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biostatistics

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

Select the best response: This type of sample selects a fixed numbers of persons who have experienced the study outcome and fixed number of individuals who have not yet experienced the study outcome from the population.(a) naturalistic sample(b) purposive cohort(c) case–control
Select the best response: This type of sample selects a fixed number of exposed and nonexposed individuals from the population.(a) naturalistic sample(b) purposive cohort(c) case–control
Select the best response: This type of sample selects a random cross section of the population.(a) naturalistic sample(b) purposive cohort(c) case–control
Thrombotic stroke in young women. A 1970s case–control study on cerebrovascular disease (stroke) and oral contraceptive use in young women matched cases to controls according to neighborhood, age, sex, and race.bb Table 18.26 displays data from this study for thrombotic stroke.(a) Calculate the
Estrogen and endometrial cancer. Table 18.25 contains data from a matched case-control study on conjugated estrogen use and endometrial cancer.TABLE 18.25 Estrogen use and endometrial cancer, case–control, matched pairs.Data from Antunes, C. M., Strolley, P. D., Rosenshein, N. B., Davies, J. L.,
Diet and adenomatous polyps. Recall the illustrative example concerning a matched case–control study of adenomatous colon polyps and diet. There were 45 matched pairs in which the case but not the control reported low fruit and vegetable consumption. There were 24 pairs in which the control but
Unadjusted and multivariate adjusted odds ratios for baldness and myocardial infarction study, Exercise 18.18.Baldness level Unadjusted OR Multivariate adjusted 1 1.0 (reference) 1.0(reference)2 1.0 0.8 3 1.4 1.1 4 1.9 2.0 52.6 (very small sample)Could not estimate
Baldness and myocardial infarction, lurking variables. This exercise is a continuation of Exercise 18.17. The investigators found differences in cases and controls besides those have to due to baldness. For example, the median age of cases was 47 years, while the median age of controls was 43
Baldness and myocardial infarction, self-assessed baldness. A case–control study was completed to learn about the relationship between male-pattern baldness and the risk of cardiovascular disease.Cases were men less than 55 years of age hospitalized for a heart attack. Cases were excluded if they
Vasectomy and prostate cancer. Table 18.20 displays data from a case–control study on vasectomy and prostate cancer. Calculate the odds ratio; include a 95% confidence interval for the OR. (Optional:Calculate a P-value to test the association.)
Doll and Hill, 1950. An important historical case–control completed by Doll and Hill found that 647 of the 649 lung cancer cases were smokers. In contrast, 622 of 649 controls smoked.u Display these data as a 2-by-2 table and calculate the odds ratio associated with smoking. Include a 95%
Cell phones and intracranial tumors, study 2. A (different) case–control study on intracranial tumors and cell phone use completed between 1994 and 1998 used a structured questionnaire to explore the relationship between cell phone use and primary brain cancer in 469 cases and 422 controls.
Cell phones and brain tumors, study 1. A case–control study examined the relationship between cellular telephone use and intracranial tumors. The study (completed between 1994 and 1998)included 782 cases with various types of intracranial tumors and 799 controls with a variety of nonmalignant
Do seatbelt laws prevent serious injury? Table 18.14 lists frequencies for severity of motor vehicle accident injuries before and after enactment of a seatbelt law. Conditional (column) percents are included in the table.(a) Comment on the conditional distributions of injuries before and after
Anger and heart disease. Exercise 18.4 found coronary heart disease risks of 1.0%, 1.3%, and 2.8% in low, intermediate, and high angertrait groups. Table 18.9 lists observed frequencies for the problem.Apply a test of trend to these data. Show all hypothesis-testing steps.
Anger and heart disease. Exercise 18.4 found a higher risk of coronary disease in individuals with higher anger levels. Table 18.9 contains the data. Test the association for statistical significance.Report the chi-square statistic, its df, and P-value.
contains observed counts.(a) Use a chi-square statistic to test the association for statistical significance.(b) Repeat the test with a z-statistic (Section 17.3). Show the relation between the chi-square statistic and the z-statistic.
