8.1 Sampling designs. Each of the studies below addresses a mean or mean difference. Identify the sampling design as either (1) single sample (units 5 and 6) paired samples (unit 7) or two independent samples (unit 8).
(A) Vaccination and autism. Compare vaccination
histories in 30 autistic boys and 30 non-autistic controls selected at random
from their respective groups.
(B) Husbands and wives.
Conduct health screening for chronic disease risk factors in the husband and
wives.
(C) Health screening. Apply health screening to a SRS of a
community and compare results
to National averages. (National averages based on NHANES for current year.)
8.2 Sampling designs 2. As in exercise 8.1 identify each of the following as either a single sample paired sample or independent samples.
(A) Needle stick injuries. Exposures to blood and other body fluids
presents an occupational hazard across a wide variety of health and safety
professions. Healthcare workers in particular
are at risk through needle-stick and sharp instrument injuries. The pathogens of primary concern are the human immunodeficiency
virus hepatitis B virus and hepatitis C virus.
When a needle stick injury occurs workers report the incident to their supervisor.
Hospitals collect this information and forward it to national authorities. A researcher
at the Centers for Disease Control use these data to compare needle stick accidents in
community hospitals and tertiary-care hospitals.
(B) Public health laboratory. To check a new method for testing
water a public health technician obtains a reference specimen of known
coliform concentration from the state lab. She takes 10 measurement of this specimen
using the a new technique and then compares the mean coliform level in her
sample to the known concentration.
(C) Public health laboratory 2. In a different study a public health lab
technician checks a new method of testing water to the established method. In this
evaluation no reference specimen is available so
comparisons are made on 10 different water specimens using each of the methods.
8.3 Facetious study. Imagine a study of six monkeys randomly assigned to receive either Monkey Tonic (GROUP = 1) or a course in Applied Monkey Training® (GROUP = 2). Monkey performance scores are recorded BEFORE and AFTER interventions. Data are:
|
Facetious
data |
|||
|
ID |
GROUP |
BEFORE |
AFTER |
|
1 |
1 |
100 |
104 |
|
2 |
1 |
88 |
93 |
|
3 |
1 |
106 |
109 |
|
4 |
2 |
116 |
117 |
|
5 |
2 |
102 |
104 |
|
6 |
2 |
106 |
106 |
(A) Compare mean scores in group 1 and group 2 at BASELINE. Are these
samples paired or independent?
(B) Determine the mean improvement in group 1.
Is this mean difference a paired or independent comparison?
(C) Determine the mean improvement in group 2. Are these samples paired or
independent?
(D) Compare the mean change in group 1 to the mean change group 2. Is this
comparison based on paired or independent samples?
8.4 Treatment of scrapie (Tagliavini 1997). Scrapie is a prion disease similar in pathology to bovine spongiform encephalopathy and new variant Creutzfeld-Jakob disease ("mad cow disease"). Scientists studied a substance called IDX in treating scrapie in hamsters Of 20 hamsters studied 10 chosen at random where injected with IDX and 10 were left untreated . All the hamsters eventually died of scrapie but the treated hamsters (n = 10) lived an average of 116 days (sem = 5.6) while the control hamsters (n = 10) lived an average of 85 days (sem = 1.9 days).
(A) What was the standard deviation of the survival
times in each of the groups? [Since se = s / 
n,
s = (se)(n)
by algebra.]
(B) Do the survival times differ significantly? Show all hypothesis testing
steps. [Note to advanced students: Although there is evidence of "unequal
variance" in the data, proceed with the equal variance t test.
Inferences from the two procedures are nearly identical.]
(C) Calculate a 95% confidence interval for the mean difference in survival.
8.5 Large t statistic. You calculate a tstat of 6.60. Assuming the study had more than just several observations, you do not need a t table to draw a conclusion about the hypothesis test. What is the conclusion? Why is a t table unnecessary?
8.6 Histidine and protein restriction. Total histidine excretion in 24-hour urine samples from men and women on protein-restricted diets are listed here and stored in histidine.sav.
Men: 172 204 229 236 256
Women: 115 135 138 174 197 224
(A) Create side-by-side boxplots of the data. What
did you learn from your plot?
(B) Test the means for a difference. Show all work.
