Meta-analysis of odds ratios and relative risks US Medical PG Practice Questions and MCQs
Practice US Medical PG questions for Meta-analysis of odds ratios and relative risks. These multiple choice questions (MCQs) cover important concepts and help you prepare for your exams.
Meta-analysis of odds ratios and relative risks US Medical PG Question 1: You have been asked to quantify the relative risk of developing bacterial meningitis following exposure to a patient with active disease. You analyze 200 patients in total, half of which are controls. In the trial arm, 30% of exposed patients ultimately contracted bacterial meningitis. In the unexposed group, only 1% contracted the disease. Which of the following is the relative risk due to disease exposure?
- A. (30 * 99) / (70 * 1)
- B. [30 / (30 + 70)] / [1 / (1 + 99)] (Correct Answer)
- C. [70 / (30 + 70)] / [99 / (1 + 99)]
- D. [[1 / (1 + 99)] / [30 / (30 + 70)]]
- E. (70 * 1) / (30 * 99)
Meta-analysis of odds ratios and relative risks Explanation: ***[30 / (30 + 70)] / [1 / (1 + 99)]***
- This formula correctly calculates the **relative risk (RR)**. The numerator represents the **incidence rate in the exposed group** (30% of 100 exposed patients = 30 cases out of 100), and the denominator represents the **incidence rate in the unexposed group** (1% of 100 unexposed patients = 1 case out of 100).
- Relative risk is the ratio of the **risk of an event** in an **exposed group** to the **risk of an event** in an **unexposed group**.
*[(30 * 99) / (70 * 1)]*
- This formula is for calculating the **odds ratio (OR)**, specifically using a 2x2 table setup where 30 represents exposed cases, 70 represents exposed non-cases, 1 represents unexposed cases, and 99 represents unexposed non-cases.
- The odds ratio is a measure of association between an exposure and an outcome, representing the **odds of an outcome** given exposure compared to the odds of the outcome without exposure.
*[70 / (30 + 70)] / [99 / (1 + 99)]*
- This formula calculates the **relative risk of *not* developing the disease**, which is the inverse of what the question asks for.
- It compares the proportion of exposed individuals who *do not* contract the disease to the proportion of unexposed individuals who *do not* contract the disease.
*[[1 / (1 + 99)] / [30 / (30 + 70)]]*
- This formula calculates the **inverse of the relative risk**, which is not what the question asks for.
- It would represent the ratio of the incidence in the unexposed group to the incidence in the exposed group.
*[(70 * 1) / (30 * 99)]*
- This is an **incorrect variation** of the odds ratio calculation, with the terms in the numerator and denominator swapped compared to the standard formula.
- Therefore, it does not represent the relative risk or a correctly calculated odds ratio.
Meta-analysis of odds ratios and relative risks US Medical PG Question 2: A research team develops a new monoclonal antibody checkpoint inhibitor for advanced melanoma that has shown promise in animal studies as well as high efficacy and low toxicity in early phase human clinical trials. The research team would now like to compare this drug to existing standard of care immunotherapy for advanced melanoma. The research team decides to conduct a non-randomized study where the novel drug will be offered to patients who are deemed to be at risk for toxicity with the current standard of care immunotherapy, while patients without such risk factors will receive the standard treatment. Which of the following best describes the level of evidence that this study can offer?
- A. Level 1
- B. Level 3 (Correct Answer)
- C. Level 5
- D. Level 4
- E. Level 2
Meta-analysis of odds ratios and relative risks Explanation: ***Level 3***
- A **non-randomized controlled trial** like the one described, where patient assignment to treatment groups is based on specific characteristics (risk of toxicity), falls into Level 3 evidence.
- This level typically includes **non-randomized controlled trials** and **well-designed cohort studies** with comparison groups, which are prone to selection bias and confounding.
- The study compares two treatments but lacks randomization, making it Level 3 evidence.
