Basic Biostatistics Statistics For Public Health Practice 2nd Edition B.Burt Gerstman - Solutions

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Given the binomial parameters and show by means of the binomial expansion given in Table 4.3.1 that g f 1x2 = 1.
Singh et al. (A-8) looked at the occurrence of retinal capillary hemangioma (RCH) in patients with von Hippel–Lindau (VHL) disease. RCH is a benign vascular tumor of the retina. Using a retrospective consecutive case series review, the researchers found that the number of RCH tumor incidents
Tubert-Bitter et al. (A-9) found that the number of serious gastrointestinal reactions reported to the British Committee on Safety of Medicine was 538 for 9,160,000 prescriptions of the anti-inflammatory drug piroxicam. This corresponds to a rate of .058 gastrointestinal reactions per 1000
If the mean number of serious accidents per year in a large factory (where the number of employees remains constant) is five, find the probability that in the current year there will be:(a) Exactly seven accidents (b) Ten or more accidents(c) No accidents (d) Fewer than five accidents
In a study of the effectiveness of an insecticide against a certain insect, a large area of land was sprayed. Later the area was examined for live insects by randomly selecting squares and counting the number of live insects per square. Past experience has shown the average number of live insects
In a certain population an average of 13 new cases of esophageal cancer are diagnosed each year.If the annual incidence of esophageal cancer follows a Poisson distribution, find the probability that in a given year the number of newly diagnosed cases of esophageal cancer will be:(a) Exactly 10 (b)
For another subject (a 29-year-old male) in the study by Diskin et al. (A-11), acetone levels were normally distributed with a mean of 870 and a standard deviation of 211 ppb. Find the probability that on a given day the subject’s acetone level is:(a) Between 600 and 1000 ppb(b) Over 900 ppb(c)
In the study of fingerprints, an important quantitative characteristic is the total ridge count for the 10 fingers of an individual. Suppose that the total ridge counts of individuals in a certain population are approximately normally distributed with a mean of 140 and a standard deviation of
One of the variables collected in the North Carolina Birth Registry data (A-3) is pounds gained during pregnancy. According to data from the entire registry for 2001, the number of pounds gained during pregnancy was approximately normally distributed with a mean of 30.23 pounds and a standard
Suppose the average length of stay in a chronic disease hospital of a certain type of patient is 60 days with a standard deviation of 15. If it is reasonable to assume an approximately normal distribution of lengths of stay, find the probability that a randomly selected patient from this group will
Given a normally distributed population with a mean of 75 and a variance of 625, find: (a) P(50x100) (b) P(x >90) (c) P(x < 60) (e) P(30x110) (d) P(x = 85)
The weights of a certain population of young adult females are approximately normally distributed with a mean of 132 pounds and a standard deviation of 15. Find the probability that a subject selected at random from this population will weigh:(a) More than 155 pounds (b) 100 pounds or less(c)
The National Health and Nutrition Examination Survey of 1988–1994 (NHANES III, A-1) estimated the mean serum cholesterol level for U.S. females aged 20–74 years to be 204 mg/dl. The estimate of the standard deviation was approximately 44. Using these estimates as the mean and standard deviation
The study cited in Exercise 5.3.1 reported an estimated mean serum cholesterol level of 183 for women aged 20–29 years. The estimated standard deviation was approximately 37. Use these estimates as the mean and standard deviation for the U.S. population. If a simple random sample of size 60 is
If the uric acid values in normal adult males are approximately normally distributed with a mean and standard deviation of 5.7 and 1 mg percent, respectively, find the probability that a sample of size 9 will yield a mean:(a) Greater than 6 (b) Between 5 and 6(c) Less than 5.2
Wright et al. (A-2) used the 1999–2000 National Health and Nutrition Examination Survey (NHANES)to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium (mg). They found in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of
In the study cited in Exercise 5.3.4, researchers found the mean sodium intake in men and women 60 years or older to be 2940 mg with a standard deviation of 1476 mg. Use these values for the mean and standard deviation of the U.S. population and find the probability that a random sample of 75
Given a normally distributed population with a mean of 100 and a standard deviation of 20, find the following probabilities based on a sample of size 16: (a) P(x100) (c) P(96 108) (b) P(X 110)
Given m = 50, s = 16 and n = 64 find: (a) P(4555) (c) P(x < 47) (b) P(x > 53) (d) P(4956)
Suppose a population consists of the following values: 1, 3, 5, 7, 9. Construct the sampling distribution of based on samples of size 2 selected without replacement. Find the mean and variance of the sampling distribution.
