Essentials of Biostatistics in Public Health Chapter 5 Answer Key

Chapter 5 Answer Key Essentials of Biostatistics for Public Health Chapter 5 Solutions to Problems 1. A recent study reported that the prevalence of hyperlipidemia (defined as total cholesterol over 200) is 30% in children 2-6 years of age. If 12 children are analyzed, a. What is the probability that at least 3 are hyperlipidemic? P(X > 3) = 1 – P(X < 3) = 1 – {P(X=0) + P(X=1) + P(X=2)} 12! 0.30 (1−0.3)12− 0= P(X = 0) = 0!(12−0 )! 0.0138 12! 1 12−1 0.3 (1−0.3) = P(X = 1) = 1!(12−1)! 0.0712 12! 2 12− 2 0.3 (1−0.3) = P(X = 2) = 2!(12−2)! 0.1678 = 1- {0.0138 + 0.0712 + 0.1678) =1- 0.2528 = 0.7472. b. What is the probability that 3 are hyperlipidemic? 12! P(X=3) = 3!(12−3)! 0.3 3 (1−0.3)12−3 = 0.2397. c. How many would be expected to meet the criteria for hyperlipidemia? =np=12(0.3) = 3.6 2. Hyperlipidemia in children has been hypothesized to be related to high cholesterol in their parents. The following data were collected on parents and children. CHILD Not Hyperlipidemic Hyperlipidemic Both Parents Hyperlipidemic 13 45 One Parent Hyperlipidemic 34 42 Neither Parent Hyperlipidemic 83 6 a. What is the probability that one or both parents are hyperlipidemic? P(one or both parents are hyperlipidemic) = (58+76)/223 = 134/223 = 0.600. b. What is the probability that the child and both parents are hyperlipidemic? P(child and both parents are hyperlipidemic) = 45/223 = 0.202. c. What is the probability that a child is hyperlipidemic IF neither of his/her parents are hyperlipidemic? P(child is hyperlipidemic | neither parent is hyperlipidemic) = 6/89 = 0.067. d. What is the probability that a child is hyperlipidemic IF both of his/her parents are hyperlipidemic? P(child is hyperlipidemic | both parents are hyperlipidemic) = 45/58 = 0.776. 3. Total cholesterol in children aged 10-15 is assumed to follow a normal distribution with a mean of 191 and a standard deviation of 22.4. a. What proportion of children 10-15 years of age have total cholesterol between 180 and 190? 180−191 P(180 < X < 190) = P( 22.4 = 0.4840 – 0.3121 = 0.1719. 190−191 < Z < 22.4 ) = P(-0.49 < Z < -0.04) b. What proportion of children 10-15 years of age would be classified as hyperlipidemic (Assume that hyperlipidemia is defined as a total cholesterol level over 200)? 200−191 P(X > 200) = P(Z > 22. 4 ) = P(Z > 0.40) = 1-0.6554 = 0.3446. c. If a sample of 20 children are selected, what is the probability that the mean cholesterol level in the sample will exceed 200? 200−191 P( X̄ > 200) = P(Z > 22. 4 / √ 20 ) = P(Z > 1.80) = 1-0.9641 = 0.0359. 4. A national survey of graduate students is conducted to assess their consumption of coffee. The following table summarizes the data. Male Female Do not drink coffee 145 80 Drink Decaffeinated Only 94 121 Drink Caffeinated Coffee 365 430 a. What proportion of students drink decaffeinated coffee only? P(decaffeinated coffee only) = 215/1235 = 0.174. b. What proportion of coffee drinkers (caffeinated and decaffeinated) are female? P(female | caffeinated or decaffeinated coffee) = (121+430)/(215 + 795) = 551/1010 = 0.546. c. What proportion of the females do not drink coffee? P (do not drink coffee | female) = 80/631 = 0.127. 5. Among coffee drinkers, men drink a mean of 3.2 cups per day with a standard deviation of 0.8 cups. Assume the number of drinking per day follows a normal distribution. a. What proportion drink 2 cups per day or more? P(X > 2) = P(Z > 2−3 .2 0 . 8 ) = P(Z > -1.5) = 1-0.0668 = 0.9332. b. What proportion drink no more than 4 cups per day? 4−3.2 P(X < 4) = P(Z < 0. 8 ) = P(Z < 1) = 0.8413. c. If the top 5% of coffee drinkers are considered heavy coffee drinkers, what is the minimum number of cups consumed by a heavy coffee drinker? X =  + Z = 3.2 + 1.645(0.8) = 4.5 drinks. d. If a sample of 20 men is selected, what is the probability that the mean number of cups per day is greater than 3? 