Cytomegalovirus infection and coronary restenosis. Exercise 18.3 found that 43% of CMV+ patients experienced arterial restenosis within 6 months of atherectomy. In contrast, 8% of CMV− patients experienced a similar outcome. Table
Drove when drinking alcohol. Table 18.7 presented cross-tabulated results for the number of adolescents who drove or rode in a car or other vehicle when the driver was drinking alcohol. Prevalence proportions varied from 11.3% (non-Hispanic white) to 4.9% (nonHispanic black). Test the data in Table
Chi-square approximation. Use Table E to find the approximate area under the curve to the right of 5.22 on a χ2 distribution with 3 degrees of freedom. (Suggestion: Draw the chi-square curve showing 5.22 on the horizontal axis. Shade the area to the right of 5.22.Bracket the shaded region with
Chi-square landmarks. Use Table E to find the area under the χ2 curve with 3 degrees of freedom to the right of:(a) 4.64(b) 6.25(c) 11.34
Response to leprosy treatment. In 1954, the prominent statistician W. G. Cochran (1909–1980) published an important article on the analysis of cross-tabulated counts. In this article, techniques were illustrated with an example from a study on the treatment of leprosy.Table 18.10 lists the
Anger and heart disease. A nonexperimental study looked at whether people who angered easily were more likely to develop coronary heart disease than those who angered less easily.Spielberger anger scale scores were classified into low, moderate, and high categories. Individuals were followed for up
Cytomegalovirus infection and coronary restenosis. Table 18.8 contains results from the cytomegalovirus (CMV) and restenosis study considered as an illustrative example of naturalistic samples in Section 18.1. The occurrence of coronary restenosis within 6 months of surgery is cross tabulated
YRBS, relative risks. Use the results from Exercise 18.1 to express the relation between driving while drinking and ethnic/racial group in the form of relative risks. Let group 1 (non-Hispanic white) serve as the reference category. Interpret the results.
YRBS, prevalence proportions. Table 18.7 is based on results from the Youth Risk Behavior Surveillance (YRBS) system. The number of adolescents who had been in a car or other vehicle when the driver had been drinking alcohol is cross tabulated according to racial/ethnic group classifications.
Drug testing student athletes. The Supreme Court of the United States ruled in 2002 that schools could require random drug testing of students who participate in after-school activities. At that time, it was not known whether random drug testing reduced use of illicit drugs. To address this
Acute otitis media. A double-blind randomized trial compared cefaclor to amoxicillin in the treatment of acute otitis media in children. There was no difference in clinical failure rates (all but four children in each group had a good clinical response), but after 14 days, 59 (55.7%) of 106
4S overall mortality. The 4S study introduced in Exercise 17.17 also tallied deaths due to any cause. The treatment group experienced 182 deaths (n1 = 2221) and the control group experienced 256 deaths (n2= 2223). Compare the mortality experience in the two groups in the form of a relative risk and
4S coronary mortality. The Scandinavian simvastatin survival study(4S) was a randomized clinical trial designed to evaluate the effects of the cholesterol-lowering agent simvastatin in patients with coronary heart disease. Over 5.4 years of follow-up, the treatment group consisting of 2221
Framingham women. Exercise 17.9 considered 6-year coronary disease risk in Framingham men. Comparable data for Framingham women revealed 30 coronary incidents in 689 women with high serum cholesterol. Among 445 women with low serum cholesterol, there were 8 incidents. Calculate the risk difference
Telephone survey completion rates. The study described in Exercise 17.14 also kept tabs of whether leaving advanced messages increased the likelihood of ultimately completing the interview. Data for this outcome are tallied in Table 17.8. Perform analyses similar to those requested in the prior
Telephone survey contact rates. Telephone surveys typically have high rates of nonresponse. This can cause bias when the variables being studied are associated with factors that determine the response rate. For mail and home surveys, it is known that advanced-warning letters letting participants
Nondifferential misclassification tends to bias ratio measures of association in what direction?Exercises 17.13 Smoking cessation trial. Exercise 17.4 considered a randomized trial of smoking cessation in which treatment with bupropion in combination with a nicotine patch was compared to use of a
Change one word in this false statement so that it becomes true:Confidence intervals protect against all errors with (1 − α)100%confidence.
List three types of systematic errors in comparative research.
“ln” represents what mathematical function?
Which RR indicates the stronger association: 0.25 or 2?
What is the value of the RR when the two groups being compared have the same risk of the outcome?
Name two exact methods that can be used for testing proportions when the samples are small.
What benchmark is used to determine whether a Normalapproximation (z) test is appropriate for testing two proportions?
What does it mean when we say a test statistic is conservative?
Which z-statistic for testing the equality of proportions provides the more conservative result, the regular z-statistic or the continuitycorrected z-statistic?
Select the best response: The function of the P-values is to protect us from making too many(a) systematic errors(b) type I errors(c) type II errors
What is wrong with the following statement? A 95% confidence interval for the risk difference has a 95% chance of capturing .
What is wrong with the statement ?
State the null hypothesis for testing the equality of proportions in three different unique but equivalent ways.
In large samples, the sampling distribution of the ______(mathematical function) of the risk ratio is Normally distributed with a mean of ln(RR) and standard deviation of .