Use the data of Example 5.3.1 to construct the sampling distribution of based on samples of size 3 selected without replacement. Find the mean and variance of the sampling distribution.
Use the data cited in Exercise 5.3.1. Imagine we take samples of size 5, 25, 50, 100, and 500 from the women in this age group.(a) Calculate the standard error for each of these sampling scenarios.(b) Discuss how sample size affects the standard error estimates calculated in part (a) and the
In the study cited in Exercises 5.3.4 and 5.3.5, the calcium levels in men and women ages 60 years or older are summarized in the following table:Use these estimates as the mean and standard deviation for the U.S. populations for these age groups. If we take a random sample of 40 men and 35 women,
Given two normally distributed populations with equal means and variances of s21= 100 andwhat is the probability that samples of size n1 = 25 n 2 = 16 will yield a value of x1 - x 2 greater than or equal to 8? 80,
Given two normally distributed populations with equal means and variances of and what is the probability that samples of size and will yield a value of as large as or larger than 12?
For a population of 17-year-old boys and 17-year-old girls, the means and standard deviations, respectively, of their subscapular skinfold thickness values are as follows: boys, 9.7 and 6.0; girls, 15.6 and 9.5. Simple random samples of 40 boys and 35 girls are selected from the populations.What is
Smith et al. (A-5) performed a retrospective analysis of data on 782 eligible patients admitted with myocardial infarction to a 46-bed cardiac service facility. Of these patients, 248 (32 percent)reported a past myocardial infarction. Use .32 as the population proportion. Suppose 50 subjects are
In the study cited in Exercise 5.5.1, 13 percent of the patients in the study reported previous episodes of stroke or transient ischemic attack. Use 13 percent as the estimate of the prevalence of stroke or transient ischemic attack within the population. If 70 subjects are chosen at random from
In the same 1999–2000 NHANES (A-3) report cited in Example 5.5.1, researchers estimated that 64 percent of U.S. adults ages 20–74 were overweight or obese (overweight: BMI 25-29, obese:BMI 30 or greater). Use this estimate as the population proportion for U.S. adults ages 20–74. If 125
Gallagher et al. (A-6) reported on a study to identify factors that influence women’s attendance at cardiac rehabilitation programs. They found that by 12 weeks post-discharge, only 64 percent of eligible women attended such programs. Using 64 percent as an estimate of the attendance percentage
Given a population in which and a random sample from this population of size 100, find: (a) P(p = .65) (c) P(.56.63) (b) P(p .58)
It is known that 35 percent of the members of a certain population suffer from one or more chronic diseases. What is the probability that in a sample of 200 subjects drawn at random from this population 80 or more will have at least one chronic disease?
According to the 2000 U.S. Census Bureau (A-8), in 2000, 9.5 percent of children in the state of Ohio were not covered by private or government health insurance. In the neighboring state of Pennsylvania, 4.9 percent of children were not covered by health insurance. Assume that these proportions are
In the report cited in Exercise 5.6.1 (A-8), the Census Bureau stated that for Americans in the age group 18–24 years, 64.8 percent had private health insurance. In the age group 25–34 years, the percentage was 72.1. Assume that these percentages are the population parameters in those age
From the results of a survey conducted by the U.S. Bureau of Labor Statistics (A-9), it was estimated that 21 percent of workers employed in the Northeast participated in health care benefits programs that included vision care. The percentage in the South was 13 percent. Assume these percentages
Herniated lumbar discs. Data from a case–control study on participation in sports and herniated lumbar discs stratified by smoking status are shown in the following tables. Quantify the effect of “no sports participation” on the risk of disc herniation.Stratum 1: Smokers Cases Controls No
Oral cancer. Data from a case–control study on oral cancer and smoking stratified by alcohol consumption are shown in the following tables. Quantify the effects of smoking on oral cancer risk while controlling for smoking. Briefly, summarize your findings.
