Epi4921 Assignments
Problem 1
Exercises 1.
No Earplugs Earplugs
Hearing-loss 40 60 100
Normal hearing 16 384 400
56 444 500
a. Risk of hearing-loss attributable to not wearing earplugs;
ARe= CI0-CI1= (a/(a+b))-(c/(c+d))
(40/100)-(16/400)= 0.36
There is a 36% higher risk for the no earplug wearing group to get hearingloss compared
to the wearing group in four years.
36% higher risk in for 4 years in the exposed group compared to the non exposed.
b. Was the proportion of hearing-loss in the non-wearers was attributable to not
wearing earplugs?
APe= CI1-CI0/CI1=( a/(a+b))-(c/(c+d))/(a/(a+b)
APe= (40/100)-(16/400)/ (40/100)= 0.9
90% of the risk to get hearingloss is associated with not wearing earplugs.
For the proportion you don’t need to know the years.
c. What proportion of the hearing-loss in all the workers was attributable to not
wearing earplugs
APt= (CIt-CIo)/CIt=(e/e+f)-(c/c+d)/(e/e+f)
Apt=(56/444)-(16/384)/(56/444)= 0.64
56/500 -16/400/ 56/500=
64% of the risk to get hearingloss for the total population is associated with not
wearing earplugs.
d. RR= (CI1/CI0)
RR= (40/100)/(16/400)= 10
The risk to get hearingloss is 10 times higher when not wearing earplugs.
e. OD= (AD/BC)=
OD= (40 x384)/(16x60)= 16
The Odds of getting hearing-loss when NOT wearing earplugs is 16 times higher.
,Exercise 2
Other P1 Trauma P2
N cases N death Case N cases N death Case
Fatality Fatality
Ratio Ratio
(x1000)
Mild 3734 37 9,9 687 3 4.3
Moderat 1887 94 49.8 1238 37 29.3
e
Serious 1665 327 198.7 1429 172 120.3
Total 7266 458 63.0 3354 212 63.2
a. What do you think of this reasoning?
Severe trauma is not equal for the other hospitals Vs trauma-center
There is an over estimation because, the increase of trauma level.
b. How would you compare the data of the trauma-center with the other hospitals
Looking at the case fatality rate, the trauma-center is doing better.
STD-pop Death Death New N death New N
rate P1 rate P2 p1 death P2
Mild 3734+637= 4371 9.9 4.3 9.9 x 4371= 18795.3
43272.9
Moderat 1887+1238=3125 49.8 29.3 155625 91562.5
e
Serious 1665+1429=3094 198.7 120.3 614777.8 372208.2
total 7266+3354=10620 63.0 63.2 669060 671184
Wmidl = 3734/687= 4427
Wmod= 3125
Wserieus= 3074
W total= 7266+3354= 10620
SR= ∑ wιRιιRιRιιRι /∑ wιRιιRι
SR= (4427 x(9,9/1000))+(3125(49.8 x1000)+ (3074x (198,8/100)/ 10620= 76.3
SRb= 45,5/1000
aιιRι aιιRι
SRR= ∑ wιRιιRι ( )/ ∑wιRιιRι ( )= Sra/SRb= 76.3/45,5 = 1.68
N 1 ιRι N 0 ιRι
,The chance of dying at the other hospitals is higher compared to the trauma hospital.
no precentages because it is an indication and standard
Step 1= create standard population
Step 2= apply rates to STD pop
Step 3= calculate new death rate
Vb:standard pop
3734+ 687= 4421
37+3=40
40/4421=
New death rate p1= 63
New death rate p2= 63.2
Exercise 3
a. Standardized mortality ratio
SMR= observed cases (O)/ Expected cases (E)) x 100
58/ 42.76= 1.36
used mortality rate
36% of the cohort higher than expected if the cohort had the same as the reference
mortality rate.
Age N death (2 x4)
1250 x 0.12
(9.9/100000
)
3423 x (17.7 0.6
x 100000)
1.46
3.06
4.68
0.06
0.65
1.94
4.98
6.70
0.000412
0.41
2.19
6.75
, 9.46
Total 42.76
b. Why indirect standardization is appropriate?
No strata- specific rates, or when the study group is to small
- The indirect method of adjustment, which has been particularly popular in the field
of occupational epidemiology, the expected number of events (e.g., deaths) in a
study group (e.g., an occupational cohort) is calculated by applying reference rates
(“standard” rates) to the number of individuals in each stratum of the study group(s)
Exercise 4
a. Incidence rate of mortality due to myocardial infection for smoking (English male
doctors 55-64 years)
IR= N/Person-time
IR= 206/28612=0.007199
b. The 10-year cumulative incidence of MI-death for smoking doctors of 55 years (until
65 years).
CI= 1-e^-(i x t)=
1-e^-(10x0.0072)
1-e^-(0.0722)= 0.0697
A smoking d of 55 has a 6,9% chance to die from MI in the next 10 years.
The changes to die from MI for the age group of 55-64 when smoking is 7 % in 10
years
c. The 40-year cumulative incidence of MI-death for doctors of 35 year, separately for
smokers and non-smokers.
Age I= smokers I= non-smokers
35-44 .407= 0.61 0.106
45-54 2.4 1.124
55-64 7.2 4.9
65-74 14.688 10.83
Total
CI= 1-e^-(i x t)+( i x t)+( i x t)
Smokers= 1-e^-(10 x 0.00061)+(10 x0.0024)+(10 x 0.0072) +(10x 0.014688)=
Smokers = 1-e^-(0.0061+0.024+0.072+0.1468)= 0.2203
Smoking d of 35 years has a 22% chance to die due to Mi over 40 years
Problem 1
Exercises 1.
