NBME CBSE PUBLIC HEALTH SCIENCES EXAM 2026/2027 |
Actual Exam 160 Questions & Correct Detailed Answers |
Verified Answers | A Grade | Pass Guaranteed
Section 1: Biostatistics & Evidence-Based Medicine (Questions 1-30)
Q1. In a randomized controlled trial of a new antihypertensive medication, the primary
outcome shows a mean systolic blood pressure reduction of 15 mmHg in the treatment
group versus 8 mmHg in the placebo group. The reported p-value is 0.03, and the 95%
confidence interval for the difference is 0.5 to 13.5 mmHg. Which statement best
describes the statistical and clinical significance of these results?
A. The results are statistically significant but may lack clinical significance given the
wide confidence interval includes values near zero.
B. The results are both statistically and clinically significant because the p-value is less
than 0.05.
C. The results are not statistically significant because the confidence interval crosses
the null value.
D. The results are clinically significant only if the number needed to treat is less than 50.
Correct Answer: A. The results are statistically significant but may lack clinical
significance given the wide confidence interval includes values near zero. [CORRECT]
Rationale: A p-value of 0.03 indicates statistical significance at α=0.05, but the 95% CI
(0.5–13.5) includes values very close to the null (0), suggesting the true effect could be
minimal; clinical significance requires consideration of effect size and confidence
precision, not just p-value alone. B incorrectly equates statistical significance with
clinical significance. C is wrong because the CI does not include the null value of 0. D
introduces an unrelated NNT threshold without data.
,Q2. A researcher is designing a study to detect a 10% absolute reduction in mortality
with 80% power and a two-sided alpha of 0.05. The control group mortality is expected
to be 25%. Which factor, if increased, would most directly reduce the required sample
size?
A. Increasing the alpha level from 0.05 to 0.10
B. Increasing the desired power from 80% to 90%
C. Increasing the minimum detectable effect size from 10% to 15%
D. Increasing the expected control group mortality from 25% to 35%
Correct Answer: C. Increasing the minimum detectable effect size from 10% to 15%
[CORRECT]
Rationale: Sample size is inversely related to effect size; a larger detectable difference
requires fewer subjects to achieve the same power. A would actually decrease sample
size but is less direct than effect size. B would increase sample size. D changes
baseline risk but does not directly reduce sample size without changing the effect size
proportion.
Q3. In a normal distribution of serum cholesterol levels in a population of 10,000 adults,
the mean is 200 mg/dL and the standard deviation is 25 mg/dL. Approximately how
many individuals would be expected to have cholesterol levels between 175 mg/dL and
225 mg/dL?
A. 2,500
B. 5,000
C. 6,800
D. 9,500
Correct Answer: C. 6,800 [CORRECT]
,Rationale: The range 175–225 mg/dL represents mean ± 1 SD; in a normal distribution,
approximately 68% of values fall within 1 SD, so 0.68 × 10,000 = 6,800. A represents 25%
(half of 1 SD). B represents 50% (mean only). D represents 95% (mean ± 2 SD).
Q4. A clinical trial reports a relative risk reduction of 30% for stroke with a new
anticoagulant. The absolute risk of stroke in the control group is 4% and 2.8% in the
treatment group. What is the number needed to treat (NNT) to prevent one stroke over
the study period?
A. 12
B. 33
C. 83
D. 120
Correct Answer: C. 83 [CORRECT]
Rationale: Absolute Risk Reduction (ARR) = 4% – 2.8% = 1.2% = 0.012; NNT = 1/ARR =
1/0.012 = 83.3, rounded to 83. A incorrectly uses relative risk. B miscalculates ARR as
3%. D inverts the control group risk.
Q5. A screening test for prostate cancer has a sensitivity of 85% and a specificity of
90%. In a population of 1,000 men with a disease prevalence of 10%, how many true
positive results would be expected?
A. 85
B. 90
C. 95
D. 100
Correct Answer: A. 85 [CORRECT]
, Rationale: True positives = Sensitivity × Total with disease = 0.85 × (1,000 × 0.10) = 0.85
× 100 = 85. B reflects specificity applied to healthy population. C and D overestimate
sensitivity or prevalence.
Q6. A study comparing three different teaching methods for medical students uses a
one-way ANOVA and reports F(2, 87) = 4.56, p = 0.013. The researcher then conducts
pairwise t-tests between all groups without adjustment. What is the primary statistical
concern with this approach?
A. The F-test was underpowered to detect differences.
B. Multiple comparisons inflate the family-wise Type I error rate.
C. ANOVA assumes equal variances, which was violated.
D. The sample size was too small for parametric testing.
Correct Answer: B. Multiple comparisons inflate the family-wise Type I error rate.
[CORRECT]
Rationale: Conducting multiple pairwise tests without correction (e.g., Bonferroni,
Tukey) increases the probability of at least one false positive; with 3 groups, 3
comparisons are made, inflating α. A is unsupported. C and D introduce assumptions
not discussed in the stem.
Q7. A new biomarker for myocardial infarction has an area under the ROC curve (AUC)
of 0.72. Which statement best characterizes this test's discriminatory ability?