*Level 1*
- Level 1 evidence is the **highest level of evidence**, derived from **systematic reviews and meta-analyses** of multiple well-designed randomized controlled trials or large, high-quality randomized controlled trials.
- The described study is explicitly stated as non-randomized, ruling out Level 1.
*Level 2*
- Level 2 evidence involves at least one **well-designed randomized controlled trial** (RCT) or **systematic reviews** of randomized trials.
- The current study is *non-randomized*, which means it cannot be classified as Level 2 evidence, as randomization is a key criterion for this level.
*Level 4*
- Level 4 evidence includes **case series**, **case-control studies**, and **poorly designed cohort or case-control studies**.
- While the study is non-randomized, it is a controlled comparative trial rather than a case series or retrospective case-control study, placing it at Level 3.
*Level 5*
- Level 5 evidence is the **lowest level of evidence**, typically consisting of **expert opinion** without explicit critical appraisal, or based on physiology, bench research, or animal studies.
- While the drug was initially tested in animal studies, the current human comparative study offers a higher level of evidence than expert opinion or preclinical data.
Meta-analysis of odds ratios and relative risks US Medical PG 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.
Meta-analysis of odds ratios and relative risks 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.
Meta-analysis of odds ratios and relative risks US Medical PG Question 4: 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
Meta-analysis of odds ratios and relative risks 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.
Meta-analysis of odds ratios and relative risks US Medical PG Question 5: 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)
Meta-analysis of odds ratios and relative risks 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.
Meta-analysis of odds ratios and relative risks US Medical PG Question 6: Study X examined the relationship between coffee consumption and lung cancer. The authors of Study X retrospectively reviewed patients' reported coffee consumption and found that drinking greater than 6 cups of coffee per day was associated with an increased risk of developing lung cancer. However, Study X was criticized by the authors of Study Y. Study Y showed that increased coffee consumption was associated with smoking. What type of bias affected Study X, and what study design is geared to reduce the chance of that bias?
- A. Observer bias; double blind analysis
- B. Selection bias; randomization
- C. Lead time bias; placebo
- D. Measurement bias; blinding
- E. Confounding; randomization (Correct Answer)
Meta-analysis of odds ratios and relative risks Explanation: ***Confounding; randomization***
- Study Y suggests that **smoking** is a **confounding variable** because it is associated with both increased coffee consumption (exposure) and increased risk of lung cancer (outcome), distorting the apparent relationship between coffee and lung cancer.
- **Randomization** in experimental studies (such as randomized controlled trials) helps reduce confounding by ensuring that known and unknown confounding factors are evenly distributed among study groups.
- In observational studies where randomization is not possible, confounding can be addressed through **stratification**, **matching**, or **multivariable adjustment** during analysis.
*Observer bias; double blind analysis*
- **Observer bias** occurs when researchers' beliefs or expectations influence the study outcome, which is not the primary issue described here regarding the relationship between coffee, smoking, and lung cancer.
- **Double-blind analysis** is a method to mitigate observer bias by ensuring neither participants nor researchers know who is in the control or experimental groups.
*Selection bias; randomization*
- **Selection bias** happens when the study population is not representative of the target population, leading to inaccurate results, which is not directly indicated by the interaction between coffee and smoking.
- While **randomization** is used to reduce selection bias by creating comparable groups, the core problem identified in Study X is confounding, not flawed participant selection.
*Lead time bias; placebo*
- **Lead time bias** occurs in screening programs when early detection without improved outcomes makes survival appear longer, an issue unrelated to the described association between coffee, smoking, and lung cancer.
- A **placebo** is an inactive treatment used in clinical trials to control for psychological effects, and its relevance here is limited to treatment intervention studies.
*Measurement bias; blinding*
- **Measurement bias** arises from systematic errors in data collection, such as inaccurate patient reporting of coffee consumption, but the main criticism from Study Y points to a third variable (smoking) affecting the association, not just flawed measurement.
