Essentials of Biostatistics in Public Health CH05 answer key PDF

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD 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 ) = 0.0712 P(X = 1) = 1!(12−1)! 12! 0. 32 (1−0 .3 )12− 2= 2!(12−2)! P(X = 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! 0 .3 3 (1−0. 3 )12−3 = 0.2397. P(X=3) = 3 !(12− 3 )! 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.

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD 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?

180−191 190−191 22.4 < Z < 22.4 ) = P(-0.49 < Z < -0.04) P(180 < X < 190) = P( = 0.4840 – 0.3121 = 0.1719. 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( X¯ > 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

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Drink Caffeinated Coffee 365 430 2

Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD 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?

2−3 .2 0 . 8 ) = P(Z > -1.5) = 1-0.0668 = 0.9332. P(X > 2) = P(Z > b. What proportion drink no more than 4 cups per day?

4−3.2 0. 8 ) = P(Z < 1) = 0.8413. P(X < 4) = P(Z < 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 P( X¯ > 3) = P(Z > 0. 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: © 2012 Jones & Bartlett Learning, LLC

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD a. Exactly half of them consume caffeine and smokes?

8! 0 . 4 4 (1−0 . 4 )8 − 4 = 4!(8−4)! 0.2322. P(X = 4) = 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 = 0.0079 P(X=7) = 7 !(8−7 )! 8! 0. 48 (1−0 . 4 )8−8= 8 !(8− 8 )! P(X=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 = 0.2322. P(X = 4) = 4!(8−4 )! 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.

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD 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 = P(X = 5) = 5 !(15−5 )! 0.2061. b. None meet the criteria for hypertension

15 ! 0 . 30 (1−0 . 3 )15−0 = P(X = 0) = 0!(15−0)! 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 Femal e Total

Optimal 22 43

Normal 73 132

Hypertension 55 65

Total 150 240

65

205

120

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. © 2012 Jones & Bartlett Learning, LLC

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD

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(Z < 0.42) = 0.6628. P(X < 90) = P(Z < 12 b. What proportion have diastolic blood pressures between 80 and 90?

80− 85 90 −85 P(80 < X < 90) = P( 12 < 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 P( X¯ > 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 ! 0. 10 (1−0 .1 )16− 0= P(X = 0) = 0 !(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)

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD

16 ! 0. 10 (1−0 .1 )16− 0= 0.1853 P(X = 0) = 0 !(16− 0 )! 16! 1 16−1 0.1 (1−0.1 ) = P(X = 1) = 1!(16−1)! 0.3294 16 ! 0 .12 (1−0 . 1)16− 2= 0.2745 P(X = 2) = 2 !(16−2)! 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 ! 0. 0550 (1−0 . 055 )16 −0 = 0 !(16− 0 )! 0.4045. P(X = 0) = 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

BMI 35+ 120 300

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 35) = 120/420 = 0.2857. d. Are BMI and preterm delivery independent? Justify. Check P(preterm | BMI > 35) ?=? P(preterm) © 2012 Jones & Bartlett Learning, LLC

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD P(preterm | BMI > 35) = 0.2857 and P(preterm) = 0.0867 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?

P (-1 < X¯

( μ−1 )−μ ( μ+1)−μ 60) = P(Z >

60−50 8 ) = P(Z > 1.25) = 1-0.8944 = 0.1056.

b. What proportion of healthy males have HDL lower than 40?

P(X < 40) = P(Z <

40 −50 8 ) = P(Z < -1.25) = 0.1056.

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.

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Chapter 5 Answer Key Essentials of Biostatistics for Public Health, 2nd Edition Lisa M. Sullivan, PhD 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 = P(X = 4) = 4 !(6−4)! (15)(0.3164)(0.0625) = 0.2966. 6! 0 . 755 (1−0 .75 )6−5 = (6)(0.2373)(0.25) = 0.3560. P(X = 5) = 5 !(6−5 )! 6! 0 .756 (1−0 . 75)6−6 = P(X = 6) = 6 !(6 − 6 )! (1)(0.1780)(1) = 0.1780.

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