Intermediate Methods in Observational Epidemiology
2008
Exercise No. 5 – Recapitulation
Instructors’ Guide
Exercise 1. The univariate analysis of a study of the relationship of atherogenic
risk factors to uterine leiomyoma (Faerstein E, Szklo M, Rosenshein NB, Am J
Epidemiol 2001;153:11-19) yielded the results for age at diabetes diagnosis
shown in the table:
Age when first diagnosed with
diabetes
Never diagnosed
<35 years
≥35 years
Cases
Controls
303
9
6
378
9
7
Odds
Ratio
1.0
a. Using the category, “never diagnosed” as reference (odds ratio= 1.0), calculate the
ratios of the absolute odds of each of the other categories to the reference odds.
Show your calculations
For <35 years, odds ratio= (9/303) ÷ (9/378)=1.25
For ≥35 years , odds ratio= (6/303) ÷ (7/378)=1.07
b. Which traditional causality criterion is not met by these odds ratios?
Dose-response, which would require that the relationship be monotonic
Exercise 2. A study to examine the relationship of inflammatory markers
(such as interleukin-6 and C-reactive protein) to incident dementia was
conducted within the Rotterdam Study cohort (n=6,713) (Engelhart MJ,
Geerlings MJ, Meijer J, et al. Inflammatory proteins in plasma and risk of
dementia. The Rotterdam Study. Arch Neurol 2004;61:668-672). A random
cohort sample (n=727) and the 188 individuals who developed dementia on
follow-up were compared. Serum inflammatory markers were measured in
cases and in the random sample.
a. Which type of study have the authors conducted?
A case-cohort study, that is, a case-control study within a defined cohort, in which the
control group was a random sample of the total cohort
b. During the cohort follow-up, 42 cases in the cohort sample developed
dementia. Under which circumstances should they not be excluded from
the cohort sample?
When a relative risk is the preferred measure of association.
c. By dividing the odds of exposure in cases to that in controls, which
measure of association is obtained?
Relative risk.
d. If the authors wished to study the relationship of inflammatory markers to
stroke, could they use the same control group? Why or why not?
Yes, the same control group (cohort sample) could be used, as for every type of case,
the exposure odds ratio would yield the relative risk.
e. The relative risk of dementia associated with an IL-6 value in the highest
quintile compared with that in the lowest quintile was found to be about
1.9. Assuming no random variability and that the relative risks of the
second, third and fourth quintiles compared with the lowest quintile were
close to 1.0, calculate the proportion of dementia explained by values in
the highest quintile.
Prevalence of exposure (IL-6 values in the highest quintile)= 0.20
Relative risk= 1.9
Population attributable risk using Levin’s formula:
Percent Population AR 
0.20 (19
.  10
. )
018
.
 100 
 100  15.2%
0.20 (19
.  10
. )  10
.
118
.
Exercise 3. The number and rate per 1000 white residents of Washington County,
Maryland State (USA) aged 45-64 years on 15 July 1963, of cases of colon cancer
diagnosed 15 July 1963 through 14 July 1975 are shown in the table below, by residence
and grades of school completed.
Initial
Population
characteristic
Total
18 125
Residence
Urban
9 351
Rural
8 774
Grades
completed
13+
2 418
<13
15 707
Cases
N
Cases/1000 Rate ratio Odds/1000
population
population
116
6.4
Odds
ratio
50
66
5.3
7.5
0.71
1.0
5.4
7.6
0.71
1.0
23
93
9.5
5.9
1.6
1.0
9.6
6.0
1.6
1.0
a. Calculate the rates, the rate ratios, the odds and the odds ratios for residence and
education.
See table.
b. The investigators selected a random sample of controls from the total population
(an actual sampling procedure was done). The results are shown in the table
below.
Initial
Characteristic
Total
Residence
Urban
Rural
Grades
completed
13+
<13
Cases
Sample
Measure of
association
116
116
50
66
57
59
0.78
1.0
23
93
9
107
2.9
1.0
c. Calculate the values of the measure of association for residence and education.
Which measures of association have you calculated? Are they the same as those
obtained in the table showing prospective data? Why?
See table. Rate ratios. No, because they are based on a random sample of controls
(and, thus, there is random variability).
Next, for each case the investigator selected an individually matched (non-case) control
(matching was done for age, gender and race). The results are shown in the table below
for residence.
Matched
controls
Urban
Rural
Urban
23
27
Cases
Rural
32
34
d. Using the Mantel-Haenszel method, calculate the odds ratio, treating each pair as
a separate stratum, and using “rural” as the reference category.
No. pairs
23
Case
Urban 1
Rural 0
Control
1
0
2
No. pairs
32
Case
Urban 0
Rural 1
Control
1
0
2
No. pairs
27
Case
Urban 1
Rural 0
Control
0
1
2
No. pairs
34
Case
Urban 0
Rural 1
Control
0
1
2
ai d i
 0
 0
 1
 0
23   32    27    34  
Ni
 2
 2
 2
 2