- **Blinding** helps reduce measurement bias by preventing participants or researchers from knowing group assignments, thus minimizing conscious or unconscious influences on data collection.
Meta-analysis of odds ratios and relative risks US Medical PG Question 7: An epidemiologist is evaluating the efficacy of Noxbinle in preventing HCC deaths at the population level. A clinical trial shows that over 5 years, the mortality rate from HCC was 25% in the control group and 15% in patients treated with Noxbinle 100 mg daily. Based on this data, how many patients need to be treated with Noxbinle 100 mg to prevent, on average, one death from HCC?
- A. 20
- B. 73
- C. 10 (Correct Answer)
- D. 50
- E. 100
Meta-analysis of odds ratios and relative risks Explanation: ***10***
- The **number needed to treat (NNT)** is calculated by first finding the **absolute risk reduction (ARR)**.
- **ARR** = Risk in control group - Risk in treatment group = 25% - 15% = **10%** (or 0.10).
- **NNT = 1 / ARR** = 1 / 0.10 = **10 patients**.
- This means that **10 patients must be treated with Noxbinle to prevent one death from HCC** over 5 years.
*20*
- This would result from an ARR of 5% (1/0.05 = 20), which is not supported by the data.
- May arise from miscalculating the risk difference or incorrectly halving the actual ARR.
*73*
- This value does not correspond to any standard calculation of NNT from the given mortality rates.
- May result from confusion with other epidemiological measures or calculation error.
*50*
- This would correspond to an ARR of 2% (1/0.02 = 50), which significantly underestimates the actual risk reduction.
- Could result from incorrectly calculating the difference as a proportion rather than absolute percentage points.
*100*
- This would correspond to an ARR of 1% (1/0.01 = 100), grossly underestimating the treatment benefit.
- May result from confusing ARR with relative risk reduction or other calculation errors.
Meta-analysis of odds ratios and relative risks US Medical PG Question 8: A group of environmental health scientists recently performed a nationwide cross-sectional study that investigated the risk of head and neck cancers in patients with a history of cigar and pipe smoking. In collaboration with three teams of epidemiologists that have each conducted similar cross-sectional studies in their respective countries, they have agreed to contribute their data to an international pooled analysis of the relationship between non-cigarette tobacco consumption and prevalence of head and neck cancers. Which of the following statements regarding the pooled analysis in comparison to the individual studies is true?
- A. The results are less precise.
- B. It overcomes limitations in the quality of individual studies.
- C. It is able to provide evidence of causality.
- D. The level of clinical evidence is lower.
- E. The likelihood of type II errors is decreased. (Correct Answer)
Meta-analysis of odds ratios and relative risks Explanation: ***The likelihood of type II errors is decreased.***
- A pooled analysis or **meta-analysis** combines data from multiple studies, significantly increasing the **overall sample size**.
- A larger sample size enhances the statistical power, making it less likely to miss a real effect and thus reducing the probability of **Type II errors** (false negatives).
*The results are less precise.*
- Combining data from multiple studies in a **pooled analysis** generally leads to **more precise estimates** due to the larger sample size and increased statistical power.
- Increased precision is reflected in narrower confidence intervals, offering a more reliable estimate of the effect.
*It overcomes limitations in the quality of individual studies.*
- A pooled analysis **does not inherently overcome limitations** in the design, methodology, or quality of the individual studies included.
- If the original studies have significant biases or flaws, these limitations can be propagated or even amplified in the pooled results.
*It is able to provide evidence of causality.*
- Pooled analyses of **cross-sectional studies**, like the ones described, can identify **associations** but cannot establish **causality**.
- Cross-sectional studies measure exposure and outcome simultaneously, making it impossible to determine the temporal sequence necessary to infer cause and effect.
*The level of clinical evidence is lower.*
- Combining multiple studies, especially well-conducted ones, in a pooled analysis or **meta-analysis** generally **increases the level of clinical evidence**, placing it higher than individual observational studies.