Fill in the blank: In large samples, the sampling distribution of a risk difference is ______ distributed with a mean of p1− p2 and standard deviation of .
What term refers to the ratio of two incidence proportions?
What term refers to the difference between two incidence proportions?
Select the best response: The term ______ is often used to refer to the explanatory variable in studies of risk.(a) outcome(b) exposure(c) dependent variable
Select the best response: This statistic is a direct estimate of risk.(a) prevalence(b) incidence proportion(c) risk ratio
This symbol denotes the population proportion in group i.
This symbol is used to denote the sample proportion in group i.
This is the symbol the text uses to denote the observed number of successes in group i.
Sample-size plan. Suppose it is twice as easy to recruit nonexposed subjects in the study considered in Exercise 17.11. How many individuals will we have to study with an allocation ratio of 2 to 1?
Sample-size plan. It is important to develop a detailed plan of study early in the development of an investigation in order to assure adequacy of the sample size. Consider a test of H0 : p1 = p2 against Ha :p1 ≠ p2 at α = 0.05 with a power that is equal to 0.80. We predict that 20% of group 1
Determinants of sample-size requirements. List the factors that determine the sample-size requirements for a test of two proportions.
Framingham Heart Study. An early publication from the Framingham Heart Study identified 51 incidences of coronary disease among 424 men with high serum cholesterol (245 mg per 100 mL and above) over a 6-year period. Among 454 men with lower cholesterol (less than 210 mg per 100 ml), there were 16
Induction of labor and meconium staining, risk difference estimate. Calculate a 95% confidence for the risk difference (induced minus noninduced) for the data in Exercise 17.7.
Induction of labor and meconium staining. Meconium staining during childbirth may be a sign of fetal distress. A randomized trial was conducted to see whether induction of labor (by pitocin and other drugs delivered near term) would reduce the risk of meconium staining. The treatment group (n1 =
Médecine d’observation. Pierre-Charles Alexandre Louis (1787–1872) is often referred to as the “father of clinical statistics.” In 1837, he wrote “I conceive that without the aid of statistics nothing like real medical science is possible.”n In his most famous study, Louis evaluated
Joseph Lister and antiseptic surgery. Joseph Lister (1827–1912)demonstrated that postoperative mortality dropped from 16 fatalities in 35 procedures to 6 fatalities in 40 procedures after adopting antiseptic surgical techniques. Determine the risks of postoperative mortality in each group. Then,
Smoking cessation trial. A randomized controlled, double-blind trial was conducted to see whether sustained-release bupropion (a pharmaceutical normally used to treat depression) provided benefit over use of the nicotine patch alone in helping people stop smoking.The control group (nicotine patch
Cytomegalovirus and coronary restenosis. Each year, cardiologists surgically repair thousands of clogged coronary arteries only to have many of these arteries narrow again (restenose) soon after surgery. A study sponsored by the NIH was conducted to help determine whether infection with a common
Hypothetical situation. Consider a fictitious study based in two populations. Twenty-five percent of the individuals in both populations have a particular risk factor (p1 = p2 = 0.25). We randomly selected 3750 subjects from each population (n1 = n2 =3750).(a) These are large samples, so the
Prevalence of cigarette use in two ethnic groups. Among U.S.adults from 1999 to 2000, American Indians/Alaska Natives had the greatest prevalence of smoking (p1 = 0.40); Chinese-Americans had the lowest prevalence (p2 = 0.12).b Suppose we take an SRS of 1000 individuals from each of these groups
Freshman binge drinking. A survey of drinking in college students found 1802 binge drinkers among 5266 U.S. freshman completing a survey.x(a) The prevalence of binge drinking in this sample is or 34.2%. Assume that the data represent an SRS of U.S. freshman. Also assume that the responders provided
Power. Determine the power of a test of H0: p = 0.10 when:(a) p is actually 0.20, α = 0.05 (two-sided), and n = 25.(b) p is actually 0.15, α = 0.01 (two-sided), and n = 100.
Familial history of breast cancer, sample size requirements.Suppose we want to test the hypothesis that women with a family history of breast cancer are at a higher risk of developing breast cancer than women who do not have this family history. Let us assume that 3% (0.03) of women who do not have
Sample size requirements. How large a sample is needed to estimate the incidence of female breast cancer in a population with 95%confidence and a margin of error that is no greater than 1% (0.01)?Assume that the expected incidence proportion in the population is 3% (0.03). How large a sample would
Incidence of improvement. Of 75 patients in a clinical study, 20 showed spontaneous improvement within a month. Calculate the 1-month incidence proportion of improvement. Include a 95%confidence interval for the proportion. Assume that the data represent an SRS of a defined clinical population.
