A researcher is investigating whether there is an association between the use of social media in teenagers and bipolar disorder. In order to study this potential relationship, she collects data from people who have bipolar disorder and matched controls without the disorder. She then asks how much on average these individuals used social media in the 3 years prior to their diagnosis. This continuous data is divided into 2 groups: those who used more than 2 hours per day and those who used less than 2 hours per day. She finds that out of 1000 subjects, 500 had bipolar disorder of which 300 used social media more than 2 hours per day. She also finds that 400 subjects who did not have the disorder also did not use social media more than 2 hours per day. Which of the following is the odds ratio for development of bipolar disorder after being exposed to more social media?
Q2
A randomized control double-blind study is conducted on the efficacy of 2 sulfonylureas. The study concluded that medication 1 was more efficacious in lowering fasting blood glucose than medication 2 (p ≤ 0.05; 95% CI: 14 [10-21]). Which of the following is true regarding a 95% confidence interval (CI)?
Q3
An investigator for a nationally representative health survey is evaluating the heights and weights of men and women aged 18–74 years in the United States. The investigator finds that for each sex, the distribution of heights is well-fitted by a normal distribution. The distribution of weight is not normally distributed. Results are shown:
Mean Standard deviation
Height (inches), men 69 0.1
Height (inches), women 64 0.1
Weight (pounds), men 182 1.0
Weight (pounds), women 154 1.0
Based on these results, which of the following statements is most likely to be correct?
Q4
A 45-year-old man comes to the clinic concerned about his recent exposure to radon. He heard from his co-worker that radon exposure can cause lung cancer. He brings in a study concerning the risks of radon exposure. In the study, there were 300 patients exposed to radon, and 18 developed lung cancer over a 10-year period. To compare, there were 500 patients without radon exposure and 11 developed lung cancer over the same 10-year period. If we know that 0.05% of the population has been exposed to radon, what is the attributable risk percent for developing lung cancer over a 10 year period after radon exposure?
Q5
An investigator is measuring the blood calcium level in a sample of female cross country runners and a control group of sedentary females. If she would like to compare the means of the two groups, which statistical test should she use?
Q6
You are reading through a recent article that reports significant decreases in all-cause mortality for patients with malignant melanoma following treatment with a novel biological infusion. Which of the following choices refers to the probability that a study will find a statistically significant difference when one truly does exist?
Q7
A medical research study is beginning to evaluate the positive predictive value of a novel blood test for non-Hodgkin’s lymphoma. The diagnostic arm contains 700 patients with NHL, of which 400 tested positive for the novel blood test. In the control arm, 700 age-matched control patients are enrolled and 0 are found positive for the novel test. What is the PPV of this test?
Q8
You submit a paper to a prestigious journal about the effects of coffee consumption on mesothelioma risk. The first reviewer lauds your clinical and scientific acumen, but expresses concern that your study does not have adequate statistical power. Statistical power refers to which of the following?
Q9
A research fellow proposes a nested case-control study within an existing cohort examining antibiotic exposure and C. difficile infection. The mentor suggests this design wastes the cohort structure and that relative risk should be calculated instead. The fellow argues that odds ratios from nested case-control studies approximate relative risk while being more efficient. Evaluate the validity of each position and synthesize the optimal approach.
Q10
Two studies examine statin therapy and stroke prevention. Study A (cohort, n=10,000) reports RR=0.75. Study B (case-control, n=2,000) reports OR=0.68. The absolute stroke rate in the general population is 2% over 5 years. Analyze these findings to determine which study provides more accurate information for clinical decision-making and why.
Adjusted odds ratios US Medical PG Practice Questions and MCQs
Question 1: A researcher is investigating whether there is an association between the use of social media in teenagers and bipolar disorder. In order to study this potential relationship, she collects data from people who have bipolar disorder and matched controls without the disorder. She then asks how much on average these individuals used social media in the 3 years prior to their diagnosis. This continuous data is divided into 2 groups: those who used more than 2 hours per day and those who used less than 2 hours per day. She finds that out of 1000 subjects, 500 had bipolar disorder of which 300 used social media more than 2 hours per day. She also finds that 400 subjects who did not have the disorder also did not use social media more than 2 hours per day. Which of the following is the odds ratio for development of bipolar disorder after being exposed to more social media?
