Heather Surcouf
NURS 802 –Epidemiology
Module 7 Cohort Studies
October 20, 2012
1. A study of bladder cancer and cigarette smoking over a 20 yr period in Boston provided the
following data:
Bladder cancer rates per 100,000 males
Cigarette smokers
48.0
non-smokers
25.4
-- Find the RR and AR%
RR= Ie/Io = 48/25.4 = 1.89 = 1.89-1 x100 = 89%
RR = 89% The RR = 1.89 (not a percent) and is described as those who smoke are 89%
more likely to develop bladder cancer than those who don’t smoke.
AR Ie-Io = 48-25.4 =22.6, as a percentage this would be :(48-25.4/48) x 100 APE = 47.083%
AR = 47.08% okay.. good
2. In a study of hormone replacement and breast cancer (JAMA, 13 Feb 2002) use of hormone
replacement therapy was compared between 700 cases of breast cancer and 700 controls (no
breast cancer). Of the breast cancer cases, 280 were exposed to hormone replacement therapy
(HRT) while only 100 of the controls were exposed to HRT. Explain the risk of breast cancer
given HRT?
Hrt
No hrt
Cases
280
420
700
Controls
100
600
700
OR= (axd)/ (bxc) = (280x600) / (100/420) = .4 = exposure increases disease risk I got a risk of
4.0.
HRT users are 4 times more likely to develop breast cancer Yes.
3. Tulane Univ. reported that 9776 subjects enrolled in the National Health and Nutrition
Examination Survey were given dietary assessment and followed for an average of 19 years.
Those with high fiber food intake experienced 1843 incident cases of CHD. (Arch Int Med, Sep.
2003). What type of study design is this?
From this information determine and describe the risk of CHD given the “exposure” to high fiber
foods.
Soluble Fiber
High
Low
CHD Yes
1843
a
2881 c
CHD No
2352 b
2700 d
RR =Ie/ Io= (a/(a+b)) /(c/(c+d) =1843/(1843 + 2352) / 2881 / (2881+2700) = .44/.52=
RR = .85 good.
(.85-1) x 100 or 15% Ok. good
High fiber diet offers a 15% decrease in CHD risk ok.
This indicates that exposure to fiber offers a weak protective effect for risk of CHD
Prospective Cohort Study
4. Of one hundred cases of oral cancer in hospitalized men, 86 were found to dip tobacco. These
were compared to 500 men with other unrelated conditions, of which 190 were found to dip. Set
up a 2 X 2 table and determine the risk of men getting oral cancer in this study.
Oral Cancer
Dip
86
No Dip
14
Total
100
RR =Ie/ Io= (a/(a+b)) /(c/(c+d)
No Oral Cancer
190
310
500
86 /(86 + 190) / 14 (14+310) = (86/276) / (14/324) = .31/.04 =
RR = 7.75
There is a strong risk associated with dip and oral cancerSince this is a case-control study,
use the OR to estimate the risk.
5. In a benzene factory 500 workers were followed over 10 years. 100 of the workers were
consistently exposed to benzene. The other 400 workers wore face masks and were not exposed.
By the end of 10 years physicians found that 40 of those exposed had early symptoms of
leukemia while only 16 mask wearers had those symptoms.
Leukemia S&S
Exposed benzene
Not exposed benzene
40
16
No Leukemia S&S
60
384
RR =Ie/ Io= (a/(a+b)) /(c/(c+d) =(40/ 100) / (16/ 400) = .4/.04 = 10 good
- What was the risk of leukemia in those exposed versus those not exposed to benzene?
Those who were exposed to Benzene had a 10 times increased risk to developing Leukemia
S&S ok.. good
- What proportion of the leukemia was attributed to not wearing a mask (AR%)?
AR = Ie-Io = (a/(a+b)) - (c/(c+d) = =(40/ 100) - (16/ 400) = .4 - .04 = .36
APE= (Ie-Io/Ie) x 100 (.36/.4) x 100 = 90%
90% of Leukemia S&S were attributed to not wearing a mask ok good
6. Your cases are 100 women with breast cancer. You also identify 100 women without breast
cancer. You found that 80% of your cases are multiparous and 90% of your controls are
multiparous. What is the exposure of interest? What is the risk of breast cancer for a nulliparous
woman in your population?
Breast Ca
No Breast Ca
Nulliparous
Multiiparous
Total
20
80
10
90
100
40
170
100
Exposure would be Nulliparous
OR = (axd)/ (bxc) = (20x90)/ (10x80) =1800/800 = 2.25
Nulliparous women are 2.25 times more likely to have breast cancer ok good
7. Diarrhea is a common cause of death among babies in developing countries. There may be a
link between diarrhea deaths and use of infant formula in these countries. A researcher found 200
families in Brazil where a death from diarrhea had occurred in first-born babies during the year.
In the same neighborhood 200 control families were found where first-born babies did not
develop diarrhea. In both groups, mothers were interviewed regarding infant formula use. Of
those with diarrhea deaths, 73 used infant formulas. Forty-six of the controls used formulas.
What type of study is this? What is the risk of dying from diarrhea given that the first-born used
infant formula? What proportion of the deaths can be attributed to using infant formulas?
Infant formula
No Formula
Diarrhea death
73
127
200
No death
46
154
200
119
281
OR = (73x154) / (127x46) = 11242/5842 = 1.92
The risk of dying from diarrhea in first born infants using formula is 1.92 times more likely
or 92% more likely ok
APE = Ie-Io/Ie x 100 = (a/(a+b)) - (c/(c+d) / Ie x 100= ( 73/ 119 )–( 127/ 281) =(( .61 - .45)
/.61) x 100 = 26
26% deaths can be attributed to using infant formula ok