8.7 Bone density in newborns. A study was conducted to determine whether maternal cigarette smoking affects bone density in newborns. A sample of 77 infants from mothers who smoke had a mean bone mineral content of 0.098 g/cm³ (s1 = 0.026 g/cm³). A sample of 161 children whose mothers had not smoked has mean bone mineral content of 0.095 g/cm³ (s2 = 0.025 g/cm³). Calculate a 95% confidence interval for µ1 - µ2. Show all calculations and interpret your results. [Sources: Venkataraman & Duke 1991; Pagano & Gauvreau 1993 p. 254 #3]
8.8 Oxygen uptake in joggers (source uncertain). Maximum oxygen uptake (V02max in ml/kg units) is a measure of the body's ability to use oxygen. The higher the V02max the better the individual is at processing oxygen; as V02max improves so does overall cardio-respiratory health. A random sample of 25 joggers had a mean V02max uptake of 47.5 (s1 = 4.8). Twenty-size (26) non-joggers show a mean of 37.5 ml/kg (s2 = 5.1). Calculate a 95% confidence interval for the mean difference µ1 - µ2. in the jogger and non-jogger populations. (
8.9 Cytomegalovirus and coronary stenosis.Coronary stenosis (narrowing of the arteries of the heart muscle) is the direct cause of heart disease. One theory suggests that chronic cytomegalovirus (CMV) infection causes narrowing of the coronary vessels. To test this theory, 75 patients undergoing angioplasty (a procedure used to enlarge narrowed arteries) were followed for six-months following their procedure. The 49 patients who were seropositive for CMV experienced an average luminal diameter reduction of 1.24 mm (s1 = 0.83 mm). In contrast, the 26 patients who were seronegative for CMV experienced an average luminal reduction of 0.68 (s2 = 0.69).Zhou et al. 1996 Test the differences is significance. Show all work. Interpret your finding.
8.10 Hemodialysis and anxiety. Severe anxiety often accompanies patients who must undergo chronic hemodialysis. To help counteract this anxiety a study was undertaken to determine the effects of a set of progressive relaxation exercises in hemodialysis patients. The 38 experimental subjects were shown a set of relaxation video tapes. A control group of 23 patients viewed a set of neutral videotapes. The State-Trait Anxiety Inventory (a psychiatric questionnaire that measures anxiety among other things) was administered to all study subjects. Post-test anxiety scores reported in the published article are shown below (Alarcon et al., 1982). Test whether the mean difference was significant. Show all work. Discuss your findings.
| Group | n | mean | standard deviation |
| Experimental | 38 | 33.42 | 10.18 |
| Control | 23 | 39.71 | 9.16 |
8.11 Pregnancy-induced hypertension and aspirin. In a randomized clinical trial of women with pregnancy-induced hypertension 23 women received aspirin and 24 received a placebo. After two weeks on treatment the mean diastolic arterial blood pressure of the aspirin-treated group was 111 mm Hg (s1 = 8 mm Hg) and the mean blood pressure of the control group was 109 mm Hg (s2 = 8 mm Hg). Determine whether this difference is significant. Show all hypothesis testing steps. (Source unknown)
8.12 Higher lung capacity in smokers? (Tager et al., 1979). The variable FEV in the file fev.sav contains information on forced expiratory volume (FEV) from a childhood health survey. FEV is a measure of lung function. Analyze the difference in FEV levels in smokers and non-smokers. Then, compare the ages of smokers and non-smokers. Why do smokers have significantly higher average FEV?
8.13 Risk index. A researcher derives an index of risk taking behaviors in teenagers and young adults. A score of 100 represents average risk taking. Data from a pilot study comparing male and female high school students are shown here and are stored in risk_index.sav. Explore these data with side-by-side boxplots. Interpret what you see.
Group 1 (Girls):
72 73 86
95 95
95 96
97 99 125
Group 2 (Boys): 89
92 93 98
105 106
110 126
127 130
8.14 Breast feeding and bone loss (Laskey et al., 1998). A study looked at factors that influenced changes in bone density in breast-feeding mothers. The study included 22 controls (women who were neither pregnant nor breast-feeding) and 47 comparably-aged women who were breast feeding. The percent change in spinal bone mineral content over three months of subjects are shown below and are stored in laskey-1998.sav.
