HLTH 501 WEEK 4 MIDTERM EXAM / HLTH501 WEEK 4 MIDTERM EXAM : LIBERTY UNIVERSITY (LATEST UPDATE 2020_2021) With ALL the information that you need Rated A

HLTH 501 WEEK 4 MIDTERM EXAM / HLTH501 WEEK 4 MIDTERM EXAM : LIBERTY UNIVERSITY (LATEST UPDATE 2020_2021) With ALL the information that you need Rated A 1. The relative risk of developing lung cancer among individuals who smoke is 15.0. Which of the following is the most accurate interpretation of this information? A) Individuals who have lung cancer are 15.0% more likely to smoke than individuals who do not have lung cancer. B) Individuals who have lung cancer are 15.0 times more likely to smoke than individuals who do not have lung cancer. C) Individuals who smoke are 15.0 times more likely to develop lung cancer than individuals who do not smoke. D) Individuals who smoke are 15.0% more likely to develop lung cancer than individuals who do not smoke. 2. A healthcare provider notices an outbreak of foodborne illness among individuals who attended a holiday cookout. Which study type would be most beneficial in identifying the source of the outbreak? A) A case series B) A retrospective cohort study C) A cross-sectional survey D) A case-control study 3. A researcher hopes to measure the exposure a large number of individuals who have already developed lung cancer had to secondhand smoke throughout their lifetime. Which study design may prove useful in helping the researcher measure the exposure the individuals had to second hand smoke? A) A case series B) A retrospective cohort study C) A prospective cohort study D) A case-control study 4. A researcher hopes to measure the exposure a large number of individuals who have already developed lung cancer had to secondhand smoke throughout their lifetime. Which study design may prove useful in helping the researcher measure the exposure the individuals had to second hand smoke? A) Randomization B) Crossover matching C) Matching D) Stratification 5. A new drug is released to market and it has been determined that the drug is responsible for causing hypertension among those individuals who take the drug. If 30,276 individuals were prescribed the drug in its first year on the market and 17,620 individuals developed hypertension, what is the point prevalence of hypertension among those individuals prescribed the drug, after the drug’s first year on the market? A) 0.00582 B) 0.05820 C) 0.58200 D) 5.82000 6. After attending a church cookout, a number of the attendees are admitted to the emergency room with nausea, vomiting, and diarrhea. After discovering all the individuals admitted to the ER consumed the egg salad at the event, the leftover egg salad was tested and found to be positive for salmonella. If 83 individuals consumed the egg salad and 49 individuals were admitted to the ER with salmonella, what was the cumulative incidence of salmonella assuming that the 49 individuals admitted to the ER were the only ones to be affected by the bacteria? A) 0.041 B) 0.059 C) 0.410 D) 0.590 7. Drug manufactures of Drug X, which was released on the market 5 years ago to treat Crohn’s disease, have noticed that many individuals treated with the drug have developed drug-induced lupus. A cohort study was conducted for post-marketing surveillance. It has documented that 6 individuals have developed drug-induced lupus over the course of the past 5 years. If 1 individual developed the disease at the 1-year mark, 3 developed the disease at the 2-year mark, 1 developed the disease at the 3-year mark, and the final individual developed the disease at the 4-year mark, what is the incidence rate of drug-induced lupus? A) 0.0429 B) 0.4290 C) 0.5710 D) 0.9571 8. In which situation would it be most appropriate to use the mean of the set of numbers to describe the data set instead of the median? A) 152, 157, 159, 163, 164, 165, 976 B) 0, 899, 906, 912, 915, 917, 919 C) 0, 57, 847, 859, 866, 866, 872 D) 165, 165, 177, 178, 186, 199, 211 9. A 95% confidence interval for relative risk of developing a bacterial infection after having a surgery is (3.71, 5.33) when comparing a non-laparoscopic surgery procedure to a laparoscopic surgery procedure. This indicates which of the following? A) The researchers are confident because the probability that these results would occur is somewhere between 3.71% and 5.33%. B) The researchers are confident that the probability that these results would occur is somewhere between 0.371% and 0.533%. C) The researchers are 95% confident that the increased risk of developing an infection is between 3.71 to 5.33 times higher among individuals who had the non-laparoscopic surgery when compared to those individuals who received the laparoscopic surgery. D) The researchers are 95% confident that the risk of developing an infection is between 3.71 to 5.33 times higher among individuals who had the laparoscopic surgery when compared to those individual who received the non-laparoscopic surgery. 