A. 1.5
B. 6 (Correct Answer)
C. 0.17
D. 0.67
Explanation: ***6***
- To calculate the odds ratio, we first construct a 2x2 table [1]:
- Bipolar Disorder (Cases): 500
- No Bipolar Disorder (Controls): 500 (1000 total subjects - 500 cases)
- Cases exposed to more social media (>2 hrs/day): 300
- Cases not exposed to more social media (≤2 hrs/day): 200 (500 - 300)
- Controls not exposed to more social media (≤2 hrs/day): 400
- Controls exposed to more social media (>2 hrs/day): 100 (500 - 400)
- The odds ratio (OR) is calculated as (odds of exposure in cases) / (odds of exposure in controls) = (300/200) / (100/400) = 1.5 / 0.25 = **6** [1].
*1.5*
- This value represents the **odds of exposure** (more than 2 hours of social media) in individuals with bipolar disorder (300 cases exposed / 200 cases unexposed = 1.5).
- It is not the odds ratio, which compares these odds to the odds of exposure in the control group.
*0.17*
- This value is close to the reciprocal of 6 (1/6 ≈ 0.166), suggesting a potential miscalculation or an inverted odds ratio.
- An odds ratio of 0.17 would imply a protective effect (lower odds of bipolar disorder with more social media), which is contrary to the calculation and typical interpretation in this context.
*0.67*
- This value is the reciprocal of 1.5 (1/1.5 ≈ 0.67) which represents the odds of *not* being exposed in cases (200/300).
- It does not represent the correct odds ratio, which compares the odds of exposure in cases to the odds of exposure in controls.
Question 2: A randomized control double-blind study is conducted on the efficacy of 2 sulfonylureas. The study concluded that medication 1 was more efficacious in lowering fasting blood glucose than medication 2 (p ≤ 0.05; 95% CI: 14 [10-21]). Which of the following is true regarding a 95% confidence interval (CI)?
A. If the same study were repeated multiple times, approximately 95% of the calculated confidence intervals would contain the true population parameter. (Correct Answer)
B. The 95% confidence interval is the probability chosen by the researcher to be the threshold of statistical significance.
C. When a 95% CI for the estimated difference between groups contains the value ‘0’, the results are significant.
D. It represents the probability that chance would not produce the difference shown, 95% of the time.
E. The study is adequately powered at the 95% confidence interval.
Explanation: ***If the same study were repeated multiple times, approximately 95% of the calculated confidence intervals would contain the true population parameter.***
- This statement accurately defines the **frequentist interpretation** of a confidence interval (CI). It reflects the long-run behavior of the CI over hypothetical repetitions of the study.
- A 95% CI means that if you were to repeat the experiment many times, 95% of the CIs calculated from those experiments would capture the **true underlying population parameter**.
*The 95% confidence interval is the probability chosen by the researcher to be the threshold of statistical significance.*
- The **alpha level (α)**, typically set at 0.05 (or 5%), is the threshold for statistical significance (p ≤ 0.05), representing the probability of a Type I error.
- The 95% confidence level (1-α) is related to statistical significance, but it is not the *threshold* itself; rather, it indicates the **reliability** of the interval estimate.
*When a 95% CI for the estimated difference between groups contains the value ‘0’, the results are significant.*
- If a 95% CI for the difference between groups **contains 0**, it implies that there is **no statistically significant difference** between the groups at the 0.05 alpha level.
- A statistically significant difference (p ≤ 0.05) would be indicated if the 95% CI **does NOT contain 0**, suggesting that the intervention had a real effect.
*It represents the probability that chance would not produce the difference shown, 95% of the time.*
- This statement misinterprets the meaning of a CI and probability. The chance of not producing the observed difference is typically addressed by the **p-value**, not directly by the CI in this manner.
- A CI provides a **range of plausible values** for the population parameter, not a probability about the role of chance in producing the observed difference.
