NR 503 Epidemiology Final Exam Version 2 Latest 2024: Population Health, Epidemiology & St, Exams of Epidemiology

Download NR 503 Epidemiology Final Exam Version 2 Latest 2024: Population Health, Epidemiology & St and more Exams Epidemiology in PDF only on Docsity! NR 503 Epidemiology Final Exam Version 2 Latest 2024: Population Health, Epidemiology & Statistical Principles: {38 Questions and Answers} 1) Which of the following is a condition which may occur during the incubation period? a. Onset of clinical illness b. Receipt of infection c. Signs & symptoms of disease d. Transmission of infection e. Isolation of disease carrier through quarantine Rationale: The incubation period is defined as the interval from receipt of infection to the time of onset of clinical illness. Accordingly, individuals may transmit infectious agents during the incubation period as they show no signs of disease that would enable the isolation of sick individuals by quarantine. 2) Chicken pox is a highly communicable disease. It may be transmitted by direct contact with a person infected with the varicella-zoster virus (VZV). The typical incubation time is between 10 to 20 days. A boy started school 2 weeks after showing symptoms of chicken pox including mild fever, skin rash, & fluid-filled blisters. One month after the boy returned to school, none of his classmates had been infected by VZV. The main reason was: a. Herd immunity b. All had been immunized prior to the school year c. Contact was after infectious period d. Subclinical infections were not yet detected e. Disease was endemic in the class Rationale: The disease is spread by contact with an infected individual who can transmit the agent (VZV) to immunologically naive persons during the incubation period & for several days after onset of clinical illness. Since the boy started school 14 days after showing signs consistent with chicken pox, it is most likely that he was no longer infectious. 3) Which of the following is characteristic of a single-exposure, common-vehicle outbreak? a. Long latency period before many illnesses develop b. There is an exponential increase in secondary cases following initial exposures c. Cases include only those who have been exposed to sick persons d. The epidemic curve has a normal distribution when plotted against the logarithm of time e. Wide range in incubation times for sick individuals Rationale: Single-exposure, common-vehicle outbreaks involve a sudden, rapid increase in cases of disease that are limited to persons not eat ice cream, each attack rate was equal to or less than 35% (14/40 & 9/30). 7) Which of the following reasons can explain why a person who did not consume the infective food item got sick? a. They were directly exposed to persons who did eat the infective food item b. Diarrhea is a general symptom consistent with a number of illnesses c. There may have been an inaccurate recall of which foods were eaten d. All of the above e. None of the above Rationale: Without knowledge as to the specific agent in this instance, it is also likely that it can be spread by direct contact with infected persons. Since diarrhea is a general disease symptom, it is possible that several infectious agents may be present at this meal or others eaten during the same time period. Further, information regarding food consumption may have been collected long after the disease episode. This may have led persons to incorrectly remember the foods that they consumed. An outbreak of gastroenteritis occurred at a boarding school with a student enrollment of 846. Fifty-seven students reported symptoms including vomiting, diarrhea, nausea, & low-grade fever between 10 p.m. on September 24 & 8 p.m. on September 25. The ill students lived in dormitories that housed 723 of the students. The table below provides information on the number of students per type of residence & the number reporting illnesses consistent with the described symptoms & onset time. Calculate the attack rate among all students at the boarding school. 1. Calculate the attack rate among all students at the boarding school. 57/846 = The answer is found by dividing the total number of cases (57) by the total number of students (846). This equals 6.7%. 10. Calculate the attack rates for boys & girls separately. a. For boys, the attack rate includes all cases (40 + 3) divided by the total number of students who are boys (380 + 46). The attack rate is 10.1%. b. For girls, the attack rate includes all cases (12 + 2) divided by the total number of students who are girls (343 + 77). The attack rate is 3.3%. 11. What is the proportion of total cases occurring in boys? Answer: The proportion of cases occurring in boys is equal to the number of cases in boys divided by the total number of cases (43/57). This equals 75.4%. 12. What is the proportion of total cases occurring in students who live in dormitories? Answer: The proportion of cases occurring in dormitory residents is equal to the number of cases in residents divided by the total number of cases (52/57). This equals 91.2%. 