Join us on Telegram : Click here : @eduwaves360 Unlocked the Medical premiums Click here : www.eduwaves360.com Medical Courses : Discussion Group : https://t.me/usmle_study_materials_2 @usmle_discussion_group eduwaves360.com | Telegram : @eduwaves360 Question # 1 A public health campaign increases vaccination rates against human papillomaviruses 16 and 18. Increased vaccination rates would have which of the following effects on the Papanicolaou test? Answer A Increased true negative rate B Decreased true positive rate C Decreased positive predictive value D Increased positive likelihood ratio E Increased negative likelihood ratio F Decreased negative predictive value Image eduwaves360.com | Telegram : @eduwaves360 Hint Increasing the rate of HPV vaccination would decrease the number of new cases of cervical cancer, thereby decreasing disease prevalence. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Increased true negative rate Explanation Why True negative rate (specificity) is not influenced by the prevalence of the disease. B - Decreased true positive rate Explanation Why True positive rate (sensitivity) is not influenced by the prevalence of a disease. C - Decreased positive predictive value Explanation Why The positive predictive value (PPV) of a test is directly proportional to prevalence. Therefore, PPV decreases as the prevalence of a disease decreases. D - Increased positive likelihood ratio Explanation Why Positive likelihood ratio is not influenced by the prevalence of a disease. eduwaves360.com | Telegram : @eduwaves360 E - Increased negative likelihood ratio Explanation Why Negative likelihood ratio is not influenced by the prevalence of a disease. F - Decreased negative predictive value Explanation Why The negative predictive value (NPV) of a test is inversely proportional to disease prevalence. In this case increasing the HPV vaccination rate would decrease the prevalence of the disease, which, in turn, would increase the NPV. eduwaves360.com | Telegram : @eduwaves360 Question # 2 An investigator is studying the frequency of polycythemia in a population of a remote, mountainous region. A representative sample of 100 men shows a normal distribution of hemoglobin concentration with a mean concentration of 17 g/dL and a standard error of 0.1 g/dL. Which of the following best represents the probability that a subject will have a hemoglobin concentration greater than 18 g/dL? Answer A 30% B 15% C 95% D 99% E 5% F 70% Image eduwaves360.com | Telegram : @eduwaves360 Hint The standard deviation can be calculated using the standard error of the mean, which is defined as: SEM = standard deviation/√(sample size). Accordingly, standard deviation = SEM x √(sample size). In this case: standard deviation = 0.1 g/dL x √(100) = 1 g/dL. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 30% Explanation Why In a normal distribution, one standard deviation above and below the mean covers approx. 68% of the sample. With the standard deviation for this population at 1 g/dL, 68% of the population would fall between 16 g/dL and 18 g/dL. The remaining 32% (∼30%) of the population would have hemoglobin concentrations that lie outside this range. B - 15% Image Explanation Why In a normal distribution, one standard deviation above and below the mean covers approximately eduwaves360.com | Telegram : @eduwaves360 68% of the sample. With the standard deviation for this population at 1 g/dL, 68% of the population will fall between 16 g/dL and 18 g/dL. The remaining 32% of the population will fall above or below one standard deviation of the mean: half (16%) would fall under 16 g/dL while the other half (16%) would be greater than 18 g/dL. 15% is the closest approximation to the probability that a subject has a hemoglobin concentration greater than 18 g/dL. C - 95% Explanation Why In a normal distribution, two standard deviations above and below the mean covers approx. 95% of the sample. With the standard deviation for this population at 1 g/dL, 95% of the population would fall between 15 g/dL and 19 g/dL. D - 99% Explanation Why In a normal distribution, three standard deviations above and below the mean cover approx. 99% of the sample. With the standard deviation for this population at 1 g/dL, 99% of the population would fall between 14 g/dL and 20 g/dL. E - 5% Explanation Why In a normal distribution, two standard deviations above and below the mean cover approx. 95% of the sample. With the standard deviation for this population calculated at 1 g/dL, 95% of the population would fall between 15 g/dL and 19 g/dL. The remaining 5% of the population would have hemoglobin concentrations that lie outside this range. eduwaves360.com | Telegram : @eduwaves360 F - 70% Explanation Why In a normal distribution, one standard deviation above and below the mean covers approx. 68% of the sample. With the standard deviation for this population at 1 g/dL, approximately 70% (68%) of the population would fall between 16 g/dL and 18 g/dL. eduwaves360.com | Telegram : @eduwaves360 Question # 3 A group of investigators is studying the association between a fire retardant chemical used on furniture and interstitial lung disease. They use hospital records to identify 50 people who have been diagnosed with interstitial lung disease. They also identify a group of 50 people without interstitial lung disease who are matched in age and geographic location to those with the disease. The participants' exposure to the chemical is assessed by surveys and home visits. Which of the following best describes this study design? Answer A Case-control study B Cross-sectional study C Case series D Retrospective cohort study E Randomized controlled trial Image eduwaves360.com | Telegram : @eduwaves360 Hint This study identifies groups of people with and without the disease of interest, and then compares the presence of risk factors between these two groups. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Case-control study Explanation Why In case-control studies, researchers select patients with (cases) and without the outcome of interest (control group). In this study, the outcome of interest is interstitial lung disease. This study design allows odds ratios to be determined, i.e. the odds of developing interstitial lung disease dependent on exposure to the fire retardant chemical. Case-control studies are retrospective studies and are prone to recall bias. B - Cross-sectional study Explanation Why Cross-sectional studies analyze data from a representative subset of a population at a specific point in time and can be used to determine prevalence or risk factors for a disease. Because this study design selects individuals based on the presence of a disease, it is not representative of the population, and therefore, not a cross-sectional study. C - Case series Explanation Why Case series are a series of reports on similar cases of a disease or exposure (i.e., to a drug). Unlike the study described here, case series do not use a matched comparative group (i.e., control group) in their design. eduwaves360.com | Telegram : @eduwaves360 D - Retrospective cohort study Explanation Why For retrospective cohort studies individuals are selected based on exposure and then it is observed if these individuals developed a specific disease. Because this study selected patients based on their known outcomes, it is not a cohort study. E - Randomized controlled trial Explanation Why In a randomized controlled trial, participants are prospectively selected and randomly assigned to control or treatment groups. Because this study retroactively selects participants and does not have an intervention of interest, it is not a randomized controlled trial. eduwaves360.com | Telegram : @eduwaves360 Question # 4 A 1-year-old boy is brought to the emergency room by his parents because of inconsolable crying and diarrhea for the past 6 hours. As the physician is concerned about acute appendicitis, she consults the literature base. She finds a paper with a table that summarizes data regarding the diagnostic accuracy of multiple clinical findings for appendicitis: Clinical finding Sensitivity Specificity Abdominal guarding (in children of all ages) 0.70 0.85 Anorexia (in children of all ages) 0.75 0.50 Abdominal rebound (in children ≥ 5 years of age) 0.85 0.65 Vomiting (in children of all ages) 0.40 0.63 Fever (in children from 1 month to 2 years of age) 0.80 0.80 Based on the table, the absence of which clinical finding would most accurately rule out appendicitis in this patient? Answer A Guarding B Fever C Anorexia D Rebound Image eduwaves360.com | Telegram : @eduwaves360 Answer E Image Vomiting eduwaves360.com | Telegram : @eduwaves360 Hint A clinical finding with a high true positive rate for appendicitis can be used to help rule out the condition. Note that this patient is only 1 year of age. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Guarding Explanation Why Guarding has the highest specificity (0.85) of all the clinical findings presented. A measure or test with high specificity is most useful for the confirmation of a diagnosis after a positive screening test. Therefore, it is used to confirm, rather than rule out, a disease. B - Fever Explanation Why In the table, fever has the highest sensitivity (0.80) for acute appendicitis in children under the age of 2 years. If a highly sensitive test (e.g., core body temperature) yields a negative result (e.g., afebrile), the disease can be more reliably ruled out. For this reason, tests with high sensitivity are often used for screening purposes. C - Anorexia Explanation Why Anorexia has the lowest specificity for acute appendicitis out of the clinical findings presented in the table. A test with a low specificity means that the feature being tested is more likely to be present in individuals who do not have the disease (i.e., high false positive rate). For this reason, even if this patient had no history of anorexia, appendicitis could not be confidently ruled out. eduwaves360.com | Telegram : @eduwaves360 D - Rebound Explanation Why In the table, rebound has the highest sensitivity (0.85) of all clinical findings studied in the > 5 year age group. Rebound would be the correct answer if the patient were over 5 years of age. However, this child is only 1 year of age, and the table does not present data on the role of rebound tenderness in the evaluation of appendicitis in those < 5 years of age. Therefore, a negative test result for rebound would not help to rule out appendicitis in this child. E - Vomiting Explanation Why Vomiting has the lowest sensitivity (0.40) for appendicitis out of all the clinical findings presented. A test with a low sensitivity means that the feature being tested is less likely to be present in individuals who have the disease. For this reason, even if this patient had no history of vomiting, appendicitis could not be confidently ruled out. eduwaves360.com | Telegram : @eduwaves360 Question # 5 A geriatric investigator is evaluating the consistency of Alzheimer dementia diagnoses based on clinical symptoms. Patients with known chart diagnoses of Alzheimer dementia were evaluated by multiple physicians during a fixed time interval. Each evaluator was blinded to the others' assessments. The extent to which the diagnosis by one physician was replicated by another clinician examining the same patient is best described by which of the following terms? Answer A Validity B Specificity C Predictive value D Precision E Sensitivity Image eduwaves360.com | Telegram : @eduwaves360 Hint The investigator wants to assess whether the diagnosis of Alzheimer dementia based on the clinical presentation produces reliable results. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Validity Explanation Why Validity describes how well the results of a test reflect what it was originally developed to measure. Rather than interrogating the validity of the Alzheimer evaluation by comparing it to other diagnostic modalities, this investigator is interested in whether repeated measures would result in the same diagnosis. B - Specificity Explanation Why Specificity refers to the ability of a specific test (e.g., Alzheimer evaluation) to correctly identify those who do not have a certain disease. As there are no standard values for this study group offered here, no conclusions about specificity can be drawn. C - Predictive value Explanation Why Predictive value reflects the likelihood that an individual who tests positive for a disease actually has the disease (positive predictive value) or the likelihood that an individual who tests negative for a disease is truly disease-free (negative predictive value). Here the investigator is instead interested if repeated measures by different examiners would yield the same result, which is assessing a different test characteristic. eduwaves360.com | Telegram : @eduwaves360 D - Precision Explanation Why Independent, repeated measurements assess a test's precision, or how reproducible the Alzheimer diagnosis is. In this case, interrater reliability is used to measure precision (e.g., a high interrater reliability means that the test yields similar results when performed by different examiners). A precise tool will have a low rate of random error and a high rate of consistency in spite of the variability between individual evaluators. E - Sensitivity Explanation Why Sensitivity (epidemiology) refers to the ability of a specific test (e.g., Alzheimer evaluation) to correctly identify those who have a certain disease (e.g., Alzheimer disease). As there are no standard values for this study group offered here, no conclusions about sensitivity can be drawn. eduwaves360.com | Telegram : @eduwaves360 Question # 6 An investigator is studying the maternal and fetal consequences of a recent spike in benzodiazepine addiction in Ireland. She is particularly interested in whether benzodiazepine use contributes to oral cleft deformities in newborns exposed to alprazolam, clonazepam, or lorazepam during the first trimester. The investigator uses statistical data from the local demographic institute to calculate the number of newly diagnosed cases of oral cleft deformities in Ireland over the past 5 years. Which of the following terms describes the investigator's statistical measure of interest? Answer A Mortality rate B Prevalence C Relative risk D Cumulative incidence E Odds ratio F Attributable risk Image eduwaves360.com | Telegram : @eduwaves360 Hint The investigator does not include pre-existing cases of cleft deformities. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Mortality rate Explanation Why Mortality rate refers to the number of deaths within a group during a period of time. The investigator is interested in cases of disease, not death. B - Prevalence Explanation Why Prevalence includes all cases of a disease in a population, both new and preexisting. This investigator, however, is only interested in new cases of cleft deformities within the 5-year span. C - Relative risk Explanation Why Relative risk compares the risk of disease in one group (exposed to a risk factor) to the risk of disease in another group (without exposure to the risk factor). Although this investigator is interested in the rate of cleft palate, she is not comparing two groups that differ by exposure to a risk factor. D - Cumulative incidence Explanation Why Cumulative incidence is the proportion of new cases of disease (here cleft deformities) over a defined eduwaves360.com | Telegram : @eduwaves360 period of time (here 5 years). It represents the likelihood of developing the disease during that time. It is calculated by dividing the number of new cases in a given time period by the population at risk in the same time interval. E - Odds ratio Explanation Why Odds ratio is a statistical measure commonly used in case-control studies. Odds ratio compares the odds of exposure to a specific event or risk factor between a group with a disease and a group without the disease. This investigator is not comparing risk exposure between groups. F - Attributable risk Explanation Why Attributable risk is the proportion of disease occurrence among exposed individuals attributable to that exposure. Here, however, the investigator is only considering the overall number of newly diagnosed cases of oral cleft deformities, irrespective of exposure status. eduwaves360.com | Telegram : @eduwaves360 Question # 7 A 22-year-old man is brought to the emergency department by his roommate 20 minutes after being discovered unconscious at home. On arrival, he is unresponsive to painful stimuli. His pulse is 65/min, respirations are 8/min, and blood pressure is 110/70 mm Hg. Pulse oximetry shows an oxygen saturation of 75%. Despite appropriate lifesaving measures, he dies. The physician suspects that he overdosed. If the suspicion is correct, statistically, the most likely cause of death is overdose with which of the following groups of drugs? Answer A Synthetic opioids B Amphetamines C Heroin D Antidepressants E Acetaminophen F Benzodiazepines Image eduwaves360.com | Telegram : @eduwaves360 Hint Overdose with this drug is the leading cause of death among adults under the age of 50 years in the United States. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Synthetic opioids Explanation Why Opioids, either alone or taken in combination with other drugs, are involved in up to 75% of overdose-related deaths. The incidence of opioid-related drug overdoses has increased dramatically over the last few decades. Deaths resulting from an overdose of synthetic opioids are far more common than those resulting from other drugs, including heroin, cocaine, psychostimulants, and benzodiazepines. Illicitly obtained synthetic opioids now outnumber prescribed synthetic opioids. Altered mental status, respiratory depression, and miosis are the classic triad of intoxication with opioid analgesics. Suspected opioid intoxication should be treated immediately with naloxone, to reverse the opioid effects, and airway management. B - Amphetamines Explanation Why Amphetamines are a commonly used street drug. Although overdose can result in serotonin syndrome, cardiac events, and cerebrovascular accidents, these adverse effects are usually not fatal. C - Heroin Explanation Why Deaths from heroin overdose have been on the rise in recent years. Statistically, more overdoserelated deaths are due to prescribed medications, not illegal drugs such as heroin. eduwaves360.com | Telegram : @eduwaves360 D - Antidepressants Explanation Why Antidepressants, such as tricyclic antidepressants, account for nearly 20% of overdose-related deaths. However, statistically, another type of drug is most likely responsible for this patient's overdoserelated death. E - Acetaminophen Explanation Why Acetaminophen is a commonly abused over-the-counter medication. Its main side effect is hepatotoxicity; acetaminophen overdose is the leading cause of acute hepatic failure in the US. Acetaminophen is involved in < 5% of overdose-related deaths. F - Benzodiazepines Explanation Why Benzodiazepines are a commonly abused prescription medication. When taken in excess, they cause CNS and respiratory depression. Benzodiazepines are the second most common drug involved in overdose-related deaths, accounting for approx. 30% of cases. eduwaves360.com | Telegram : @eduwaves360 Question # 8 A pulmonologist is analyzing the vital signs of patients with chronic obstructive pulmonary disease (COPD) who presented to an emergency room with respiratory distress and subsequently required intubation. The respiratory rates of 7 patients with COPD during their initial visit to the emergency room are shown: Patient 1 22 breaths per minute Patient 2 32 breaths per minute Patient 3 23 breaths per minute Patient 4 30 breaths per minute Patient 5 32 breaths per minute Patient 6 32 breaths per minute Patient 7 23 breaths per minute Which of the following is the mode of these respiratory rates? Answer A 30 breaths per minute B 32 breaths per minute C 10 breaths per minute D 27.7 breaths per minute Image eduwaves360.com | Telegram : @eduwaves360 Answer E Image 26 breaths per minute eduwaves360.com | Telegram : @eduwaves360 Hint The mode is least affected by outliers. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 30 breaths per minute Explanation Why 30 breaths per minute is the median of this set of values. To determine the median, the data set needs to be ordered. The values of this set, in ascending order, are 22, 23, 23, 30, 32, 32, 32. Since there is an uneven number of values in the set, the median is the middle value, which is 30. If there were an even number of values in the set, the median value would be the average of the two middle values. B - 32 breaths per minute Explanation Why 32 breaths per minute is the mode (i.e., the most common value) of this set of values because it appears three times, whereas the other values appear only once or twice. The mode is most resistant to outliers, the mean is least resistant to them, and the median lies somewhere in between. C - 10 breaths per minute Explanation Why 10 breaths per minute is the range of this set of values: 32 - 22 = 10. eduwaves360.com | Telegram : @eduwaves360 D - 27.7 breaths per minute Explanation Why 27.7 breaths per minute is the approximate mean of the respiratory rates: (22 + 32 + 23 + 30 + 32 + 32 + 23)/7 ≈ 27.7. E - 26 breaths per minute Explanation Why 26 breaths per minute is neither the median, mean, nor mode of this set of values. eduwaves360.com | Telegram : @eduwaves360 Question # 9 A researcher is investigating the risk of symptomatic intracerebral hemorrhage associated with tissue plasminogen activator (tPA) treatment in severe ischemic stroke. The outcomes of a large randomized controlled trial of ischemic stroke patients, some of whom were randomized to tPA, is shown: Symptomatic intracerebral hemorrhage No symptomatic intracerebral hemorrhage Received tPA 12 188 Did not receive tPA 25 475 Based on this data, how many patients with severe ischemic stroke would need to be treated with tPA, on average, to contribute to one case of symptomatic intracerebral hemorrhage? Answer A 6 B 0.01 C 188 D 13 E 1.2 Image eduwaves360.com | Telegram : @eduwaves360 Answer F Image 100 eduwaves360.com | Telegram : @eduwaves360 Hint The value of interest is the number needed to harm (NNH). It can be calculated as the inverse of attributable risk. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A-6 Explanation Why This calculation is based on the risk of symptomatic intracranial hemorrhage in the tPA group (12/ 200 = 0.06). In order to compare the two groups, the correct equation must also consider the risk of symptomatic intracranial hemorrhage in the control group. B - 0.01 Explanation Why This calculation is the attributable risk, which is the inverse of the number needed to harm - the statistical measure of interest. Logically, it is impossible for 0.01 patients to be treated with tPA in order for one tPA recipient to experience a symptomatic intracerebral hemorrhage, meaning that more patients will have an intracerebral hemorrhage than receive tPA. C - 188 Explanation Why This is the number of patients in the tPA group that did not experience a hemorrhage. The correct equation focuses on those afflicted with an intracerebral hemorrhage and also account for both the tPA and the control groups. eduwaves360.com | Telegram : @eduwaves360 D - 13 Explanation Why This is the difference in symptomatic intracerebral hemorrhage cases between the tPA group and control group. This calculation is incorrect because it doesn't account for the size difference between the tPA group compared to the control group. A much greater number of patients did not receive tPA. E - 1.2 Explanation Why This value approximates the odds ratio. Although an odds ratio is useful in case-control studies and quantifies risk, the resulting calculation is the odds of intracerebral hemorrhage between two groups. Instead, the intended result for this question is in units of people. F - 100 Explanation Why Number needed to harm (NNH) = 1 / (attributable risk) = 1 / (intracerebral hemorrhage risk among tPA group - intracerebral hemorrhage risk in control group) = 1 / [(12/200) - (25/500)] = 1/.01 = 100 100 patients treated with tPA will result in 1 patient treated with tPA to suffer from a symptomatic intracerebral hemorrhage. eduwaves360.com | Telegram : @eduwaves360 Question # 10 A rheumatologist is evaluating the long-term risk of venous thromboembolism in patients with newly diagnosed rheumatoid arthritis by comparing two retrospective cohort studies. In study A, the hazard ratio for venous thromboembolism was found to be 1.7 with a 95% confidence interval of 0.89–2.9. Study B identified a hazard ratio for venous thromboembolism of 1.6 with a 95% confidence interval of 1.1–2.5. Which of the following statements about the reported association in these studies is most accurate? Answer A The results of study B are less likely to be accurate than the results of study A. B The HR of study B is less likely to be statistically significant than the HR of study A. C Study A likely had a larger sample size than study B. D The p-value of study A is likely larger than the p-value of study B. E The power of study B is likely smaller than the power of study A. eduwaves360.com | Telegram : @eduwaves360 Image Hint A hazard ratio equal to 1.0 means there is no difference in the long-term VTE risk between the rheumatoid arthritis and control groups. eduwaves360.com | Telegram : @eduwaves360 Correct Answer AThe results of study B are less likely to be accurate than the results of study A. Explanation Why Accuracy refers to how well a study reflects the truth of the association between the variables of interest. It is impossible to make a statement about this study's accuracy without information regarding its methodology (inclusion/exclusion criteria, randomization strategy, group demographics, etc.), which would allow sources of systematic error to be properly evaluated. BThe HR of study B is less likely to be statistically significant than the HR of study A. Explanation Why Because the 95% confidence interval of the hazard ratio (HR) excludes 1.0 for study B but includes 1.0 for study A, it can be concluded that the HR for study B is statistically significant, while the HR for study A is not. C - Study A likely had a larger sample size than study B. Explanation Why Assuming that all aspects of study design remain the same, a larger sample size (n) will decrease standard deviation, which results in a narrower confidence interval. Since study B has a narrower confidence interval than study A, it can be inferred that study B in fact had a larger sample size than study A. However, because no other information about the study populations is provided, it cannot be definitively concluded that this is the case. eduwaves360.com | Telegram : @eduwaves360 D - The p-value of study A is likely larger than the p-value of study B. Explanation Why The 95% confidence interval of the hazard ratio (HR) for study A includes 1.0 and therefore is not statistically significant (p-value > 0.05). In contrast, the 95% confidence interval of the HR for study B does not include 1.0 and therefore is statistically significant (p-value < 0.05). Although the exact p-values are not given here, we can use this significance criteria to infer that (p-value of study B) < 0.05 < (p-value of study A). E - The power of study B is likely smaller than the power of study A. Explanation Why Statistical power is increased by several factors, including a large sample size, a large magnitude of effect between groups, increased precision of measurement, and the stringency of the statistical significance criteria chosen for the study (a p-value of 0.03 will fulfill a significance criteria of p < 0.05 before a significance criteria of p < 0.01). Because study B's 95% confidence interval of the hazard ratio (HR) excludes 1.0, this study had sufficient statistical power to detect a difference between the groups. In contrast, study A's 95% confidence interval of the HR includes 1.0; study A is therefore less powerful than study B. eduwaves360.com | Telegram : @eduwaves360 Question # 11 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? Answer A It overcomes limitations in the quality of individual studies. B It is able to provide evidence of causality. C The results are less precise. D The likelihood of type II errors is decreased. E The level of clinical evidence is lower. Image eduwaves360.com | Telegram : @eduwaves360 Hint The teams are conducting a meta-analysis, which increases statistical power by pooling data across different studies. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - It overcomes limitations in the quality of individual studies. Explanation Why The quality of a meta-analysis entirely depends on the individual studies it is composed of. Therefore, the analysis' results are only as valid as the results of each individual study. Merely pooling several studies in itself will not add any benefit in terms of quality. B - It is able to provide evidence of causality. Explanation Why Pooling data from various cross-sectional studies may increase the probability that an observed association reaches statistical significance. However, it does not change the fact that observational studies cannot imply causality no matter how strong the statistical power is. C - The results are less precise. Explanation Why Increasing the sample size through pooling not only improves precision and accuracy of the results but also adds the benefit of greater generalizability, and thus more external validity. eduwaves360.com | Telegram : @eduwaves360 D - The likelihood of type II errors is decreased. Explanation But Type I errors are not affected by statistical power as they depend on the significance level. Explanation Why A type II error is the acceptance of the null hypothesis although it is actually false. Increasing sample size by pooling data from several studies translates into increased statistical power, which is the main determinant for type II errors. The higher the statistical power, the higher the probability of rejecting the null hypothesis when it is truly false. Hence, the likelihood of type II error is decreased. E - The level of clinical evidence is lower. Explanation Why Assuming that each individual contributing study is accurate and valid, a meta-analysis exhibits the highest level of clinical evidence of all study types, even exceeding that of randomized clinical trials or systematic reviews. eduwaves360.com | Telegram : @eduwaves360 Question # 12 A group of researchers studying the relationship between major depressive disorder and unprovoked seizures identified 36 patients via chart review who had been rehospitalized for unprovoked seizures following discharge from an inpatient psychiatric unit and 105 patients recently discharged from the same unit who did not experience unprovoked seizures. The results of the study show: Unprovoked seizure No seizure Major depressive disorder 20 35 No major depressive disorder 16 70 Based on this information, which of the following is the most appropriate measure of association between history of major depressive disorder (MDD) and unprovoked seizures? Answer A 0.36 B 1.95 C 0.19 D 2.5 E 0.17 Image eduwaves360.com | Telegram : @eduwaves360 Hint The researchers began the study by identifying patients with or without the outcome (unprovoked seizures) and then determined the presence or absence of the potential risk factor. This study design is consistent with a case-control study. The appropriate measure of association in a case-control study would be the odds ratio. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 0.36 Explanation Why [20/(20 +35)] = 0.36 is the risk of unprovoked seizures among patients with MDD. The appropriate statistical measure in a case-control study compares odds, not risks. B - 1.95 Explanation Why [20/(20 +35)] / [16/(16 +70)] = 1.95 is the relative risk of seizures among psychiatric inpatients with MDD against those without MDD. The relative risk is commonly used in cohort studies. Casecontrol studies compare groups using an odds ratio. C - 0.19 Explanation Why [16/(16 +70)] = 0.19 is the risk of unprovoked seizures among patients without MDD. The appropriate statistical measure in a case-control study compares odds, not risks. D - 2.5 Explanation Why The odds of an unprovoked seizure in a patient with major depressive disorder (MDD) is the ratio of eduwaves360.com | Telegram : @eduwaves360 the number of patients with MDD who develop an unprovoked seizure (a = 20) to the number of patients without MDD who developed an unprovoked seizure (c = 16). Similarly, the odds of no seizure in a patient with MDD is 35 (= b) divided by 70 (= d). The odds ratio is calculated as: (a/c) / (b/d) = ad/bc = (20*70)/(35*16) = 2.5. Based on this study, the odds of an unprovoked seizure after discharge is 2.5 times higher in patients with MDD compared to patients without MDD. E - 0.17 Explanation Why [20/(20 +35)] - [16/(16 +70)] = 0.17 is the difference between the risks of unprovoked seizures of patients with MDD and without MDD (attributable risk). The appropriate way to compare odds is through a ratio (division) rather than a difference (subtraction). eduwaves360.com | Telegram : @eduwaves360 Question # 13 A randomized controlled trial is conducted to evaluate the relationship between the angiotensin receptor blocker losartan and cardiovascular death in patients with congestive heart failure (diagnosed as ejection fraction < 30%) who are already being treated with an angiotensin-converting enzyme (ACE) inhibitor and a beta blocker. Patients are randomized either to losartan (N = 1500) or placebo (N = 1400). The results of the study show: Cardiovascular death No cardiovascular death Losartan + ACE inhibitor + beta blocker 300 1200 Placebo + ACE inhibitor + beta blocker 350 1050 Based on this information, if 200 patients with congestive heart failure and an ejection fraction < 30% were treated with losartan in addition to an ACE inhibitor and a beta blocker, on average, how many cases of cardiovascular death would be prevented? Answer A 20 B 10 C 0.25 D 50 E 0.05 Image eduwaves360.com | Telegram : @eduwaves360 Hint To answer this question correctly, first calculate the number needed to treat (NNT) (i.e., the number of individuals that must be treated for one person to benefit from treatment), which is the inverse of absolute risk reduction (1/ARR). Then, as the second step, consider that the question asks for the number of persons who would benefit if 200 patients were treated! eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 20 Explanation Why This value is the number needed to treat, which is calculated by taking the inverse of the ARR. In this study, the ARR is the risk in the placebo group, 350 / (350 + 1050) = 0.25, minus the risk in the losartan group, 300 / (300 + 1200) = 0.20. Thus, the ARR = 0.25 - 0.20 = 0.05. The NNT is therefore 1/0.05 = 20 patients. However, the question asks how many deaths will be prevented if 200 patients with congestive heart failure and an ejection fraction < 30% were treated with losartan in addition to an ACE inhibitor and a beta blocker. B - 10 Explanation Why The absolute risk reduction in this study is the risk in the placebo group, 350 / (350 + 1050) = 0.25, minus the risk in the losartan group, 300 / (300 + 1200) = 0.20. Thus, the ARR = 0.25 - 0.20 = 0.05. Therefore, the number needed to treat is 1/0.05 = 20 patients. This means that for every 20 patients treated, one death will be prevented. For every 200 patients treated, 10 deaths will be prevented. C - 0.25 Explanation Why This value is the risk of cardiovascular death in the placebo group, which is calculated by dividing the number of deaths in the placebo group by the number at risk in the placebo group: 350 / (350 + 1050) = 0.25. eduwaves360.com | Telegram : @eduwaves360 D - 50 Explanation Why This value is the difference in the number of cardiovascular deaths in the placebo group (350) and the losartan group (300). This does not relate to the number of deaths that would be prevented if 200 patients with congestive heart failure and an ejection fraction < 30% were treated with losartan in addition to an ACE inhibitor and a beta blocker. E - 0.05 Explanation Why This value is the absolute risk reduction. In this study, the ARR is the risk in the placebo group, 350 / (350 + 1050) = 0.25, minus the risk in the losartan group, 300 / (300 + 1200) = 0.20. Thus, the ARR = 0.25 - 0.20 = 0.05. eduwaves360.com | Telegram : @eduwaves360 Question # 14 An investigator is studying the efficacy of a new bisphosphonate analog in preventing hip fractures in patients above 60 years of age with risk factors for osteoporosis but no confirmed diagnosis. Participating patients were randomized to either pharmacologic therapy with the new bisphosphonate analog or a placebo. The results show: Hip fracture No hip fracture Pharmacologic therapy 3 97 No pharmacologic therapy 10 190 Based on this information, which of the following best represents the proportionate reduction in the risk of hip fractures brought about due to pharmacologic therapy, in comparison to the control group? Answer A 3% B 5% C 40% D 2% E 60% Image eduwaves360.com | Telegram : @eduwaves360 Hint The value of interest is the relative risk reduction (RRR), which is calculated by subtracting the relative risk (RR) from 1 (i.e., RRR = 1 - RR). eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 3% Explanation Why This value is the risk of hip fracture in the treatment group, which is calculated as 3/(97 + 3) = 0.03. B - 5% Explanation Why This value is the risk of hip fracture in the control group, which is calculated as 10/(190 + 10) = 0.05. C - 40% Explanation But Drug advertisements often use measures of relative risk because they tend to overestimate the effectiveness of a treatment or intervention. In the case of this new bisphosphonate analog, the drug company is likely to promote the 40% reduction in relative risk rather than the 2% reduction in absolute risk. Explanation Why The relative risk (RR) of hip fracture in the treatment group compared to the control group is calculated as the risk of hip fracture in the treatment group [3/(97 + 3) = 0.03] divided by the risk of hip fracture in the control group [10/(190 + 10) = 0.05]. 0.03/0.05 = 0.6 = 60%. The relative risk reduction, RRR = 1 - RR, thus equals 1 - 0.6 = 0.4, which indicates a 40% reduction in the risk of hip fracture due to the pharmacologic therapy compared to control. eduwaves360.com | Telegram : @eduwaves360 D - 2% Explanation Why This value is the absolute risk reduction, which is the risk in the control group minus the risk in the treatment group. In this study, the ARR equals [10/(190 + 10) = 0.05] minus [3/(97 + 3) = 0.03]. 0.05 - 0.03 = 0.02 = 2%. E - 60% Explanation Why This value is the relative risk of hip fracture in the treatment group compared to the control group, which is calculated as the risk of hip fracture in the treatment group [3/(97 + 3) = 0.03] divided by the risk of hip fracture in the control group [10/(190 + 10) = 0.05]. 0.03/0.05 = 0.6 = 60%. However, this value is not the reduction in risk due to the pharmacologic therapy. eduwaves360.com | Telegram : @eduwaves360 Question # 15 An investigator is studying the efficacy of preventative measures to reduce pesticide poisonings among Central American farmers. The investigator evaluates the effect of a ban on aldicarb, an especially neurotoxic pesticide of the carbamate class. The ban aims to reduce pesticide poisonings attributable to carbamates. The investigator followed 1,000 agricultural workers residing in Central American towns that banned aldicarb as well as 2,000 agricultural workers residing in communities that continued to use aldicarb over a period of 5 years. The results show: Pesticide poisoning No pesticide poisoning Total Aldicarb ban 10 990 1000 No aldicarb ban 100 1900 2000 Which of the following values corresponds to the difference in risk attributable to the ban on aldicarb? Answer A 0.04 B 0.2 C 25 D 0.19 E 90 Image eduwaves360.com | Telegram : @eduwaves360 Answer F Image 0.8 eduwaves360.com | Telegram : @eduwaves360 Hint The value of interest is the absolute risk reduction, which is the difference between the event rate (pesticide poisoning) in the group without exposure to the intervention (no aldicarb ban) and the group with exposure to the intervention (aldicarb ban). eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 0.04 Explanation Why The event rate in the group with exposure to the intervention is the number of pesticide poisonings in the group with the aldicarb ban, which is 10/1000. The event rate in the group without exposure to the intervention is the number of pesticide poisonings in the group that did not ban aldicarb, which is 100/2000. Therefore, the absolute risk reduction = (100/2000) - (10/1000) =0.05 - 0.01 =0.04 or 4%. B - 0.2 Explanation Why This value is the relative risk of exposure to the intervention (aldicarb ban), which is calculated by taking the event rate in the group with exposure to the intervention divided by the event rate in the group without exposure to the intervention (aldicarb ban) = (10/1000) / (100/2000) = 0.01/0.0.5 = 0.2. C - 25 Explanation Why This value is the number needed to treat, which is the reciprocal of the absolute risk reduction (the reciprocal of the difference of the event rate in the group without exposure to the intervention and the group with exposure to the intervention), which = 1/ [(100/2000) - (10/1000)] = 1/(0.05 - 0.01) = 25. eduwaves360.com | Telegram : @eduwaves360 D - 0.19 Explanation Why This value is the odds ratio, which is calculated as the ratio between the odds of the event in the group with exposure to the intervention, which is (10/1000) / (990/1000) = (10/990), and the odds of the event in the group without exposure to the intervention, which is (100/2000) / (1900/2000) = (100/1900). The odds ratio is therefore (10/990) / (100/1900) = (10 x 1900) / (990 x100), which is approximately 0.19. E - 90 Explanation Why This value is the difference in the number of pesticide poisonings between the group without exposure to the intervention (aldicarb ban) and the group with exposure to the intervention, which is 100 - 10 = 90. F - 0.8 Explanation Why This value is the relative risk reduction, which is calculated with the formula 1 - (relative risk). The relative risk of exposure to the intervention (aldicarb ban) is calculated by taking the event rate in the group with exposure to the intervention divided by the event rate in the group without exposure to the intervention = (10/1000) / (100/2000) = 0.01/0.05 =0.2. Thus, the relative risk reduction = 1 - 0.2 = 0.8. eduwaves360.com | Telegram : @eduwaves360 Question # 16 After learning in a lecture that cesarean section rates vary from < 0.5% to over 30% across countries, a medical student wants to investigate if national cesarean section rates correlate with national maternal mortality rates worldwide. For his investigation, the student obtains population data from an international registry that contains tabulated cesarean section rates and maternal mortality rates from the last 10 years for a total of 119 countries. Which of the following best describes this study design? Answer A Survival analysis B Case series C Meta-analysis D Case-control study E Cross-sectional study F Retrospective cohort study G Ecological study H Prospective cohort study Image eduwaves360.com | Telegram : @eduwaves360 Hint This study investigates data on the population level rather than the individual level. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Survival analysis Explanation Why Survival analyses investigate the time interval for a particular outcome to manifest. Since this student is not analyzing the time between exposure to the risk factor (i.e., cesarean section) and the event (i.e., maternal mortality), survival analysis does not describe the proposed study. B - Case series Explanation Why A case series is a quantitative compilation of similar individual cases. Here the student collected data for various countries rather than considering individual cases. C - Meta-analysis Explanation Why In meta-analyses, the results of multiple similar studies are pooled and analyzed in order to improve the statistical power of the results. This student, however, is proposing a single study pooling the data from different countries, not combining the results of previous individual studies. eduwaves360.com | Telegram : @eduwaves360 D - Case-control study Explanation Why Case-control studies compare a group selected for the presence of a disease or outcome of interest (cases) to a group without that disease or outcome (controls). The presence of a specific factor (e.g., cesarean section) is then retrospectively considered for in each individual of both groups in order to determine the existence and strength of an association between the factor and the disease/outcome. In this study, the student collected data for various countries rather than considering individual cases for an exposure to the risk factor. E - Cross-sectional study Explanation Why Cross-sectional studies are used to assess the prevalence of a condition or disease and probable risk factors at a single point in time for a specific population. Cross-sectional studies analyze neither prevalence over time nor prevalence in multiple populations, as proposed in this study. F - Retrospective cohort study Explanation Why Retrospective cohort studies identify individuals with and without certain risk factors, such as a prior cesarean section, and then analyze their medical records for evidence of progression to a certain disease or condition, such as maternal mortality. However, this student is using aggregated population-level data of each country's maternal mortality rates and cesarean section rates separately. He is not working at the level of individuals. eduwaves360.com | Telegram : @eduwaves360 G - Ecological study Explanation Why An ecological study is an observational study that assesses data collected at the population level for exposures and/or outcomes. It is used to investigate causal relationships between a specific factor (cesarean section in this case) and its outcome (maternal mortality rate in this case) at the population level. This student collected and studied the data for cesarean section and maternal mortality rates for many countries. As each unit of observation was an entire country (i.e., large population), this study design is an ecological study. Ecological studies cannot be employed to draw conclusions on an individual level (e.g., personal risk of maternal death in a patient with or without a cesarean section). H - Prospective cohort study Explanation Why Prospective cohort studies identify individuals with and without certain risk factors, such as a prior cesarean section, and then follow these two groups for a period of time to see if a certain disease or condition develops. This student, however, is using aggregated population-level data of each country's maternal mortality rates and cesarean section rates separately. He is not working at the level of individuals. eduwaves360.com | Telegram : @eduwaves360 Question # 17 A group of investigators are studying the effects of transcranial direct current stimulation (tDCS) on cognitive performance in patients with Alzheimer disease. A cohort of 50 patients with mild Alzheimer disease were randomized 1:1 to either tDCS or sham tDCS over the temporoparietal cortex. Both procedures were conducted so that patients experienced the same sensations while receiving treatment. After 1 week of observation during which no treatments were delivered, the two groups were switched. Neuropsychiatric testing was subsequently conducted to assess differences in recognition memory between the two groups. Which of the following best describes the study design? Answer A Pretest-posttest B Case-control C Crossover D Meta-analysis E Parallel group F Factorial Image eduwaves360.com | Telegram : @eduwaves360 Hint In this study, the same subject is receiving different treatments during different time periods of the study, which helps to reduce confounding. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Pretest-posttest Explanation Why In a pretest-posttest study design, data (e.g., a test score, a laboratory study) is collected before and after an intervention (e.g., a new teaching method, administration of a new drug) in order to study the effect of the intervention. All the participants of a pretest-posttest study receive the same single intervention, unlike in this study, where one group received tDCS and another received sham tDCS. B - Case-control Explanation Why A case-control design is an observational study in which a particular risk factor (e.g., smoking) is retroactively assessed in two groups, which are selected according to the presence/absence of a particular outcome (e.g., lung cancer). The investigator does not intervene in an observational study, unlike in the experimental study described here, where the participants received tDCS/sham tDCS. C - Crossover Explanation But In a crossover study, statistical power is increased by having each participant serve as his/her own control, reducing the number of participants required for a study to be statistically significant. Explanation Why In a crossover study design, every participant receives all the interventions but at different times and/ or in a different sequence. In the study described in this vignette, all the participants sequentially received tDCS and sham tDCS (i.e., the groups crossed over). A washout period between treatments is commonly included in crossover studies to enable any potentially confounding effects of the first eduwaves360.com | Telegram : @eduwaves360 treatment to wear off prior to starting the sequential treatment. D - Meta-analysis Explanation Why In a meta-analysis, the results of multiple similar studies are pooled and analyzed in order to improve the statistical power of the results. The study described in this vignette is a single epidemiological study. E - Parallel group Explanation Why In a parallel group design, participants are randomized into different groups (treatment vs. placebo) and remain within that group until the end of the study. The participants in the study here switched groups after a week to experience both interventions, so this is not a parallel group study. F - Factorial Explanation Why A factorial study tests the effects of two or more interventions simultaneously, each with two or more independent variables. The study described is testing the effect of only one intervention (tDCS on cognitive performance in a certain group of patients), so it is not a factorial study. eduwaves360.com | Telegram : @eduwaves360 Question # 18 A team of epidemiologists is investigating an outbreak of hemolytic uremic syndrome (HUS) caused by Shiga toxin-producing E. coli O104:H4. In Europe, multiple episodes of illness were reported in May 2017 within a large extended family of 16 family members, who all attended a family reunion in late April where they ate sprouts contaminated with E. coli. In the ensuing weeks, multiple family members were admitted to local hospitals for treatment of HUS. A graph depicting the course of the disease is shown. Each row represents a patient. The gray bars represent the duration of the disease. Based on the graph, which of the following is the attack rate among the individuals at risk in the month of May? Answer A 6/15 B 6/8 Image eduwaves360.com | Telegram : @eduwaves360 Answer C 7/8 D 8/16 E 5/8 F 7/16 Image eduwaves360.com | Telegram : @eduwaves360 Hint The attack rate is the ratio of new cases to the number of individuals at risk. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 6/15 Explanation Why Attack rate (cumulative incidence) is calculated by dividing the number of individuals who contracted an illness in a given period of time by the number of individuals who were at risk of contracting that illness during this time. According to this graph, 6 people contracted HUS in May (patient 1 was ill before May; patient 6 fell ill in June). The number of people at risk of contracting HUS in May is 15, as patient 1 was already ill in April. Patient 6 was at risk of contracting the illness throughout May and is included as an exposed individual. However, since the illness did not manifest until June, it is not counted toward the attack rate in May. Therefore, the attack rate of HUS for May is 6/15. B - 6/8 Explanation Why This is the ratio of those who contracted HUS in May (6 people) to all the individuals who contracted HUS in this family (8 people). This value does not represent the attack rate of HUS in May. C - 7/8 Explanation Why This is the ratio of those who contracted HUS (7 people) after April to all the individuals who contracted HUS in this family (8 people). This value does not represent the attack rate of HUS in May. eduwaves360.com | Telegram : @eduwaves360 D - 8/16 Explanation Why This would be the attack rate (cumulative incidence) of HUS among the family members during April, May, and June, not just May. Eight members in a family of 16 members contracted HUS during these months. Thus, the attack rate of HUS in this family during this time is 8/16. E - 5/8 Explanation Why This is the ratio of those who fell ill in May and recovered that same month (5 people; patient 1 was ill before May, patient 6 fell ill in June, and patient 2 remained ill in June) to all the family members who contracted HUS (8 people). This value does not represent the attack rate of HUS in May. F - 7/16 Explanation Why This would be the prevalence of HUS among the family members in May, not the attack rate. Prevalence is calculated by dividing the total number of individuals with the disease at a particular point in time or over a time period by the total number of individuals in the group. Seven people had HUS in May (patient 6 fell ill in June but was disease-free in May) in a family of 16. Thus, the prevalence of HUS among these family members in May is 7/16. eduwaves360.com | Telegram : @eduwaves360 Question # 19 An infectious disease investigator is evaluating the diagnostic accuracy of a new interferon-gammabased assay for diagnosing tuberculosis in patients who have previously received a Bacillus CalmetteGuérin (BCG) vaccine. Consenting participants with a history of BCG vaccination received an interferon-gamma assay and were subsequently evaluated for tuberculosis by sputum culture. Results of the study are summarized in the table below. Tuberculosis, confirmed by culture No tuberculosis Total Positive interferon-gamma assay 90 6 96 Negative interferon-gamma assay 10 194 204 Total 100 200 300 Based on these results, what is the sensitivity of the interferon-gamma-based assay for the diagnosis of tuberculosis in this study? Answer A 194/200 B 90/100 C 90/96 D 100/300 Image eduwaves360.com | Telegram : @eduwaves360 Answer E Image 194/204 eduwaves360.com | Telegram : @eduwaves360 Hint The sensitivity of a test is the true positive rate (i.e., the proportion of people with a condition who test positive). eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 194/200 Explanation Why This is the specificity of this test, which is calculated using the formula: true negatives/(true negatives + false positives). Of all the participants who do not have tuberculosis (6 + 194), the test was negative in 194 (i.e., 194 true negative results). Therefore, the specificity of this test is 194/200. B - 90/100 Explanation Why The sensitivity (epidemiology) of this test is calculated using the formula: true positives/(true positives + false negatives). Of all the participants who were confirmed to have tuberculosis by culture (90 + 10), the test was positive in 90 (i.e., 90 true positive results). Therefore, the sensitivity of this test is 90/100. C - 90/96 Explanation Why This is the positive predictive value (PPV) of this test, which is calculated using the formula: true positives/(true positives + false positives). Of all the participants who tested positive (i.e., 90 + 6), 90 have confirmed tuberculosis. Therefore, the PPV of this test is 90/96. eduwaves360.com | Telegram : @eduwaves360 D - 100/300 Explanation Why This is the prevalence of tuberculosis in the cohort of all participants in this study. Prevalence is calculated using the formula: total number of cases/total population at a given point in time. In this study with 300 participants, 100 have confirmed tuberculosis. Therefore, the prevalence of tuberculosis in this cohort is 100/300. E - 194/204 Explanation Why This is the negative predictive value (NPV) of this test, which is calculated using the formula: true negatives/(true negatives + false negatives). Of all the participants who tested negative (i.e., 194 + 10), 194 do not have tuberculosis. Therefore, the NPV of this test is 194/204. eduwaves360.com | Telegram : @eduwaves360 Question # 20 A group of investigators is evaluating the diagnostic properties of a new blood test that uses two serum biomarkers, dityrosine and Nε-carboxymethyl-lysine, for the clinical diagnosis of autism spectrum disorder (ASD) in children. The test is considered positive only if both markers are found in the serum. 