*The study is adequately powered at the 95% confidence interval.*
- **Statistical power** is the probability of correctly rejecting a false null hypothesis, typically set at 80% or 90%. It is primarily determined by sample size, effect size, and alpha level.
- A 95% CI is a measure of the **precision** of an estimate, while power refers to the **ability of a study to detect an effect** if one exists. They are related but distinct concepts.
Question 3: An investigator for a nationally representative health survey is evaluating the heights and weights of men and women aged 18–74 years in the United States. The investigator finds that for each sex, the distribution of heights is well-fitted by a normal distribution. The distribution of weight is not normally distributed. Results are shown:
Mean Standard deviation
Height (inches), men 69 0.1
Height (inches), women 64 0.1
Weight (pounds), men 182 1.0
Weight (pounds), women 154 1.0
Based on these results, which of the following statements is most likely to be correct?
A. 86% of heights in women are likely to fall between 63.9 and 64.1 inches.
B. 99.7% of heights in women are likely to fall between 63.7 and 64.3 inches. (Correct Answer)
C. 68% of weights in women are likely to fall between 153 and 155 pounds.
D. 95% of heights in men are likely to fall between 68.85 and 69.15 inches.
E. 99.7% of heights in men are likely to fall between 68.8 and 69.2 inches.
Explanation: ***99.7% of heights in women are likely to fall between 63.7 and 64.3 inches.***
* For a **normal distribution**, approximately 99.7% of values fall within **±3 standard deviations** of the mean.
* For women's height: Mean = 64 inches, Standard Deviation = 0.1 inches. Therefore, 3 SD = 0.3 inches. The range is 64 ± 0.3, which is **63.7 to 64.3 inches**.
*86% of heights in women are likely to fall between 63.9 and 64.1 inches.*
* The range 63.9 to 64.1 inches represents **±1 standard deviation** (64 ± 0.1 inches).
* For a normal distribution, approximately **68%** (not 86%) of values fall within ±1 standard deviation of the mean.
*68% of weights in women are likely to fall between 153 and 155 pounds.*
* While 153 to 155 pounds represents **±1 standard deviation** (154 ± 1 pound), the problem states that the **distribution of weight is not normally distributed**.
* The **68-95-99.7 rule** (empirical rule) only applies to data that follows a normal distribution.
*95% of heights in men are likely to fall between 68.85 and 69.15 inches.*
* For a normal distribution, 95% of values fall within **±2 standard deviations**.
* For men's height: Mean = 69 inches, Standard Deviation = 0.1 inches. Therefore, 2 SD = 0.2 inches. The range for 95% should be 69 ± 0.2, which is **68.8 to 69.2 inches**, not 68.85 to 69.15 inches.
*99.7% of heights in men are likely to fall between 68.8 and 69.2 inches.*
* For a normal distribution, 99.7% of values fall within **±3 standard deviations**.
* For men's height: Mean = 69 inches, Standard Deviation = 0.1 inches. Therefore, 3 SD = 0.3 inches. The range for 99.7% should be 69 ± 0.3, which is **68.7 to 69.3 inches**, not 68.8 to 69.2 inches.
Question 4: A 45-year-old man comes to the clinic concerned about his recent exposure to radon. He heard from his co-worker that radon exposure can cause lung cancer. He brings in a study concerning the risks of radon exposure. In the study, there were 300 patients exposed to radon, and 18 developed lung cancer over a 10-year period. To compare, there were 500 patients without radon exposure and 11 developed lung cancer over the same 10-year period. If we know that 0.05% of the population has been exposed to radon, what is the attributable risk percent for developing lung cancer over a 10 year period after radon exposure?
A. 3.8%
B. 0.31%
C. 2.2%
D. 6.0%
E. 63.3% (Correct Answer)
Explanation: ***63.3%***
- The **attributable risk percent (ARP)** quantifies the proportion of disease in the exposed group that is attributable to the exposure. It is calculated as [(Incidence in exposed - Incidence in unexposed) / Incidence in exposed] * 100.