13. Which proportion is more informative for the purpose of the outbreak investigation? Answer: Both proportions are useful. Dormitory residents account for over 90% of the cases indicating an outbreak of an infectious agent that was transmitted at the school. Furthermore, over 75% of the cases were boys indicating that the responsible agent was more likely to have been transmitted in the boys’ dormitory. A group of researchers are interested in conducting a clinical trial to determine whether a new cholesterol-lowering agent was useful in preventing coronary heart disease (CHD). They identified 12,327 potential participants for the trial. At the initial clinical exam, 309 were discovered to have CHD. The remaining subjects entered the trial & were divided equally into the treatment & placebo groups. Of those in the treatment group, 505 developed CHD after 5 years of follow-up while 477 developed CHD during the same period in the placebo group. 14. What was the prevalence of CHD at the initial exam? Answer: The prevalence of CHD at the initial exam was 309 cases of CHD divided by 12,327 participants. This equals a prevalence of 25.1 cases of CHD per 1,000 persons. 15. What was the incidence of CHD during the 5-year study? Answer: The incidence rate reflects the number of new cases developing in the population at risk. Since prevalent CHD cases were excluded from the study, the population at risk was 12,018 (12,327 persons less 309 cases of CHD). During the 5-year study period, 982 incident cases of CHD developed. This equals an incidence rate of 81.7 cases of CHD per 1,000 persons. 16. Which of the following are examples of a population prevalence rate? a. The number of ear infections suffered by 3-year-old children in March, 2006 b. The number of persons with hypertension per 100,000 population c. The number of cases of skin cancer diagnosed in a dermatology clinic d. b & c e. All of the above Rationale: Prevalence is the number of affected persons in a specified population size at a given time. Only answer (b) fits this definition. Example (a) is more consistent with an incident rate while answer (c) is a selected group of persons who may not be representative of a general population. 17. What would be the effect on age-specific incidence rates of uterine cancer if women with hysterectomies were excluded from the denominator of incidence calculations assuming that most women who have had hysterectomies are older than 50 years of age. A. The rates in all age groups would remain the same. B. Only rates in women older than 50 years of age would tend to decrease. C. Rates in women younger than 50 years would increase compared to women older than 50 years of age. D. Rates would increase in women older than 50 years of age but may decrease in younger women as they get older. E. It cannot be determined whether the rates would increase or decrease. Rationale: Women who have had hysterectomies (i.e., removal of the uterus) are no longer at risk for uterine cancer. For women older than 50 years of age, this would increase the age-specific incidence rate as there would be the same number of uterine cancers occurring among fewer women at risk. Further, rates may decrease among younger women who have had hysterectomies as they are no longer at risk for uterine cancer & thus may decrease the number of potential cases occurring in their age group over time. A survey was conducted among 1,000 r&omly sampled adult males in the United States in 2005. The results from this survey are shown below. 18. The researchers stated that there was a doubling of risk of hypertension in each age group younger than 60 years of age. You conclude that the researchers’ interpretation: a. Is correct b. Is incorrect because prevalence rates are estimated c. Is incorrect because it was based on proportions of the population sample d. Is incorrect because incidence rates do not describe risk e. Is incorrect because the calculations do not include adult females Rationale: The survey reports the disease status of a population at a specific point in time. In this case, a r&om sample of adult males in 2005 provides a reliable estimate of the prevalence of hypertension. Since there is no information on duration of hypertension in these men, incidence cannot be calculated. Therefore, the researchers are not able to make a statement concerning risk of hypertension in the population. 19. The incidence & prevalence rates of a chronic childhood illness for a specific community are given below. • Incidence rates do not vary by age, but older adults have pneumonia for a longer duration compared to younger adults • None of the above Rationale: For prevalence to be higher among older adults, either incidence or duration of pneumonia must be increased in this age group. 