10. Which of the following accurately represents a 95% confidence interval for a sample of 5 individuals whose blood glucose levels were 98, 111, 105, 120, and 147? A) (153.5, 78.9) B) (78.9, 153.5) C) (63.4, 169.0) D) (169.0, 63.4) 11. With a binomial distribution with n = 25 and p = 0.48, which is larger? A) P(12 successes) B) P(9 successes) C) P(20 successes) D) P(10 successes) True/False 12. True or False? Biostatistics is integral to the practice of public health because it allows public health professionals to accurately monitor and track the prevalence of disease within a population. 13. True or False? The margin of error in a research study is indicative of the precision and accuracy of the research results obtained in the study. 14. True or False? An active-controlled trial allows researchers to offer participants a placebo in place of an alternative treatment method that has been previously proven to be effective. 15. True or False? Incidence reflects the likelihood of developing a disease among a group of participants free of the disease who are considered at risk of developing the disease over a specified observation period; prevalence involves estimating the proportion of people who have disease at a point in time. 16. True or False? If the odds of catching the flu among individuals who take vitamin C is 0.0342 and the odds of catching the flu among individuals not taking vitamin C is 0.2653, then the individuals not taking vitamin C are 7.7573 times more likely to catch the flu than individuals taking vitamin C. 17. True or False? Histograms are often used to display categorical data, while bar charts are commonly used to display variables that fall on a measurable continuum. 18. True or False? Variability within a sample is oftentimes measured using sample variance. 19. True or False? A newly proposed endoscopy method used to screen for colon cancer is tested among a random sample of the population. Given that 16 individuals from a test population of 100,000 individuals has colon cancer and 14 individuals test positive for colon cancer, the researchers should determine the sensitivity of the newly proposed screening method is .875. 20. True or False? A newly developed medical procedure to prevent death in women who give birth and experience eclampsia has a 97.5% success rate of saving the woman’s life. If a doctor uses this medical procedure in 6 women who experience eclampsia and have no other comorbid conditions, the probability that all 6 women will survive is approximately 0.859. 21. True or False? The mean measure of high density lipoprotein (HDL), or good cholesterol, is 54 with a standard deviation of 17 in patients over age 50. If the HDL values are obtained from a sample of 32 individuals, the probability of getting a mean HDL value over 62 is 0.049. 22. True or False? If there are outliers, then the mean will be greater than the median. 23. True or False? The 90th percentile of the standard normal distribution is 1.645. 24. True or False? The mean is the 50th percentile of any normal distribution. 25. True or False? The mean is a better measure of location when there are no outliers. 26. True or False? Case-control study, prospective cohort study, cross-over trial, and retrospective cohort study are all examples of observational study designs. 27. True or False? For the standard normal distribution, Q3 = 0.675. Essay 28. The risk of hepatoma among alcoholics without cirrhosis of the liver is 24%. Suppose we observe 7 alcoholics without cirrhosis. What is the probability that exactly one of these 7 people have a hepatoma? Answer P(X=1) =7C1*(0.24^1)*((1-0.24)^(7-1)) =0. 29. Glucose levels in patients free of diabetes are assumed to follow a normal distribution with a mean of 120 and a standard deviation of 16. What proportion of patients have glucose levels exceeding 115? Answer P(X 115) = P(z (115 - 120)/16) = P(z -0.3125) = 0.6227 30. The following are body mass index (BMI) scores measured in 12 patients who are free of diabetes and participating in a study of risk factors for obesity. Body mass index is measured as the ratio of weight in kilograms to height in meters squared. Using the following data, compute the mean BMI, the standard deviation of BMI, the median BMI and the Q1 and Q3. 25 27 31 33 26 28 38 41 24 32 35 40 Answer Mean:- 31.67 Standard D- 5.9 Median- 31.5 Q1/Q3- Q1- 26.5, Q2- 36.5 31. The following table shows the numbers of patients classified as underweight, normal weight, overweight and obese according to their diabetes status. What is the probability that a patient selected at random is overweight? Underweight Normal Weight Overweight Obese Diabetes 8 34 65 43 No Diabetes 12 85 93 40 Answer 1) Probability that they are overweight = total overweight patients / total number of patients = 158/380 = 0.4157 32. A new non-invasive screening test is proposed that is claimed to be able to identify patients with impaired glucose tolerance based on a battery of questions related to health behaviors. The new test is given to 75 patients. Based on each patient’s responses to the questions they are classified as positive or negative for impaired glucose tolerance. Each patient also submits a blood sample and their glucose tolerance status is determined. The results are tabulated below. What is the sensitivity of the screening test? Screening Test Impaired Glucose Tolerance Not Impaired Positive 17 13 Negative 8 37 Answer given this, for a, patient having impaired glucose tolerance in the screening test required probabilty = 25/75 = 0.3 for (b) 13/75 = -.173 answer 33.A study is conducted to test a new drug claimed to reduce diastolic blood pressure in adults with a history of coronary heart disease. What is the most efficient study to test whether the drug reduces diastolic blood pressure? Answer: A randomized clinical trial. It will be the best and most efficient study to test whether the drug reduces diastolic blood pressure. 