- In this case, **Incidence in exposed (radon)** = 18/300 = 0.06 or 6%. **Incidence in unexposed** = 11/500 = 0.022 or 2.2%. Therefore, ARP = [(0.06 - 0.022) / 0.06] * 100 = (0.038 / 0.06) * 100 = **63.3%**.
*3.8%*
- This value represents the difference in the **absolute risk** or incidence between the exposed and unexposed groups (6% - 2.2% = 3.8%).
- It does not represent the proportion of disease in the exposed group that is due to the exposure.
*0.31%*
- This value is not derived from the given data using standard epidemiological formulas for attributable risk percent.
- It is possibly a miscalculation or an irrelevant measure in this context.
*2.2%*
- This value represents the **incidence of lung cancer in the unexposed group** (11/500 = 0.022 or 2.2%).
- It is a component of the ARP calculation but not the ARP itself.
*6.0%*
- This value represents the **incidence of lung cancer in the radon-exposed group** (18/300 = 0.06 or 6%).
- It is used in the numerator and denominator for calculating the attributable risk percent but is not the final ARP.
Question 5: An investigator is measuring the blood calcium level in a sample of female cross country runners and a control group of sedentary females. If she would like to compare the means of the two groups, which statistical test should she use?
A. Chi-square test
B. Linear regression
C. t-test (Correct Answer)
D. ANOVA (Analysis of Variance)
E. F-test
Explanation: ***t-test***
- A **t-test** is appropriate for comparing the means of two independent groups, such as the blood calcium levels between runners and sedentary females.
- It assesses whether the observed difference between the two sample means is statistically significant or occurred by chance.
*Chi-square test*
- The **chi-square test** is used to analyze categorical data to determine if there is a significant association between two variables.
- It is not suitable for comparing continuous variables like blood calcium levels.
*Linear regression*
- **Linear regression** is used to model the relationship between a dependent variable (outcome) and one or more independent variables (predictors).
- It aims to predict the value of a variable based on the value of another, rather than comparing means between groups.
*ANOVA (Analysis of Variance)*
- **ANOVA** is used to compare the means of **three or more independent groups**.
- Since there are only two groups being compared in this scenario, a t-test is more specific and appropriate.
*F-test*
- The **F-test** is primarily used to compare the variances of two populations or to assess the overall significance of a regression model.
- While it is the basis for ANOVA, it is not the direct test for comparing the means of two groups.
Question 6: You are reading through a recent article that reports significant decreases in all-cause mortality for patients with malignant melanoma following treatment with a novel biological infusion. Which of the following choices refers to the probability that a study will find a statistically significant difference when one truly does exist?
A. Type II error
B. Type I error
C. Confidence interval
D. p-value
E. Power (Correct Answer)
Explanation: ***Power***
- **Power** is the probability that a study will correctly reject the null hypothesis when it is, in fact, false (i.e., will find a statistically significant difference when one truly exists).
- A study with high power minimizes the risk of a **Type II error** (failing to detect a real effect).
*Type II error*
- A **Type II error** (or **beta error**) occurs when a study fails to reject a false null hypothesis, meaning it concludes there is no significant difference when one actually exists.
- This is the **opposite** of what the question describes, which asks for the probability of *finding* a difference.
*Type I error*
- A **Type I error** (or **alpha error**) occurs when a study incorrectly rejects a true null hypothesis, concluding there is a significant difference when one does not actually exist.
- This relates to the **p-value** and the level of statistical significance (e.g., p < 0.05).
*Confidence interval*
- A **confidence interval** provides a range of values within which the true population parameter is likely to lie with a certain degree of confidence (e.g., 95%).
- It does not directly represent the probability of finding a statistically significant difference when one truly exists.
*p-value*
- The **p-value** is the probability of observing data as extreme as, or more extreme than, that obtained in the study, assuming the null hypothesis is true.
- It is used to determine statistical significance, but it is not the probability of detecting a true effect.
Question 7: A medical research study is beginning to evaluate the positive predictive value of a novel blood test for non-Hodgkin’s lymphoma. The diagnostic arm contains 700 patients with NHL, of which 400 tested positive for the novel blood test. In the control arm, 700 age-matched control patients are enrolled and 0 are found positive for the novel test. What is the PPV of this test?