25. Which of the following statements are true? More than one answer may be correct. a. Prevalence rates are always larger than incidence rates b. In a steady state, the prevalence of disease is equal to the attack rate c. Diagnostic criteria rarely impact estimates of disease prevalence & incidence d. Prevalence rates are useful for public health planning e. Incidence rates can be used to estimate prevalence when the mean duration of the disease is known 26. A disease has an incidence of 10 per 1,000 persons per year, & 80% of those affected will die within 1 year. Prior to the year 2000, only 50% of cases of the disease were detected by physician diagnosis prior to death. In the year 2000, a lab test was developed that identified 90% of cases an average of 6 months prior to symptom onset; however, the prognosis did not improve after diagnosis. Comparing the epidemiology of the disease prior to 2000 with the epidemiology of the disease after the development of the lab test, which statement is true concerning the disease in 2000? a. Incidence is higher & prevalence is higher than in 1999 b. Incidence is higher in 2000 but prevalence remains the same c. Incidence is the same in 2000 but prevalence is higher than in 1999 d. Both incidence & prevalence remain the same as in 1999 e. Incidence is the same in 2000 but prevalence is lower than in 1999 Rationale: With increased ability to detect cases of the disease at earlier times, both the number of incidence & prevalent cases will increase through better detection. 27. Which statement is true concerning the duration of the disease after the development of the lab test? • Mean duration of a case of the disease is shorter in 2000 • Mean duration of a case of the disease is the same in 2000 • Mean duration of a case of the disease is longer in 2000 • No inference about mean duration can be made since the lab test has only been available for 1 year Rationale: Though the prognosis is similar after the development of the lab test, the duration of new cases identified by the test can be increased by up to 6 months due to earlier detection. 28. Which statement is true concerning the disease-specific mortality rate after the development of the lab test? • The mortality rate for the disease is decreased in 2000 • The mortality rate for the disease is the same in 2000 • The mortality rate for the disease is increased in 2000 • No inference about the mortality rate can be made since the lab test has only been available for 1 year Rationale: With the implementation of the lab test, the increase in early detection of cases will increase incidence, duration, & prevalence; however, since the prognosis is still the same, at least 80% of patients will die during the year 2000. This should result in a similar mortality rate as the previous year given no change in transmission, prevention, or medical care of the disease. 29. In a coastal area of a country in which a tsunami struck, there were 100,000 deaths in a population of 2.4 million for the year ending December 31, 2005. What was the all-cause crude mortality rate per 1,000 persons during 2005? Answer: The answer is 41.7 per 1,000 persons. The rate is calculated by dividing 100,000 deaths by the population of 2,400,000 persons. To express as a rate per 1,000 persons, the rate is multiplied by 1,000. 30. In an industrialized nation, there were 192 deaths due to lung diseases in miners ages 20 to 64 years. The expected number of deaths in this occupational group, based on age-specific death rates for lung diseases in all males ages 20 to 64 years, was 238 during 1990. What was the st&ardized mortality ratio (SMR) for lung diseases in miners? Answer: The answer is 81. The ratio is calculated by dividing 192 observed deaths by the 238 expected deaths for this age group. To express it as an SMR, the ratio is often multiplied by 100. 31. In 2001, a state enacted a law that required the use of safety seats for all children under 7 years of age & m&atory seatbelt use for all persons. The table below lists the number of deaths due to motor vehicle accidents (MVAs) & the total population by age in 2000 (before the law) & in 2005 (4 years after the law was enacted). What is the age-specific mortality rate due to MVAs for children ages 0 to 18 years in 2000? Answer: 6.1 per 1,000 Rationale: The rate is found by combining the MVA deaths & total population size for the two age groups under 7 years & 7 to 18 years during the year 2000. This equals (44 + 105) divided by (3,500 + 21,000). Multiplying this rate by 1,000 persons gives the answer indicated. 