Download epidemiology exam 2 practice question answers

1. Which one of the following would NOT be expected to increase
the statistical power of a clinical trial to test a new medical
treatment?
a) treating a more clinically homogeneous group of patients
b) lengthening the trial to allow enrollment of more patients
c) adopting a more stringent standard of evidence to prove the
new treatment works
d) adopting stronger measures to ensure compliance with
therapy
e) testing only a subgroup of patients for whom the new drug
would be expected to have the greatest advantage
2. Two researchers conduct separate randomized clinical trials comparing the
same new therapy and publish their results in consecutive New England
Journal of Medicine papers. The first study found a difference in favor of the
new therapy and reported p=.02 to support their contention that this
difference was real. The second study used twice as many patients as the
first and also observed a difference in favor of the experimental therapy, but
reported that p=.21 and therefore that observed difference was not
statistically significant. Which statement below is NOT a possible
explanation of the different results?
a) One of the reported p-values must be incorrect, since larger sample sizes
generate smaller p-values.
b) The observed difference favoring the new therapy was smaller in the
second clinical trial than in the first.
c) Because of differences in designs of the two clinical trials, measurement
error and biological variability were greater in the second trial than in the
first.
d) The first trial included an adjustment for confounding variables, while the
second examined the crude difference between therapies.
e) The first study was not blinded, and diagnostic suspicion bias contaminated
the results.
3. You conduct a case-control study examining the relationship between
drinking soda and colon cancer and find that among 1500 who have colon
cancer, 400 drink soda, while among the 3000 controls who don’t have
colon cancer, 450 drink soda.
a) Draw a 2x2 table and calculate the crude OR.
Colon cancer +
Colon cancer -
Drinks soda +
400
450
Drinks soda -
1100
2550
Crude OR = (400*2550)/(450*1100) = 2.06
3. (continued) Now you stratify by gender and find the following:
• Among women, 200 of 1000 who have colon cancer drink soda, while
among the 2000 who don’t have colon cancer, 300 drink soda.
• Among men, 200 of the 500 who have colon cancer drink soda, but only
150 of the 1000 who don’t have colon cancer drink soda.
b) Draw out the stratified 2x2 tables and calculate their respective ORs.
WOMEN
Colon cancer +
Colon cancer Drinks soda +
200
300
Drinks soda 800
1700
OR = (200*1700)/(300*800) = 1.42
MEN
Colon cancer +
Colon cancer Drinks soda +
200
150
Drinks soda 300
850
OR = (200*850)/(150*300) = 3.78
c) Is this an example of effect modification or confounding?
The stratum-specific ORs are substantially different (> 20% different).
Therefore this is an example of Effect Modification.
d) Explain in lay terms what this conclusion means.
There is a natural, inherent difference between how drinking soda affects
risk of developing colon cancer in men versus women. Men who drink soda
have a greater increase in risk of colon cancer than women do because of
something intrinsic to the male body’s response to the soda.
e) What do you do now?
We cannot report the crude (combined) OR because it falsely elevates the
risk in women and falsely reduces the risk in men. It is misleading. We
instead report the separate, stratified ORs that we calculated.
4. You conduct a case-control study to examine the relationship between
eating margarine and depression. You find that among the 185 patients
who suffer from depression, 65 eat margarine, while 50 of the 230 controls
eat margarine.
a) Draw a 2x2 table and calculate the crude OR.
Depression +
Depression -
Eats margarine +
65
50
Eats margarine -
120
180
Crude OR = (65*180)/(50*120) = 1.95
4. (continued) Among the 100 women in this group who suffer from
depression, 25 eat margarine. Among the 50 female controls, 5 eat
margarine. Among the 85 male cases, 40 eat margarine. Among the 180
male controls, 45 eat margarine.
b) Draw out the stratified 2x2 tables and calculate their respective ORs.
WOMEN
Eats margarine +
Eats margarine OR = (25*45)/(5*75) = 3
Depression +
25
75
Depression 5
45
MEN
Depression +
Depression Eats margarine +
40
45
Eats margarine 45
135
OR = (40*135)/(45*45) = 2.67
c) Is this an example of effect modification or confounding?
The stratum-specific ORs are very similar (they don’t differ by >20%), but
they differ from the crude OR. This is an example of confounding.
d) Explain in lay terms what this conclusion means.
The relationship between eating margarine and depression is distorted by
gender. Gender is a variable associated with both eating margarine and
with developing depression. For example, maybe more men burn their
toast, and burnt toast is also associated with an increase in depression.
e) What do you do now?
We cannot report the crude (combined) OR because it falsely represents
the risk. We instead need to conduct a Mantel-Haenszel adjustment to
calculate a combined OR that controls for the effect of gender. Good news
for you guys is that you do not need to know how to do this!
5. Ahh, the fun of the matched case-control adjustment...
Consider the following matched case-control study examining the
relationship between eating peanut butter for breakfast and developing
peptic ulcer disease:
Pair:
1 2 3 4 5 6 7 8 9 10
PUD (+): + + + - - - + + + +
Controls: + + - + + - + - - Draw a regular 2x2 table and calculate a summary OR.
PUD +
PUD -
Peanut butter +
7
5
Peanut butter -
3
5
Crude OR = (7*5)/(3*5) = 2.33
5. (continued) Let’s approach the same question differently.
Consider the following matched case-control study examining the
relationship between eating peanut butter for breakfast and developing
peptic ulcer disease:
Pair:
1 2 3 4 5 6 7 8 9 10
PUD (+): + + + - - - + + + +
Controls: + + - + + - + - - Calculate the matched OR.
To approach this type of question, first select only the discordant pairs.
Now, we make one ratio:
Discordant pairs with case exposed
Discordant pairs with case unexposed
= 4/2 = 2.0
The strength of association between exposure and disease changes when
we use matched cases and controls. Remember that this method increases
statistical power (a more homogenous population) but may decrease
generalizability to the population at large.
6. In an "outcomes" analysis of coronary bypass surgery, health services
researchers identify charts of all patients diagnosed with three vessel
disease at three major clinical centers during the past ten years. These
patients are separated into those who initially were treated surgically and
those who were initially treated medically, with surgery used if medical
treatment was unsuccessful. Aggregate results for mortality and a variety of
other outcome variables were compiled for each group, to produce
prognostic profiles for those initially treated medically vs. those who
received immediate surgery. This study was a:
a)
b)
c)
d)
e)
prospective cohort study
retrospective cohort study
cross-sectional survey
hospital-based case-control study
controlled clinical trial
7. Consider the following study:
In a study examining the relationship between oral contraceptives and
bacteriuria, you follow women who do and do not use oral contraceptives
over a three-year period, and find that 70 of the 500 individuals who use
OC acquired bacteriuria, while 150 of 3000 individuals who don’t use OC
acquired bacteriuria.
a) Is this a cohort or case-control study?
Cohort
b) What is the incidence among the exposed?
Ie = 70/500 = 0.14
c) What is the incidence among the unexposed? Io = 150/3000 = 0.05
d) What is the relative risk?
RR = Ie/Io = 2.8
e) What is the attributable risk?
AR = Ie-Io = 0.09
f) What is the attributable risk %?
AR% = (Ie-Io)/Ie = 0.64
Now suppose that the prevalence of use of OC in the population is 20%.
g) What is the expected incidence of bacteriuria in our population?
It = (Pe)(Ie) + (1-Pe)(Io) = (0.2)(0.14) + (0.8)(.05) = 0.068
h) What is our population attributable risk?
PAR = (It-Io) = 0.018
i) What is our population attributable risk %?
PAR% = (It-Io)/It = 0.26