Infant survival. Data on the survival of 715 infants in relation to the amount of medical care received are as follows:Crude data (clinics 1 and 2 combined)Did not survive Survived Less care 20 373 More care 6 316 Data from Whittaker, J. (1990). Graphical Models in Applied Multivariate Statistics.
presents a hypothetical example comparing survival rates in victims by evacuation method.TABLE 19.11 Data for Exercise 19.6. Survival following motor vehicle accidents by evacuation method and seriousness of accident. Data are hypothetical.Data from Oppe, S., & De Charro, F. T. (2001). The effect
Helicopter evacuation and survival following trauma. Accident victims may be transported to the hospital by helicopter or, more typically, by road ambulance. Does the use of helicopter actually save lives? Table
Data for Exercise 19.5. Acceptance to UC Berkeley Graduate programs by gender and major.Data from Bickel, P. J., Hammel, E. A., & O’Connell, W. (1975). Sex bias in graduate admission: Data from Berkeley. Science, 187, 398–404. Data reported on p. 17 of Freedman, D. A., Pisani, R., Purves, R., &
Why should we not use a Mantel–Haenszel summary measure of effect when interaction is present?Exercises 19.5 Sex bias in graduate school admissions? Table 19.3 contains data on the acceptance of applications to graduate programs at the University of California Berkeley for 1973. Assuming men
Provide the alternative hypothesis when testing for an interaction in the odds ratios from three strata.
Provide the null hypothesis when testing for an interaction in odds ratios from three strata.
Select the best response: Which of the following is associated with a statistical test for interaction?(a) H0: RR1= RR2(b) H0: RR = 1(c) H0: RR = 0
Select the best response: A synonym for “statistical interaction” is(a) confounding.(b) bias.(c) effect measure modification.
Select the best response: Mantel–Haenszel summary measures of effect are weighted averages of _________________ measures of effect.(a) crude(b) strata-specific(c) adjusted
Select the best response: Which of the following is addressed by the Mantel–Haenszel test statistic for risk ratios?(a) H0: RR1= RR2(b) H0: RR = 1(c) H0: RR = 0
What is a Mantel–Haenszel summary measure of effect?
What is a strata-specific measure of effect?
What is a crude measure of effect?
What is a measure of effect?
What is Simpson’s paradox?
List five statistical methods that are used to mitigate confounding.
List the three properties of a confounder.
Select the best response: The Latin term ceteris paribus means(a) to mix together.(b) with other things equal.(c) bias.
Select the best response: The Latin term confundere means(a) to mix together.(b) with all things equal.(c) bias.
Select the best response: Confounding is(a) any error in inference.(b) any systematic error in inference.(c) any systematic error in inference due to the influence of extraneous variables.
Smoking and cervical cancer. Table 19.9 displays data from a case–control study on smoking and invasive cervical cancer with data stratified by number of sexual partners.(a) Calculate the odds ratios within each stratum. Comment on your findings.(b) Test the odds ratios for interaction. Report
Is participating in a follow-up survey associated with having medical aid? Exercise 19.1 considered whether participation in a follow-up survey was associated with having medical aid. Race was a confounding variable in this analysis. Table 19.5 lists both crude and race-specific cross
Is participating in a follow-up survey associated with having medical aid? Use the strata-specific data in Table 19.5B and C to calculate the Mantel–Haenszel summary prevalence proportion ratio(“relative risk”) for the relationship between follow up while adjusting for race. Include a
Is participating in a follow-up survey associated with having medical aid? A South African study compared whether participating in a 5-year follow-up interview was associated with having medical aid (a type of health insurance). Based on the setup of the problem in the original source, whether an
Hemorrhagic stroke in young women. Exercise 18.21 addressed data from a matched-pair case–control study on oral contraceptive use and thrombotic stroke in young women. The investigators also collected information on hemorrhagic stroke. Table 18.34 lists data for this outcome.(a) Calculate the
Practice with chi-square. Calculate a chi-square test of association statistic and P-value for the case–control data in Table 18.21. Justify that it is acceptable to use a chi-square statistic in this situation even though two of the cells have small numbers.