32. Using the pooled total of the 2000 & 2005 populations as the st&ard rate, calculate the age- adjusted mortality rate due to MVAs in 2005. Answer: 2.3 MVA deaths per 1,000 persons. The key to calculating the age-adjusted rate is to pool the observed numbers for both time periods & to calculate the expected numbers of deaths in the 2005 population assuming that a common rate applied to the population. For example, for those under 7 years, the pooled rate equals (44 + 20) divided by (3,500 + 4,000). The pooled rate for this group is 8.5 per 1,000 persons. When this rate is multiplied by the 4,000 children under 7 years of age in 2005, the expected number of deaths is 34.13. Performing the same calculation for each age group results in 111.7 deaths in those 7 to 18 years of age, 175.8 deaths in those 19 to 49 years, & 237.35 deaths for those 50 years or more. The total number of deaths expected in 2005 based on this pooled rate is 558.98. Therefore, the age-adjusted overall rate for 2005 is 558.98 deaths divided by 240,000 persons. 33. Based on the information in the table, it was reported that there was an increased risk of death due to MVAs in the state after the law was passed. These conclusions are: Answer: Correct, because both the total & the age-adjusted mortality rates are higher in 2005 than in 2000 Rationale: The overall crude (unadjusted) mortality rate is 2.6 per 1,000 persons in 2005. This is found by dividing 640 deaths by a population of 240,000 persons. This rate is then multiplied by 1,000. The overall adjusted mortality rate is 2.3 per 1,000 persons as calculated in question 34. Both of these rates are higher than the overall crude mortality rate of 2.0 per 1,000 persons for the year 2000. 34. For colorectal cancer diagnosed at an early stage, the disease can have 5-year survival rates of greater than 80%. Which answer best describes early stage colorectal cancer? • Incidence rates & mortality rates will be similar • Mortality rates will be much higher than incidence rates • Incidence rates will be much higher than mortality rates • Incidence rates will be unrelated to mortality rates • None of the above Rationale: For diseases with a long duration as indicated by high 5-year survival rates for early stage colorectal cancer, the incidence will be much higher than the mortality rate since more persons are being diagnosed with the disease than are dying of it. . The following table gives the mean annual age-specific mortality rates from measles during the first 25 years of life in successive 5-year periods. You may assume that the population is in a steady state (i.e., migrations out are equal to migrations in). 35. The age-specific mortality rates for the cohort born in 1915-1919 are: Answer= 2.4 3.3 2.0 0.6 0.1 Rationale: This is found by tracking the cohort of children born between 1915 & 1919 by each 5-year age group. For example, this group would be 0 to 4 years of age in 1915 to 1919 with a rate of measles mortality of 2.4. In 1920 to 1924, this group of children would be 5 to 9 years of age & have a rate of measles mortality of 3.3. Continuing in a diagonal manner, the remaining three rates can be found in the table. a. The PPV of the school nurse’s exam is equal to the number of true positive cases divided by the total number of those that the school nurse labels as positive. In this exam, the PPV is 34.6% (90 true myopic children divided by 260 children labeled as myopic by the school nurse). 4. How many children will be labeled myopic following the optometrist’s exam? a. Since the optometrist will only test children who have been labeled as myopic by the school nurse, the testing group for this sequential exam is 260 children. The optometrist labels 105 children as myopic. Among the 90 myopic children correctly referred by the school nurse, the optometrist identifies 88 of them as myopic (98% sensitivity multiplied by 90 true cases of myopia). Further, the optometrist will incorrectly identify 17 false positive cases among the 170 children referred by the school nurse who do not have myopia. The sum of the cases labeled as positive by the optometrist equals 105 children (89 true cases plus 17 false positive cases). 5. What is the positive predictive value (PPV) of the optometrist’s exam? a. The PPV of the optometrist’s exam is equal to the number of true positive cases divided by the total number that the optometrist labels as positive. The optometrist will only test 260 children referred by the school nurse. Of these children, the optometrist will correctly identify 89 cases of myopia among 105 children labeled as positive for the condition. The PPV equals 83.8% (89 true myopic children divided by 105 children labeled as positive). 6. What is the negative predictive value (NPV) of the optometrist’s exam? a. The NPV of the optometrist’s exam is 98.7%. The NPV equals the number of true negative cases divided by all negative cases indicated by the exam. In this instance, the optometrist correctly identifies 153 children as negative for myopia; however, there are 2 false negative cases following the optometrist’s exam (90 true cases referred by the school nurse less the 88 cases detected by the optometrist). The NPV equals 153 divided by 155, or 98.7%. 7. What is the overall sensitivity of the sequential examinations? a. The overall sensitivity of the sequential exams is 58.7%; 88 true positive cases of myopia are found following the optometrist’s exam among the 150 prevalent cases in the school population. 