Binge drinking in U.S. colleges. Alcohol abuse is a serious problem on college campuses. A nationwide survey of students at 4-year colleges found that 3314 of the 17,096 student respondents met the criterion for being a “frequent binge drinker” (five or more drinks in a row three or more times
Perinatal growth failure. Among the 33 perinatal growth failure cases discussed in the prior exercise, eight (24.2%) had very-low intelligence test scores when tested at 8 years of age. In normal birthweight babies, we’d expect 2.5% of the population to exhibit this trait. Perform a one-sided
Perinatal growth failure. Failure to grow normally during the first year of life (perinatal growth failure) was observed in 33 of 249 verylow birth-weight babies.v Calculate a 90% confidence interval for p based on this information. Assume data were derived by an SRS.
Alternative medicine. According to an April 29, 1998, New York Times article, a nationwide telephone survey conducted for a managed alternative care company found that of 1500 adults interviewed, 660 said they would use alternative medicine if traditional medical care failed to produce the desired
Sample-size requirement. As in the prior exercise, you are planning a study that intends to estimate a population proportion with 95%confidence. How many individuals do you need if you intend to cut your margin of error in half (i.e., m = 0.03)? Why is the sample size requirement for this exercise
Sample-size requirement. You are planning a study that wants to estimate a population proportion with 95% confidence. How many individuals do you need to study to achieve a margin of error of no greater than 6%? A reasonable estimate for the population proportions is not available before the study,
Leukemia gender preference. A simple random sample of 262 leukemia cases consisted of 150 males and 112 females. Does this provide evidence of a gender preference for the disease? (Observed proportion, male = 150/(150 + 112) = 0.5725. Test H0: p = 0.5.)
Kidney cancer survival. An oncologist treats 40 kidney cancer cases.Sixteen of the cases survive at least 5 years. Historically, one in five cases were expected to survive this long. Test whether there has been a significant improvement in survival.
AIDS-related risk factor. In a study of AIDS-related risk factors, 5 of 2673 heterosexual respondents reported a history of receiving a blood transfusion or having a sexual partner from a high-risk group.t Assume this is an SRS of U.S. heterosexuals. Provide a 95%confidence interval for the
Insulation workers. Twenty-six cancer deaths were observed in a cohort of 556 insulation workers. Based on national statistics, a cohort of this size and age distribution was expected to experience 14.4 incident cases during the observation period. Therefore, under the null hypothesis, p = p0 =
BRCA1 mutations in familial breast cancer cases. Of 169 women having breast cancer and a familial risk factor, 27 had an inherited BRCA1 mutation.s Based on this information, estimate the prevalence of BRCA1 mutation in women with familial breast cancer. Include a 95% confidence interval for the
Cerebral tumors and cell phone use. In a case-controlled study on cerebral tumors and cell phone use, tumors occurred more frequently on the same side of the head where cellular telephones had been used in 26 of 41 cases.r Test the hypothesis that there is an equal distribution of contralaterial
True or false? P-values address random sampling error.Exercises 16.12 Drove when drinking alcohol. The Youth Risk Behavior Surveillance survey for 2005 estimated that, within a 30-day period, 10% of the adolescent population had driven or ridden in a car or other vehicle when the driver had been
True or false? P-values address information bias.
True or false? Confidence intervals compensate for random sampling error.
True or false? Confidence intervals compensate for selection bias.
What is the distinction between selection bias and information bias?
What factors determine the sample size requirements for estimating population proportion p?
In determining the sample size requirements for estimating p, we specify p* as an educated guess for population proportion p. When no such educated guess is available, we let p* = 0.5. What is the justification for doing this?
Fill in the blanks: For Normal-based methods, the margin of error m is equal to approximately ______ the confidence interval width.
The plus-four confidence interval for p adds ______ (a number)imaginary successes to the numerator of the proportion estimate and ______ (a number) imaginary successes to the denominator to derive.
Select the best response: The plus-four method of calculating a confidence interval for p is based on ______ score method.(a) Student’s(b) Fisher’s(c) Wilson’s
Exact inferential procedures for counts and proportions are based on what probability mass function?
Where does the value of p0 come from when stating H0?
What does the symbol p0 represent in the expression H0: p = p0?
Fill in the blanks: The random number of successes in n independent Bernoulli trials has a binomial distribution with parameters n and p.When the sample is large, the random number of successes can also be described by a ______ distribution with μ = ______ and σ =______.
Fill in the blank: A proportion is an ______ of ones and zeros, where“one” represents presence or the condition and 0 represents its absence.

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