No Earplugs Earplugs
Hearing-loss 40 60 100
Normal hearing 16 384 400
56 444 500
a. Risk of hearing-loss attributable to not wearing earplugs;
ARe= CI0-CI1= (a/(a+b))-(c/(c+d))
(40/100)-(16/400)= 0.36
There is a 36% higher risk for the no earplug wearing group to get hearingloss compared
to the wearing group in four years.
36% higher risk in for 4 years in the exposed group compared to the non exposed.
b. Was the proportion of hearing-loss in the non-wearers was attributable to not
wearing earplugs?
APe= CI1-CI0/CI1=( a/(a+b))-(c/(c+d))/(a/(a+b)
APe= (40/100)-(16/400)/ (40/100)= 0.9
90% of the risk to get hearingloss is associated with not wearing earplugs.
For the proportion you don’t need to know the years.
c. What proportion of the hearing-loss in all the workers was attributable to not
wearing earplugs
APt= (CIt-CIo)/CIt=(e/e+f)-(c/c+d)/(e/e+f)
Apt=(56/444)-(16/384)/(56/444)= 0.64
56/500 -16/400/ 56/500=
64% of the risk to get hearingloss for the total population is associated with not
wearing earplugs.
d. RR= (CI1/CI0)
RR= (40/100)/(16/400)= 10
The risk to get hearingloss is 10 times higher when not wearing earplugs.
e. OD= (AD/BC)=
OD= (40 x384)/(16x60)= 16
The Odds of getting hearing-loss when NOT wearing earplugs is 16 times higher.
,Exercise 2
Other P1 Trauma P2
N cases N death Case N cases N death Case
Fatality Fatality
Ratio Ratio
(x1000)
Mild 3734 37 9,9 687 3 4.3
Moderat 1887 94 49.8 1238 37 29.3
e
Serious 1665 327 198.7 1429 172 120.3
Total 7266 458 63.0 3354 212 63.2
a. What do you think of this reasoning?
Severe trauma is not equal for the other hospitals Vs trauma-center
There is an over estimation because, the increase of trauma level.
b. How would you compare the data of the trauma-center with the other hospitals
Looking at the case fatality rate, the trauma-center is doing better.
STD-pop Death Death New N death New N
rate P1 rate P2 p1 death P2
Mild 3734+637= 4371 9.9 4.3 9.9 x 4371= 18795.3
43272.9
Moderat 1887+1238=3125 49.8 29.3 155625 91562.5
e
Serious 1665+1429=3094 198.7 120.3 614777.8 372208.2
total 7266+3354=10620 63.0 63.2 669060 671184
Wmidl = 3734/687= 4427
Wmod= 3125
Wserieus= 3074
W total= 7266+3354= 10620
SR= ∑ wιRιιRιRιιRι /∑ wιRιιRι
SR= (4427 x(9,9/1000))+(3125(49.8 x1000)+ (3074x (198,8/100)/ 10620= 76.3
SRb= 45,5/1000
aιιRι aιιRι
SRR= ∑ wιRιιRι ( )/ ∑wιRιιRι ( )= Sra/SRb= 76.3/45,5 = 1.68
N 1 ιRι N 0 ιRι
,The chance of dying at the other hospitals is higher compared to the trauma hospital.
no precentages because it is an indication and standard
Step 1= create standard population
Step 2= apply rates to STD pop
Step 3= calculate new death rate
Vb:standard pop
3734+ 687= 4421
37+3=40
40/4421=
New death rate p1= 63
New death rate p2= 63.2
Exercise 3
a. Standardized mortality ratio
SMR= observed cases (O)/ Expected cases (E)) x 100
58/ 42.76= 1.36
used mortality rate
36% of the cohort higher than expected if the cohort had the same as the reference
mortality rate.
Age N death (2 x4)
1250 x 0.12
(9.9/100000
)
3423 x (17.7 0.6
x 100000)
1.46
3.06
4.68
0.06
0.65
1.94
4.98
6.70
0.000412
0.41
2.19
6.75
, 9.46
Total 42.76
b. Why indirect standardization is appropriate?
No strata- specific rates, or when the study group is to small
- The indirect method of adjustment, which has been particularly popular in the field
of occupational epidemiology, the expected number of events (e.g., deaths) in a
study group (e.g., an occupational cohort) is calculated by applying reference rates
(“standard” rates) to the number of individuals in each stratum of the study group(s)
Exercise 4
a. Incidence rate of mortality due to myocardial infection for smoking (English male
doctors 55-64 years)
IR= N/Person-time
IR= 206/28612=0.007199
b. The 10-year cumulative incidence of MI-death for smoking doctors of 55 years (until
65 years).
CI= 1-e^-(i x t)=
1-e^-(10x0.0072)
1-e^-(0.0722)= 0.0697
A smoking d of 55 has a 6,9% chance to die from MI in the next 10 years.
The changes to die from MI for the age group of 55-64 when smoking is 7 % in 10
years
c. The 40-year cumulative incidence of MI-death for doctors of 35 year, separately for
smokers and non-smokers.
Age I= smokers I= non-smokers
35-44 .407= 0.61 0.106
45-54 2.4 1.124
55-64 7.2 4.9
65-74 14.688 10.83
Total
CI= 1-e^-(i x t)+( i x t)+( i x t)
Smokers= 1-e^-(10 x 0.00061)+(10 x0.0024)+(10 x 0.0072) +(10x 0.014688)=
Smokers = 1-e^-(0.0061+0.024+0.072+0.1468)= 0.2203
Smoking d of 35 years has a 22% chance to die due to Mi over 40 years