A. Excellent discrimination; the test is suitable as a standalone diagnostic.
B. Fair discrimination; better than chance but limited clinical utility alone.
C. Poor discrimination; the test performs no better than flipping a coin.
D. Perfect discrimination; the test distinguishes all cases from non-cases.
Correct Answer: B. Fair discrimination; better than chance but limited clinical utility
alone. [CORRECT]
Actual Exam 160 Questions & Correct Detailed Answers |
Verified Answers | A Grade | Pass Guaranteed
Section 1: Biostatistics & Evidence-Based Medicine (Questions 1-30)
Q1. In a randomized controlled trial of a new antihypertensive medication, the primary
outcome shows a mean systolic blood pressure reduction of 15 mmHg in the treatment
group versus 8 mmHg in the placebo group. The reported p-value is 0.03, and the 95%
confidence interval for the difference is 0.5 to 13.5 mmHg. Which statement best
describes the statistical and clinical significance of these results?
A. The results are statistically significant but may lack clinical significance given the
wide confidence interval includes values near zero.
B. The results are both statistically and clinically significant because the p-value is less
than 0.05.
C. The results are not statistically significant because the confidence interval crosses
the null value.
D. The results are clinically significant only if the number needed to treat is less than 50.
Correct Answer: A. The results are statistically significant but may lack clinical
significance given the wide confidence interval includes values near zero. [CORRECT]
Rationale: A p-value of 0.03 indicates statistical significance at α=0.05, but the 95% CI
(0.5–13.5) includes values very close to the null (0), suggesting the true effect could be
minimal; clinical significance requires consideration of effect size and confidence
precision, not just p-value alone. B incorrectly equates statistical significance with
clinical significance. C is wrong because the CI does not include the null value of 0. D
introduces an unrelated NNT threshold without data.
,Q2. A researcher is designing a study to detect a 10% absolute reduction in mortality
with 80% power and a two-sided alpha of 0.05. The control group mortality is expected
to be 25%. Which factor, if increased, would most directly reduce the required sample
size?
A. Increasing the alpha level from 0.05 to 0.10
B. Increasing the desired power from 80% to 90%
C. Increasing the minimum detectable effect size from 10% to 15%
D. Increasing the expected control group mortality from 25% to 35%
Correct Answer: C. Increasing the minimum detectable effect size from 10% to 15%
[CORRECT]
Rationale: Sample size is inversely related to effect size; a larger detectable difference
requires fewer subjects to achieve the same power. A would actually decrease sample
size but is less direct than effect size. B would increase sample size. D changes
baseline risk but does not directly reduce sample size without changing the effect size
proportion.
Q3. In a normal distribution of serum cholesterol levels in a population of 10,000 adults,
the mean is 200 mg/dL and the standard deviation is 25 mg/dL. Approximately how
many individuals would be expected to have cholesterol levels between 175 mg/dL and
225 mg/dL?
A. 2,500
B. 5,000
C. 6,800
D. 9,500
Correct Answer: C. 6,800 [CORRECT]
,Rationale: The range 175–225 mg/dL represents mean ± 1 SD; in a normal distribution,
approximately 68% of values fall within 1 SD, so 0.68 × 10,000 = 6,800. A represents 25%
(half of 1 SD). B represents 50% (mean only). D represents 95% (mean ± 2 SD).
Q4. A clinical trial reports a relative risk reduction of 30% for stroke with a new
anticoagulant. The absolute risk of stroke in the control group is 4% and 2.8% in the
treatment group. What is the number needed to treat (NNT) to prevent one stroke over
the study period?
A. 12
B. 33
C. 83
D. 120
Correct Answer: C. 83 [CORRECT]
Rationale: Absolute Risk Reduction (ARR) = 4% – 2.8% = 1.2% = 0.012; NNT = 1/ARR =
1/0.012 = 83.3, rounded to 83. A incorrectly uses relative risk. B miscalculates ARR as
3%. D inverts the control group risk.
Q5. A screening test for prostate cancer has a sensitivity of 85% and a specificity of
90%. In a population of 1,000 men with a disease prevalence of 10%, how many true
positive results would be expected?
A. 85
B. 90
C. 95
D. 100
Correct Answer: A. 85 [CORRECT]
, Rationale: True positives = Sensitivity × Total with disease = 0.85 × (1,000 × 0.10) = 0.85
× 100 = 85. B reflects specificity applied to healthy population. C and D overestimate
sensitivity or prevalence.
Q6. A study comparing three different teaching methods for medical students uses a
one-way ANOVA and reports F(2, 87) = 4.56, p = 0.013. The researcher then conducts
pairwise t-tests between all groups without adjustment. What is the primary statistical
concern with this approach?
A. The F-test was underpowered to detect differences.
B. Multiple comparisons inflate the family-wise Type I error rate.
C. ANOVA assumes equal variances, which was violated.
D. The sample size was too small for parametric testing.
Correct Answer: B. Multiple comparisons inflate the family-wise Type I error rate.
[CORRECT]
Rationale: Conducting multiple pairwise tests without correction (e.g., Bonferroni,
Tukey) increases the probability of at least one false positive; with 3 groups, 3
comparisons are made, inflating α. A is unsupported. C and D introduce assumptions
not discussed in the stem.
Q7. A new biomarker for myocardial infarction has an area under the ROC curve (AUC)
of 0.72. Which statement best characterizes this test's discriminatory ability?
A. Excellent discrimination; the test is suitable as a standalone diagnostic.
B. Fair discrimination; better than chance but limited clinical utility alone.
C. Poor discrimination; the test performs no better than flipping a coin.
D. Perfect discrimination; the test distinguishes all cases from non-cases.
Correct Answer: B. Fair discrimination; better than chance but limited clinical utility
alone. [CORRECT]