34. Approximately 30% of obese patients develop diabetes. Answer the following questions if a physician sees 10 patients who are obese. What is the probability that half of them will develop diabetes? Answer PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k Where k = number of successes in trials n = is the number of independent trials p = probability of success on each trial P( X = 5 ) = ( 10 5 ) * ( 0.3^5) * ( 1 - 0.3 )^5 = 0.1029 35. A study is run to investigate body mass index (BMI) in children living in urban neighborhoods. Based on the following data: compute the sample mean, the sample standard deviation, and the median. 29 26 24 32 30 22 27 28 31 Answer mean = total sum / n mean = (29+26+24+32+30+19+27+28+31)/9 mean =27.3333 36. The following table shows the distribution of BMI in children living in United States and European urban neighborhoods. (The data are in millions.) What is the probability that a child living in a U.S. urban neighborhood is overweight? Neighborhood Normal Weight Overweight Obese United States 125 50 40 Europe 101 42 21 37. Suppose that the probability that a child living in an urban area in the United States is obese is 20%. If a social worker sees 15 children living in urban areas, answer the following: What is the probability that none are obese? Answer PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k Where k = number of successes in trials n = is the number of independent trials p = probability of success on each trial P( X = 0 ) = ( 15 0 ) * ( 0.2^0) * ( 1 - 0.2 )^15 = 0.03518 38. Body mass index (BMI) in children is approximately normally distributed with a mean of 24.5 and a standard deviation of 6.2. In a random sample of 10 children, what is the probability that their mean BMI exceeds 25? Answer P(xbar25) =P((xbar-mean)/(s/vn) (25-24.5)/(6.2/sqrt(10))) =P(Z0.26) =0.3974 (check standard normal table) 39. An investigator wants to test whether exposure to secondhand smoke before 1 year of life is associated with development of childhood asthma (defined as asthma diagnosed before 5 years of age). Give two possible study designs. ANSWER cohort study which is astudy that involves a group of individuals who meeta set of criteria and is used to make an association between a certain risk factor and developing a disease in this case secondhand smoke and asthma a con in this study design is the possibility that the participants have other risk factors that could cause asthma that would distort the data * case-control studies also link a risk factor to a specific outcome this study however is typically more useful when the condition or outcome is rare this type of study is cost and time efficient this study however has to include participants whom already have asthma and the investigators must rely on the memory of the participants caregiver to determine the level of secondhand smoke exposure prior to diagnoses 40. A study is run to estimate the incidence of atrial fibrillation (AF) in men and women over the age of 60. Development of atrial fibrillation was monitored over a 10-year follow-up period. The data are summarized below. Using this data, compute the cumulative incidence of AF in men and in women. Developed AF Did not Develop AF Men 120 6453 Women 86 7074 Answer a = 120 , b = 6453, c = 86 , d = 7074 The relative risk or risk ratio is given by RR /(a b) c/(c+ d) =1.5200 The odds ratio is given by Odds ratio = 1.5296 41. The following data were collected in a survey of 8th graders and summarize their cell phone status. What proportion of the 8th graders have cell phones? No cell phone Conventional cell phone user Smart phone user Boys 55 65 35 Girls 31 78 27 Answer Having phones = Total- (55 + 31) = 291 - 86 = 205 Total students = 86 + 65 + 78 + 35 + 27 = 291 proportion of the 8th graders has cell phones = 205/291 = 0.7045 or 70.45% 42. A ferritin test is a popular test to measure a person’s current iron stores. In women, ferritin is approximately normally distributed with a mean of 89 ng/mL and a standard deviation of 23 ng/mL.If 50 women are tested, what is the probability that the mean ferritin exceeds 90? Answer 142.5 89 - 2.3261 23 From standard normal tables, we get: P(Z 2.3261) = 0.9900 b) For 50 women, the distribution of the mean value is given using central limit theorem as: Χ.Ν(μ 89, σχ - νn 23 Χ Ν μ-89, σχ ν50 Χ.