3−3 .2 X̄ 0. P( > 3) = P(Z > 8 /√ 20 ) = P(Z > -1.12) = 1-0.1314 = 0.8686. 6. A study is conducted to assess the impact of caffeine consumption, smoking, alcohol consumption and physical activity on cardiovascular disease. Suppose that 40% of participants consume caffeine and smoke. If 8 participants are evaluated, what is the probability that: a. Exactly half of them consume caffeine and smokes? 8! 0.4 4 (1−0.4)8−4 = 4!(8−4)! P(X = 4) = 0.2322. b. More than 6 consume caffeine and smoke? P(X > 6) = P(X=7) + P(X=8) 8! 0.4 7 (1−0.4)8−7 = 7!(8−7)! P(X=7) = 0.0079 8! 8 8−8 0. 4 (1−0.4) = P(X=8) = 8!(8−8)! 0.0007 P(X > 6) = P(X=7) + P(X=8) = 0.0079 + 0.0007 = 0.0086. c. Exactly 4 do NOT consume caffeine or smoke? 8! 0.6 4 (1−0.6 )8−4 = 4!(8−4 )! P(X = 4) = 0.2322. 7. As part of the study described in Problem 6, investigators wanted to assess the accuracy of self-reported smoking status. Participants are asked whether they currently smoke or not. In addition, laboratory tests are performed on hair samples to determine presence or absence of nicotine. The laboratory assessment is considered the gold standard, or truth about nicotine. The data are as follows: Self-Reported Non-Smoker Self-Reported Smoker Nicotine Absent 82 12 Nicotine Present 14 52 a. What is the sensitivity of self-reported smoking status? Sensitivity = P(Smoker | Nicotine Present) = 52/66 = 0.7878. b. What is the specificity of self-reported smoking status? Specificity = P(Non Smoker | Nicotine Absent) = 82/94 = 0.8723. 8. A recent study of cardiovascular risk factors reported that 30% of adults meet criteria for hypertension. If 15 adults are assessed, what is the probability that a. Exactly 5 meet the criteria for hypertension? 15! 0.3 5 (1−0.3)15−5 = 5 !(15− 5 )! P(X = 5) = 0.2061. b. None meet the criteria for hypertension 15! 0.30 (1−0.3)15−0 = 0!(15−0 )! P(X = 0) = 0.0047. c. How many would you expect to meet the criteria for hypertension? =np=15(0.3) = 4.5. 9. The following table displays blood pressure status by sex. Male Female Total Optimal 22 43 65 Normal 73 132 205 Hypertension 55 65 120 Total 150 240 390 a. What proportion of the participants have optimal blood pressure? P(Optimal) = 65/390 = 0.1667. b. What proportion of men have optimal blood pressure? P(Optimal | Male) = 22/150 = 0.1467. c. What proportion of participants with hypertension are male? P(Male | Hypertension) = 55/120 = 0.458. d. Are hypertensive status and male gender independent? Check P(Male | Hypertension) ?=? P(Male) P(Male | Hypertension) = 0.458 and P(Male) = 150/390 = 0.385 0.458 ≠ 0.385, not independent. 10. Diastolic blood pressures are assumed to follow a normal distribution with a mean of 85 and a standard deviation of 12. a. What proportion of people have diastolic blood pressure less than 90? 90 −85 P(X < 90) = P(Z < 12 ) = P(Z < 0.42) = 0.6628. b. What proportion have diastolic blood pressures between 80 and 90? 80− 85 90 −85 12 P(80 < X < 90) = P( < Z < 12 ) = P(-0.42 < Z < 0.42) = 0.6628 – 0.3372 = 0.3256. c. If someone has a diastolic blood pressure of 100, what percentile is he/she in? 100−85 P(X < 100) = P(Z < 12 ) = P(Z < 1.25) = 0.8944, 89.44th percentile. 11. Consider the data described in Problem 10. If a sample of 15 participants are sampled, what is the probability that their mean diastolic blood pressure exceeds 90? 90 −85 X̄ P( > 90) = P(Z > 12/ √ 15 ) = P(Z > 1.61) = 1-0.9463 = 0.0537. 12. Data from a large national study reports that 10% of pregnant women deliver prematurely. A local obstetrician is seeing 16 pregnant women in his next clinic session. a. What is the probability that none will deliver prematurely? 16! P(X = 0) = 0!