bi ci
 0
 1
 0
 0
 N 23 2   32 2   27 2   34 2 
i

OR MH
27

2  27  2  27  0.84
32
2 32 32
2
e. Calculate the ratio of discrepant pairs.
OR= 27 ÷ 32 = 0.84
f. Now pretend that cases and controls are not individually matched. Calculate the
“unmatched” odds ratio.
Cases
Controls
Urban
50
55
Rural
66
61
OR= (50/66) ÷ (55/61)= 0.84
g. Why do you think the “unmatched” odds ratio is similar/same/different from the
“matched” odds ratio?
The “matched” and “unmatched” odds ratios are the same because age, gender and
race are likely not confounders of the association between residence and colon cancer.
Exercise 4. The authors of the paper on melanoma and tanning ability mentioned
in the lecture (Weinstock et al. Am J Epidemiol 1991;133:240-5) provided the
following information for cases only:
Pre-melanoma diagnosis
(gold standard)
Exposed
Unexposed
8
7
1
18
9
25
Post-melanoma diagnosis information
No tan or light tan (exposed)
Medium, average, deep or dark tan (unexposed)
Total (true classification) 
Total defined according to the
melanoma diagnosis informa
15
19
34
a. Using the data on the table, calculate the sensitivity and specificity of the postmelanoma diagnosis information for cases.
Answer
Sensitivity: 8/9 (89%); Specificity: 18/25 (72%)
b. Assuming that the misclassification was non-differential for cases and controls,
use the table below showing the “gold standard” results to calculate the biased
(post-melanoma diagnosis) odds ratio.
“Gold standard results” (pre-melanoma diagnosis)
Tanning
Cases
Controls
ability
Exposed
9
79
Unexposed
25
155
Odds Ratio: [9 x 155] / [79 x 25] = 0.71
Answer
In cases
Exposed
Unexposed
Exposed
Unexposed
total
8
1
9
7
18
25
In
controls
Exposed
Unexposed
total
Exposed
Unexposed
Misclassified
total
70
9
79
43
112
155
113
121
234
Misclassified
total
15
19
34
Study (post-melanoma diagnosis, biased) results based on the assumption
of non-differential misclassification
Tanning
Cases
Controls
ability
Exposed
15
113
Unexposed
19
121
Odds Ratio= 0.84
c. The post-melanoma diagnosis (“misclassified”) odds ratio was found by
Weinstock et al to be 1.6. On the basis of your calculations, was the
misclassification in Weinstock et al’s study non-differential?
Answer
No. If the misclassification had been non-differential, the above odds ratio would not
have been found to differ from the “misclassified” odds ratio (1.6) observed in the
study.
Exercise 5. The association between presence of factor V Leiden allele (heterozygotes
and homozygotes), oral contraceptives (OC) and risk of venous thromboembolism was
examined in a case-control study conducted by Vanderbroucke and colleagues (Lancet
1994;344:1453-7). Factor V is activated in the coagulation process, where it acts as a cofactor for other factors involved in thrombus formation. Factor V Leiden (FV Leiden) is
a mutation of factor V that renders factor Va resistant to cleavage by activated protein C.
This mutation predisposes the patient to thrombosis. Results are presented in the
following table.
Factor V
OC use
Cases
Controls
+
+
25
2
+
10
4
+
84
63
36
100
Total
155
169
interaction between Factor V and OC use.
OR stratified
1.0
1.0
Ignore
random
variability as
well as the
possibility of
a. Calculate the odds ratios using as reference categories those indicated by 1.0 in the
table. Answer:
Factor V
+
+
Total
use.
OC use
+
+
-
Cases
25
10
84
36
155
Controls
2
4
63
100
169
OR stratified
5.0
1.0
3.7
1.0
b. Ignoring
Factor V,
calculate the
(crude) odds
ratio for OC
OC use
+
-
Cases
Controls
OR stratified
109
65
3.8
46
104
1.0
c.
Using
155
169
the MantelHaenszel method, calculate the factor V-adjusted odds ratio for the relationship of oral
contraceptive use to thromboembolism. Show your calculations.
OR MH
25  4 84  100

41
283  3.8

2  10 63  36

41
283
d. Based on the comparison between factor V-adjusted odds ratio and the crude odds
ratio, do you conclude that factor V is a confounder?
No: the crude and the adjusted odds ratios are identical.
e. Why is the factor-V-adjusted OR closer to one of the factor V strata than it is to the
other?
Because the study sample is weighted towards the Factor V negative stratum
f.
In general, is significance testing indicated to evaluate a potential confounding
effect? Why?
No. A small difference between groups under comparison (cases vs. controls or, in
cohort studies, exposed vs. unexposed) in the prevalence (or level) of the confounder
may confound the association if, in addition, the confounder is strongly related to the
exposure (in case-control studies) or the disease (in cohort studies)
Exercise 6. A recent case-control study examined the relationship of alcohol drinking to
cancer of the liver (Donato F, et al. Am Journal of Epidem 2002; 155: 323-331).
Unadjusted results are shown below. (Adjustment for age, residence, alcohol drinking,
hepatitis B surface antigen and Hepatitis C virus by using unconditional logistic
regression analysis did not change the patterns). For this exercise, assume that the data
are valid and that there is no random variability.
Distribution of liver cancer cases and controls according to alcohol drinking of men and
women, Brescia, Italy, 1995–2000
Men
Women
Alcohol
Cases
Controls
Cases
Controls
drinking No.
%
No.
%
OR*
No.
%
No.
%
OR*
Never
8
2.1
42
6.1
24
28.6
54
39.1
1.0
1.0
Former
151
39.7
89
13.0
31
36.9
19
13.8
8.9
3.7
Current
281
58.2
555
80.9
29
34.5
65
47.1
2.1
1.0
*OR- odds ratio
a. Calculate the odds ratios separately for men and women in the above table using
“never” drinkers as the reference category.
See table.
b. What kind of interaction can be inferred by inspecting the above table?
Multiplicative.
c. What are some possible reasons why results are heterogeneous?
True (biologic) interaction; residual confounding and/or misclassification of different
magnitude between gender strata; random variability; former or current drinking
heavier in men than in women.
d. State a plausible explanation for why odds ratios are higher for former than for
current drinking.
Individuals with liver disease symptoms stop drinking.