8. What is the overall specificity of the sequential examinations? a. The overall specificity of the sequential exams is 98%; 833 children will be correctly labeled as negative for myopia among the 850 true negative cases. This is found by summing the number of true positives after each exam (680 following that of the school nurse plus 153 following the optometrist) & dividing by the true negative children in the population. This equals 833 divided by 850, or 98%. 9. What would be the positive predictive value (PPV) of the exam for myopia if the optometrist tested all 1,000 children? a. The PPV of the optometrist’s exam would be equal to the number of true positive cases divided by all children labeled positive by the optometrist. Applying the sensitivity & specificity of the optometrist’s exam to the 1,000 children would indicate that 147 true positive cases are labeled positive by the optometrist. Additionally, the optometrist would find 85 false positive cases (850 true negative cases multiplied by 90% specificity). The PPV would be 63.4% (147 true positive cases divided by 232 total positives indicated by the optometrist 10. Which of the following improves the reliability of diabetes screening tests? a. Having the same lab analyze all samples b. Taking more than one sample for each subject & averaging the results c. Insuring that the instrument is st&ardized before each sample is analyzed d. a & c only e. All of the above Rationale: Reliability is improved by consistency of analyses, especially when multiple samples are taken for a subject & the analytic instrument is routinely st&ardized. 11. A prostate specific antigen (PSA) test is a quick screening test for prostate cancer. A researcher wants to evaluate it using two groups. Group A consists of 1,500 men who had biopsy-proven adenocarcinoma of the prostate while group B consists of 3,000 age- & race-matched men all of whom showed no cancer at biopsy. The results of the PSA screening test in each group is shown in the table. What is the sensitivity of the PSA screening test in the combined groups? Answer: The sensitivity equals the number of true positives detected among all true positives. Since a biopsy is the gold st&ard test for prostate cancer, all 1,500 men in group A are positive for prostate cancer. The PSA test indicated that 1,155 of these men had prostate cancer, a sensitivity of 77%. 12. What is the specificity of the screening test in the combined groups? a. Answer: 85 b. Rationale: The specificity equals the number of true negatives detected among all true negatives. Among the 3,000 men who did not have prostate cancer, the test correctly identified 2,760 men as negative for prostate cancer (3,000 minus 240 false positives). This gives a sensitivity of 92%. 13. What is the positive predictive value (PPV) of the screening test in the combined groups? a. The PPV is 83%. b. rationale: This value is found by dividing 1,155 true positives by the total number of all positives indicated by the PSA test (1,155 plus 24). 14. The PSA screening test is used in the same way in two equal-sized populations of men living in different areas of the United States, but the proportion of false positives among those who have a positive PSA test in the first population is lower than that among those who have a positive PSA test in the second population. What is the likely explanation for this finding? a. It is impossible to determine what caused the difference b. The prevalence of disease is higher in the first population c. The specificity of the test is lower in the first population d. The specificity of the test is higher in the first population e. The prevalence of the disease is lower in the first population Rationale: We can assume that the specificity of the test will be similar in each population. Therefore the proportion of false positives found among the true negatives should be the same in each population. However, the proportion of false positives among all positives on the PSA screening test will be influenced by the number of true positives detected by the test. Since the sensitivity of the test will also be the same, we can assume that more true positives exist in the population of men with a lower proportion of false positive tests due to an increase in the PPV. 15. Test A has a sensitivity of 95% & a specificity of 90%. Test B has a sensitivity of 80% & a specificity of 98%. In a community of 10,000 people with 5% prevalence of the disease, Test A has always been given before Test B. What is the best reason for changing the order of the tests? a. The net sensitivity will be increased if Test B is given first b. The total number of false positives found by both tests is decreased if Test B is given first c. The net specificity will be decreased if Test B is given first d. The total number of false negatives found by both tests is decreased if Test B is given first e. There is no good reason to change the order of the tests Rationale: A sequential testing process would only refer those with positive results to the second test. Since Test B has a higher specificity, then fewer false positives will be referred for Test A, thereby decreasing the number of false positives found. This can be shown by calculation if we assume that 500 persons have the disease among the 10,000 in the population. Test B will find only 190 false positives for referral (9,500 true negatives less the number of true negatives multiplied by 98% specificity). Performing Test A first results in 950 false positives referred for the second test (9,500 true negative less the number of true negatives multiplied by 90% specificity). 