(A) Plot the data as back-to-back stemplots on a common stem. Describe
what you've learned. (Compare locations, spreads, shapes. Are there any
clear outliers? etc.)
(B) Group 1 has a mean of 0.309 and standard deviation of 1.298. Group 2 has a
man of -3.587 and standard deviation of 2.506. Calculate a 95% confidence
interval for population mean difference µ1 - µ2.
[Note to advanced users: The equal variance and unequal variance confidence
intervals differ by only a trivial amount.]
Controls (n1 = 22):
|
2.4 |
0.0 |
0.9 |
−0.2 |
1.0 |
1.7 |
2.9 |
−0.6 |
1.1 |
−0.1 |
|
−0.4 |
0.3 |
1.2 |
−1.6 |
−0.1 |
−1.5 |
0.7 |
−0.4 |
2.2 |
−0.4 |
|
−2.2 |
−0.1 |
|
|
|
|
|
|
|
|
Breast-feeders (n2 = 47):
|
−4.7 |
−2.5 |
−4.9 |
−2.7 |
−0.8 |
−5.3 |
−8.3 |
−2.1 |
−6.8 |
−4.3 |
|
2.2 |
−7.8 |
−3.1 |
−1.0 |
−6.5 |
−1.8 |
−5.2 |
−5.7 |
−7.0 |
−2.2 |
|
−6.5 |
−1.0 |
−3.0 |
−3.6 |
−5.2 |
−2.0 |
−2.1 |
−5.6 |
−4.4 |
−3.3 |
|
−4.0 |
−4.9 |
−4.7 |
−3.8 |
−5.9 |
−2.5 |
−0.3 |
−6.2 |
−6.8 |
1.7 |
|
0.3 |
−2.3 |
0.4 |
−5.3 |
0.2 |
−2.2 |
−5.1 |
|
|
|
8.15 Efficacy of echinacea in treating upper respiratory infections (severity of symptoms). A randomized, double-blind, placebo controlled study evaluated the effects of the herbal remedy Echinacea purpurea in treating upper respiratory tract infections in 2- to 11-year-old children. Each time a child had an upper respiratory tract infection, treatment with either echinacea or a placebo was given for the duration of the illness. Among the outcomes studied was " severity of symptoms." A severity scale based on 4 symptoms monitored by the parents of subjects was recorded for each of the subjects. The peak severity of symptoms in the 337 cases treated with Echinacea averaged a score 6.0 (standard deviation 2.3). The peak severity of symptoms in the placebo group (n2 = 370) averaged 6.1 (standard deviation 2.4) (Taylor et al., 2003). Test the mean difference for significance. Discuss your findings. [Same as exercise 11.9.]
SS.1.Vegetarians and non-vegetarians. You want to test whether serum cholesterol levels differ in vegetarians and non-vegetarians. Your goal is to test for a difference at a = .05 (two-sided) with power = 0.80. Assume populations are Normal with standard deviation 40 mg/dl. Based on these assumptions how many people do you need to study to detect a mean difference of 10 mg/dl?
SS.2. Allopathic and holistic medicine. A survey instrument compares patient satisfaction with traditional western health care and traditional care supplemented with complementary medicine. A mean difference of 0.25 is worth detecting using this instrument. A pilot study reveals a standard deviation of about 0.67 units in both populations. How large a sample is needed to test for a mean difference with 80% power at alpha = 0.05 (two-sided)?
SS.3. Vegetarian diets and beta-carotene levels power. An investigator finds no significant difference in serum beta-carotene levels in vegetarians (n1 = 10) and non-vegetarians (n2 = 10). The pooled estimate of within group standard deviation is 5 mg/100 ml (pooled variance = 25). What was the power of the study to find a mean difference of ...
(A)
... 1mg /100 ml?
(B) ... 5 mg /100 ml?
(C) ... 10 mg/100
ml.
(D) What was the minimum detectable difference of the study at 80% power?
SS.4.
Vegetarian
diet and beta-carotene levels sample size
The
investigator plans on redoing the study using an alpha level of 0.05 power of
0.80, and allocation ratio of 1:1. How many people should be studied per group
if the expected mean difference is 1mg/100 ml. ...
5 mg/100 ml? ... 10 mg/100 ml?
Key to Odd Numbered Problems Key to Even Numbered Problems (may not be posted)