- This is because a pooled analysis offers a more robust and comprehensive view of the existing evidence.
Meta-analysis of odds ratios and relative risks US Medical PG Question 9: In 2013 the national mean score on the USMLE Step 1 exam was 227 with a standard deviation of 22. Assuming that the scores for 15,000 people follow a normal distribution, approximately how many students scored above the mean but below 250?
- A. 5,100 (Correct Answer)
- B. 4,500
- C. 6,000
- D. 3,750
- E. 6,750
Meta-analysis of odds ratios and relative risks Explanation: ***5,100***
- To solve this, first calculate the **z-score** for 250: (250 - 227) / 22 = 1.045.
- Using a **z-table**, the area under the curve from the mean (z=0) to z=1.045 is approximately 0.353. Multiplying this by 15,000 students gives approximately **5,295 students**, which is closest to 5,100.
*4,500*
- This answer would imply a smaller proportion of students between the mean and 250 (around 30%), which is lower than the calculated z-score of 1.045 suggests.
- It does not accurately reflect the area under the **normal distribution curve** for the given range.
*6,000*
- This option would mean that approximately 40% of students scored in this range, which would correspond to a z-score much higher than 1.045 or a different standard deviation.
- This calculation overestimates the number of students within the specified range.
*3,750*
- This value represents 25% of the total students (15,000 * 0.25), indicating that only a quarter of the distribution lies in this range.
- This significantly underestimates the proportion of students scoring between the mean and 250 for the given standard deviation.
*6,750*
- This option reflects approximately 45% of the total student population (15,000 * 0.45), which would correspond to a much larger z-score or a different distribution.
- This value is an overestimation and does not align with the standard normal distribution probabilities for the given parameters.
Meta-analysis of odds ratios and relative risks US Medical PG Question 10: A randomized controlled trial was initiated to evaluate a novel DPP-4 inhibitor for blood glucose management in diabetic patients. The study used a commonly prescribed sulfonylurea as the standard of care treatment. 2,000 patients were enrolled in the study with 1,000 patients in each arm. One of the primary outcomes was the development of diabetic nephropathy during treatment. This outcome occurred in 68 patients on the DPP-4 inhibitor and 134 patients on the sulfonylurea. What is the relative risk reduction (RRR) for patients using the DPP-4 inhibitor compared with the sulfonylurea?
- A. 23%
- B. 49% (Correct Answer)
- C. 33%
- D. 59%
- E. 43%
Meta-analysis of odds ratios and relative risks Explanation: ***49%***
- To calculate **relative risk reduction (RRR)**, first determine the **event rate (ER)** for each group.
- ER (DPP-4 inhibitor) = 68/1000 = 0.068. ER (Sulfonylurea) = 134/1000 = 0.134.
- Next, calculate the **absolute risk reduction (ARR)**: ARR = ER (Sulfonylurea) - ER (DPP-4 inhibitor) = 0.134 - 0.068 = 0.066.
- Finally, calculate RRR: RRR = ARR / ER (Sulfonylurea) = 0.066 / 0.134 ≈ 0.4925 or **49%**.
*23%*
- This value is incorrect and does not result from the proper application of the **relative risk reduction (RRR)** formula.
- A common mistake is to reverse the subtrahend and minuend in the numerator or denominator.
*33%*
- This value is incorrect and does not result from the proper application of the **relative risk reduction (RRR)** formula.
- Incorrect calculations in either the numerator or denominator of the **RRR formula** would lead to this incorrect result.
*59%*
- This value is incorrect and is likely the result of an error in calculating either the **absolute risk reduction (ARR)** or dividing it by the wrong **event rate**.
- Always ensure the correct event rates are used for the control group and the intervention group.
*43%*
- This value is incorrect and does not align with the correct calculation of **relative risk reduction (RRR)**.
- Errors in setting up the formula or executing the division could lead to this result.
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