A. 400 / (400 + 0) (Correct Answer)
B. 700 / (700 + 300)
C. 400 / (400 + 300)
D. 700 / (700 + 0)
E. 700 / (400 + 400)
Explanation: ***400 / (400 + 0) = 1.0 or 100%***
- The **positive predictive value (PPV)** is calculated as **True Positives / (True Positives + False Positives)**.
- In this scenario, **True Positives (TP)** are the 400 patients with NHL who tested positive, and **False Positives (FP)** are 0, as no control patients tested positive.
- This gives a PPV of 400/400 = **1.0 or 100%**, indicating that all patients who tested positive actually had the disease.
*700 / (700 + 300)*
- This calculation does not align with the formula for PPV based on the given data.
- The denominator `(700+300)` suggests an incorrect combination of various patient groups.
*400 / (400 + 300)*
- The denominator `(400+300)` incorrectly includes 300, which is the number of **False Negatives** (patients with NHL who tested negative), not False Positives.
- PPV focuses on the proportion of true positives among all positive tests, not all diseased individuals.
*700 / (700 + 0)*
- This calculation incorrectly uses the total number of patients with NHL (700) as the numerator, rather than the number of positive test results in that group.
- The numerator should be the **True Positives** (400), not the total number of diseased individuals.
*700 / (400 + 400)*
- This calculation uses incorrect values for both the numerator and denominator, not corresponding to the PPV formula.
- The numerator 700 represents the total number of patients with the disease, not those who tested positive, and the denominator incorrectly sums up values that don't represent the proper PPV calculation.
Question 8: You submit a paper to a prestigious journal about the effects of coffee consumption on mesothelioma risk. The first reviewer lauds your clinical and scientific acumen, but expresses concern that your study does not have adequate statistical power. Statistical power refers to which of the following?
A. The probability of detecting an association when no association exists.
B. The probability of not detecting an association when an association does exist.
C. The probability of detecting an association when an association does exist. (Correct Answer)
D. The first derivative of work.
E. The square root of the variance.
Explanation: ***The probability of detecting an association when an association does exist.***
- **Statistical power** is defined as the probability that a study will correctly reject a false null hypothesis, meaning it will detect a true effect or association if one exists.
- A study with **adequate statistical power** is less likely to miss a real effect.
*The probability of detecting an association when no association exists.*
- This describes a **Type I error** or **false positive**, often represented by **alpha (α)**.
- It is the probability of incorrectly concluding an effect or association exists when, in reality, there is none.
*The probability of not detecting an association when an association does exist.*
- This refers to a **Type II error** or **false negative**, represented by **beta (β)**.
- **Statistical power** is calculated as **1 - β**, so this option describes the complement of power.
*The first derivative of work.*
- The first derivative of work with respect to time represents **power** in physics, which is the rate at which work is done.
- This option is a **distractor** from physics and is unrelated to statistical power in research.
*The square root of the variance.*
- The **square root of the variance** is the **standard deviation**, a measure of the dispersion or spread of data.
- This is a statistical concept but is not the definition of statistical power.
Question 9: A research fellow proposes a nested case-control study within an existing cohort examining antibiotic exposure and C. difficile infection. The mentor suggests this design wastes the cohort structure and that relative risk should be calculated instead. The fellow argues that odds ratios from nested case-control studies approximate relative risk while being more efficient. Evaluate the validity of each position and synthesize the optimal approach.
A. Calculate RR from full cohort since all data are available and RR is more interpretable
B. Use nested case-control only if computational resources are limited, otherwise use full cohort
C. Use nested case-control with OR since it's mathematically equivalent to cohort RR with proper sampling (Correct Answer)
D. Perform both analyses and compare results to validate the nested case-control approach
E. Calculate hazard ratios using Cox regression as a compromise between efficiency and accuracy
Explanation: ***Use nested case-control with OR since it's mathematically equivalent to cohort RR with proper sampling***
- In a **nested case-control study** using **incidence density sampling** (risk-set sampling), the **Odds Ratio (OR)** provides an unbiased estimate of the **Rate Ratio** or **Relative Risk (RR)** without requiring the rare disease assumption.
