk
AR = ERexposed − ERunexposed
Extra risk caused by a harmful exposure. Number needed to harm = 1 / AR.
Attributable risk percent
ARP = RR − 1RR
Trigger words: "what percent of cases can be attributed to…"
Worked example · NNT from percentages

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.

Worked example · NNT from a 2×2 table

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.

Worked example · attributable risk percent

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%.

Worked example · working backwards from ARR

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.
Case fatality rate = deaths from the condition ÷ people with that condition. Stay inside the one category the question names and ignore the rest of the table.
Worked example

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
p-value: the probability that the result happened by chance. A study is statistically significant when p < 0.05. So p = 0.04 is significant and p = 0.06 is not.
Power = 1 − β: the probability of detecting an association that really exists. The exam describes β in words, so learn the words in each box.
EXTRA Confidence intervals say the same thing as the p-value. A 95% CI for an RR or OR that includes 1 is not significant; a 95% CI for a difference between two means that includes 0 is not significant. A larger sample size raises power and narrows the CI.
Worked example

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?

TestComparesUse when
t-testMeans2 groups
ANOVAMeans3 or more groups
Chi-square (χ²)Outcomes as percentages or categoriesLarge sample
Fisher exactOutcomes as percentages or categoriesSmall sample

Correlation

Draw the line that best fits the dots. Its direction gives the sign of r.

5Distributions

The 68, 95, 99.7 rule, and the tail trap.

34%34% 13.5%13.5% 2.35%2.35% 0.15%0.15% −3 SD−2 SD−1 SD mean +1 SD+2 SD+3 SD 68%95%99.7%
Within ±1 SD: 68%. Within ±2 SD: 95%. Within ±3 SD: 99.7%.
Exam trap · split the leftover Whatever falls outside the range is shared between two tails. Outside ±2 SD is 5% in total, so only 2.5% lies above +2 SD. Outside ±1 SD is 32%, so 16% lies above +1 SD.
Worked example

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.

BiasWhat happensKeywords in the stem
Lead-timeThe 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-timeScreening picks up a disproportionate number of slow, harmless cases, so outcomes look better than they are.Benign, slowly progressive
Hawthorne effectPeople change their behaviour because they know they are being studied.Participants stop smoking while being followed in a cohort
Recall biasData depend on memory, which is unreliable and differs between sick and healthy people.Retrospective study, case-control, "asked to remember"
Sampling biasThe people studied do not represent everyone, so the result may not apply to other populations.Studied in one specific population only
Worked example

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

LevelGoalExamples
PrimaryStop the disease before it happensVaccines, education
SecondaryCatch the disease early to reduce its impactScreening: blood pressure check at a health fair, Pap smear, mammogram
Tertiary EXTRATreat established disease to limit complicationsCardiac rehabilitation after MI, chemotherapy

8Hardy-Weinberg

Two equations and one move: find q², take the square root, work backwards.

Alleles (genes)
p + q = 1
p = frequency of the dominant allele. q = frequency of the recessive allele.
People (phenotypes)
p² + 2pq + q² = 1
p² = homozygous dominant. 2pq = heterozygous (carriers, look dominant). q² = homozygous recessive (show the recessive trait).
The method The stem almost always hands you q²: the share of people who show the recessive trait or have the autosomal recessive disease. Then: √q² = q → p = 1 − q → plug into whatever is asked.

EXTRA For a rare disease p ≈ 1, so the carrier frequency 2pq ≈ 2q.

Worked example · problem 1

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%.

Worked example · problem 2

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 +TPFP
Test −FNTN
Reality H1Reality H0
Study H1Power = 1 − βα (Type I)
Study H0β (Type II)Correct
MeasureFormulaCue
SensitivityTP / (TP + FN)TP, down. SnNOut.
SpecificityTN / (TN + FP)TN, up. SpPIn.
PPVTP / (TP + FP)↑ with prevalence ↑
NPVTN / (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 / bcCase-control. One number over one.
ARRERcontrol − ERtreatmentDifference in risk
RRRARR / ERcontrol= 1 − RR
NNT1 / ARRRound up
Attributable riskERexposed − ERunexposedNNH = 1 / AR
ARP(RR − 1) / RR"What percent…"
Case fatalitydeaths / cases of that conditionStay in the one category
Power1 − βSignificant if p < 0.05
Normal curve±1 SD 68% · ±2 SD 95% · ±3 SD 99.7%Halve the leftover for one tail
Hardy-Weinbergp + q = 1 · p² + 2pq + q² = 1Stem gives q². Take the root.