Bayesian alternatives to p-values — MCQs

10 questions

A scientist in Chicago is studying a new blood test to detect Ab to EBV with increased sensitivity and specificity. So far, her best attempt at creating such an exam reached 82% sensitivity and 88% specificity. She is hoping to increase these numbers by at least 2 percent for each value. After several years of work, she believes that she has actually managed to reach a sensitivity and specificity much greater than what she had originally hoped for. She travels to China to begin testing her newest blood test. She finds 2,000 patients who are willing to participate in her study. Of the 2,000 patients, 1,200 of them are known to be infected with EBV. The scientist tests these 1,200 patients' blood and finds that only 120 of them tested negative with her new exam. Of the patients who are known to be EBV-free, only 20 of them tested positive. Given these results, which of the following correlates with the exam's specificity?

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 25-year-old man with a genetic disorder presents for genetic counseling because he is concerned about the risk that any children he has will have the same disease as himself. Specifically, since childhood he has had difficulty breathing requiring bronchodilators, inhaled corticosteroids, and chest physiotherapy. He has also had diarrhea and malabsorption requiring enzyme replacement therapy. If his wife comes from a population where 1 in 10,000 people are affected by this same disorder, which of the following best represents the likelihood a child would be affected as well?

A 26-year-old medical student comes to the physician with a 3-week history of night sweats and myalgias. During this time, he has also had a 3.6-kg (8-lb) weight loss. He returned from a 6-month tropical medicine rotation in Cambodia 1 month ago. A chest x-ray (CXR) shows reticulonodular opacities suggestive of active tuberculosis (TB). The student is curious about his likelihood of having active TB. He reads a study that compares sputum testing results between 2,800 patients with likely active TB on a basis of history, clinical symptoms, and CXR pattern and 2,400 controls. The results are shown: Sputum testing positive for TB Sputum testing negative for TB Total Active TB likely on basis of history, clinical symptoms, and CXR pattern 700 2100 2,800 Active TB not likely on basis of history, clinical symptoms, and CXR pattern 300 2100 2,400 Total 1000 4200 5,200 Which of the following values reflects the probability that a patient with a diagnosis of active TB on the basis of history, clinical symptoms, and CXR pattern actually has active TB?

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 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 student health coordinator plans on leading a campus-wide HIV screening program that will be free for the entire undergraduate student body. The goal is to capture as many correct HIV diagnoses as possible with the fewest false positives. The coordinator consults with the hospital to see which tests are available to use for this program. Test A has a sensitivity of 0.92 and a specificity of 0.99. Test B has a sensitivity of 0.95 and a specificity of 0.96. Test C has a sensitivity of 0.98 and a specificity of 0.93. Which of the following testing schemes should the coordinator pursue?

A prospective cohort study was conducted to assess the relationship between LDL and the incidence of heart disease. The patients were selected at random. Results showed a 10-year relative risk of 2.3 for people with elevated LDL levels compared to individuals with normal LDL levels. The 95% confidence interval was 1.05-3.50. This study is most likely to have which of the following p values?

A pharmaceutical company has modified one of its existing antibiotics to have an improved toxicity profile. The new antibiotic blocks protein synthesis by first entering the cell and then binding to active ribosomes. The antibiotic mimics the structure of aminoacyl-tRNA. The drug is covalently bonded to the existing growing peptide chain via peptidyl transferase, thereby impairing the rest of protein synthesis and leading to early polypeptide truncation. Where is the most likely site that this process occurs?

Q10

A first-year medical student is conducting a summer project with his medical school's pediatrics department using adolescent IQ data from a database of 1,252 patients. He observes that the mean IQ of the dataset is 100. The standard deviation was calculated to be 10. Assuming that the values are normally distributed, approximately 87% of the measurements will fall in between which of the following limits?

Bayesian alternatives to p-values US Medical PG Practice Questions and MCQs

Question 1: A scientist in Chicago is studying a new blood test to detect Ab to EBV with increased sensitivity and specificity. So far, her best attempt at creating such an exam reached 82% sensitivity and 88% specificity. She is hoping to increase these numbers by at least 2 percent for each value. After several years of work, she believes that she has actually managed to reach a sensitivity and specificity much greater than what she had originally hoped for. She travels to China to begin testing her newest blood test. She finds 2,000 patients who are willing to participate in her study. Of the 2,000 patients, 1,200 of them are known to be infected with EBV. The scientist tests these 1,200 patients' blood and finds that only 120 of them tested negative with her new exam. Of the patients who are known to be EBV-free, only 20 of them tested positive. Given these results, which of the following correlates with the exam's specificity?