Ν(μ-89, στ-3.2527) P(X90) Converting it to a standard normal variable, we get: 90 89 P(Z 3.2527 P(Z 0.3074 Getting it from the standard normal tables, we get: P(Z 0.3074) 0.3793 Therefore 0.3793 is the required probability here. 43. A pilot study is run to investigate the effect of a lifestyle intervention designed to increase medication adherence in patients with HIV. Medication adherence is measured as the percentage of prescribed pills that are taken over a one-week observation period. Ten patients with HIV agree to participate and their medication adherence before and after the intervention are shown below. Compute the standard deviation of the difference in adherence before versus after intervention. Participant ID Before Intervention After Intervention 1 75% 80% 2 82% 84% 3 66% 70% 4 74% 70% 5 88% 90% 6 66% 75% 7 51% 60% 8 93% 90% 9 88% 90% 10 91% 95% Answer: The mean difference score is M = 3% Estimated standard deviation of difference mean score is S= Σ(X-M)2 η-1 ν = 0.0166 = 0.043 The standard deviation of the difference in adherence before versus after intervention is 0.043 44. The following table shows the results of a screening test hypothesized to detect persons at risk for side effects of a new cosmetic surgery. Compute the sensitivity of the test, the specificity of the test and the false negative fraction. Side effects present Side effects absent Screen positive 12 6 Screen negative 85 204 Answer: Total population sampled is 307 (=12+6+85+204). Compute sensitivity: Sensitivity of the test = TP TP + FN) X 100 = 12 (12 + 85) X 100 = 12.37 %. Compute specificity: Specificity of the test = TN / (FP + TN) X 100 = 204 (6 + 204) X 100 = 97.14 %. Compute false positive fraction: False positive fraction = FP (FP + TP) = 6 (6 + 12) = 1/3. Compute false negative fraction: False negative fraction = FN (FN + TN) = 85 (85 + 204) = 5/17. 45. A clinical trial designed to show the efficacy of a new drug in reducing progression to hypertension reports the following data. Using this information, compute the relative risk of progression to hypertension among patients receiving the new drug as compared to the placebo. New Drug Placebo Sample size 200 200 Progression to hypertension 14% 19% ANSWER=(28X200)I(38X 200)= (0.14/0.19)=0.7368 46. The following are grade point averages measured in a sample of 8 undergraduate students who are applying to graduate schools in public health. Compute the sample mean, the sample standard deviation, and the sample median. 3.28 2.97 3.05 3.61 3.39 2.95 3.00 3.10 Answer: mean - 3.53 Median :3.08 STD- 47. The following table shows the results of a screening test hypothesized to identify persons at risk for a rare blood disease. Compute the specificity of the test. No Disease Disease Screen Negative 1274 28 Screen Positive 51 45 Answer: specificity=True negative rate=true negative/(true negative+false positive)=1274/(1274+28)=0.9785 48. The data shown below describe children in four countries in terms of adequate dental care. What is the probability that a child from the United States has adequate dental care? Adequate Dental Care Inadequate Dental Care United Kingdom 2539 176 United States 4563 209 France 1298 307 Mexico 2874 256 Answer: proportion of children with inadequate dental care are from the US=209/948=0.2205 49. The mean total serum cholesterol level in women free of cardiovascular disease is 205 with a standard deviation of 19.2. If values between the 5th and 95th percentiles are considered “normal,” what is the upper limit of normal? Answer Z-score of 95th percentile =1.645 The upper limit of normal,small X=mu +Zsigma small X=205 +(1.645)(19.2) 236.584 small =236.6 50. Four hundred melanoma patients were diagnosed according to the type of skin cancer and the location of the skin cancer. This data is presented below. What proportion of patients had superficial spreading melanoma? Location Type Head and Neck Trunk Extremities Total Hutchinson’s melanomic freckle 22 2 10 34 Indeterminate 11 17 28 56 Nodular 19 33 73 125 Superficial spreading melanoma 16 54 115 185 Total 68 106 226 400 51. The gestation period for human births can be taken as normally distributed with a mean of 266 days and a standard deviation of 16 days. What is the probability that a randomly chosen baby had a gestation period of more than 280 days? If a gestation period is 276 days, what percentile among human births is this? x=gestation period for human births Answer X~Normal(y = 266,0= 164) What is the probability that a randomly chosen baby had a gestation period of more than 280 days? ie P(X 280) P(X 280) = 1- P(X280) = 1 - ( 4 - =1-P(Z0.875) =1-0.8092 =0.1908 P(X 280) = 0.1908 2)If a gestation period is 276 days, what percentile among human births is this? ie P(x 276) = P(X - H 276 – 266 16 =P(Z0.625) =0.7340 P(X 276) = 0.7340 If a gestation period is 276 days, what percentile among human births is 0.7340 52. The faculty and staff at Dana-Farber Cancer Institute in Boston are required to undergo a tuberculosis test each year. Use the following table to answer the questions below. If a Dana-Farber faculty or staff worker goes in to have a TB test, what is the chance of him/her getting a positive test result? Test Result Tuberculosis Yes No Total Positive 4995 99,950 104,945 Negative 5 9,895,050 9,895,055 Total 5,000 9,995,000 10,000,000 Answer Chance of Getting Positive Result = 104,945 /10,000,000 = 0.00104 What is the positive predictive value of the test? = 4995/ (104,945) = 0.0475