Wynder and Graham’s case–control study of smoking and lung cancer. A historically important case–control study of smoking and lung cancer compared smoking histories of 605 lung cancer cases to 780 noncancer controls. Table 18.33 displays data from this study.(a) Calculate the odds ratio
Baldness and myocardial infarction, interviewer assessments.Exercises 18.17 and 18.18 considered a case–control study of baldness and heart attacks in men using self-assessments of baldness.This same study also had the interviewers assess the extent of baldness of study subjects using a Hamilton
Brain tumors and electric blanket use. A case–control study assessed risks of brain tumors in children. Table 18.31 cross tabulates the data according to case status and electric blanket exposure.Calculate the odds ratio and its 95% confidence interval. Comment on the results.TABLE 18.31 Data for
Esophageal cancer and tobacco use. The case–control study on esophageal cancer and alcohol use that we used to illustrate case–control methods in this chapter also collected data on tobacco use as a potential risk factor. Table 18.30 shows data from the study with tobacco use dichotomized at 20
IUDs and infertility. A case–control study found a history of intrauterine device use in 89 of 283 infertile women. In contrast, 640 of the 3833 controls had used IUDs.ee Calculate the odds ratio and its 95% confidence interval for IUD use and infertility. (Note: Data must first be put into a
Yates, 1934 (three-by-two). Table 18.29 displays results from a historically important methodological paper written by Frank Yates on the analysis of cross-tabulated counts.dd Explain why a chi-square statistic should not be used to test these data. Then use WinPepi >Compare2.exe > Program F1 to
Data for Exercise 18.23. 18.24 University Group Diabetes Program. A multicentered clinical trial completed in the 1960s assessed the efficacy of various oral hypoglycemics, insulin, and diet in preventing vascular complications of diabetes.cc It found that 26 of 204 patients (13%)treated with the
Don’t sweat the small stuff. Consider a study in which 40 of 320 individuals (12.5%) in the treatment group experience an outcome.In contrast, 26 of 336 (7.7%) of the individuals in the control group experience the outcome. Table 18.28 displays the data in crosstabulated form. Calculate
Binge drinking by gender. Wechsler and coworkers (1994) found that 19.4% of students at 4-year U.S. colleges engaged in frequent binge drinking. Table 18.27 reports the data from this study by gender.TABLE 18.27 Data for Exercise 18.22. Frequency of binge drinking by gender. Binge drinking is
Select the best response: These types of samples match individuals to one other based on factors thought to have the potential to confound the relationship being studied.(a) independent samples(b) paired samples(c) case–control samples
Select the best response: When the outcome is rare (incidence less than 5%), the odds ratio has the same interpretation as the(a) risk(b) risk ratio(c) risk difference
Select the best response: This measure of association is used in case–control studies.(a) risk(b) risk ratio(c) odds ratio
State the null hypothesis for the Mantel test for trend in words.
State the null hypothesis for the chi-square tests of association in words.
Fill in the blank: The expected value in table cell is equal to the (row total) × (column total) divided by __________.
Select the best response: Expected values in R-by-C table cells assume(a) null hypothesis is true.(b) null hypothesis is false.(c) alternative hypothesis is true.
True or false? Say whether each of these statements about chi-square probability density functions is true or false.(a) have a mean equal to 0(b) have a mean equal to their degrees of freedom(c) have a mode of 0(d) have a mode of 0 when df = 1 or 2(e) have a variance of 1(f) have a variance of two
Select the best response: Chi-square distributions __________ 0.(a) are centered on(b) start on(c) end on
Fill in the blanks: Chi-square probability density functions are ______________ (symmetrical or asymmetrical) continuous distributions that become increasingly ____________ (symmetrical or asymmetrical) as their degrees of freedom increase.
A chi-square statistic for testing for an association between the row variable and the column variable in an R-by-C table has this many degrees of freedom.
This is the most common statistic used to test for associations among categorical variables.
This is the ratio of two odds.
This is the ratio of “successes” to “failures.”
Select the best response: A risk ratio of 0.56 indicates ___________ association between the study exposure and the outcome.(a) a positive(b) no(c) a negative
Select the best response: A risk difference of 0.56 per 1000 indicates ____________ association between the exposure and the outcome.(a) a positive(b) no(c) a negative
This statistic represents the incidence proportion in the exposed group divided by the incidence proportion in the nonexposed group.
This statistic represents the incidence proportion in the exposed group minus the incidence proportion in the nonexposed group.
Select the best response: This statistic is an estimate of the risk of an outcome.(a) prevalence(b) incidence proportion(c) risk ratio
This symbol represents the sample proportion in group i.
How do you calculate a conditional column percent?
How do you calculate a conditional row percent?
What is a conditional distribution?
How do you calculate a marginal percent?
What is a marginal distribution?
What does the C stand for in an R-by-C table?
What does the R stand for in an R-by-C table?

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Basic Biostatistics Statistics For Public Health Practice 2nd Edition B.Burt Gerstman - Solutions

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