A. 82%
B. 90%
C. 84%
D. 86%
E. 98% (Correct Answer)

Explanation: ***98%*** - **Specificity** measures the proportion of **true negatives** among all actual negatives. - In this case, 800 patients are known to be EBV-free (actual negatives), and 20 of them tested positive (false positives). This means 800 - 20 = 780 tested negative (true negatives). Specificity = (780 / 800) * 100% = **98%**. *82%* - This value represents the *original sensitivity* before the scientist’s new attempts to improve the test. - It does not reflect the *newly calculated specificity* based on the provided data. *90%* - This value represents the *newly calculated sensitivity* of the test, not the specificity. - Out of 1200 EBV-infected patients, 120 tested negative (false negatives), meaning 1080 tested positive (true positives). Sensitivity = (1080 / 1200) * 100% = 90%. *84%* - This percentage is not directly derived from the information given for either sensitivity or specificity after the new test results. - It does not correspond to any of the calculated values for the new test's performance. *86%* - This percentage is not directly derived from the information given for either sensitivity or specificity after the new test results. - It does not correspond to any of the calculated values for the new test's performance.

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

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.

Question 3: A 25-year-old man with a genetic disorder presents for genetic counseling because he is concerned about the risk that any children he has will have the same disease as himself. Specifically, since childhood he has had difficulty breathing requiring bronchodilators, inhaled corticosteroids, and chest physiotherapy. He has also had diarrhea and malabsorption requiring enzyme replacement therapy. If his wife comes from a population where 1 in 10,000 people are affected by this same disorder, which of the following best represents the likelihood a child would be affected as well?

A. 0.01%
B. 2%
C. 0.5%
D. 1% (Correct Answer)
E. 50%

Explanation: ***Correct Option: 1%*** - The patient's symptoms (difficulty breathing requiring bronchodilators, inhaled corticosteroids, and chest physiotherapy; diarrhea and malabsorption requiring enzyme replacement therapy) are classic for **cystic fibrosis (CF)**, an **autosomal recessive disorder**. - For an autosomal recessive disorder with a prevalence of 1 in 10,000 in the general population, **q² = 1/10,000**, so **q = 1/100 = 0.01**. The carrier frequency **(2pq)** is approximately **2q = 2 × (1/100) = 1/50 = 0.02**. - The affected man is **homozygous recessive (aa)** and will always pass on the recessive allele. His wife has a **1/50 chance of being a carrier (Aa)**. If she is a carrier, she has a **1/2 chance of passing on the recessive allele**. - Therefore, the probability of an affected child = **(Probability wife is a carrier) × (Probability wife passes recessive allele) = 1/50 × 1/2 = 1/100 = 1%**. *Incorrect Option: 0.01%* - This percentage is too low and does not correctly account for the carrier frequency in the population and the probability of transmission from a carrier mother. *Incorrect Option: 2%* - This represents approximately the carrier frequency (1/50 ≈ 2%), but does not account for the additional 1/2 probability that a carrier mother would pass on the recessive allele. *Incorrect Option: 0.5%* - This value would be correct if the carrier frequency were 1/100 instead of 1/50, which does not match the given population prevalence. *Incorrect Option: 50%* - **50%** would be the risk if both parents were carriers of an autosomal recessive disorder (1/4 chance = 25% for affected, but if we know one parent passes the allele, conditional probability changes). More accurately, 50% would apply if the disorder were **autosomal dominant** with one affected parent, which is not the case here.

Question 4: A 26-year-old medical student comes to the physician with a 3-week history of night sweats and myalgias. During this time, he has also had a 3.6-kg (8-lb) weight loss. He returned from a 6-month tropical medicine rotation in Cambodia 1 month ago. A chest x-ray (CXR) shows reticulonodular opacities suggestive of active tuberculosis (TB). The student is curious about his likelihood of having active TB. He reads a study that compares sputum testing results between 2,800 patients with likely active TB on a basis of history, clinical symptoms, and CXR pattern and 2,400 controls. The results are shown: Sputum testing positive for TB Sputum testing negative for TB Total Active TB likely on basis of history, clinical symptoms, and CXR pattern 700 2100 2,800 Active TB not likely on basis of history, clinical symptoms, and CXR pattern 300 2100 2,400 Total 1000 4200 5,200 Which of the following values reflects the probability that a patient with a diagnosis of active TB on the basis of history, clinical symptoms, and CXR pattern actually has active TB?