- This design is highly efficient as it preserves the **temporal sequence** of the cohort while significantly reducing the need to process exposure data for the entire **at-risk population**.
*Calculate RR from full cohort since all data are available and RR is more interpretable*
- While **Relative Risk** is intuitive, calculating it requires data for the **entire denominator**, which may be prohibitively expensive or time-consuming if additional exposure processing (e.g., biomarker testing) is needed.
- The mentor's insistence ignores the **efficiency gains** of the nested design, which provides the same statistical inference with a fraction of the data processing.
*Perform both analyses and compare results to validate the nested case-control approach*
- Performing both analyses is redundant and contradicts the primary goal of using a **nested case-control** design, which is to **save resources** by not analyzing the full cohort.
- Validation is unnecessary because the **mathematical validity** of the nested case-control method is already well-established in epidemiological theory.
*Use nested case-control only if computational resources are limited, otherwise use full cohort*
- The primary limitation addressed by nested designs is usually **resource-intensive exposure assessment** (e.g., expensive lab assays) rather than mere **computational power**.
- Even if resources are available, the nested approach is often preferred in large cohorts to maintain a manageable **sub-sample** while achieving nearly identical statistical power.
*Calculate hazard ratios using Cox regression as a compromise between efficiency and accuracy*
- **Cox regression** is typically performed on the **full cohort** data, whereas the fellow is specifically proposing a sampling method to reduce the data set size.
- While a **Hazard Ratio** is a valid measure of effect, it does not solve the resource issue unless used in conjunction with a **case-cohort** or nested design, which leads back to the fellow's original point.
Question 10: Two studies examine statin therapy and stroke prevention. Study A (cohort, n=10,000) reports RR=0.75. Study B (case-control, n=2,000) reports OR=0.68. The absolute stroke rate in the general population is 2% over 5 years. Analyze these findings to determine which study provides more accurate information for clinical decision-making and why.
A. Study B because case-control designs eliminate confounding
B. Study B because odds ratios are more generalizable across populations
C. Study A because relative risk is always more accurate than odds ratio
D. Study A because the cohort design allows calculation of absolute risk reduction (Correct Answer)
E. Both are equally valid since stroke is a rare outcome making OR approximate RR
Explanation: ***Study A because the cohort design allows calculation of absolute risk reduction***
- **Study A (Cohort)** is superior for clinical decision-making because it directly measures **incidence**, enabling the calculation of **Absolute Risk Reduction (ARR)** and **Number Needed to Treat (NNT)**.
- **Cohort studies** establish a clear **temporal relationship** between exposure and outcome, providing stronger evidence for causality compared to retrospective designs.
*Study A because relative risk is always more accurate than odds ratio*
- **Relative Risk (RR)** is not inherently more "accurate," but it is mathematically more appropriate for describing risk in populations where **incidence** is known.
- The term accuracy refers to the lack of **bias** and **random error**, whereas RR and OR are simply different measures of association depending on study design.
*Study B because case-control designs eliminate confounding*
- **Case-control designs** are actually more prone to **selection and recall bias** and do not inherently eliminate **confounding** better than cohort studies.
- Confounding is typically controlled through **matching**, **stratification**, or **multivariate analysis** regardless of the study design used.
*Study B because odds ratios are more generalizable across populations*
- **Odds Ratios (OR)** are not more generalizable; they are frequently used in case-control studies because **incidence** cannot be directly calculated in those samples.
- Generalizability depends on the **representativeness** of the study sample to the target population, not on the mathematical measure (OR vs RR) used.
*Both are equally valid since stroke is a rare outcome making OR approximate RR*
- While the **Rare Disease Assumption** states that OR approximates RR when outcome prevalence is low (<10%), their clinical utility still differs.
- Even if the values are similar, the **Cohort study (Study A)** remains more robust for clinical decisions as it provides the **denominator** needed to calculate absolute benefits.
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