16.16. Two neurologists, Drs. J & K, independently examined 70 magnetic resonance images (MRIs) for evidence of brain tumors. As shown in the table below, the neurologists read each MRI as either “positive” or “negative” for brain tumors. Based on the above information, the overall percent agreement between the two doctors including all observations is: a. 62.9% b. Rationale: The two doctors agree on 44 of the 70 MRI readings. This includes the 26 that they both labeled as positive for brain tumors & the 18 that they both agreed were negative for brain tumors. 17. What is the estimate of kappa for the reliability of the two doctors’ test results? a. 24.9% b. Rationale: The estimate of kappa expresses the observed agreement of two testers in excess of chance alone. It is found by applying the expected agreement rates for both testers. In this case, Dr. K labeled 38 of the 70 MRIs as positive (54.3% of all MRIs) & 32 as negative (45.7% of all slides). Dr. J labeled 57.1% of the MRIs as positive (40 of 70) & 42.9% as negative. We would expect that if Dr. K had the same rate of positive & negative findings as Dr. J then they would agree by chance on 21.7 of the 38 positive MRIs that were found (38 multiplied by 0.571). Further, they would agree by chance on 13.7 of the 32 negative MRIs that were found (32 multiplied by 0.429). Therefore, we would expect the two doctors to agree by chance on 50.6% of the MRIs (21.7 positive agreements plus 13.7 negative agreements equals 35.4, then divide this by the total of 70 to get an expected overall agreement of 50.6%). Now, kappa can be calculated as the observed agreement less expected divided by 100% less the expected agreement— in factor were to differentially influence survival during a portion of the follow-up time then we would not be able to assume a cumulative survival that is consistent during the entire study period. 27.27. Complete the table. What is the probability that a person enrolled in the study will survive to the end of the third year? a. The answer is 48.6%. b. Completing the table gives the following results for each column: c. Column 5 from top to bottom: 350, 255, 184 d. Column 6 from top to bottom: 0.229, 0.228, 0.185 e. Column 7 from top to bottom: 0.771, 0.772, 0.815 f. Column 8 from top to bottom: 0.771, 0.596, 0.486 Rationale: The cumulative survival at the end of the follow-up period equals the probability of survival during each of the years of follow-up. In this example, multiplying 0.771 by 0.772, then multiplying this product by 0.815 equals the cumulative survival rate of 0.486. 28. Before reporting the results of this survival analysis, the investigators compared baseline characteristics of the 38 people who withdrew from the study before its end to those who had complete follow-up. This was done for which of the following reasons: a. To check whether those remaining in the study represent the total study population b. Rationale: A key assumption for the use of survival analysis is that persons who are lost to follow-up have the same mortality experience as those remaining in the study. The failure to satisfy this assumption introduces a bias in the survival estimates since the observed population has different attributes that are associated with survival compared to the population that is lost to follow-up. 29. Which of the following is a key assumption involved in the use of life-table analysis? a. The risk of disease does not change within each interval over the period of observation b. There are no losses to follow-up in the study population c. The frequency of exposure is similar in treatment & comparison groups d. The disease is common e. The study subjects are representative of the population from which they were drawn Rationale: Life-table analysis depends upon a consistent rate of survival during all periods of the study. Changes in the rate of survival may be due to external influences that are operating at later times on only a portion of the initial population. Since those who have died earlier in the study period will not experience these external influences, the comparison between periods is rendered invalid. 