A. 1.4
B. 0.50
C. 0.70
D. 0.88
E. 0.25 (Correct Answer)

Explanation: ***0.25*** - This value represents the **positive predictive value (PPV)** for active TB based on the initial clinical assessment criteria (history, symptoms, CXR). - PPV is calculated as the number of true positives (700) divided by the total number of individuals with a positive clinical diagnosis (700 + 2100 = 2800). So, 700 / 2800 = 0.25. - **This answers the question**: the probability that someone with a clinical diagnosis of active TB actually has the disease. *Incorrect 1.4* - This value is not a valid probability, as probabilities must be between 0 and 1.0. - It might arise from an incorrect calculation or misinterpretation of the provided data. *Incorrect 0.50* - This value does not correspond to any standard diagnostic metric calculated from the provided data. - The actual prevalence of TB (based on positive sputum) is 1000/5200 = 0.19, not 0.50. - This is likely a distractor with no meaningful interpretation in this context. *Incorrect 0.70* - This value represents the **sensitivity** of the sputum test for detecting active TB. - Sensitivity is calculated as true positives (700) divided by total with disease (700 + 300 = 1000). So, 700 / 1000 = 0.70. - Sensitivity tells us how good the test is at detecting disease when present, not the probability of having disease given a positive clinical diagnosis. *Incorrect 0.88* - This value represents the **specificity** of the clinical assessment. - Specificity is calculated as true negatives (2100) divided by total without disease (2100 + 300 = 2400). So, 2100 / 2400 = 0.875 ≈ 0.88. - Specificity tells us how good the assessment is at ruling out disease in those without it, not the probability of disease given a positive assessment.

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: 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 7: A student health coordinator plans on leading a campus-wide HIV screening program that will be free for the entire undergraduate student body. The goal is to capture as many correct HIV diagnoses as possible with the fewest false positives. The coordinator consults with the hospital to see which tests are available to use for this program. Test A has a sensitivity of 0.92 and a specificity of 0.99. Test B has a sensitivity of 0.95 and a specificity of 0.96. Test C has a sensitivity of 0.98 and a specificity of 0.93. Which of the following testing schemes should the coordinator pursue?

A. Test A on the entire student body followed by Test B on those who are positive
B. Test A on the entire student body followed by Test C on those who are positive
C. Test C on the entire student body followed by Test B on those who are positive
D. Test C on the entire student body followed by Test A on those who are positive (Correct Answer)
E. Test B on the entire student body followed by Test A on those who are positive

Explanation: ***Test C on the entire student body followed by Test A on those who are positive*** - To "capture as many correct HIV diagnoses as possible" (maximize true positives), the initial screening test should have the **highest sensitivity**. Test C has the highest sensitivity (0.98). - To "capture as few false positives as possible" (maximize true negatives and confirm diagnoses), the confirmatory test should have the **highest specificity**. Test A has the highest specificity (0.99). *Test A on the entire student body followed by Test B on those who are positive* - Starting with Test A (sensitivity 0.92) would miss more true positive cases than starting with Test C (sensitivity 0.98), failing the goal of **capturing as many cases as possible**. - Following with Test B (specificity 0.96) would result in more false positives than following with Test A (specificity 0.99). *Test A on the entire student body followed by Test C on those who are positive* - This scheme would miss many true positive cases initially due to Test A's lower sensitivity compared to Test C. - Following with Test C would introduce more false positives than necessary, as it has a lower specificity (0.93) than Test A (0.99). *Test C on the entire student body followed by Test B on those who are positive* - While Test C is a good initial screen for its high sensitivity, following it with Test B (specificity 0.96) is less optimal than Test A (specificity 0.99) for minimizing false positives in the confirmation step. - This combination would therefore yield more false positives in the confirmatory stage than using Test A. *Test B on the entire student body followed by Test A on those who are positive* - Test B has a sensitivity of 0.95, which is lower than Test C's sensitivity of 0.98, meaning it would miss more true positive cases at the initial screening stage. - While Test A provides excellent specificity for confirmation, the initial screening step is suboptimal for the goal of capturing as many diagnoses as possible.

Question 8: A prospective cohort study was conducted to assess the relationship between LDL and the incidence of heart disease. The patients were selected at random. Results showed a 10-year relative risk of 2.3 for people with elevated LDL levels compared to individuals with normal LDL levels. The 95% confidence interval was 1.05-3.50. This study is most likely to have which of the following p values?