50 children who have been diagnosed with ASD based on established clinical criteria and 50 children without the disorder undergo testing. The results show: Diagnosis of ASD No diagnosis of ASD Test positive 45 15 Test negative 5 35 Which of the following is the specificity of this new test? Answer A 30% B 10% C 88% D 70% E 90% F 75% Image eduwaves360.com | Telegram : @eduwaves360 Hint The specificity of a test refers to the true negative rate. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 30% Explanation Why This is the false positive rate of this test, which is calculated with the following formula: (1 – specificity). The specificity of this test is 35/(35 + 15), which is 0.7. Therefore, the false positive rate of this test is 1 – 0.7, which is 0.3 or 30%. B - 10% Explanation Why This is the false negative rate of this test, which is calculated with the following formula: (1 – sensitivity). The sensitivity of this test is 45/(45 + 5), which is 0.9. Therefore, the false negative rate of this test is 1 – 0.9, which is 0.1 or 10%. C - 88% Explanation Why This is the negative predictive value (NPV) of this test, which is calculated with the following formula: true negatives/(true negatives + false negatives). Of all the children who had a negative test (i.e., true negatives + false negatives: 35 + 5), only 35 do not have ASD. Therefore, the NPV of this test is 35/(35 + 5), which is 35/40 or ∼ 88%. eduwaves360.com | Telegram : @eduwaves360 D - 70% Explanation Why The specificity (epidemiology) of a test is calculated with the following formula: true negatives/(true negatives + false positives). Of the children who do not have ASD (35 + 15), 35 tested negative (i.e., 35 true negative test results). Therefore, the specificity of this test is 35/(35 + 15), which is 35/50 or 70%. E - 90% Explanation Why This is the sensitivity (epidemiology) of this test, which is calculated with the following formula: true positives/(true positives + false negatives). Of all the children who have ASD (45 + 5), 45 tested positive (i.e., 45 true positive test results). Therefore, the sensitivity of this test is 45/(45 + 5), which is 45/50 or 90%. F - 75% Explanation Why This is the positive predictive value (PPV) of this test, which is calculated with the following formula: true positives/(true positives + false positives). Of all the children who had a positive test (i.e., true positive + false positive: 45 + 15), only 45 have ASD. Therefore, the PPV of this test is 45/(45 + 15), which is 45/60 or 75%. eduwaves360.com | Telegram : @eduwaves360 Question # 21 A 54-year-old man comes to the physician for an annual health maintenance examination. He reports that he feels well. He has smoked one pack of cigarettes daily for 22 years and drinks three 12-oz bottles of beer each night. He works as an accountant and says he does not have time to exercise regularly. He is 178 cm (5 ft 10 in) tall and weighs 98 kg (216 lb); BMI is 31 kg/m2. His blood pressure is 146/90 mm Hg. Physical examination shows no abnormalities. His serum cholesterol concentration is 252 mg/dL and hemoglobin A1C is 6.9%. Which of the following preventative measures is likely to have the greatest impact on this patient's all-cause mortality risk? Answer A Increased physical activity B Weight loss C Lipid-lowering therapy D Antidiabetic medication E Reduced alcohol intake F Blood pressure reduction G Smoking cessation Image eduwaves360.com | Telegram : @eduwaves360 Hint This man has several modifiable risk factors that increase his mortality risk. Although he should be encouraged to make several important lifestyle changes, only one of the patient's risk factors is the single greatest preventable cause of death. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Increased physical activity Explanation Why Increased physical activity provides a protective effect against cardiovascular and all-cause mortality with a dose-response relationship. However, the magnitude of this protection for all-cause mortality is less than the magnitude of protection from another behavioral modification that this patient could make. B - Weight loss Explanation Why Weight loss would be beneficial in this patient who meets the criteria for World Health Organization Class I obesity (BMI > 30 kg/m2). However, for patients in this BMI category, the all-cause mortality benefits of weight loss are less than the magnitude of protection from another behavioral modification that this patient could make. C - Lipid-lowering therapy Explanation Why Lipid-lowering therapy, most commonly achieved with the use of statins, provides a protective effect against cardiovascular and all-cause mortality with a dose-response relationship. Even with high dose statin therapy, the all-cause mortality risk reduction is less than the magnitude of protection from another behavioral modification that this patient could make. eduwaves360.com | Telegram : @eduwaves360 D - Antidiabetic medication Explanation Why Antidiabetic medication would be helpful in this patient who meets diagnostic criteria for type 2 diabetes mellitus given his hemoglobin A1C > 6.5%. Glycemic control can prevent some adverse events such as limb amputation or progression of diabetic nephropathy and diabetic retinopathy. However, there is little evidence to suggest that these interventions lead to a significant reduction of all-cause mortality. E - Reduced alcohol intake Explanation Why Reduced alcohol intake will likely enhance this patient's overall health. His current alcohol intake is about 40 grams per day. Although the exact effect of reduced alcohol intake on all-cause mortality is still actively debated, most experts agree that if a benefit exists, it is small. F - Blood pressure reduction Explanation Why Blood pressure reduction can reduce the risk of all-cause mortality and cardiovascular disease in this patient. Tight blood pressure control (< 130/90 mm Hg) is the single most important intervention to reduce the risk of stroke. However, the magnitude of this protection for all-cause mortality is less than the magnitude of protection from another behavioral modification that this patient could make. eduwaves360.com | Telegram : @eduwaves360 G - Smoking cessation Explanation Why Smoking cessation will confer the greatest protection against all-cause mortality for this patient. Smoking is the single greatest preventable cause of death worldwide, regardless of age at quitting or the number of previous pack years. eduwaves360.com | Telegram : @eduwaves360 Question # 22 A group of investigators is examining the effect of the drug orlistat as an adjunct therapy to lifestyle modification on weight loss in obese volunteers. 800 obese participants were randomized to receive orlistat in addition to counseling on lifestyle modification and 800 obese participants were randomized to receive counseling on lifestyle modification alone. At the conclusion of the study, the investigators found that patients who underwent combined therapy lost a mean of 8.2 kg (18.1 lb), whereas patients counseled on lifestyle modification alone lost a mean of 4.3 kg (9.5 lb) (p < 0.001). The investigators also observed that of the 120 participants who did not complete the study, 97 participants were in the lifestyle modification group and 23 participants were in the combination group. Based on this information, the investigators should be most concerned about which of the following? Answer A Error in randomization B Lead-time bias C Recall bias D Attrition bias E Nonresponse bias F Confounding bias Image eduwaves360.com | Telegram : @eduwaves360 Hint Longitudinal studies with a large number of participants lost to follow-up are prone to this type of issue. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Error in randomization Explanation Why Error in randomization is possible in any study that randomizes participants to different arms. For example, one treatment arm could have a greater number of younger participants who might be more motivated or able to lose weight compared to older participants. However, the disproportionate number of dropouts in the lifestyle modification group does not directly suggest an error in randomization. B - Lead-time bias Explanation Why Lead-time bias is often a concern in studies that involve early screening for disease. This study does not evaluate the detection of a condition and its usual clinical presentation and thus is not likely to be affected by this type of bias. C - Recall bias Explanation Why Recall bias is often a concern in retrospective studies that involve patient recall of an exposure. This type of bias is not a concern in prospective studies like this one. eduwaves360.com | Telegram : @eduwaves360 D - Attrition bias Explanation Why Attrition bias is a major concern in this study due to unequal loss of participants from the different arms. The lifestyle modification group had significantly more dropouts than the medication group, which raises concerns about whether the reason for dropout is related to poorer outcomes. Other reasons why participants might not complete a study include adverse events and death. To reduce attrition bias, investigators should aim for high follow-up rates. E - Nonresponse bias Explanation Why Nonresponse bias, also known as participation bias, is a type of selection bias that is of concern in written surveys, polls, or telephone questionnaires with high rates of nonresponse. It occurs if the responders differ significantly from the nonresponders, which makes the results nonrepresentative. The individuals in this study volunteered to participate; therefore, nonresponse bias is not a concern. The discrepancy observed between the number of participants in the lifestyle modification and combination therapy lost to follow-up is more concerning for a different type of bias. F - Confounding bias Explanation Why Confounding bias is an issue when a third variable that has not been factored into a study (the confounder) is associated with both the dependent variable (the outcome) as well as the independent variable (e.g., exposure to a risk factor). It does not affect the number of dropouts in a study. eduwaves360.com | Telegram : @eduwaves360 Question # 23 An investigator is studying the relationship between suicide and unemployment using data from a national health registry that encompasses 10,000 people who died by suicide, as well as 100,000 matched controls. The investigator finds that unemployment was associated with an increased risk of death by suicide (odds ratio = 3.02; p < 0.001). Among patients with a significant psychiatric history, there was no relationship between suicide and unemployment (p = 0.282). Likewise, no relationship was found between the two variables among patients without a psychiatric history (p = 0.32). These results are best explained by which of the following? Answer A Matching B Selection bias C Observer bias D Effect modification E Stratification F Confounding Image eduwaves360.com | Telegram : @eduwaves360 Hint Upon stratifying for psychiatric history, the relationship between suicide and unemployment has changed such that it is no longer significant for both groups. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Matching Explanation Why In epidemiology, matching is a study design method to reduce bias due to confounding. It is often used in case-control studies, as it is used here, to control for known confounders. Matching does not explain why the association between suicide and unemployment was no longer significant in the stratified analysis. B - Selection bias Explanation Why Selection bias occurs if the study population does not reflect the target population. For example, it could have occurred here if the study had only recruited participants who owned firearms but then applied the results to a population where the prevalence of gun ownership was low. Selection bias does not explain why the association between suicide and unemployment was no longer significant in the stratified analysis. C - Observer bias Explanation Why Observer bias occurs when the measurement of a variable is influenced by the experimenter's knowledge or expectations; there is no evidence to suggest that the different odds ratios from the crude analysis and the stratified analysis were influenced by the experimenter. eduwaves360.com | Telegram : @eduwaves360 D - Effect modification Explanation Why Effect modification occurs when the relationship between an exposure and an outcome differs in the presence and absence of a third variable. When a stratified analysis is conducted, the stratum-specific effects will be different from each other. Here, the stratum-specific effects are the same: there was no relationship between suicide and unemployment in the group with a psychiatric history or the group without a psychiatric history. E - Stratification Explanation Why Stratification is an analytical tool in which a group is divided into subgroups to determine if a relationship that is true for the entire group holds within each of the subgroups. It is the technique that was used here to determine that the crude analysis odds ratios differed from the stratum-specific odds ratios. However, stratification does not explain why odds ratios differed. F - Confounding Explanation But To minimize confounding during study design, stratification, randomization, and/or matching can be used. Good study design includes anticipating potential confounders. Explanation Why Confounding may occur when the relationship between an exposure and an outcome is distorted by the effect of a third variable that has an association with both the exposure and the outcome. Here, the crude analysis showed a significant relationship between suicide and unemployment. However, when the sample was stratified by psychiatric history, there was no relationship between suicide and unemployment in the group with a psychiatric history or the group without a psychiatric history. eduwaves360.com | Telegram : @eduwaves360 Because a psychiatric history is associated with both unemployment and suicide, it is a potential confounder that may account for the observed results in the crude analysis. eduwaves360.com | Telegram : @eduwaves360 Question # 24 An investigator has conducted a prospective study to evaluate the relationship between asthma and the risk of myocardial infarction (MI). She stratifies her analyses by biological sex and observed that among female patients, asthma was a significant predictor of MI risk (hazard ratio = 1.32, p < 0.001). However, among male patients, no relationship was found between asthma and MI risk (p = 0.23). Which of the following best explains the difference observed between male and female patients? Answer A Confounding B Measurement bias C Stratified sampling D Random error E Effect modification Image eduwaves360.com | Telegram : @eduwaves360 Hint Upon stratifying for biological sex, the investigator found that the relationship between asthma and MI changed such that it is significant for females, but not for males. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Confounding Explanation Why Confounding occurs when a third variable is correlated with both the exposure variable and the outcome variable. If biological sex was a confounder, stratifying the groups by this variable would show that asthma is not a significant predictor of MI risk in either group. B - Measurement bias Explanation Why Measurement bias occurs when key study variables (e.g., exposure and outcome) are inaccurately measured or classified. There is no evidence to suspect this type of bias here. C - Stratified sampling Explanation Why While a stratified analysis was used in this study to further explore the influence of biological sex on the relationship between the exposure and the outcome, it is distinct from stratified sampling, which is a term that describes a component of certain experimental designs. eduwaves360.com | Telegram : @eduwaves360 D - Random error Explanation Why While all epidemiological studies are susceptible to random error, this does not explain why the relationship between asthma and MI differs between biologically male and biologically female patients. Furthermore, the p-value of p < 0.001 implies that there is a very small probability that the difference observed between the subgroups in this study is due to random error. E - Effect modification Explanation But Effect modification is distinct from confounding, but they are often confused. Stratification by the variable in question would not reveal a difference between groups if the variable was a simple confounder. Explanation Why Effect modification occurs when the magnitude of the relationship between the exposure (asthma) and the outcome of interest (MI) is modified by a third variable (biological sex). The standard test to assess for the presence of effect modification is to perform a stratified analysis of the data by the variable in question. In this case, stratification by biological sex revealed that asthma was a significant predictor of MI risk in females, but not in males. Thus, there is an interaction between asthma and biological sex that may have gone unnoticed if males and females were analyzed together. eduwaves360.com | Telegram : @eduwaves360 Question # 25 A group of researchers recently conducted a meta-analysis of twenty clinical trials encompassing 10,000 women with estrogen receptor-positive breast cancer who were disease-free following adjuvant radiotherapy. After an observation period of 15 years, the relationship between tumor grade and distant recurrence of cancer was evaluated. The results show: Distant recurrence No distant recurrence Well differentiated 500 4500 Moderately differentiated 375 2125 Poorly differentiated 550 1950 Based on this information, which of the following is the 15-year risk for distant recurrence in patients with high-grade breast cancer? Answer A 500/1425 B 550/1425 C 500/5000 D 550/2500 E 2500/10000 Image eduwaves360.com | Telegram : @eduwaves360 Answer F Image 1950/8575 eduwaves360.com | Telegram : @eduwaves360 Hint Absolute risk is the probability of developing disease over a certain period of time, in this case the risk of distant recurrence in patients with high-grade (i.e., poorly differentiated) breast cancer during the observation period of 15 years. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 500/1425 Explanation Why 500/(550 + 375 + 550) = 500/1425 is the number of participants with low-grade (i.e., welldifferentiated) breast cancer among the participants who were diagnosed with distant recurrence over the observation period of 15 years. B - 550/1425 Explanation Why 550/(550 + 375 + 550) = 550/1425 is the number of participants with high-grade breast cancer among the participants diagnosed with distant recurrence over the observation period of 15 years. C - 500/5000 Explanation Why 500/5000 = 0.1, or 10% is the absolute risk of distant recurrence in participants with low-grade (i.e., well-differentiated) breast cancer over the observation period of 15 years. D - 550/2500 Explanation Why The probability of distant recurrence in participants with high-grade breast cancer over the 15-year eduwaves360.com | Telegram : @eduwaves360 time period is 550/(550 + 1950) = 550/2500 = 0.22, or 22%. The absolute risk is equivalent to the incidence rate (the number of new cases per population at risk in a given time period). Here, there were 550 new cases of distant recurrence in 2500 participants with poorly differentiated breast cancer over the observation period of 15 years. E - 2500/10000 Explanation Why (550 + 1950)/10000 = 2500/10000 is the prevalence of high-grade cancer in the entire cohort of participants with estrogen receptor-positive breast cancer. F - 1950/8575 Explanation Why 1950/(1950 + 2125 + 4500) = 1950/8575 is the number of patients with high-grade breast cancer among the participants who did not have distant recurrence after the observation period of 15 years. eduwaves360.com | Telegram : @eduwaves360 Question # 26 An investigator studying the epidemiology of breast cancer finds that prevalence of breast cancer has increased significantly in the United States since the 1980s. After analyzing a number of large epidemiological surveillance databases, the epidemiologist notices that the incidence of breast cancer has remained relatively stable over the past 30 years. Which of the following best explains these epidemiological trends? Answer A Increased awareness of breast cancer among clinicians B Increased average age of population at risk for breast cancer C Improved screening programs for breast cancer D Improved treatment of breast cancer E Increased exposure to risk factors for breast cancer Image eduwaves360.com | Telegram : @eduwaves360 Hint The investigator found that more people are being documented as having breast cancer, while the number of new cases has remained the same. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Increased awareness of breast cancer among clinicians Explanation Why Increased awareness of breast cancer among clinicians would likely lead to a higher index of suspicion for breast cancer, increased testing, and, subsequently, an increase in the number of new diagnoses of breast cancer. In this case, however, the number of new diagnoses (i.e., incidence) has remained stable for the past 30 years. B - Increased average age of population at risk for breast cancer Explanation Why Because the risk of breast cancer increased with age, an aging population would put more individuals at risk of developing breast cancer and therefore lead to an increase in the number of new cases. This would, however, increase both the incidence and prevalence of the disease. In this case, the incidence of breast cancer has remained stable. C - Improved screening programs for breast cancer Explanation Why If screening programs for breast cancer are improved, more people are diagnosed with the disease and, therefore, both incidence and prevalence would rise. In this case, incidence has remained stable for the past 30 years. eduwaves360.com | Telegram : @eduwaves360 D - Improved treatment of breast cancer Explanation Why In a stable population, the approximate prevalence is calculated as the incidence multiplied by the average duration of the disease. An increased prevalence of disease with a stable incidence can be explained by factors that result in increased survival and prolonged duration of the disease, such as improved treatment of patients with breast cancer. E - Increased exposure to risk factors for breast cancer Explanation Why An increase in exposure to risk factors for breast cancer would likely result in an increased incidence and prevalence of breast cancer. In this case, however, incidence has remained stable for the past 30 years. eduwaves360.com | Telegram : @eduwaves360 Question # 27 A prospective cohort study is conducted to evaluate the risk of pleural mesothelioma in construction workers exposed to asbestos in Los Angeles. Three hundred construction workers reporting current occupational asbestos exposure were followed alongside 300 construction workers without a history of asbestos exposure. After 8 years of follow-up, no statistically significant difference in the incidence of pleural mesothelioma was observed between the two groups (p = 0.13), even after controlling for known mesothelioma risk factors such as radiation, age, and sex. Which of the following is the most likely explanation for the observed results of this study? Answer A Length-time bias B Lead-time bias C Latency period D Observer effect E Confounding bias F Berkson bias Image eduwaves360.com | Telegram : @eduwaves360 Hint A 35-year follow-up would likely reveal a statistically significant difference in mesothelioma rates between the workers exposed to asbestos and those not exposed to asbestos. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Length-time bias Explanation Why Length-time bias is a concern in screening studies, not prospective cohort studies, and refers to an apparent benefit in survival. This apparent benefit results from a higher detection rate of disease (rather than a true delay of death) that is longer in duration but lower in severity. Patients with aggressive forms of a disease develop the disease and die within a timespan that is shorter than the screening interval and may never be screened. Length-time bias thus affects the measured survival, not incidence, of disease. B - Lead-time bias Explanation Why Lead-time bias refers to an apparent benefit in survival due to earlier diagnosis of disease, rather than a true delay of death (e.g., often discussed in the context of cancer screening). The lead time of a diagnostic test is the average length of time between the initial detection of disease and the predetermined outcome (e.g., death or onset of clinical symptoms). The focus of this prospective cohort study, however, is mesothelioma incidence rather than survival of those diagnosed with mesothelioma. C - Latency period Explanation Why The mean latency period for the development of mesothelioma after asbestos exposure is over 30 years. An observation period of 8 years is unlikely to be long enough to observe a statistically significant difference in rates of mesothelioma between those exposed and those not exposed to asbestos. Malignancies and other chronic diseases typically have very long latency periods. It can therefore take years to study disease pathogenesis and the influence of factors that may modify the risk (e.g., a new treatment). Infectious diseases, on the other hand, have a very short latent period eduwaves360.com | Telegram : @eduwaves360 (incubation period). D - Observer effect Explanation Why The observer effect refers to the tendency of subjects to alter their behavior when they know they are being observed. It is unlikely that the act of observation by the investigators influenced the development of mesothelioma in those exposed or not exposed to asbestos. E - Confounding bias Explanation Why Residual confounding is possible even after controlling for known mesothelioma risk factors (i.e., that there are confounders other than radiation, age, and sex that mask the true difference in incidence between exposed and non-exposed groups). However, there is a more likely reason that no significant differences in incidence were observed between groups. F - Berkson bias Explanation Why Berkson bias can result from a sample being taken from a subpopulation rather than the general population. Berkson bias is of particular concern in case-control studies in which cases and controls are selected from a hospital inpatient population. In these cases, the relationship between exposure and outcome may be different than the relationship between exposure and outcome in the general population. There is no reason to suspect that the relationship between exposure and outcome in the participants from this prospective cohort study would be different from the general population. eduwaves360.com | Telegram : @eduwaves360 Question # 28 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 of 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? Answer A 1.4 B 0.25 C 0.50 D 0.70 Image eduwaves360.com | Telegram : @eduwaves360 Answer E Image 0.88 eduwaves360.com | Telegram : @eduwaves360 Hint The positive predictive value (PPV) reflects the likelihood that a patient with a certain test result or clinical feature (in this case, high probability of active TB on the basis of history, clinical symptoms, and CXR) truly has the condition based on gold standard testing (in this case, sputum testing). eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 1.4 Explanation Why 1.4 is the positive likelihood ratio of the TB test. This ratio reflects the likelihood that a person with a given disease will have a positive test result compared to a person without the disease having a positive test result. Thus, a positive likelihood ratio is the true positive rate/false positive rate. The first step to calculating the positive likelihood ratio is to calculate the sensitivity (true positive rate) and the specificity (true negative rate) of this TB test. In this study, TP = 700, FN = 300, FP = 2100, and TN = 2100: Sensitivity = TP/(FN + TP) = 700/(300 + 700) → 700/1000 = 0.70, or 70% Specificity = TN/(FP + TN) = 2100/(2100 + 2100) → 2100/4200 = 0.50, or 50% Remember that the positive likelihood ratio (+LR) is true positive rate/false positive rate. As specificity is the true negative rate, the false positive rate is yet to be calculated. The formula for calculating the positive likelihood ratio is sensitivity/(1 - specificity), where 1 - specificity represents the false positive rate. +LR = 0.70 /(1 - 0.50) → 0.70/0.50 = 1.4 B - 0.25 Explanation But The PPV increases as the prevalence of a disease increases in a population. Explanation Why The PPV is the proportion of positive tests that are true positives. The PPV is calculated by dividing the number of true positives (TP) by the total number of positives (i.e. true positives plus false positives, FP). In this study, TP = 700 and FP = 2100: eduwaves360.com | Telegram : @eduwaves360 PPV = TP/(TP + FP) PPV = 700/(700 + 2100) → 700/2800 = 0.25,or 25% C - 0.50 Explanation Why 0.50 or 50% is the specificity of this TB test. This test is also known as the true negative rate because it represents the probability of a test being negative in the case of a patient without the disease. As a result, highly specific tests are most commonly used as confirmatory tests because they have low false positive rates. In this study, FP = 2100 and TN = 2100: Specificity = TN/(FP + TN) Specificity = 2100/(2100 + 2100) → 2100/4200 = 0.50, or 50% D - 0.70 Explanation Why 0.70 or 70% is the sensitivity of this TB test. This value is also known as the true positive rate because it is the probability of the test being positive in cases of true disease. High sensitivity tests, therefore, are useful for ruling out diseases, as they rarely have false negative results. In this study, TP = 700 and FN = 300: Sensitivity = TP/(FN + TP) Sensitivity = 700/(300 + 700) → 700/1000 = 0.70, or 70% eduwaves360.com | Telegram : @eduwaves360 E - 0.88 Explanation But The NPV varies inversely with the prevalence of a disease. Explanation Why The negative predictive value (NPV) is the probability that an individual who tested negative is actually disease-free. The NPV is calculated by dividing the number of true negatives (TN) by the total number of negatives, i.e. true negatives plus false negatives (FN). In this study, TN = 2100 and FN = 300: NPV = TN/(FN + TN) NPV = 2100/(300 + 2100) → 2100/2400 = 0.88, or 88% eduwaves360.com | Telegram : @eduwaves360 Question # 29 A study is conducted to investigate the relationship between the development of type 2 diabetes mellitus and the use of atypical antipsychotic medications in patients with schizophrenia. 300 patients who received the atypical antipsychotic clozapine and 300 patients who received the typical antipsychotic haloperidol in long-acting injectable form were followed for 2 years. At the end of the observation period, the incidence of type 2 diabetes mellitus was compared between the two groups. Receipt of clozapine was found to be associated with an increased risk of diabetes mellitus relative to haloperidol (RR = 1.43, 95% p < 0.01). Developed type 2 diabetes mellitus Did not develop type 2 diabetes mellitus Clozapine 30 270 Haloperidol 21 279 Based on these results, which of the following best represents the difference in the probability of developing type 2 diabetes mellitus in patients receiving clozapine and those taking a typical antipsychotic? Answer A 33.3 B 0.3 C 0.03 D 1.48 E 0.43 Image eduwaves360.com | Telegram : @eduwaves360 Hint The attributable risk is the excess risk in the clozapine group compared to the haloperidol group. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 33.3 Explanation Why 33.3 is the number needed to harm (NNH), meaning the number of people that have to be exposed to the risk factor clozapine before one person develops diabetes mellitus. The NNH is calculated as one divided by the attributable risk: 1/[(30/300) - (21/300)] = 1/(0.1-0.07) ≈ 33.3. B - 0.3 Explanation Why 0.3 is the attributable risk percent (ARP), not the attributable risk. The ARP is defined as the proportion of cases attributable to exposure to a risk factor, which can be easily calculated based on the relative risk (RR): ARP = (RR - 1)/RR. Since the relative risk of developing diabetes among those exposed to clozapine compared to those in the haloperidol group (not exposed) is (30/300) / (21/300) ≈ 1.43 , the attributable risk percent is thus approximately (1.43-1)/1.43 = 0.30, or 30%. This means that approximately 30% of type 2 diabetes mellitus cases in the clozapine group can be attributed to exposure to clozapine. C - 0.03 Explanation Why 0.03 is the attributable risk due to clozapine. Attributable risk is defined as the difference in disease occurrence attributable to an exposure. It is calculated as the incidence in the exposed group (clozapine group, 30/300 = 0.1) minus the incidence in the unexposed group (haloperidol group, 21/ 300 = 0.07): 0.1 - 0.07 = 0.03. This means that the absolute risk increase of developing type 2 diabetes mellitus attributable to clozapine is 3%. eduwaves360.com | Telegram : @eduwaves360 D - 1.48 Explanation Why 1.48 is the odds ratio of developing diabetes among those exposed to clozapine (exposed) compared to those in the haloperidol group (unexposed), not the attributable risk. The odds ratio is calculated as the odds of disease in the exposed group divided by the odds of disease in the unexposed group. In this case: OR = (30/21) / (270/279) ≈ 1.48. E - 0.43 Explanation Why Those exposed to clozapine have a 43% increased risk of developing diabetes, given that the relative risk of developing diabetes in the clozapine group compared to the haloperidol group is 1.43. This is not equivalent, however, to the attributable risk of an exposure, which is calculated as the difference in risks rather than from the ratio of one risk to another. eduwaves360.com | Telegram : @eduwaves360 Question # 30 A researcher is investigating the relationship between interleukin-1 (IL-1) levels and mortality in patients with end-stage renal disease (ESRD) on hemodialysis. In 2017, 10 patients (patients 1–10) with ESRD on hemodialysis were recruited for a pilot study in which IL-1 levels were measured (mean = 88.1 pg/mL). In 2018, 5 additional patients (patients 11–15) were recruited. Results are shown: Patient IL-1 level (pg/mL) Patient IL-1 level (pg/mL) Patient 1 (2017) 84 Patient 11 (2018) 91 Patient 2 (2017) 87 Patient 12 (2018) 32 Patient 3 (2017) 95 Patient 13 (2018) 86 Patient 4 (2017) 93 Patient 14 (2018) 90 Patient 5 (2017) 99 Patient 15 (2018) 81 Patient 6 (2017) 77 Patient 7 (2017) 82 Patient 8 (2017) 90 Patient 9 (2017) 85 Patient 10 (2017) 89 Which of the following statements about the results of the study is most accurate? Answer A Image Systematic error was introduced by the five new patients who joined the study in 2018. eduwaves360.com | Telegram : @eduwaves360 Answer B The median of IL-1 measurements is now larger than the mode. C The mean of IL-1 measurements is now larger than the mode. D The standard deviation was decreased by the five new patients who joined the study in 2018. E The median of IL-1 measurements is now larger than the mean. F The range of the data set is unaffected by the addition of five new patients in 2018. Image eduwaves360.com | Telegram : @eduwaves360 Hint Patient 12 has a much lower interleukin-1 (IL-1) levels (an outlier) than the other patients. This will result in a negatively skewed distribution. eduwaves360.com | Telegram : @eduwaves360 Correct Answer ASystematic error was introduced by the five new patients who joined the study in 2018. Explanation Why Systematic errors (bias) are reproducible inaccuracies resulting from an inappropriate experimental design or inaccurate measuring or classification of variables. There is no evidence of either in this case. B - The median of IL-1 measurements is now larger than the mode. Explanation Why The addition of a significant outlier in the five new patients in 2018 results in a more negativelyskewed distribution, and, as such, the new median (87 pg/mL) is actually less than the mode (90 pg/ mL). Prior to the addition of the five new patients in 2018, this data had no mode, as no value was repeated. C - The mean of IL-1 measurements is now larger than the mode. Explanation Why The mean value has decreased with the addition of the five new patients (to 84.1 pg/mL). In a negatively-skewed distribution, the mean is smaller, not larger, than the mode (which is in this case: 90 pg/mL). D- eduwaves360.com | Telegram : @eduwaves360 The standard deviation was decreased by the five new patients who joined the study in 2018. Explanation Why With the addition of a significant outlier in the five new patients in 2018, the variability of the data greatly increased. In this case, the standard deviation has now risen from around 6.5 pg/mL to over 15 pg/mL. E - The median of IL-1 measurements is now larger than the mean. Image Explanation But In a perfect normal distribution of data, mean, median, and mode are all equal. The mode is the most resistant measure of central tendency to outliers. eduwaves360.com | Telegram : @eduwaves360 Explanation Why The mean is inherently more susceptible to outliers when compared to the median. The mean value of the IL-1 levels following the addition of the new patients in 2018 changes to 84.1 pg/mL, compared to the previous mean of 88.1 pg/mL. The median IL-1 level, however, only changes from 88 pg/mL to 87 pg/mL over the same time frame. In this case the addition of Patient 12, who had an IL-1 level of only 32 pg/mL, both increased the dispersion of data and resulted in a more negativelyskewed distribution. FThe range of the data set is unaffected by the addition of five new patients in 2018. Explanation Why The range of data has increased to 67 pg/mL (Patient 5 (99) – Patient 12 (32)) from 22 pg/mL (Patient 5 (99) – Patient 6 (77)). Outliers typically result in the widening of the range of data. eduwaves360.com | Telegram : @eduwaves360 Question # 31 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? Answer A 68% of weights in women are likely to fall between 153 and 155 pounds. B 95% of weights in men are likely to fall between 180 and 184 pounds. C 99.7% of heights in women are likely to fall between 63.7 and 64.3 inches. D 86% of heights in women are likely to fall between 63.9 and 64.1 inches. E 99.7% of heights in men are likely to fall between 68.8 and 69.2 inches. Image eduwaves360.com | Telegram : @eduwaves360 Answer F Image 95% of heights in men are likely to fall between 68.85 and 69.15 inches. eduwaves360.com | Telegram : @eduwaves360 Hint For a standard normal distribution, a range that includes 1 standard deviation (SD) from the mean includes 68% of the sample. 2 SDs ≅ 95% of the sample; 3 SDs ≅ 99.7% of the sample. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 68% of weights in women are likely to fall between 153 and 155 pounds. Explanation Why For a standard normal distribution, a range that includes 1 SD from the mean includes 68% of the sample. However, the weight distribution in women was not normally distributed. B - 95% of weights in men are likely to fall between 180 and 184 pounds. Explanation Why For a standard normal distribution, a range that includes 2 SD from the mean includes 95% of the sample. However, the weight distribution in men was not normally distributed. C - 99.7% of heights in women are likely to fall between 63.7 and 64.3 inches. Explanation Why The sampled heights of women follow a standard normal distribution. Thus, a range that includes 3 SD (≅ 0.3 inches) from the mean includes 99.7% of the sampled heights. D - 86% of heights in women are likely to fall between 63.9 and 64.1 inches. Explanation Why The sampled heights of women follow a standard normal distribution. Thus, a range that includes 1 SD (≅ 0.1 inches) from the mean includes 68% of the sampled heights, not 86%. eduwaves360.com | Telegram : @eduwaves360 E - 99.7% of heights in men are likely to fall between 68.8 and 69.2 inches. Explanation Why The sampled heights of men follow a standard normal distribution. Thus, a range that includes 2 SD (≅ 0.2 inches) from the mean includes 95% of the sampled heights, not 99.7%. F - 95% of heights in men are likely to fall between 68.85 and 69.15 inches. Explanation Why The sampled heights of men follow a standard normal distribution. Thus, 95% of the sampled heights fall within 2 SD (≅ 0.2 inches) of the mean, not 0.15 inches. The stated range includes approximately 87% of the sampled heights, not 95%. eduwaves360.com | Telegram : @eduwaves360 Question # 32 A group of researchers conducted a study to determine whether there is an association between folic acid supplementation before pregnancy and autism spectrum disorder (ASD) in offspring. The researchers retrospectively surveyed 200 mothers with children diagnosed with ASD during the first 4 years of life and 200 mothers with healthy children. All participants were interviewed about their prenatal consumption of folic acid using standardized questionnaires. A 94% response rate was obtained from the surveys. The study ultimately found that folic acid supplementation was associated with lower rates of ASD in offspring (OR = 0.3, p < 0.01). Which of the following type of bias is most likely to have influenced these results? Answer A Non-response bias B Interviewer bias C Latency period D Recall bias E Survival bias F Length-time bias Image eduwaves360.com | Telegram : @eduwaves360 Hint A significant period of time (4 years) has passed between the prenatal exposure being studied here and the administration of the questionnaire. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Non-response bias Explanation Why It is unlikely than non-response bias influenced the results of the study, as the response rate was high (94%). B - Interviewer bias Explanation Why This study used standardized questionnaires, making interviewer bias unlikely. There is no indication that there were any differences in the way that the questionnaires here were administered to the two groups. C - Latency period Explanation Why The latency period would be a concern if there were insufficient time between prenatal folic acid supplementation and ASD. However, most symptoms of ASD appear within the first 18 months of life. The study identified cases as those with symptoms within the first 4 years of life, a more than adequate period of time between the exposure and potential outcome. eduwaves360.com | Telegram : @eduwaves360 D - Recall bias Explanation Why Recall bias is a major concern in case-control studies, as subjects with a particular outcome may unintentionally or intentionally recall past exposures in more detail than those without the outcome. This difference in recall may lead to misclassification of exposure and false conclusions about the relationship between exposure and outcome. E - Survival bias Explanation Why It is unlikely that survival bias is at play here. While it is possible that the children in the study have more severe disease than those whose mothers opted not to participate, it is not likely that this is because the true population of those with ASD had different rates of survival or resolution of disease. F - Length-time bias Explanation Why Length-time bias takes place in screening studies, not case-control studies. It affects the measured survival of those with the outcome, not the exposures in those with the outcome compared to those without the outcome. eduwaves360.com | Telegram : @eduwaves360 Question # 33 A 47-year-old man comes to the physician for a routine health maintenance examination. He has no complaints and has no history of serious illness. He works as a forklift operator in a factory. His brother died of malignant melanoma. He smokes occasionally and drinks a glass of wine once a week. His pulse is 79/min and blood pressure is 129/84 mm Hg. Which of the following causes of death is this patient most at risk for over the next 15 years? Answer A Lung cancer B Coronary artery disease C Malignant melanoma D Industrial accident E Prostate cancer F Suicide Image eduwaves360.com | Telegram : @eduwaves360 Hint What is the leading cause of death in men in the United States? eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Lung cancer Explanation Why Lung cancer is the second most common cancer among men as well as women and the leading cause of cancer deaths worldwide. Although this patient's occasional smoking slightly increases his risk of lung cancer-related death, his risk of death due to another cause is higher. B - Coronary artery disease Explanation Why Coronary heart disease (CHD) is the leading cause of death in the US and worldwide, and the death rate is higher in men than in women. This patient's risk for coronary heart disease is higher than a non-smoker due to his occasional smoking. Malignancy is the most common cause of death among patients 45–64 years of age. However, heart disease is a more common cause of death than any single type of cancer in the United States. C - Malignant melanoma Explanation Why Malignant melanoma is a life-threatening dermatologic disease and the fifth most common cancer in men. While this patient's family history increases his risk of developing melanoma (around 10% of melanoma cases are familial), his risk of death due to another cause is higher. eduwaves360.com | Telegram : @eduwaves360 D - Industrial accident Explanation Why Industrial accidents are the third most common cause of death in American men. Although this patient's occupation as a forklift operator increases his risk of accident-related injuries, his risk of death due to another cause is higher. E - Prostate cancer Explanation Why Prostate cancer is the most common cause of cancer in men in the US and the second most common cause of cancer-related death. Although the fact that this patient is male means he is at risk of developing prostate cancer, his risk of death due to another cause is higher. F - Suicide Explanation Why Suicide or intentional self-harm is currently the tenth leading cause of death in the US. This patient does not express any suicidal intention and has no signs of depression; his risk of death due to another cause is higher. eduwaves360.com | Telegram : @eduwaves360 Question # 34 A clinical trial is conducted to determine the efficacy of ginkgo biloba in the treatment of Parkinson disease. A sample of patients with Parkinson disease is divided into two groups. Participants in the first group are treated with ginkgo biloba, and participants in the other group receive a placebo. A change in the Movement Disorder Society-Unified Parkinson Disease Rating Scale (MDS-UPDRS) score is used as the primary endpoint for the study. The investigators, participants, and data analysts were meant to be blinded throughout the trial. However, while the trial is being conducted, the patients' demographics and their allocated treatment groups are mistakenly disclosed to the investigators, but not to the participants or the data analysts, because of a technical flaw. The study concludes that there is a significant decrease in MDS-UPDRS scores in patients treated with ginkgo biloba. Which of the following is most likely to have affected the validity of this study? Answer A Recall bias B Pygmalion effect C Hawthorne effect D Effect modification E Procedure bias F Ascertainment bias Image eduwaves360.com | Telegram : @eduwaves360 Hint In this scenario, the investigators' belief in the efficacy of ginkgo biloba could have influenced the outcome of the analysis. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Recall bias Explanation Why Recall bias occurs when the subject's awareness of a condition alters their recall of exposure to risk factors. This type of bias most commonly occurs in retrospective studies; there is no evidence of recall bias in this scenario. B - Pygmalion effect Explanation Why Pygmalion effect (observer-expectancy bias) occurs when the experimenter's expectation influences the outcome of the study. In this scenario, patient information was mistakenly disclosed to the investigators during the trial. The investigators' belief in the efficacy of treatment with ginkgo biloba may have influenced data collection and analysis. To avoid this bias, study designs have to include placebo groups and experimenters need to be blinded throughout the study. C - Hawthorne effect Explanation Why Hawthorne effect occurs when individuals in test groups change their behavior once they are aware that they are being observed. The Hawthorne effect is difficult to eliminate in study designs that adhere to informed consent. However, this effect would have occurred in both groups alike and would not have affected the results of the study. eduwaves360.com | Telegram : @eduwaves360 D - Effect modification Explanation Why Effect modification occurs when an exposure has a different effect on different subgroups because of an additional factor internal to the respective subgroups that affects the outcome of the exposure. No such internal factor has been described in this scenario. E - Procedure bias Explanation Why Procedure bias occurs when individuals in different study groups receive different treatment because of their affiliation with their respective groups. There is no evidence that the two groups were treated differently in this scenario. F - Ascertainment bias Explanation Why Ascertainment bias (sampling bias) occurs when certain individuals are more likely to be selected for a study group, which results in a nonrandomized, unrepresentative sample of a population. Though patient information was mistakenly disclosed to the investigators, this occurred after treatment groups were established; there is no evidence that the study group was subject to sampling bias. eduwaves360.com | Telegram : @eduwaves360 Question # 35 During a clinical study on an island with a population of 2540 individuals, 510 are found to have fasting hyperglycemia. Analysis of medical records of deceased individuals shows that the average age of onset of fasting hyperglycemia is 45 years, and the average life expectancy is 70 years. Assuming a steady state of population on the island with no change in environmental risk factors, which of the following is the best estimate of the number of individuals who would newly develop fasting hyperglycemia over 1 year? Answer A 50 B 10 C 30 D 40 E 20 Image eduwaves360.com | Telegram : @eduwaves360 Hint The prevalence is 0.2 (510/2540) and the average duration of the disease is 25 years (70 - 45). A population is in a "steady state" when the birth rate equals the death rate, the prevalence is fairly constant, and the incidence of the disease remains constant. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 50 Explanation Why In populations with a steady state, incidence rates can be calculated from prevalence and the average duration of the disease. This answer is not the correct value. B - 10 Explanation Why In populations with a steady state, incidence rates can be calculated from prevalence and the average duration of the disease. This answer is not the correct value. C - 30 Explanation Why In populations with a steady state, incidence rates can be calculated from prevalence and the average duration of the disease. This answer is not the correct value. D - 40 Explanation Why In populations with a steady state, incidence rates can be calculated from prevalence and the average duration of the disease. This answer is not the correct value. eduwaves360.com | Telegram : @eduwaves360 E - 20 Image Explanation Why If the population is in a steady state, the relationship between incidence rate (IR), prevalence (P), and the average duration of the disease (T) can be described mathematically as P/(1-P) = IR x T. Since the average duration of the disease is 25 years and the prevalence is 0.2, the incidence rate is about 0.01 per year. The predicted number of new cases of fasting hyperglycemia after one year would, therefore, be IR × population at risk = 0.01 × (2540 - 510) ≈ 20 cases. eduwaves360.com | Telegram : @eduwaves360 Question # 36 An investigator is studying determinants of childhood obesity by observing a cohort of pregnant women with obesity. After delivery, he regularly records the height and weight of the cohort's children. The results of the correlation analysis between mean childhood BMI at 4 years of age and mean maternal BMI before pregnancy are shown. Based on these findings, which of the following is the most likely correlation coefficient? Answer A -0.45 B 1.80 C 0.95 Image eduwaves360.com | Telegram : @eduwaves360 Answer D 0.45 E 0 F -1.80 Image eduwaves360.com | Telegram : @eduwaves360 Hint In the descriptive analysis of data, a scatter plot can be helpful to ascertain whether there is an association between two variables. The correlation coefficient (r) represents the strength of the linear relationship between two variables (i.e., how close the data points are to a straight line of regression) as well as the polarity of the relationship. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - -0.45 Explanation Why A correlation coefficient (r) of -0.45 reflects a moderately strong relationship between two variables, which is shown by this scatter plot. However, a negative correlation suggests that one variable tends to decrease as the other increases, which is not the case here. B - 1.80 Explanation Why The correlation coefficient (r) must be positive since the plot is ascending (positive correlation). However, r ranges from -1 to 1; 1.8 is therefore not a meaningful value. C - 0.95 Explanation Why A correlation coefficient (r) of 0.95 reflects a nearly perfect positive relationship between two variables. As seen in this graph, one variable tends to increase as the other variable increases. While there is clearly a positive trend, the points on the scatter plot do not come close to clustering into a single straight line. This scatter plot suggests that the relationship between the two variables is moderately strong but not the nearly perfect relationship expected with an r of 0.95. eduwaves360.com | Telegram : @eduwaves360 D - 0.45 Explanation Why A correlation coefficient (r) of 0.45 reflects a moderately strong positive linear correlation between maternal BMI and childhood BMI. While there is a clear positive trend (i.e., as one variable increases, the other variable also tends to increase), the points on the scatter plot do not cluster into a straight line, indicating that the relationship is moderately strong but not a perfect correlation. Correlation does not imply causation! It merely shows whether a linear relationship exists between the two variables and how strong it is. E-0 Explanation Why A correlation coefficient (r) of 0 suggests that there is no relationship between the two variables. However, this graph shows a clear trend, where one variable tends to increase as the other variable increases; there does appear to be a relationship between the two variables. F - -1.80 Explanation Why The correlation coefficient (r) must be positive because the plot is ascending (positive correlation). Moreover, r ranges from -1 to 1; -1.8 is therefore not a meaningful value. eduwaves360.com | Telegram : @eduwaves360 Question # 37 A clinical trial is conducted to determine the role of cerebrospinal fluid (CSF) beta-amyloid levels as a biomarker in the early detection and prognosis of Alzheimer disease. A total of 100 participants are enrolled and separated into three groups according to their Mini-Mental State Examination (MMSE) score: mild dementia (20–24 points), moderate dementia (13–20 points), and severe dementia (< 13 points). Participants' CSF level of beta-amyloid 42 is measured using an immunoassay. It is found that participants with severe dementia have a statistically significantly lower mean CSF level of beta-amyloid 42 compared to the other two groups. Which of the following statistical tests was most likely used to compare measurements between the study groups? Answer A Chi-square test B Fishers exact test C Two-sample t-test D Pearson correlation analysis E Analysis of variance Image eduwaves360.com | Telegram : @eduwaves360 Hint The statistical test was conducted to determine if there are statistically significant differences in the mean of a continuous variable (CSF beta-amyloid 42 level) between three groups of individuals (mild, moderate, and severe dementia). eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Chi-square test Explanation Why The Chi-square test is used to determine if there is a relationship between ≥ 2 categorical variables. It tests whether the difference between two proportions is statistically significant. Here, the investigators are comparing the means of a continuous variable (CSF beta-amyloid 42 level) between three groups, so the Chi-square test is not applicable. The Chi-square test could be used, for example, to assess if there is a statistically significant difference in the proportion of individuals with abnormal CSF beta-amyloid 42 levels in the dementia (MMSE ≤ 24) and non-dementia (MMSE > 24) group. B - Fishers exact test Explanation Why Fishers exact test is used to determine if there is a relationship between ≥ 2 categorical variables. Like the Chi-square test, it tests whether the difference between two proportions is statistically significant, but Fishers is usually used when the study sample is small. Here, the investigators are comparing the means of a continuous variable (CSF beta-amyloid 42 level) between three groups, so Fishers exact test is not applicable. C - Two-sample t-test Explanation Why A two-sample t-test can be used to determine if there is a statistically significant difference in the means of a continuous variable between two categorical variables (i.e., groups). Here, the investigators are comparing the means of three groups, so a two-sample t-test is not appropriate. A ttest could be used, for example, if the investigators were comparing mean CSF beta-amyloid 42 levels in individuals with (MMSE ≤ 24) and without (MMSE > 24) dementia. eduwaves360.com | Telegram : @eduwaves360 D - Pearson correlation analysis Explanation Why In Pearson correlation analysis, the Pearson correlation coefficient is determined. The Pearson correlation coefficient describes the strength and polarity (positive, negative) of a linear relationship between two continuous variables. The variables in this scenario include a continuous variable (CSF beta-amyloid 42 levels) and three categorical variables (mild, moderate, and severe dementia), so Pearson correlation analysis is not applicable. The correlation coefficient would be appropriate, for example, to test for a linear relationship between the CSF beta-amyloid 42 levels and the MMSE score (continuous variable ranging from 0–30). E - Analysis of variance Explanation Why Analysis of variance (ANOVA) is the appropriate statistical test in this scenario because it compares differences in means in a continuous outcome among 3 or more levels of a categorical variable (i.e., groups). ANOVA tests the null hypothesis (i.e., there is no difference in means between the groups) by comparing the variation among group means relative to the variation within groups. If two or more group means are statistically significantly different from one another, the null hypothesis can be rejected. eduwaves360.com | Telegram : @eduwaves360 Question # 38 A recent study examined trends in incidence and fatality of ischemic stroke in a representative sample of Scandinavian towns. The annual incidence of ischemic stroke was calculated to be 60 per 2,000 people. The 1-year case fatality rate for ischemic stroke was found to be 20%. The health department of a town in southern Sweden with a population of 20,000 is interested in knowing the 1-year mortality conferred by ischemic stroke. Based on the study's findings, which of the following estimates the annual mortality rate for ischemic stroke per 20,000? Answer A 12 people B 120 people C 400 people D 60 people E 600 people Image eduwaves360.com | Telegram : @eduwaves360 Hint The annual mortality rate for ischemic stroke (i.e., number of deaths due to ischemic stroke per year) can be directly calculated based on annual incidence of ischemic stroke (i.e., number of new cases of ischemic stroke per year) and the 1-year case fatality rate (i.e., proportion of deaths due to ischemic stroke during the first year after diagnosis). eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 12 people Explanation Why This is the mortality rate for a population of 2,000 people, as given in the question (incidence of 60 cases per 2,000 people with a 20% fatality rate). The question is asking for a mortality rate in a sample population of 20,000 people. B - 120 people Explanation Why The annual incidence for ischemic stroke equals 60 per 2,000 people in this study. Based on the 1-year case fatality rate of 20%, 12 of those 60 new cases each year die. Thus, 12 out of 2,000 people die every year due to ischemic stroke. Therefore, the annual mortality rate for 20,000 people is 120 (12 per 2,000, multiplied by 10). C - 400 people Explanation Why An annual mortality rate of 400 per 20,000 people is incorrect based on the annual incidence of 60 per 2,000 people and a 1-year case fatality rate of 20%. An annual mortality rate of 400 per 20,000 people occurs if one calculates 20% of 2,000 people (sample population of the study), which is incorrect as 20% of the 60 diseased people (those with an ischemic stroke) die, not all 2,000. eduwaves360.com | Telegram : @eduwaves360 D - 60 people Explanation Why An annual mortality rate of 60 per 20,000 people is incorrect based on the annual incidence of 60 per 2,000 people and a 1-year case fatality rate of 20%. This means that each year, 60 out of 2,000 healthy individuals sustain an ischemic stroke and 20% of these 60 people die. Thus, an annual mortality rate of 60 per 20,000 would only be correct in the setting of a 1-year case fatality rate of 10% instead of the 20% found in the study. E - 600 people Explanation Why This number represents the number of people who will suffer an ischemic stroke in a population of 20,000 people. Alternatively, this figure would also describe the mortality rate if ischemic stroke was found to be fatal in 100% of cases. However, the study indicates that the fatality rate is only 20% in 1 year so this value overestimates the number of deaths. eduwaves360.com | Telegram : @eduwaves360 Question # 39 A group of investigators conducted a randomized controlled trial to compare the effectiveness of rivaroxaban to warfarin for ischemic stroke prevention in patients with atrial fibrillation. A total of 14,000 participants were enrolled and one half was assigned to each of the cohorts. The patients were followed prospectively for 3 years. At the conclusion of the trial, the incidence of ischemic stroke in participants taking rivaroxaban was 1.7% compared to 2.2% in participants taking warfarin. The hazard ratio is calculated as 0.79 and the 95% confidence interval is reported as 0.64 to 0.97. If the study was conducted with a total of 7,000 participants, which of the following changes would most be expected? Answer A Increased risk of selection bias B Decreased hazard ratio C Increased confidence interval range D Decreased type I error rate E Increased risk of confounding bias F Decreased type II error rate Image eduwaves360.com | Telegram : @eduwaves360 Hint Decreasing the number of participants (i.e., the sample size) also decreases the power of the study. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Increased risk of selection bias Explanation Why Bias is a flaw in a study's methodology that may lead to inaccurate results. Selection bias is the establishment of a sample that is poorly representative of the population or disease being studied, and may therefore skew findings. As this is a methodologic or randomization flaw, the risk of selection bias does not change with the reduced sample size proposed here. It must be managed prior to recruitment. B - Decreased hazard ratio Explanation Why Hazard ratio is a measure of efficacy, such as the ability of a drug to treat (or be toxic to) a patient. As a change in the sample size in this study would have no impact on the relative efficacies of rivaroxaban and warfarin, its hazard ratio would not be expected to change. C - Increased confidence interval range Explanation Why Confidence intervals are defined as the mean ± standard error of the mean, which is calculated by multiplying a Z-score (for 95% confidence intervals this is always 2) by the standard deviation (SD) divided by the square root of the sample size (Mean +/- Z-score (SD/√sample size)). A larger sample size or a decreased SD (based on data precision) will decrease the standard error. Including more disparate data or reducing the sample size, as was done here, will increase it. eduwaves360.com | Telegram : @eduwaves360 D - Decreased type I error rate Explanation Why Type I error rejects the null hypothesis (accepting a different study outcome), when it is in fact true. This value is used to state what threshold should be used to consider the difference between two values as statistically significant. The inverse of type I error is the p-value. This value is fixed regardless of study power, sample size, or standard deviation. Therefore, it would not change if the sample size of this study were reduced. E - Increased risk of confounding bias Explanation Why Confounding is the ability for an unconsidered variable to influence a relationship being studied. For example, a study attempting to associate weight with lung cancer incidence may not consider if more overweight patients smoke. This is a methodologic flaw and must be mitigated using subgroup analyses (i.e., looking at smokers who are overweight as their own group), or by performing the same study on multiple populations. A decrease in sample size, as done here, will not change the risk of confounding. F - Decreased type II error rate Explanation Why Type II error is accepting a null hypothesis (disproving alternative hypotheses) when it is actually false. It is measured as 1 - power. When the sample size is reduced, as was done in this case, the power of the study will be reduced. Type II error rate is therefore increased, not decreased. eduwaves360.com | Telegram : @eduwaves360 Question # 40 A study is designed to investigate the relationship between type 2 diabetes mellitus and the development of pancreatic cancer. After a review of medical records, the investigators identify a group of patients who had a diagnosis of type 2 diabetes mellitus but did not have a history of pancreatic cancer as of 10 years ago. Another group of similar patients with no history of type 2 diabetes mellitus or pancreatic cancer is also identified. The investigators examine the medical charts to determine whether the group of patients who had been diagnosed with type 2 diabetes mellitus was more likely to have developed pancreatic cancer over the 10-year period. The investigators ultimately found that the risk of developing pancreatic cancer was 25% higher in patients with type 2 diabetes mellitus compared to patients without this condition. This study is best described as which of the following? Answer A Ecological study B Prospective cohort study C Case-control study D Meta-analysis E Cross-sectional study F Retrospective cohort study Image eduwaves360.com | Telegram : @eduwaves360 Hint The investigator identified patients with an exposure (diabetes mellitus) and assessed whether they developed an outcome (pancreatic cancer) via chart review. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Ecological study Explanation Why An ecological study is an observational study that assesses aggregated data with at least one variable collected at the population level (e.g., health statistics from different countries, cities, or socioeconomic groups). For instance, rather than collecting data from individual patients on their history of diabetes mellitus and development of pancreatic cancer, an ecological study would compare the mean, prevalence, incidence, or rates of pancreatic cancer between cities or communities to identify any associated risk factors. B - Prospective cohort study Explanation Why A prospective cohort study also begins by identifying individuals with or without exposure to a particular risk factor (e.g., diabetes mellitus) and determining whether the risk factor is associated with a particular outcome (e.g., pancreatic cancer). However, in a prospective cohort study, the outcome would have occurred during the course of the study, unlike this study in which the exposure and the outcome have already occurred. C - Case-control study Explanation Why Case-control studies begin by identifying individuals with or without a particular outcome (e.g., pancreatic cancer) and the investigator then compares these two groups for exposure to one or more risk factors (e.g., diabetes mellitus) and determines the disease odds ratio. In this study, the investigator began by identifying the risk factor first and then assessed the risk of the outcome. eduwaves360.com | Telegram : @eduwaves360 D - Meta-analysis Explanation Why In a meta-analysis, data from multiple studies is systematically assessed to increase statistical power and to identify discrepancies between and/or common effects among studies. Here, the results of an individual study were presented. E - Cross-sectional study Explanation Why Cross-sectional studies measure the prevalence of potential risk factors (e.g., diabetes mellitus) and outcomes (e.g., pancreatic cancer) at a single point in time. Unlike the study described here, a crosssectional study does not measure a temporal change in outcomes. F - Retrospective cohort study Explanation Why The investigator conducted a retrospective cohort study, selecting a cohort of individuals with exposure to a risk factor (e.g., diabetes mellitus) and determined the risk of a specific outcome (pancreatic cancer) by following up over a period of time. Cohort studies are well suited to determining the incidence of a disease and allow for the relative risk (RR) of developing a disease of interest to be calculated. Here, the RR of developing pancreatic cancer is 1.25 in patients with a previous diagnosis of diabetes mellitus compared to patients without this risk factor. Retrospective cohort studies are time-saving and relatively inexpensive when compared to prospective cohort studies because they involve the analysis of data that has already been collected. eduwaves360.com | Telegram : @eduwaves360 eduwaves360.com | Telegram : @eduwaves360 Join us on Telegram : Click here : @eduwaves360 Unlocked the Medical premiums Click here : www.eduwaves360.com Medical Courses : Discussion Group : https://t.me/usmle_study_materials_2 @usmle_discussion_group eduwaves360.com | Telegram : @eduwaves360 Join us on Telegram : Click here : @eduwaves360 Unlocked the Medical premiums Click here : www.eduwaves360.com Medical Courses : Discussion Group : https://t.me/usmle_study_materials_2 @usmle_discussion_group eduwaves360.com | Telegram : @eduwaves360 Question # 1 A research group wants to assess the relationship between childhood diet and cardiovascular disease in adulthood. A prospective cohort study of 500 children between 10 to 15 years old is conducted in which the participants' diets are recorded for 1 year and then the patients are assessed 20 years later for the presence of cardiovascular disease. A statistically significant association is found between childhood consumption of vegetables and decreased risk of hyperlipidemia, as well as greater exercise tolerance. When these findings are submitted to a scientific journal, a peer reviewer comments that the researchers did not discuss the study's validity. Which of the following additional analyses would most likely address the concerns about this study's design? Answer A Blinding B Crossover C Matching D Stratification E Randomization Image eduwaves360.com | Telegram : @eduwaves360 Hint To address the validity of this prospective cohort study, the researchers should address any confounding variables. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Blinding Explanation Why Blinding refers to the practice of not revealing to patients whether they are a control or treatment candidate (e.g., receiving a placebo or not). Blinding is used in cohort studies (e.g., double-blind studies) to decrease the likelihood of bias. This practice does not decrease confounding bias. B - Crossover Explanation Why In a crossover study design, participants receive all interventions but at different times and/or in a different sequence, allowing them to act as their own study control. Crossover designs can help mitigate the effect of confounders in participant selection. However, additional analysis cannot convert the described prospective cohort study into a crossover study, which is an entirely different study design. Furthermore, a crossover design would not be appropriate to assess the relationship between childhood diet and cardiovascular disease in adulthood because this relationship does not involve any intervention and the observation period of 20 years would be impractical for this study design. C - Matching Explanation Why Matching is the process of grouping study participants based on possible or known confounding variables (e.g., gender). This process, which occurs during the design phase of a study to ensure study and control groups have similar baseline characteristics, could have improved this study's validity. However, this study has already been completed, and matching cannot be retrospectively applied to data that has already been collected. eduwaves360.com | Telegram : @eduwaves360 D - Stratification Explanation Why Stratification, the division of a population into subpopulations based on a particular characteristic, would address concerns regarding the validity of this study by mitigating susceptibility to confounding bias. For example, in this study, higher socioeconomic status could be associated with a higher consumption of vegetables and therefore a lower risk of hyperlipidemia and greater exercise tolerance. The effect of confounding can be reduced by stratifying subjects by their socioeconomic status. E - Randomization Explanation Why Randomization occurs when study participants are assigned to different groups (e.g., to treatment and control groups) by chance. Randomization, a central component of clinical trials, ensures that the sample group is representative of the population being studied and can control for both known and unknown confounders. However, randomization is a tactic that is used in randomized controlled trials, not in cohort studies, which are a type of observational study. eduwaves360.com | Telegram : @eduwaves360 Question # 2 A case-control study is conducted to investigate the association between the use of phenytoin during pregnancy in women with epilepsy and the risk for congenital malformations. The odds ratio of congenital malformations in newborns born to women who were undergoing treatment with phenytoin is 1.74 (P = 0.02) compared to newborns of women who were not treated with phenytoin. Which of the following 95% confidence intervals is most likely reported for this association? Answer A 1.75 to 2.48 B 0.56 to 1.88 C 0.36 to 0.94 D 1.34 to 2.36 E 0.83 to 2.19 Image eduwaves360.com | Telegram : @eduwaves360 Hint A P value < 0.05 corresponds to a 95% confidence interval that does not contain the null hypothesis value (i.e., an odds ratio of 1). In this scenario, the null hypothesis is that there is no difference in the odds of congenital malformations in newborns born to women who received phenytoin compared to those who did not. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 1.75 to 2.48 Explanation Why A 95% confidence interval of 1.75–2.48 indicates that there is a 95% chance that the true odds ratio lies between 1.75–2.48. Though this range does not contain the null hypothesis value (i.e., an odds ratio of 1), it does not contain the calculated odds ratio of 1.74. B - 0.56 to 1.88 Explanation Why A 95% confidence interval of 0.56–1.88 indicates that there is a 95% chance that the true odds ratio lies between 0.56–1.88. Though this range contains the calculated odds ratio of 1.74, it also contains the null hypothesis value (i.e., an odds ratio of 1) and is, therefore, not statistically significant (i.e., P ≥ 0.05). C - 0.36 to 0.94 Explanation Why A 95% confidence interval of 0.36–0.94 indicates that there is a 95% chance that the true odds ratio lies between 0.36–0.94. Though this range does not contain the null hypothesis value (i.e., an odds ratio of 1), it does not contain the calculated odds ratio of 1.74. eduwaves360.com | Telegram : @eduwaves360 D - 1.34 to 2.36 Explanation Why A 95% confidence interval of 1.34–2.36 indicates that there is a 95% chance that the true odds ratio lies between 1.34–2.36 This range contains the calculated odds ratio of 1.74; in addition, it does not contain the null hypothesis value (i.e., an odds ratio of 1) and is, therefore, statistically significant (i.e., P < 0.05). E - 0.83 to 2.19 Explanation Why A 95% confidence interval of 0.83–2.19 indicates that there is a 95% chance that the true odds ratio lies between 0.83–2.19. Though this range contains the calculated odds ratio of 1.74, it also contains the null hypothesis value (i.e., an odds ratio of 1) and is, therefore, not statistically significant (i.e., P ≥ 0.05). eduwaves360.com | Telegram : @eduwaves360 Question # 3 A large, multicenter study is conducted to assess the prevalence and mortality rate of pneumonia requiring hospitalization caused by various pathogens in 3 states over the past year. The study encompassed 250,000 adults between 45–79 years of age. The results for five different pathogens are shown: All cases Fatal cases Pathogen Number Percentage Number Percentage S. aureus 460 34.8 69 50.3 Influenza virus 288 21.8 12 8.8 P. aeruginosa 136 10.3 17 12.4 S. pneumoniae 338 25.5 27 19.7 Klebsiella spp. 101 7.6 12 8.8 Total 1323 100 137 100 A journalist would like to report the yearly case fatality rate of the pneumonia pathogen responsible for most hospitalizations in these states. Based on the study results, which of the following responses by the investigator is most accurate? Answer A 10.4% B 50.3% C 34.8% Image eduwaves360.com | Telegram : @eduwaves360 Answer D 15.0% E 12.5% Image eduwaves360.com | Telegram : @eduwaves360 Hint Of the given pathogens, S. aureus is the pathogen responsible for most hospitalizations, as it is responsible for 34.8% of all pneumonia cases requiring hospital admission in this cohort. The case fatality rate is the proportion of fatal cases of a specified disease within a reference period. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 10.4% Explanation Why This value is the overall mortality rate for all patients of pneumonia that were admitted to the hospital: 137 deaths/1323 cases = 0.104 (or 10.4%). However, this journalist is asking about the mortality rate of pneumonia caused by one specific pathogen. B - 50.3% Explanation Why This value is the proportion of fatalities attributable to S. aureus (69/137 ≈ 0.503, or 50,3%). Although S. aureus is indeed the most common pathogen causing pneumonia that leads to hospital admission, this value does not represent the case fatality rate. C - 34.8% Explanation Why This value is the proportion of total cases caused by S. aureus, otherwise known as incidence: 460 cases/1323 total cases = 0.348 (or 34.8%). While this value correctly identifies S. aureus as the most common pathogenic causing pneumonia that required hospital admission, the journalist is most interested in identifying the mortality rate of this infection, not the incidence. eduwaves360.com | Telegram : @eduwaves360 D - 15.0% Explanation Why The case fatality rate for the most common organism causing pneumonia that requires hospitalization is 15.0%. There were 460 cases of pneumonia caused by S. aureus in the study cohort, of which 69 were fatal. The case fatality rate, therefore, is 69/460 = 0.15, or 15.0%. E - 12.5% Explanation Why This value is the case fatality rate from P. aeruginosa, 17 deaths/136 cases = 0.125 (or 12.5%). However, the journalist is interested in the case fatality rate of pneumonia caused by a different pathogen. eduwaves360.com | Telegram : @eduwaves360 Question # 4 The height of American adults is expected to follow a normal distribution, with a typical male adult having an average height of 69 inches with a standard deviation of 0.1 inches. An investigator has been informed about a community in the American Midwest with a history of heavy air and water pollution in which a lower mean height has been reported. The investigator plans to sample 30 male residents to test the claim that heights in this town differ significantly from the national average based on heights assumed be normally distributed. The significance level is set at 10% and the probability of a type 2 error is assumed to be 15%. Based on this information, which of the following is the power of the proposed study? Answer A 0.85 B 0.15 C 0.10 D 0.90 E 0.05 Image eduwaves360.com | Telegram : @eduwaves360 Hint Statistical power is the probability that the study will detect a statistically significant difference between the two experimental groups if one truly exists. The statistical power can be calculated directly using the probability of incorrectly rejecting the alternative hypothesis. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 0.85 Explanation But The statistical power of any study can be increased by increasing the sample size, expected effect size, or precision of measurement. Explanation Why The power of the proposed study refers to the probability of correctly rejecting the null hypothesis. By convention, most studies aim to achieve 80% statistical power. The statistical power can be calculated as follows: Power = 1 - β, where β is the probability of a type 2 error. In this question, β is 15%. Therefore, the power of this study would be 0.85 (Power = 1 - 0.15 = 0.85 or 85%). B - 0.15 Explanation Why This value represents the probability of a type 2 error (β). In this study, it is assumed to be 15%. Although β can be used to calculate the statistical power of a study, it does not equate to it. C - 0.10 Explanation Why This value represents the significance level (α), which is the probability of rejecting the null hypothesis when it is true. This is also the probability of a type 1 error. A study is considered statistically significant if the p-value is less than the prespecified alpha (α). The significance level, however, is not equivalent to statistical power. eduwaves360.com | Telegram : @eduwaves360 D - 0.90 Explanation Why This value represents the confidence level of the study. Confidence level = 1 - α, where α is the significance level. The significance level of this study is set at 10%. Therefore, the confidence level would be 90%, meaning that there is 90% confidence that the range of values in the study (confidence interval) contains the true sample measurement. The confidence level is not equivalent to statistical power. E - 0.05 Explanation Why This value represents the significance level (α) used for the majority of medical and epidemiological studies. The α for this study was prespecified as 0.10. Additionally, the significance level of a study is not equivalent to the statistical power of that study. eduwaves360.com | Telegram : @eduwaves360 Question # 5 A group of investigators is evaluating the efficacy of intranasal ketamine in decreasing acute suicidality in teenagers admitted to inpatient psychiatric units. They conducted a pilot study in which 15 participants were randomized to receive intranasal ketamine while 15 participants were randomized to receive placebo. The investigators ultimately found no statistically significant difference in suicidality after 30 days of follow up with the level of significance fixed at 5%. They suspect inadequate statistical power. Assuming the investigators intend to keep the level of significance at 5%, which of the following changes would be most effective to increase the statistical power? Answer A Decrease alpha B Decrease the sample size C Decrease the type II error rate D Increase beta E Increase the type I error rate Image eduwaves360.com | Telegram : @eduwaves360 Hint Statistical power is the probability that a study will detect a true difference, i.e., rejecting the null hypothesis when the alternative hypothesis is true. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Decrease alpha Explanation Why Decreasing α (otherwise known as the type I error rate) is an aspect of experiment design that is intended to offset the risk of concluding that an effect or difference exists when there is, in fact, no effect or difference. The α is set by the researcher, with the most traditional value being 0.05. These researchers also used 0.05 (i.e., 5%) as the significance threshold. Changing this value will have no direct effect on the statistical power of the experiment. B - Decrease the sample size Explanation Why Decreasing the sample size will cause a reduction in statistical power. A smaller sample is more subject to the effects of random chance, so the probability of committing a type II error will necessarily increase in this condition. C - Decrease the type II error rate Explanation Why Decreasing the type II error rate (symbolized as β) reduces the chance of accepting a false null hypothesis and rejecting a viable alternative hypothesis. Reducing β can be achieved by increasing the sample size in the study or by making more precise measurements of the outcome of interest. Either of these two actions will directly increase the statistical power. In mathematical terms, power = 1 – β so it easily follows that reducing β will lead to increased statistical power. eduwaves360.com | Telegram : @eduwaves360 D - Increase beta Explanation Why Beta, or the probability of a type II error, is the complement of power [power = 1 – β]. Increasing β would cause a decrease in statistical power, the opposite of the desired effect. E - Increase the type I error rate Explanation Why Increasing the type I error rate (also known as alpha), will have no direct effect on the statistical power. Alpha is the significance level that is chosen by the researcher. In this experiment, it has been set at 0.05 (i.e., 5%), which is the most common threshold for this value. eduwaves360.com | Telegram : @eduwaves360 Question # 6 A researcher is examining the relationship between socioeconomic status and IQ scores. The IQ scores of young American adults have historically been reported to be distributed normally with a mean of 100 and a standard deviation of 15. Initially, the researcher obtains a random sampling of 300 high school students from public schools nationwide and conducts IQ tests on all participants. Recently, the researcher received additional funding to enable an increase in sample size to 2,000 participants. Assuming that all other study conditions are held constant, which of the following is most likely to occur as a result of this additional funding? Answer A Increase in range of the confidence interval B Decrease in standard deviation C Decrease in standard error of the mean D Increase in risk of systematic error E Increase in probability of type II error Image eduwaves360.com | Telegram : @eduwaves360 Hint Increasing the sample size increases the likelihood that the measured value corresponds to the true mean of the underlying population. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Increase in range of the confidence interval Explanation Why An increase in the range of the confidence interval would be expected if there was a decrease in the size of the sample population. The width of the confidence interval is an indicator of the uncertainty in the estimator statistic (in this case, the mean IQ). Smaller study populations are more subject to the effects of chance, leading to less precision in the estimate of the mean that manifests as a wider confidence interval. In contrast, increasing the sample size will mitigate the effects of chance, providing a more precise estimate of the true mean that manifests as a narrower confidence interval (decreased range). B - Decrease in standard deviation Explanation Why Standard deviation measures the dispersion or spread in data and is an intrinsic property of the population from which the sample is drawn. Increasing the sample size may increase the accuracy of estimating the standard deviation, but it will not change the standard deviation itself. C - Decrease in standard error of the mean Explanation But It is important to distinguish between the standard deviation and the SEM. The standard deviation is a measure of the variability of a sample, whereas the SEM is a measure of the standard deviation of the sampling distribution of the mean. Explanation Why The standard error of the mean (SEM) is a measure of the dispersion of a random set of sample eduwaves360.com | Telegram : @eduwaves360 means around the true population mean. It is dependent on the variability (i.e., standard deviation) of the measured values (in this case, IQ scores) and the sample size (SEM = SD/√n). By increasing the sample size, the sample means approach the true population mean, resulting in a smaller SEM. In contrast, a greater standard deviation and/or a smaller sample size will increase the SEM, resulting in a less accurate estimate. The SEM is commonly used to calculate confidence intervals that indicate, with a given level of confidence, the range of values within which the true mean of the underlying population falls. D - Increase in risk of systematic error Explanation Why Systematic error is a form of bias in which a consistent, repeatable error occurs, most often as a result of faulty equipment. A simple example would be trying to estimate the mean body weight of a population using an incorrectly calibrated scale (e.g., one that consistently shows a weight 3 pounds less than the true weight). A change in sample size would not predispose to this type of error. Random error, on the other hand, is inherent in any sample statistic that estimates a true population parameter. Random error can be reduced by increasing the sample size but systematic error cannot. E - Increase in probability of type II error Explanation Why A type II error occurs when the null hypothesis is accepted despite it being false (false-negative error). The probability of type II error (denoted by β) is inversely related to the statistical power of a study (statistical power = 1 - β). Since the statistical power of a study increases with sample size, the probability of type II error will decrease, not increase, under the new study conditions. eduwaves360.com | Telegram : @eduwaves360 Question # 7 An investigator conducts a case-control study to evaluate the relationship between benzodiazepine use among the elderly population (older than 65 years of age) that resides in assisted-living facilities and the risk of developing Alzheimer dementia. Three hundred patients with Alzheimer dementia are recruited from assisted-living facilities throughout the New York City metropolitan area, and their rates of benzodiazepine use are compared to 300 controls. Which of the following describes a patient who would be appropriate for the study's control group? Answer A An 80-year-old man with well-controlled hypertension and mild benign prostate hyperplasia who lives in an independent-living community B A 64-year-old man with well-controlled hypertension and mild benign prostate hyperplasia who lives in an assisted-living facility C A 73-year-old woman with coronary artery disease who was recently discharged to an assisted-living facility from the hospital after a middle cerebral artery stroke D A 86-year-old man with well-controlled hypertension and mild benign prostate hyperplasia who lives in an assisted-living facility E A 68-year-old man with hypercholesterolemia, mild benign prostate hyperplasia, and poorly-controlled diabetes who is hospitalized for pneumonia eduwaves360.com | Telegram : @eduwaves360 Image Hint An appropriate control group should be representative of the study's source population (elderly individuals who reside in assisted-living facilities). Benzodiazepine use among patients in the control group should be similar to patients in the case group. eduwaves360.com | Telegram : @eduwaves360 Correct Answer AAn 80-year-old man with well-controlled hypertension and mild benign prostate hyperplasia who lives in an independent-living community Explanation Why This man comes from a different source population (a community center in an independent-living community, not an assisted-living facility), which poses a risk for selection bias. Thus, he is not an appropriate candidate for the control group. BA 64-year-old man with well-controlled hypertension and mild benign prostate hyperplasia who lives in an assisted-living facility Explanation Why This man's age is outside the age of interest (> 65 years) for this study. Thus, he is not an appropriate candidate for the control group. CA 73-year-old woman with coronary artery disease who was recently discharged to an assisted-living facility from the hospital after a middle cerebral artery stroke Explanation Why Selecting an individual with a recent history of hospitalization poses a risk for Berkson bias because this woman could have increased risk of benzodiazepine exposure, which distorts the odds ratio. Moreover, this woman has a comorbid condition (stroke) that might affect her cognitive status. Thus, she is not an appropriate candidate for the control group. eduwaves360.com | Telegram : @eduwaves360 DA 86-year-old man with well-controlled hypertension and mild benign prostate hyperplasia who lives in an assisted-living facility Explanation Why The control group and case group should have similar baseline characteristics and be recruited from the same study population. This study should recruit individuals > 65 years of age who reside in an assisted-living facility such as this 86-year-old man who has no history of Alzheimer dementia. EA 68-year-old man with hypercholesterolemia, mild benign prostate hyperplasia, and poorly-controlled diabetes who is hospitalized for pneumonia Explanation Why Selecting an individual who is hospitalized poses a risk for Berkson bias because this man could have an increased risk of benzodiazepine exposure, which distorts the odds ratio. Thus, he is not an appropriate candidate for the control group. eduwaves360.com | Telegram : @eduwaves360 Question # 8 A group of investigators who are studying individuals infected with Trypanosoma cruzi is evaluating the ELISA absorbance cutoff value of serum samples for diagnosis of infection. The previous cutoff point is found to be too high, and the researchers decide to lower the threshold by 15%. Which of the following outcomes is most likely to result from this decision? Answer A Increased negative predictive value B Unchanged true positive results C Decreased sensitivity D Increased true negative results E Increased positive predictive value F Increased specificity Image eduwaves360.com | Telegram : @eduwaves360 Hint Lowering the threshold for this test will lead to a decrease in false negative results and an increase in false positive results. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Increased negative predictive value Explanation But Lowering the threshold for this test will also result in increased sensitivity (epidemiology), decreased specificity (epidemiology), and decreased positive predictive value. Explanation Why The negative predictive value (NPV) of a test is the likelihood that an individual with a negative test result does not have the condition that is being tested. NPV is calculated as TN/(TN + FN). For this test, the ELISA absorbance value must meet or exceed a specified threshold to establish a diagnosis of Trypanosoma cruzi infection. Lowering the threshold of this test will lead to a decreased number of negative test results among infected individuals (i.e., a decrease in false negative results). If the test produces fewer false negative results, then its NPV will increase (i.e., a negative test result is more likely to reflect that the individual does not have the infection). B - Unchanged true positive results Explanation Why A true positive result is a positive test result in an individual who has the condition that is being tested. For this test, the ELISA absorbance value must meet or exceed a specified threshold to establish a diagnosis of Trypanosoma cruzi infection. Lowering the threshold of this test will lead to an increased number of positive test results among infected individuals (i.e., an increase in true positive results). eduwaves360.com | Telegram : @eduwaves360 C - Decreased sensitivity Explanation Why Sensitivity (epidemiology) is calculated as 1 – FN rate. For this test, the ELISA absorbance value must meet or exceed a specified threshold to establish a diagnosis of Trypanosoma cruzi infection. Lowering the threshold of this test will lead to a decreased number of negative test results among infected individuals (i.e., a decrease in false negative results). If the test produces fewer false negative results, then its sensitivity will increase, not decrease. D - Increased true negative results Explanation Why A true negative result is a negative test result in an individual who does not have the condition that is being tested. For this test, the ELISA absorbance value must meet or exceed a specified threshold to establish a diagnosis of Trypanosoma cruzi infection. Lowering the threshold of this test will lead to a decreased number of negative test results among uninfected individuals (i.e., a decrease, not increase, in true negative results). E - Increased positive predictive value Explanation Why The positive predictive value (PPV) of a test is the likelihood that an individual with a positive test result has the condition that is being tested. PPV is calculated as TP/(TP + FP). For this test, the ELISA absorbance value must meet or exceed a specified threshold to establish a diagnosis of Trypanosoma cruzi infection. Lowering the threshold of this test will lead to an increased number of positive test results among uninfected individuals (i.e., an increase in false positive results). If the test produces more false positive results, then its PPV will decrease, not increase. eduwaves360.com | Telegram : @eduwaves360 F - Increased specificity Explanation Why Specificity (epidemiology) is calculated as 1 – FP rate. For this test, the ELISA absorbance value must meet or exceed a specified threshold to establish a diagnosis of Trypanosoma cruzi infection. Lowering the threshold of this test will lead to an increased number of positive test results among uninfected individuals (i.e., an increase in false positive results). If the test produces more false positive results, then its specificity will decrease, not increase. eduwaves360.com | Telegram : @eduwaves360 Question # 9 An investigator studying the effects of dietary salt restriction on atrial fibrillation compares two published studies, A and B. In study A, nursing home patients without atrial fibrillation were randomly assigned to a treatment group receiving a low-salt diet or a control group without dietary salt restriction. When study B began, dietary sodium intake was estimated among elderly outpatients without atrial fibrillation using 24-hour dietary recall. In both studies, patients were reevaluated at the end of one year for atrial fibrillation. Which of the following statements about the two studies is true? Answer A Study B allows for better control over selection bias B Study B is better at inferring causality C Study A allows for better control of confounding variables D Study B results can be analyzed using a chi-square test E Study A results can be analyzed using a t-test Image eduwaves360.com | Telegram : @eduwaves360 Hint The design of study A is consistent with a randomized controlled clinical trial, and study B is consistent with a prospective cohort study. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Study B allows for better control over selection bias Explanation Why Study B does not allow for better control of selection bias because the patients were not randomized. Randomization of patients in a study best mitigates this systematic error. B - Study B is better at inferring causality Explanation Why Study B may infer causality, but randomized controlled trials, such as study A, have the highest form of evidence for causality. C - Study A allows for better control of confounding variables Explanation Why As a randomized controlled trial, study A allows for better control of confounding variables. Randomization also allows for better control of systematic errors (e.g., selection bias, allocation bias). D - Study B results can be analyzed using a chi-square test Explanation Why A chi-square test is a statistical test of association between two categorical variables. Study B only eduwaves360.com | Telegram : @eduwaves360 has one continuous variable (amount of salt intake), and as such, it should be analyzed using another kind of test. E - Study A results can be analyzed using a t-test Explanation Why Study A has one independent nominal variable (salt restriction) and one dependent nominal variable (atrial fibrillation) and is therefore best analyzed using a chi-square test, not a t-test. Study B could be analyzed with a t-test or a one-way ANOVA (a single factor analysis of variance) because it had only one independent variable. eduwaves360.com | Telegram : @eduwaves360 Question # 10 A 52-year-old man comes to the physician for a follow-up examination 1 year after an uncomplicated liver transplantation. He feels well but wants to know how long he can expect his donor graft to function. The physician informs him that from the time of transplantation, the probability of graft survival is 90% at 1 year, 78% at 5 years, and 64% at 10 years. At this time, the probability of the patient's graft surviving to 10 years after transplantation is closest to which of the following? Answer A 64% B 82% C 71% D 58% E 45% Image eduwaves360.com | Telegram : @eduwaves360 Hint The given probabilities represent the expected graft survival immediately following transplant. This patient's graft is still functioning 1 year after transplantation. To determine the expected 10-year graft survival in this patient, it is necessary to calculate the conditional probability. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 64% Explanation Why 64% is the probability of graft survival to 10 years for all patients undergoing liver transplantation. However, this patient's graft has survived the first interval of 0–1 years, meaning it is part of the 90% survivor subset. Graft survival to 10 years is dependent on graft survival to 1 year, indicating that these are not independent events. B - 82% Explanation Why 82% is the probability of graft survival to 10 years given that the graft has already survived 5 years (i.e., 0.64/0.78 = 82%). This patient's graft has only survived 1 year so far, making this answer incorrect. C - 71% Explanation Why Conditional probability is the probability of event “A” occurring given that event “B” has occurred: P(A|B). It can be calculated as P(A|B) = P(A and B)/P(B). This patient's graft has survived the first interval of 0–1 years, meaning it is part of the 90% survivor subset (probability of event “B”). All cases in which the graft survives past this point are part of this first group. The probability of survival to 10 years is given as 64% (probability of event “A”). The conditional probability is therefore 64 divided by 90 (i.e., 64/90), which is approx. 0.71, or 71%. eduwaves360.com | Telegram : @eduwaves360 D - 58% Explanation Why 58% is the product of the probability of 10-year graft survival and the probability of 1-year graft survival. If the calculation aimed at finding the probability of 2 independent events occurring, as in 2 coin tosses, it would be correct. However, in this scenario, graft survival to 10 years is dependent on the probability of graft survival to 1 year, making this answer incorrect. E - 45% Explanation Why 45% is the product of all three given probabilities (90%, 78%, 64%). If the calculation aimed at finding the probability of 3 independent events occurring, as in 3 coin tosses, it would be correct. However, in this scenario, graft survival to 10 years is dependent on the probability of graft survival to 1 year, making this answer incorrect. eduwaves360.com | Telegram : @eduwaves360 Question # 11 A 2-month study is conducted to assess the relationship between the consumption of natural licorice and the development of hypokalemia. A total of 100 otherwise healthy volunteers are enrolled. Half of the volunteers are asked to avoid licorice and the other half are asked to consume licorice daily, along with their regular diet. All volunteers are monitored for the duration of the study and their serum potassium concentration is measured each week. No statistically significant difference in mean serum potassium concentrations is found between the volunteers who consumed licorice regularly and those avoiding licorice. The serum potassium concentrations remained within the range of 3.5–5.0 mEq/L in all volunteers from both groups. Two patients were excluded from the study after their baseline serum potassium concentrations were found to be 3.1 mEq/L and 3.3 mEq/L. If these patients had been included in the analysis, which of the following values would most likely have been unaffected? Answer A Median B Mode C Standard error D Variance E Mean Image eduwaves360.com | Telegram : @eduwaves360 Hint The two excluded values are outliers, which would cause a negative skew of the sample distribution. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Median Explanation Why The median represents the middle value of a data set when the set is arranged in order of the magnitude of the values. The median is usually equivalent to the 50th percentile and divides the upper half of the data set from the lower half. Although the median is relatively resistant to outliers, including the two outliers in the lower end of the data set would cause a slight decrease to the median. B - Mode Explanation But Chronic ingestion of large doses of licorice can cause pseudohyperaldosteronism with hypokalemia, hypernatremia, and hypertension. Licorice inhibits 11-beta hydroxylase, an enzyme involved in cortisol degradation, resulting in high levels of cortisol stimulating mineralocorticoid receptors. Explanation Why The mode is the most common value of a data set. Because the mode is not based on a calculated average but on the frequency of data points occurring, it is most resistant against single outliers. If a mode exists, one value of the data set appears more frequently than all other values (that is, at least twice). Therefore, including two additional values would not change this frequency distribution (as long as the two outliers are not equal). C - Standard error Explanation Why The standard error of the mean (SEM) estimates the deviation of the sample mean from the eduwaves360.com | Telegram : @eduwaves360 population mean. It is directly proportional to the standard deviation (SD) and inversely proportional to the sample size (n): SEM = SD/√n. In this case, including the two outliers would increase √n (denominator), which would decrease the SEM. At the same time, the SD (numerator) would increase because the two values are outliers, thereby increasing the SEM. Thus, adding the two outliers to the data set would not leave the SEM unaffected (even though a prediction on the direction of change is not possible without actual numbers). D - Variance Explanation Why Variance is a measure of dispersion (like standard deviation) indicating how far data is spread out. Including the two outliers would increase the variance. E - Mean Explanation Why The mean is the arithmetic average of a data set (i.e., the sum of all the data divided by the number of values in the data set) and is easily affected by outliers. In this case, including the two outliers would decrease the mean. eduwaves360.com | Telegram : @eduwaves360 Question # 12 A group of investigators have conducted a randomized clinical trial to evaluate the efficacy of adding a novel adenosine A1 receptor agonist to the standard anti-epileptic treatment in reducing the frequency of focal seizures. It was found that patients taking the combination regimen (n = 200) had a lower seizure frequency compared to patients taking the standard treatment alone (n = 200; p < 0.01). However, several participants taking the novel drug reported severe drowsiness. The investigators administered a survey to both the combination treatment group and standard treatment group to evaluate whether the drowsiness interfered with daily functioning using a yes or no questionnaire. Results are shown: Interference with daily functioning Yes (number of patients) No (number of patients) Combination treatment group 115 85 Standard treatment group 78 122 Which of the following statistical methods would be most appropriate for assessing the statistical significance of these results? Answer A Multiple linear regression B Chi-square test C Unpaired t-test D Paired t-test E Analysis of variance Image eduwaves360.com | Telegram : @eduwaves360 Answer F Image Pearson correlation coefficient eduwaves360.com | Telegram : @eduwaves360 Hint The variables being compared are both categorical variables: medication regimen group (combination treatment vs. standard treatment alone) and interference with daily activity (yes vs no). eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Multiple linear regression Explanation Why Multiple linear regression is used to predict ≥ 2 outcomes of a single exposure/intervention in a study group in which the outcomes, as well as the exposure/intervention, are continuous variables (e.g., predicting an increase in BMI and blood glucose levels with increased calorie intake). This study aims to identify the statistical significance in the difference of one outcome (interference with daily activity: yes or no) from the exposure (i.e., standard treatment vs. compound treatment). Moreover, the outcome and exposures in this study are both categorical variables. B - Chi-square test Explanation Why The Chi-square test is used to identify if a relationship between two categorical variables is statistically significant, i.e., whether the observations differ significantly from what would be expected by chance. This table shows that those receiving the combination treatment report interference with daily functioning more frequently than those receiving the standard treatment. The null hypothesis is rejected if there is a statistically significant difference in the frequency of interference with daily functioning between the two groups. C - Unpaired t-test Explanation Why An unpaired t-test compares the difference between the means of a continuous variable (e.g., serum glucose level, BMI) that is measured at one point of time in two unmatched sample groups (e.g., a study comparing the difference in BMI of 20-year-old males and females). In this study, a categorical variable (i.e., interference with daily activity: yes or no) is the outcome being compared between two groups. The unpaired t-test would therefore not be appropriate for evaluating the statistical significance of the differences observed in this table. eduwaves360.com | Telegram : @eduwaves360 D - Paired t-test Explanation Why A paired t-test compares the difference between the means of a continuous variable (e.g., serum glucose level, BMI) that is measured at two points in time within one sample group (e.g., a study comparing serum glucose levels at baseline and after 2 months of initiating a drug therapy within the same group). In this study, a categorical variable (i.e., interference with daily activity: yes or no) is the outcome being compared between the two sample groups. The paired t-test would therefore not be appropriate for evaluating the statistical significance of the differences observed in this table. E - Analysis of variance Explanation Why The analysis of variance test (ANOVA) is a parametric test that is used to identify if there is a statistically significant difference among the means of ≥ 3 continuous variables (e.g., identifying if there is a relationship between blood pressure, serum glucose levels, and BMI). However, this study is comparing two categorical variables. An ANOVA test would therefore not be appropriate for evaluating the statistical significance of the differences observed in this table. F - Pearson correlation coefficient Explanation Why The Pearson correlation coefficient describes the strength and polarity of a linear relationship between two continuous variables (e.g., to assess the relationship between the dose of an antiepileptic and its peak serum concentration). The variables in this study are categorical variables, and the aim of this study is to test the statistical significance of the differences observed in this table. Consequently, the Pearson correlation coefficient cannot be used in this case. eduwaves360.com | Telegram : @eduwaves360 Question # 13 An investigator conducts a study to determine whether earlier detection of glioblastoma multiforme (GBM) in patients increases survival time. One subset of study participants consists of asymptomatic individuals who were diagnosed with GBM after undergoing a screening MRI of the brain. The other subset of study participants was diagnosed with GBM only after they developed symptoms. Results from the study show that the asymptomatic patients who were diagnosed with screening MRI had an average survival time that was 6 weeks longer than that of the patients who were diagnosed after symptom onset. Which of the following statistical biases is most likely to have occurred as a result of the endpoint selected for this study? Answer A Selection bias B Observer-expectancy bias C Length-time bias D Confounding bias E Surveillance bias F Lead-time bias Image eduwaves360.com | Telegram : @eduwaves360 Hint Studies of screening tests that use survival times as an endpoint are susceptible to this form of bias. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Selection bias Explanation Why Selection bias is a collective term for a group of biases that occur when a sample population does not represent the general population from which the sample is drawn. There is no evidence of inappropriate sampling within the two groups selected for this experiment. Moreover, the use of the survival time as an endpoint for this screening study is more likely to result in a different bias. B - Observer-expectancy bias Explanation Why Observer bias is a form of cognitive bias where prior knowledge or expectations of the investigator affect the group assignment of subjects. As the selection of patients was conducted based on patient family histories and clinical symptoms, it is unlikely that this bias occurred. Moreover, the use of the survival time as an endpoint for this screening study results in a different bias. C - Length-time bias Explanation Why Some diseases are slowly progressive and will therefore be detectable by screening for longer periods of time than rapidly progressive diseases that are more likely to manifest clinically before being diagnosed by screening. Length-time bias refers to the overestimation of such slowly progressive disease cases, leading to the incorrect assumption that a screening test results in better outcomes or prognosis. A classic example is a cancer screening test that is assumed to improve outcomes but actually detects a higher number of asymptomatic, slow-growing tumors that would have been less dangerous or lethal irrespective of detection. eduwaves360.com | Telegram : @eduwaves360 D - Confounding bias Explanation Why A confounding bias occurs when causality is established between the dependent (outcome) and independent variables (e.g., exposure to a risk factor), but the association is, in fact, due to a third variable that has not been factored into the study (the confounder). While confounding biases could have affected this study, the use of the survival time as an endpoint for this screening study results in a different bias. E - Surveillance bias Explanation Why Surveillance bias occurs when there is a higher probability of detecting an outcome in one exposed group than another, usually due to increased screening or diagnostic testing. In this case, only patients with the outcome are being studied, making surveillance bias irrelevant. The use of survival time as an endpoint in this study is more likely to result in a different type of bias. F - Lead-time bias Explanation Why Lead-time bias occurs when the earlier detection of a disease (in this case, the diagnosis of GBM due to MRI screening) is misinterpreted as increased survival time and occurs in screening studies where survival times are used as endpoints. Earlier detection of a preexisting condition may only increase the time of awareness of the disease, but does not necessarily equate with a significant increase in actual survival time. A better endpoint measurement for screening tests is mortality rates between the screening groups, which minimizes this bias. eduwaves360.com | Telegram : @eduwaves360 Question # 14 A research group wants to assess the safety and toxicity profile of a new drug. A clinical trial is conducted with 20 volunteers to estimate the maximum tolerated dose and monitor the apparent toxicity of the drug. The study design is best described as which of the following phases of a clinical trial? Answer A Phase II B Phase III C Phase 0 D Phase IV E Phase V F Phase I Image eduwaves360.com | Telegram : @eduwaves360 Hint This phase aims to determine the safety, toxicity, pharmacokinetics, and pharmacodynamics of a drug. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Phase II Explanation Why Phase II trials are the first trials that evaluate drug efficacy among a small group of patients with the disease of interest. The drug dosing of a phase II trial would be based on the toxicity and side effect profile developed by the research group. B - Phase III Explanation Why Phase III trials generally require large numbers of volunteers taking part in a randomized control trial rather than the small number of volunteers studied here. Moreover, the primary goal of a phase III trial is to evaluate drug efficacy against the “gold standard” treatment rather than to study the safety, toxicity, pharmacokinetics, and pharmacodynamics of a drug. C - Phase 0 Explanation Why Phase 0 trials are meant to assess the pharmacokinetics and pharmacodynamics of a drug in a small group of healthy volunteer patients. Phase 0 trials do not assess safety or toxicity. eduwaves360.com | Telegram : @eduwaves360 D - Phase IV Explanation Why Phase IV trials are long-term “surveillance” trials. Often commenced and continued even after FDA approval, phase IV trials are meant to identify the long-term risks of drug treatment beyond those discovered during initial trials. Phase IV trials involve thousands of patients, many more than the small group of healthy volunteers included in the study here. E - Phase V Explanation Why Phase V trials are conducted after drug approval to compare data on drug efficacy gathered from the community with the results of previous clinical studies. These studies involve large sample sizes, in contrast to the study here. F - Phase I Explanation Why Following the initial evaluation of the drug to assess for its pharmacodynamic and pharmacokinetic properties (phase 0 trial), phase I clinical trials assess drug safety and determine tolerable drug dosages. The results of phase I trials are used to organize trials with a larger sample size that assess drug efficacy (phase II trials and phase III trials). eduwaves360.com | Telegram : @eduwaves360 Question # 15 A 50-year-old woman comes to the physician for a routine health maintenance examination. She has no personal or family history of serious illness. She smoked one pack of cigarettes daily for 5 years during her 20s. Her pulse is 70/min, and blood pressure is 120/78 mm Hg. Serum lipid studies and glucose concentration are within the reference ranges. Which of the following health maintenance recommendations is most appropriate at this time? Answer A Perform DEXA scan B Perform colonoscopy C Perform 24-hour ECG D Measure serum TSH levels E Perform BRCA gene test F Perform abdominal ultrasound Image eduwaves360.com | Telegram : @eduwaves360 Hint This screening test should be recommended for every patient ≥ 50 years of age, regardless of whether symptoms and/or risk factors are present. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Perform DEXA scan Explanation Why A routine DEXA scan to screen for osteoporosis is typically recommended in older female patients. The US Preventive Services Task Force recommends that DEXA scans are only offered to women who are 65 years of age or older. Exceptions to this are patients with premature menopause, who should be offered a DEXA scan at a younger age. Premature menopause is not described in this patient. B - Perform colonoscopy Explanation Why As part of routine colorectal cancer screening, the US Preventive Services Task Force recommends that a colonoscopy is performed every 10 years in patients between the ages of 50 and 75. This recommendation has been assigned a grade of “A”, which indicates that a large number of highquality studies associate this preventive screening with improved colon cancer treatment and survival. Generally accepted alternatives to colonoscopy include annual fecal occult blood testing, or a sigmoidoscopy every 5 years with fecal occult blood testing every 3 years. However, the supporting evidence for these alternative screening procedures is weaker. C - Perform 24-hour ECG Explanation Why The US Preventive Services Task Force has concluded that routine ECG screenings do not decrease the risk of myocardial infarction or stroke. The current evidence regarding routine ECG screening for atrial fibrillation is also insufficient to recommend this service. Therefore, an ECG would not be offered to this patient. eduwaves360.com | Telegram : @eduwaves360 D - Measure serum TSH levels Explanation Why The US Preventive Services Task Force has concluded that there is insufficient evidence to support routine thyroid function screening (e.g., serum TSH) in nonpregnant patients. Pregnancy (particularly during the first trimester) can interfere with normal circulating levels of TSH. Women who are planning pregnancy and are at increased risk for thyroid disorders (e.g., family history of autoimmune thyroid disease, presence of a goiter) can be screened, but universal screening is not generally recommended. E - Perform BRCA gene test Explanation Why The US Preventive Services Task Force recommends that screening for breast cancer with the BRCA gene test (detecting BRCA1 and/or BRCA2) is only performed in patients who have a family history of breast, ovarian, tubal, and/or peritoneal cancer. A family history of these cancers is not described in this patient. F - Perform abdominal ultrasound Explanation Why An abdominal ultrasound is recommended by the US Preventive Services Task Force to rule out an abdominal aortic aneurysm. Although this patient used to smoke, ultrasound is only recommended for male smokers between 65–75 years of age. Therefore, this patient would not be offered this service. eduwaves360.com | Telegram : @eduwaves360 Question # 16 An investigator is studying nosocomial infections in hospitals. The weekly incidence of hospital-acquired pulmonary infections within the pediatric wards of eight different hospitals is recorded. The results are shown. Which of the following values best represents the median value of these incidence rates? Answer A 7.0 B 5.5 C 2.73 Image eduwaves360.com | Telegram : @eduwaves360 Answer D 6.0 E 8.0 Image eduwaves360.com | Telegram : @eduwaves360 Hint The median represents the middle value of the data set when it is arranged by order of magnitude. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 7.0 Explanation Why 7.0 represents the mode, the value that occurs most in the data set (twice, in this case). B - 5.5 Explanation Why 5.5 represents the mean of the data set. This value is obtained by dividing the sum of all the data by the number of values in the data set. When added, the data set in this question will read: 1 + 3 + 4 + 5 + 7 + 7 + 8 + 9 = 44. There are a total of eight values in this data set. Thus, the mean will be → 44 / 8 = 5.5. C - 2.73 Explanation Why 2.73 represents the standard deviation of this data set. This is a measure of the variability of the data points, not a measure of central tendency like the median. Standard deviation is calculated by first summing the squares of the difference between each individual data point and the mean of the data set: (-1.5)2 + (-2.5)2 + (1.5)2 + (-4.5)2 + (1.5)2 + (-0.5)2 + (2.5)2 + (3.5)2 = 52. This value is divided by the total number of values in the data set minus one (8-1=7) to calculate the variance → (52 / 7) = 7.43. Standard deviation is then calculated as the square root of the variance: √7.43 = 2.73. eduwaves360.com | Telegram : @eduwaves360 D - 6.0 Explanation But The median value is not strongly affected by outliers or skewed data. Explanation Why The median divides the upper half of the data set from the lower half. That is, 50% of the data set lies below the median value and 50% lies above it. When arranged in order of magnitude, the data set in this question will read: 1, 3, 4, 5, 7, 7, 8, and 9 (a total of eight values). The median for this even data set will be the average of the two middle values, 5 and 7 → (5 + 7) / 2 = 6. E - 8.0 Explanation Why 8.0 represents the range of the data set. The range is obtained by subtracting the lowest value (Hospital D - 1) from the highest value (Hospital H - 9) of the data set. eduwaves360.com | Telegram : @eduwaves360 Question # 17 A graduate student in public health is conducting a study on population health and is comparing different demographic models. He is particularly interested in investigating health care interventions in societies with the demographic distribution shown. Which of the following measures is most likely to ensure a healthy demographic transition in this population? Answer A Image Invest in workplace health and safety measures eduwaves360.com | Telegram : @eduwaves360 Answer B Invest in childhood immunization programs C Invest in type II diabetes research programs D Invest in prostate cancer screening programs E Invest in long-term care facilities Image eduwaves360.com | Telegram : @eduwaves360 Hint The illustration shows an expansive pyramid, reflecting a youthful population with high birth rates and low life expectancy. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Invest in workplace health and safety measures Explanation Why Investing in workplace health and safety measures would have an impact on populations in which a large proportion of individuals are employed, i.e., those with a significant middle-aged population. However, this population is predominantly youthful, with a large proportion of individuals under the age of 20. Investing in workplace health and safety measures would, therefore, impact only a small proportion of the population. B - Invest in childhood immunization programs Image eduwaves360.com | Telegram : @eduwaves360 Explanation Why Investing in interventions for children such as childhood immunization programs is most likely to ensure a healthy demographic transition in a youthful population. The goal of healthy demographic transitions is to lower death rates, lower birth rates, and ensure a healthy aging population. In the given population, the majority of individuals are young. Consequently, interventions that reduce child mortality and reproduction (e.g., by improving access to contraceptives) are likely to have a positive impact on the demographic distribution. C - Invest in type II diabetes research programs Explanation Why Investing in type II diabetes research would be beneficial in populations with a significant middleaged and elderly population. However, this population is predominantly youthful, with the majority of individuals under the age of 20. In individuals under the age of 20, type II diabetes is rare. Investing in diabetes research would therefore not benefit a large proportion of the population. D - Invest in prostate cancer screening programs Explanation Why Investing in prostate cancer screening programs might lower the death rate of elderly men and would therefore be beneficial in communities with a significant elderly population. However, this population is predominantly youthful, with the majority of individuals under the age of 20. Investing in prostate cancer screening programs would therefore not benefit a large proportion of the population. E - Invest in long-term care facilities Explanation Why Investing in long-term care facilities would be beneficial in communities with a significant elderly eduwaves360.com | Telegram : @eduwaves360 population. However, this population is predominantly youthful, with the majority of individuals under the age of 20. Investing in long-term care facilities would therefore not benefit a large proportion of the population. eduwaves360.com | Telegram : @eduwaves360 Question # 18 An investigator is studying the association between exclusive breastfeeding and body weight in infants. The body weights of 15 exclusively breastfed infants at the age of 6 months are measured. Results are shown: Patient Body weight (kg) 1 7.0 2 6.0 3 6.1 4 6.8 5 7.2 6 6.4 7 6.2 8 6.8 9 6.5 10 7.3 11 6.3 12 8.5 13 6.9 14 6.6 15 5.2 One of the computed measures of central tendency is 6.8 kg. Which of the following characteristics is generally true about this measurement? eduwaves360.com | Telegram : @eduwaves360 Answer A It is the 50th percentile of a set of values. B It is not applicable for qualitative data analysis. C It is resistant to outliers. D Its value only occurs once in a data set. E It is useful to assess the extent of data variability. Image eduwaves360.com | Telegram : @eduwaves360 Hint The value of 6.8 represents the mode in this data set. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - It is the 50th percentile of a set of values. Explanation Why The median represents the 50thpercentile of a data set. The mode represents the most frequent observation in a data set, and it is not necessarily correlated with the 50th percentile. B - It is not applicable for qualitative data analysis. Explanation Why The only measure of central tendency that can be used in the analysis of qualitative data (i.e., categorical variables) is the mode. Quantitative data can be described by all three measures (i.e., mean, median, mode). C - It is resistant to outliers. Explanation Why The mode (epidemiology) of a sample does not change due to the presence of outliers, regardless of their magnitude, because it represents the most frequent value in a given sample. This is also true of the median. In contrast, the mean is a measure of central tendency that is sensitive to the presence of outliers. eduwaves360.com | Telegram : @eduwaves360 D - Its value only occurs once in a data set. Explanation Why The mode represents the most frequently measured value(s) in a given sample. In samples in which each value occurs only once, there is no mode. E - It is useful to assess the extent of data variability. Explanation Why Measures such as the range (statistics), standard deviation, and interquartile range are used to assess the extent of data variability. However, the mode is a measure of central tendency, which provides a point around which data is clustered. eduwaves360.com | Telegram : @eduwaves360 Question # 19 The World Health Organization suggests the use of a new rapid diagnostic test for the diagnosis of malaria in resource-limited settings. The new test has a sensitivity of 70% and a specificity of 90% compared to the gold standard test (blood smear). The validity of the new test is evaluated at a satellite health center by testing 200 patients with a positive blood smear and 150 patients with a negative blood smear. How many of the tested individuals are expected to have a false negative result? Answer A 155 B 15 C 195 D 60 E 140 F 135 Image eduwaves360.com | Telegram : @eduwaves360 Hint Sensitivity = (true positives) / (true positives + false negatives) eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - 155 Explanation Why This value is the number of all expected positive results of the new test (i.e., the sum of the true positive results and false positive results). B - 15 Explanation Why This value is the number of expected false positive results of the new test, which is calculated as the number of people without malaria (i.e., 150, the number of people with a negative blood smear) multiplied by the false positive rate of the new test, which is defined as 1 – specificity (epidemiology). Thus, the number of expected false positive results of the new test is 150 × (1 – 0.9) = 15. C - 195 Explanation Why This value is the number of all expected negative results of the new test (i.e., the sum of the true negative results and false negative results). eduwaves360.com | Telegram : @eduwaves360 D - 60 Explanation Why The number of expected false negative results of the new test is calculated as the number of people with malaria (i.e., 200, the number of people with a positive blood smear) multiplied by the false negative rate of the new test, which is defined as 1 – sensitivity. Thus, the number of expected false negative results of the new test is 200 × (1 – 0.7) = 60. E - 140 Explanation Why This value is the number of expected true positive results of the new test, which is calculated as the number of people with malaria (i.e., 200, the number of people with a positive blood smear) multiplied by the sensitivity (epidemiology) of the new test, which is given as 70%. Thus, the number of expected true positive results of the new test is 200 × 0.7 = 140. F - 135 Explanation Why This value is the number of expected true negative results of the new test, which is calculated as the number of people without malaria (i.e., 150, the number of people with a negative blood smear) multiplied by the specificity (epidemiology) of the new test, which is given as 90%. Thus, the number of expected true negative results of the new test is 150 × 0.9 = 135. eduwaves360.com | Telegram : @eduwaves360 Question # 20 A multicenter study was performed with 5000 women over 45 years of age to evaluate a new mammography technique for the diagnosis of breast cancer. A core-needle biopsy of the breast was obtained in all recruited women with abnormal mammography findings. Of the 29 women who underwent biopsy, 26 were found to have breast cancer; the remaining women were found to have benign breast changes. Which of the following is required to calculate the specificity of this new mammography technique? Answer A Probability that a woman with abnormal mammography findings has breast cancer B Incidence of breast cancer in the general population C Number of women with normal mammography who do not have breast cancer D Prevalence of breast cancer in the general population E Number of women with benign breast changes and normal mammography findings eduwaves360.com | Telegram : @eduwaves360 Image Hint The specificity of a test is the probability that a given test is negative when the disease is absent. Specificity can be calculated using the formula specificity = TN/(TN + FP), in which TN is the number of true negatives and FP is the number of false positives. eduwaves360.com | Telegram : @eduwaves360 Correct Answer AProbability that a woman with abnormal mammography findings has breast cancer Explanation But On the other hand, it would be possible to calculate TN in this scenario if the negative predictive value (NPV) was known because NPV = TN/(TN + FN), and the total number of patients who tested negative (i.e., TN + FN) is already known to be 4971. Explanation Why This probability is also known as the positive predictive value (PPV). The formula for calculating PPV is PPV = TP/(TP + FP), in which TP and FP are the numbers of true and false positives, respectively. In this scenario, the TP and FP are already known, and TN, which is the missing variable to calculate specificity, cannot be gleaned from the PPV alone. B - Incidence of breast cancer in the general population Explanation Why The incidence of breast cancer denotes the number of new breast cancer cases within a specific time frame (e.g., one year). This parameter is not needed for the calculation of specificity. CNumber of women with normal mammography who do not have breast cancer Explanation Why TN and FP are necessary to calculate the specificity of this new mammography technique. FP can be calculated from the data set (FP = 29 - 26 = 3). TN in this case would be the number of women with eduwaves360.com | Telegram : @eduwaves360 normal mammography findings in conjunction with no evidence of breast cancer on biopsy (i.e., no changes or only benign changes). D - Prevalence of breast cancer in the general population Explanation But If the prevalence of breast cancer among women over 45 years of age was known and the sample population is representative of women over 45 years of age in the general population (i.e., no selection bias), it would be possible to calculate the total number of individuals with breast cancer by multiplying the prevalence with the number of individuals in the study population (5000) and, subsequently, glean the total number of individuals in the study population without breast cancer (i.e., FP + TN). Since the FP is known in this scenario, TN, which is the missing variable for calculating specificity, could then be obtained. The prevalence of breast cancer in the general population, however, is not equivalent to the prevalence of breast cancer in women above 45 years of age; women younger than 45 years of age also develop breast cancer. Explanation Why The prevalence of breast cancer denotes the number of total breast cancer cases in the general population at specific point in time. This parameter is not needed for the calculation of specificity. ENumber of women with benign breast changes and normal mammography findings Explanation Why The number of women with benign breast changes and normal mammography findings constitutes only a part of TN and, therefore, would not suffice for the calculation of specificity. The other component of TN would be the number of women with no breast changes and normal mammography findings. eduwaves360.com | Telegram : @eduwaves360 Question # 21 An investigator is conducting a study to identify potential risk factors for post-transplant hypertension. The investigator selects post-transplant patients with hypertension and gathers detailed information regarding their age, gender, preoperative blood pressure readings, and current medications. The results of the study reveal that some of the patients had been treated with cyclosporine. This study is best described as which of the following? Answer A Retrospective cohort study B Meta-analysis C Ecological study D Cross-sectional study E Prospective cohort study F Case-control study G Case series Image eduwaves360.com | Telegram : @eduwaves360 Hint This is a type of descriptive observational study that lacks a control group. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Retrospective cohort study Explanation Why A retrospective cohort study is an analytical observational study that begins after the study groups develop an outcome of interest. Individuals (e.g., post-transplant patients) who are either exposed or not exposed to a particular risk factor (e.g., cyclosporine) are selected and the outcome status (e.g., hypertension or no hypertension) of these two groups is determined using pre-existing data (e.g., patient records). The strength of association can then be measured using the relative risk. The above study is retrospective, as the outcome of interest (post-transplant hypertension) is already present prior to the start of the study. However, it is not a cohort study, as the investigators started by selecting the outcome rather than the exposure (cyclosporine), and there is no control group. B - Meta-analysis Explanation Why A meta-analysis aggregates summary data (e.g., means, relative risk) from multiple studies and statistical analysis of the reviewed data is performed. It aims to increase statistical power, provide more precise estimates of the magnitude of an effect, and to identify differences between and/or common effects in individual studies. C - Ecological study Explanation Why An ecological study is a descriptive observational study in which the unit of observation is a group (e.g., population) rather than separate individuals. It is used to measure the prevalence and incidence of disease at the population level, particularly when the disease is rare. For example, an ecological study could be used to determine the incidence of post-transplant hypertension based on specific locations (e.g., the US versus China). This data can then be used to identify the exposure (e.g., physicians in the US tend to prescribe cyclosporine more often than physicians in China). However, eduwaves360.com | Telegram : @eduwaves360 the above study is conducted at the level of the individual patient rather than the population. D - Cross-sectional study Explanation Why A cross-sectional study is a descriptive observational study that simultaneously assesses the frequency of disease and exposure at a particular point in time (i.e., a snapshot of the population). It is used to determine the prevalence of disease and other variables (e.g., risk factors). For example, an investigator sends out a survey to post-transplant patients asking whether they received cyclosporine and whether they currently have a diagnosis of hypertension. The chi-squared test can be used to measure the significance of the association between a risk factor and disease. In the above study, the exposure and outcome are not measured simultaneously. Therefore, it is not a cross-sectional study. E - Prospective cohort study Explanation Why A prospective cohort study is an analytical observational study that begins before the study groups develop an outcome of interest. Individuals (e.g., post-transplant patients) are divided into groups according to exposure to a particular risk factor (e.g., cyclosporine) and are followed to see if they develop the outcome of interest (e.g., hypertension). The strength of association can then be measured using the relative risk. The above study, however, is not prospective, as the outcome of interest (post-transplant hypertension) is already present prior to the start of the study. Moreover, it is not a cohort study; the investigators started by selecting the outcome rather than the exposure (cyclosporine), and there is no control group. F - Case-control study Explanation Why A case-control study is an analytical observational study that compares a group of individuals with a disease of interest (case; e.g., hypertension) to a group without a disease of interest (control) from the same source population (e.g., post-transplant patients). This study type can be used to determine if an eduwaves360.com | Telegram : @eduwaves360 exposure is associated with an outcome. In the described scenario, investigators could perform a chart review to look for cyclosporine use in cases versus controls. The strength of association can then be measured using the odds ratio. The above study, however, is not a case-control study as it only looks at patients with post-transplant hypertension and lacks a control group (i.e., posttransplant patients without hypertension). G - Case series Explanation Why This study, which was compiled by aggregating several similar patient cases (i.e., all patients with post-transplant hypertension), represents a case (or clinical) series. A case series has no control group and samples individuals based on either exposure (e.g., all patients treated with cyclosporine) or outcome (as seen here). Without a control group, no measures of association between exposure and outcome (e.g., relative risk, odds ratio) can be calculated. Therefore, case series do not provide evidence of cause and effect. eduwaves360.com | Telegram : @eduwaves360 Question # 22 A group of investigators is planning a phase III clinical trial to assess whether the histone deacetylase 3 inhibitor BG45 should be used in the treatment of multiple myeloma (MM). Which of the following sets of characteristics best describes this clinical trial phase? Study purpose Study population A Evaluate the safety of BG45 Small number of healthy volunteers B Evaluate the efficacy and safety of BG45 against the standard of care Large number of patients with MM C Evaluate the efficacy and safety of BG45 Small number of patients with MM D Evaluate the safety of BG45 Large number of patients with MM E Evaluate the efficacy and safety of BG45 Small number of mice with MM F Evaluate the efficacy and safety of BG45 against the standard of care Small number of patients with MM Answer A A B B C C Image eduwaves360.com | Telegram : @eduwaves360 Answer D D E E F F Image eduwaves360.com | Telegram : @eduwaves360 Hint Phase III clinical trials are randomized controlled trials, in which enrolled patients are randomly allocated to either a treatment or control group. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A-A Explanation Why This set of characteristics describes a phase I clinical trial, the main objective of which is to assess the safety and toxicity of a new treatment. Phase I trials include the assessment of pharmacokinetics and pharmacodynamics. This phase involves a small number of healthy volunteers or patients with the disease that the drug is intended to treat. Phase I clinical trials are only started if cell and animal studies suggest that the new treatment is likely to be effective and safe in humans. B-B Explanation Why This set of characteristics describes a phase III clinical trial in which a new treatment is compared with the current standard of care or placebo in terms of safety and efficacy. This phase involves a large number of patients, who are randomly divided into two groups: One group is treated with the current standard of care or placebo and the other is given the new treatment. Outcomes are then analyzed. If the new treatment proves to be more effective and/or safer than the current standard treatment, a new drug application is submitted to regulatory agencies (the Food and Drug Administration in the United States) for approval. C-C Explanation Why This set of characteristics describes a phase II clinical trial, the main objective of which is to assess the efficacy of a drug. In this phase, a small number of patients are treated with the dose found to be the safest in phase I clinical trials. eduwaves360.com | Telegram : @eduwaves360 D-D Explanation Why This set of characteristics describes a phase IV clinical trial. In this phase, the main objective is to evaluate safety over time, assessing for long-term and rare adverse effects in a large number of patients. Treatments that are currently in phase IV clinical trials have already received approval from regulatory agencies (the Food and Drug Administration in the United States) and are already being used in the clinical setting. Phase IV trials are used for postmarketing surveillance. E-E Explanation Why This set of characteristics describes a preclinical trial in which in vitro experiments and animal models are used. The main objective of preclinical trials is to assess whether the new treatment is likely to be effective and safe in humans. F-F Explanation Why This set of characteristics does not describe any phase of a clinical trial. Although there is a phase of clinical trials that aims to evaluate the efficacy and safety of a drug against the standard of care, this type of clinical trial is carried out with a large number of patients. eduwaves360.com | Telegram : @eduwaves360 Question # 23 A group of investigators conducts an unblinded, randomized controlled trial to investigate the effect of single-agent pomalidomide on the survival of patients with smoldering multiple myeloma (SMM). Three-hundred eligible patients with SMM are randomized to receive 6 cycles of either pomalidomide or a placebo. At the 24-month follow-up, the investigators find that a significant number of patients were not adherent with the treatment assigned or lost to follow-up. Nonetheless, results for the patients who completed the protocol showed that pomalidomide improves progression-free survival. The investigators are worried about potential biases resulting from the significant loss to follow-up and nonadherence in the treatment group. Which of the following is the most appropriate technique to minimize selection bias and provide the most valid estimate of the real effect of pomalidomide on patients with SMM? Answer A Subgroup analysis B Noninferiority trial C Intention-to-treat analysis D Per-protocol analysis E Crossover trial Image eduwaves360.com | Telegram : @eduwaves360 Hint This technique can be used to preserve the benefits of randomization and provide a more conservative estimate of the effect of pomalidomide in patients with SMM. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Subgroup analysis Explanation Why A subgroup analysis (or stratified data analysis) is a statistical method that can be used when a third variable seems to either influence a study outcome (effect modifier) or correlate with the exposure and the outcome (confounder). Study groups are divided into subgroups according to the third variable and measures of association can be calculated for each subgroup. A subgroup analysis adjusts for an effect modifier or a confounder. In the described study, this method can improve the estimate of the real effect of pomalidomide on patients with SMM. However, this technique does not minimize selection bias because the original randomization is lost with this method. B - Noninferiority trial Explanation Why A noninferiority trial is a type of clinical trial that aims to show that the efficacy of a tested treatment is not substantially smaller than the efficacy of the established control treatment. Noninferiority trials are most commonly used when placebo control would be unethical. The described study is an unblinded, randomized controlled trial with a treatment group and a placebo group. By definition, noninferiority trials do not compare treatment to placebo, but rather to control treatment. Furthermore, noninferiority is a study design and not a method of statistical analysis, so it cannot be applied in hindsight. Noninferiority trials are also susceptible to the same biases as superiority trials. C - Intention-to-treat analysis Explanation But The converse of intention-to-treat analysis is per-protocol analysis. eduwaves360.com | Telegram : @eduwaves360 Explanation Why An intention-to-treat analysis is a statistical method in which the data of every patient enrolled in a study is used for data analysis (including data from drop-out patients), and in which each patient is analyzed as part of the treatment group that they were randomly assigned to at the beginning of the study (regardless of the treatment they eventually receive). This analysis includes data from all patients and preserves the original randomization, giving a more accurate estimate of the efficacy of the tested treatment and minimizing selection bias when compared to data analysis that excludes drop-out patients. In the described study, the large drop-out rate as a result of nonadherence and loss to follow-up may have created selection bias and the shown improvement of progression-free survival under pomalidomide may be falsely skewed in favor of the treatment. An intention-to-treat analysis is an appropriate technique to address this bias. D - Per-protocol analysis Explanation Why A per-protocol analysis is a statistical method in which the study groups are compared using data only from the patients who finished the treatment that they were originally assigned. This analysis is used to evaluate the treatment effect under optimal conditions. Using a per-protocol analysis in the described study improves the estimate of the real effect of pomalidomide on patients with SMM. However, this method tends to overestimate the effects of the tested treatment because it does not take into account when patients drop out of the study. Furthermore, this technique does not minimize selection bias because the original randomization is lost with this method. E - Crossover trial Explanation Why A crossover trial is a type of clinical study in which each patient serves as their own control. Crossover studies improve the estimates of the real effect of a treatment by reducing confounding bias. However, the described study is an unblinded, randomized controlled trial with a defined treatment group and a placebo group. The patients do not switch from the treatment group to the control group or vice versa during the study. Crossover is a study design and not a method of statistical analysis and cannot be applied in hindsight. eduwaves360.com | Telegram : @eduwaves360 Question # 24 Two researchers are investigating the prevalence of hypertension in a metropolitan city. The first researcher measured the blood pressure from 1032 inpatients at a tertiary care hospital in the city (study A). The second researcher performed community blood pressure screening on 1003 people (study B) through nursing health fairs. Both studies are conducted using calibrated sphygmomanometers from the same manufacturer. Compared to study A, the result of study B is most likely to have which of the following properties? Answer A Higher accuracy B Higher precision C Lower risk of a type I error D Lower risk of measurement bias E Higher statistical power Image eduwaves360.com | Telegram : @eduwaves360 Hint Study A exhibits Berkson bias because its patients are selected from a subpopulation (tertiary care hospital) rather than the city's general population. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - Higher accuracy Explanation Why A study's accuracy (validity) depends on systematic errors and bias. The two forms of validity are internal and external. Internal validity is likely unaffected by the blood pressure measurement, as both studies use calibrated sphygmomanometers from the same manufacturer. External validity refers to how applicable the study is to the general population; study B is more likely to exhibit higher external validity compared to study A, which involves Berkson bias. In study A, the actual prevalence of hypertension in the general population is less likely to be accurately reflected due to this bias. Therefore, study B is expected to have higher accuracy. B - Higher precision Explanation Why Precision (reliability) is the reproducibility of test results on the same sample under similar conditions. Precision depends on random errors that occur by chance (e.g., unpredictable fluctuations in blood pressure). These errors can be reduced by repeating measurements and/or increasing sample size. Precision is unlikely to be affected here, as the sample size of study B is similar to the one of study A. In addition, the measurement of blood pressure is performed similarly in both studies. C - Lower risk of a type I error Explanation Why A type I error occurs in hypothesis testing when the null hypothesis is incorrectly rejected and the alternative one is accepted, although the observed effect is due to chance. As these are independent cross-sectional studies without a comparison arm, there is no null hypothesis to reject. eduwaves360.com | Telegram : @eduwaves360 D - Lower risk of measurement bias Explanation Why Measurement bias is caused by inconsistencies in measurement or classification between sample groups. Measurement bias can be caused by human error or instrument error. In this case, an example of measurement bias would be using a defective sphygmomanometer. There is no indication to suspect measurement bias in the described studies. Instead, this study exhibits Berkson bias, which is a selection, not a measurement, bias. E - Higher statistical power Explanation Why Statistical power is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. Because these are two independent cross-sectional studies without a comparison arm, there is no null hypothesis to reject. eduwaves360.com | Telegram : @eduwaves360 Question # 25 A research group investigates the safety and toxicity profile of a new drug. Data about the length of time from the administration of the drug until the onset of adverse effects is collected for a period of 20 days. Administration of the drug is terminated for any patient experiencing adverse effects. The collected data is shown. Which of the following statements about this data set is most accurate? Answer A Image The data set follows a right-skewed distribution eduwaves360.com | Telegram : @eduwaves360 Answer B The data set follows a normal distribution C The median is greater than the mean D The mean and mode are equal E The data set follows a bimodal distribution F The data set follows a left-skewed distribution Image eduwaves360.com | Telegram : @eduwaves360 Hint The histogram shows an asymmetrical distribution with a long tail pointing to the positive direction of the scale. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A - The data set follows a right-skewed distribution Image Explanation Why A data set that follows a right-skewed (positively skewed) distribution is clustered on the left side of the scale with a long tail on the right (positive direction of the scale), as seen here. Such a distribution has the following inequality between the measures of central tendency: mean > median > mode. B - The data set follows a normal distribution Explanation Why A data set following a normal distribution would form a symmetrical, bell-shaped curve with a peak eduwaves360.com | Telegram : @eduwaves360 located at the center of the distribution. In this case, however, data is clustered on the left side of the scale, and the data set forms an asymmetrical curve with a peak on the left side of the distribution. C - The median is greater than the mean Explanation Why The median is expected to be greater than the mean in data distribution that shows data points clustered on the right side of a histogram with a long tail to the left (negative direction of the scale), unlike what is seen in this histogram. For the data set shown here, the mean is greater than the median. D - The mean and mode are equal Explanation Why The mean and mode are expected to be equal in a data set that forms a symmetrical, bell-shaped curve with a peak located at the center of the distribution. In the histogram shown here, the mean is greater than the mode. E - The data set follows a bimodal distribution Explanation Why A data set following a bimodal distribution would form a histogram with two different peaks, which suggests the existence of two groups within the study population. In this case, however, the histogram shows only one peak on the left side of the data distribution. eduwaves360.com | Telegram : @eduwaves360 F - The data set follows a left-skewed distribution Explanation Why A data set following a left-skewed (negatively skewed) distribution would form an asymmetrical curve, which is seen here, but the data would cluster on the right rather than the left side of the scale, forming a long tail to the left (negative direction of the scale). The histogram shown here follows a different type of distribution. eduwaves360.com | Telegram : @eduwaves360 Question # 26 A new rapid test for the diagnosis of a viral respiratory infection is evaluated in 200 study participants. At a given cutoff, the test shows a true-negative rate of approximately 70% and a true-positive rate of approximately 65%. Which of the following labeled graphs is most likely to represent this diagnostic test? Answer A Image A eduwaves360.com | Telegram : @eduwaves360 Answer B B C C D D E E Image eduwaves360.com | Telegram : @eduwaves360 Hint The false-positive (FP) rate is calculated from the true-negative rate (specificity) using the formula FP rate = 1 - specificity. The curve that is obtained by plotting the true-positive rate (TP rate or sensitivity) against the FP rate is called a receiving operating characteristic (ROC) curve. The area under this curve (AUC) positively correlates with the diagnostic value of the test. eduwaves360.com | Telegram : @eduwaves360 Correct Answer A-A Explanation Why This graph, which shows an ROC curve close to the left upper corner and an AUC almost equal to 1, represents the ideal diagnostic test with almost 100% sensitivity and specificity. In the test described above, with 65% sensitivity and 70% specificity, the ROC curve is not expected to be that close to the left upper corner. B-B Image Explanation Why This graph represents a diagnostic test with sensitivity and specificity between 50% and 100%, eduwaves360.com | Telegram : @eduwaves360 corresponding to the described diagnostic test and an AUC > 0.5. A test with an AUC > 0.5 has diagnostic value because it can differentiate between individuals with and without disease with better accuracy than a test in which the result is randomly determined with a 50% chance for each outcome (e.g., a coin flip). The higher the AUC, the more accurate the test; the ideal diagnostic test has an AUC of 1. C-C Explanation Why This graph represents a diagnostic test that has an FP rate equal to the TP rate and an AUC of 0.5, indicating that the test performs just as well as a test in which the result is randomly determined with a 50% chance for each outcome (e.g., a coin flip). A test that detects the same number of truepositive and false-positive cases has no predictive diagnostic value. D-D Explanation Why This graph represents a diagnostic test that has an FP rate higher than the TP rate and an AUC < 0.5, indicating that the test performs worse than a test in which the result is randomly determined with a 50% chance for each outcome (e.g., a coin flip). Such a curve for a test usually indicates a mistake in plotting the data for the ROC. E-E Explanation Why This graph represents a diagnostic test that has an FP rate of almost 100%, a TP rate of almost 0%, and an AUC of 0, meaning that the test has no sensitivity or specificity for the diagnosis of the condition. eduwaves360.com | Telegram : @eduwaves360 eduwaves360.com | Telegram : @eduwaves360 Join us on Telegram : Click here : @eduwaves360 Unlocked the Medical premiums Click here : www.eduwaves360.com Medical Courses : Discussion Group : https://t.me/usmle_study_materials_2 @usmle_discussion_group eduwaves360.com | Telegram : @eduwaves360 Join us on Telegram : Click here : @eduwaves360 Unlocked the Medical premiums Click here : www.eduwaves360.com Medical Courses : Discussion Group : https://t.me/usmle_study_materials_2 @usmle_discussion_group