30. Which of the following is a measure of disease prognosis? a. Prevalence b. Median survival time c. Age-adjusted mortality rates d. St&ardized mortality ratio e. Proportionate mortality ratio Rationale: Disease prognosis indicates the likelihood of survival once a disease has become manifest. The median survival time reflects the length of time that the 50th percentile of affected persons has. It differs from the mean survival time in that the mean survival time is an average that may be influenced by extremely low or high survival times. The median survival time consists of an ordering of all survival times with the midpoint of the distribution taken as the duration of survival. 31.31. In 2003, Sudden Acute Respiratory Syndrome (SARS) appeared in several countries, mainly in Asia. The disease was determined to have been caused by a virus that could be spread from person –to person from the index case occurring in mainl& China. This table reflects the total number of reported cases of SARS & deaths among those cases as best as can be determined. What is the overall case-fatality rate for the worldwide epidemic of SARS? a. 9.5% b. Rationale: This can be found by dividing the total number of deaths due to SARS by the total number of cases. This equals a case-fatality rate of 9.5%. 32. In 2003, Sudden Acute Respiratory Syndrome (SARS) appeared in several countries, mainly in Asia. The disease was determined to have been caused by a virus that could be spread from person –to person from the index case occurring in mainl& China. This table reflects the total number of reported cases of SARS & deaths among those cases as best as can be determined. Based on the table, we can conclude that the case-fatality rate (CFR) in Vietnam: a. Is the same as the case-fatality rate in Singapore b. Is twice as great as the case-fatality rate in Singapore c. Is almost one half that of the case-fatality rate in Singapore d. Cannot be determined because the data are not age-adjusted e. Depends on the number of secondary cases Rationale: The CFR in Vietnam equals 5 divided by 63, or 7.9%, while that of Singapore equals 15%. This is approximately one half the rate. 33. What happened to the case-fatality rate (CFR) following this reclassification? a. It was decreased b. Rationale: The increase in prevalent cases with no change in mortality would decrease the CFR since the numerator, number of deaths due to SARS, would stay the same while the denominator, number of cases, increased. 34.34. What is the probability of surviving the second year of the study given that a person survived the first year? a. The independent probability of surviving the second year for all persons who survived the first year is found by dividing the number of survivors at the end of the period by the total number present at the beginning of the period. In addition, for those who withdraw during the interval, only 50% of these persons should be counted as being present during the interval. The table should be completed with the following values: b. Column (B) from top to bottom: 248, 124, 55 c. Column (E) from top to bottom: 0.410, 0.470, 0.296 d. Column (F) from top to bottom: 0.590, 0.530, 0.704 e. Column (G) from top to bottom: 0.590, 0.313, 0.220 f. Therefore, the second year survival probability among all those surviving in the study past the first year is 53%. The probability of dying during the second year equals the number of deaths during the interval (55) divided by the total number of persons alive at the start of the interval less one half of those withdrawing from the study (117). Subtracting this value from 100% results in a survival rate of 53% for the interval. 35. For all people in the study, what is the probability of surviving to the end of the second year? a. The cumulative probability of survival through the second year equals the probability of survival for the first year multiplied by the probability for the second year. This equals 59% multiplied by 53%, or 31.3%. 36. What is the probability chance of surviving 3 years after diagnosis? a. The cumulative survival probability for all 3 years equals the product of the independent interval survival probabilities. In this example, 59% multiplied by 53% multiplied by 70.4% gives a cumulative survival probability of 22%. 37. What is the total number of person-years of follow-up for patients in the study assuming a median survival time of one half of the year for all persons dying during an interval & an observation time of one half of the year for all persons withdrawing from the study? a. This calculation involves attributing the correct amounts of person-years to each group during an interval. For the first year of the study, 96 deaths occur. Using the median survival time, we can calculate that these persons contributed 48 person-years of observation. Additionally, 28 persons withdraw from the study. Again, allocating one half of the year to each of these patients results in 14 person-years. Of the remaining 124 persons who survive for the full year, they contribute 124 person-years of observation.