A. 0.20
B. 0.06
C. 0.08
D. 0.04 (Correct Answer)
E. 0.10

Explanation: ***0.04*** - A 95% confidence interval that **does not include 1 (one)** suggests a **statistically significant** association, meaning the p-value is likely to be **less than 0.05**. - The given CI of 1.05-3.50 for the relative risk (RR) is entirely above 1, indicating a significant positive association, and therefore, a p-value less than 0.05. *0.20* - A p-value of 0.20 is **greater than 0.05**, which would imply the finding is **not statistically significant**. - If the p-value were 0.20, the 95% confidence interval would likely **include 1**, suggesting no significant difference in risk. *0.06* - A p-value of 0.06 is **greater than 0.05**, indicating that the association is **not statistically significant at the conventional alpha level**. - If the p-value were 0.06, the 95% confidence interval would likely **include 1**, or be very close to including it, contradicting the given CI of 1.05-3.50. *0.08* - A p-value of 0.08 is **greater than 0.05**, indicating that the finding is **not statistically significant**. - If the p-value were 0.08, the 95% confidence interval would almost certainly **include 1**, which is inconsistent with the provided interval. *0.10* - A p-value of 0.10 is **greater than 0.05**, which signifies that the finding is **not statistically significant**. - If the p-value were 0.10, the 95% confidence interval for the relative risk would typically **include 1**, contradicting the given confidence interval.

Question 9: A pharmaceutical company has modified one of its existing antibiotics to have an improved toxicity profile. The new antibiotic blocks protein synthesis by first entering the cell and then binding to active ribosomes. The antibiotic mimics the structure of aminoacyl-tRNA. The drug is covalently bonded to the existing growing peptide chain via peptidyl transferase, thereby impairing the rest of protein synthesis and leading to early polypeptide truncation. Where is the most likely site that this process occurs?

A. E site
B. 30S small subunit
C. A site (Correct Answer)
D. 40S small subunit
E. P site

Explanation: ***A site*** - The **A (aminoacyl) site** is where incoming aminoacyl-tRNAs bind during translation, bringing new amino acids to the ribosome. Since the antibiotic mimics **aminoacyl-tRNA** and is covalently bonded to the peptide chain by **peptidyl transferase**, its action must occur at the A site. - Binding at the A site and subsequent peptide bond formation with the antibiotic would lead to premature polypeptide truncation, as no further amino acids can be added. *E site* - The **E (exit) site** is where deacylated tRNAs are released from the ribosome after having delivered their amino acid to the growing peptide chain in the P site. - The antibiotic's mechanism of action, involving binding and covalent incorporation into the peptide, does not align with the function of the E site. *30S small subunit* - The **30S small ribosomal subunit** in prokaryotes is primarily involved in mRNA binding and decoding, ensuring the correct aminoacyl-tRNA binds to the mRNA codon. - While the antibiotic binds to active ribosomes, its key action described as mimicking aminoacyl-tRNA and being incorporated by peptidyl transferase points to a specific binding site within the ribosome rather than the entire subunit's general function. *40S small subunit* - The **40S small ribosomal subunit** is found in **eukaryotic ribosomes**, not prokaryotic ones, and is involved in mRNA binding during initiation. - The question implies an antibiotic targeting bacterial protein synthesis (given its discussion of modifying an existing antibiotic), making eukaryotic ribosomal subunits an unlikely target. *P site* - The **P (peptidyl) site** holds the tRNA carrying the growing polypeptide chain. Peptidyl transferase activity forms a peptide bond between the amino acid in the A site and the peptide in the P site. - While peptidyl transferase is involved, the antibiotic *mimics* aminoacyl-tRNA, which is delivered to the A site for peptide bond formation, rather than the P site which already holds the growing chain.

Question 10: A first-year medical student is conducting a summer project with his medical school's pediatrics department using adolescent IQ data from a database of 1,252 patients. He observes that the mean IQ of the dataset is 100. The standard deviation was calculated to be 10. Assuming that the values are normally distributed, approximately 87% of the measurements will fall in between which of the following limits?

A. 85–115 (Correct Answer)
B. 95–105
C. 65–135
D. 80–120
E. 70–130

Explanation: ***85–115*** - For a **normal distribution**, approximately 87% of data falls within **±1.5 standard deviations** from the mean. - With a mean of 100 and a standard deviation of 10, the range is 100 ± (1.5 * 10) = 100 ± 15, which gives **85–115**. *95–105* - This range represents **±0.5 standard deviations** from the mean (100 ± 5), which covers only about 38% of the data. - This is a much narrower range and does not encompass 87% of the observations as required. *65–135* - This range represents **±3.5 standard deviations** from the mean (100 ± 35), which would cover over 99.9% of the data. - Thus, this interval is too wide for 87% of the measurements. *80–120* - This range represents **±2 standard deviations** from the mean (100 ± 20), which covers approximately 95% of the data. - While a common interval, it is wider than necessary for 87% of the data. *70–130* - This range represents **±3 standard deviations** from the mean (100 ± 30), which covers approximately 99.7% of the data. - This interval is significantly wider than required to capture 87% of the data.

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