(16−0)! 0.10 (1−0.1 )16−0= 0.1853. b. What is the probability that fewer than 3 will deliver prematurely? P(X < 3) = P(X=0) + P(X=1) + P(X=2) 16! P(X = 0) = 0!(16−0)! 0.10 (1−0.1 )16−0= 0.1853 16! 0.11 (1−0.1)16−1 = 1!(16−1)! P(X = 1) = 0.3294 16 ! 2 16− 2 0 .1 (1−0.1) = P(X = 2) = 2 !(16−2)! 0.2745 P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 0.1853 + 0.3294 + 0.2745 = 0.7892. c. What is the probability that none will deliver prematurely if in fact the true percentage who deliver prematurely is 5.5%? 16 ! P(X = 0) = 0!(16−0)! 0.0550 (1−0.055)16 −0 = 0.4045. d. If the true percentage is 10% and this obstetrician has 146 pregnant women under his care, how many would be expected to deliver prematurely? =np=146(0.1) = 14.6 13. The following table cross classifies women in the study by their body mass index (BMI) at 16 weeks gestation and whether they had preterm delivery. Preterm Full Term BMI < 30 320 4700 BMI 30-34.9 80 480 a. What is the probability that a woman delivers preterm? P(preterm) = 520/6000 = 0.0867. b. What is the probability a women has BMI <30 and delivers preterm? P(BMI < 30 and preterm) = 320/6000 = 0.0533. c. What proportion of women with BMI 35+ deliver preterm? P(preterm | BMI > 35) = 120/420 = 0.2857. d. Are BMI and preterm delivery independent? Justify. Check P(preterm | BMI > 35) ?=? P(preterm) P(preterm | BMI > 35) = 0.2857 and P(preterm) = 0.0867 BMI 35+ 120 300 0.2857 ≠ 0.0867, not independent. 14. In the study described in Problem 13, suppose BMI at 16 weeks gestation has a mean 28.5 with a standard deviation of 3.6 and BMI is assumed to follow a normal distribution. Find the following: a. The proportion of women with BMI> 30. P(X > 30) = P(Z > 30−28 . 5 3 . 6 ) = P(Z > 0.42) = 1-0.6628 = 0.3372. b. The proportion of women with BMI>40. P(X > 40) = P(Z > 40−28 .5 3.6 ) = P(Z > 3.19) = 0. c. The BMI that separates the top 10% from the rest. X =  + Z = 28.5 + 1.282(3.6) = 33.11. 15. Suppose we want to estimate the mean BMI for women in pregnancy at 20 weeks gestation. If we have a sample of 100 women and measure their BMI at 20 weeks gestation, what is the probability that the sample mean is within 1 unit of the true BMI if the standard deviation in BMI is taken to be 3.6? ( μ−1 )−μ P (-1 < X̄ < +1) = P( 3 . 6/ √ 100 1.25) = 1-0.8944 = 0.1056. P(X > 60) = P(Z > b. What proportion of healthy males have HDL lower than 40? 40 −50 8 ) = P(Z < -1.25) = 0.1056. P(X < 40) = P(Z < c. What is the 90th percentile of HDL in healthy males? X =  + Z = 50 + 1.282(8) = 60.3. 20. The following table summarizes data collected in a study to evaluate a new screening test for ovarian cancer. A total of 200 women were involved in the study – 50 had ovarian cancer and 150 did not. The results are tabulated below. Ovarian Cancer Screening Test Positive Negative 28 22 50 Free of Ovarian Cancer 23 127 150 a. Find the sensitivity of the screening test. Sensitivity = P(Test Positive | Ovarian Cancer) = 28/50 = 0.56. b. Find the false positive fraction of the screening test. False Positive = P(Test Positive | Free of Ovarian Cancer) = 23/150 = 0.15. c. What proportion of women who screen positive actually have ovarian cancer? P(Ovarian Cancer | Test Positive) = 28/51 = 0.55. 21. An experimental drug has been shown to be 75% effective in eliminating symptoms of allergies in animal studies. A small human study involving 6 participants is conducted. What is the probability that the drug is effective on more than half of the participants? P(X > 3) = P(4) + P(5) + P(6) = 0.2966 + 0.3560 + 0.1780 = 0.8306. 6! 0 .75 4 (1−0.75)6−4 = 4 !(6−4)! P(X = 4) = (15)(0.3164)(0.0625) = 0.2966. 6! 5 6−5 0.75 (1−0.75) = P(X = 5) = 5 !(6−5 )! (6)(0.2373)(0.25) = 0.3560. 6! 6 6−6 0 .75 (1−0.75) = P(X = 6) = 6 !(6−6)! (1)(0.1780)(1) = 0.1780.