Question 1
1. 15
2. 728
3. 31
4. 1988
5. c. Incidence rate
Question 2
They are both ways to compare measures of disease frequency in order to assess the impact of an exposure on a disease.
The ratio measure gives information on the strength of the relationship between an exposure and disease; the difference measure describes the excess number of cases of disease that are associated with the exposure.
Question 3
You would want to measure risk if you are following a fixed cohort of disease-free persons all of whom are followed for the same amount of time other than those who develop the health outcome of interest.
Question 4
You would want to measure rate if you were following a cohort of disease-free persons and those who do not develop the health outcome of interest have differing amounts of disease-free follow-up time.
Question 5
Both provide information on the absolute effect of the exposure or the excess risk of disease. However, the risk difference gives the number of cases of disease among the exposed that may be attributable to the exposure, and the population risk difference gives the number of cases of disease in the total population that may be attributable to the exposure.
Question 6
When everyone in the population is exposed.
Question 7
This is measured by the risk difference
correct
This is measured by the attributable proportion among the exposed
This is measured by the attributable proportion in the total population
Question 8
RD = (27/482) - (77/1908) = 1566 per 100,000
Question 9
30/100,000 person-years/45/100,000 person-years = 0.67
Women who engage in regular physical activity have 0.67 times the risk of ovarian cancer (or a 33% reduced risk of ovarian cancer) compared with women who do not engage in regular physical activity.
30/100,000 person-years - 45/100,000 person years = - 15/100,000 person-years. Note that the incidence rate difference is negative.
The excess rate of ovarian cancer among women who do not engage in regular physical activity is 15/100,000 person-years. Or, if regular physical activity prevents ovarian cancer, then 15 cases per 100,000 person-years of follow-up would be eliminated if the women engaged in regular physical activity.
Question 10
A.
APe = [(Re – R0) / Re ] * 100
APe = [(110/399- 1073/16969) / (110/399)] * 100 = [(27.6 – 6.3) / 27.6] * 100 = 77%
Alternative formula for APe = [(RR – 1) / RR] * 100
APe = [(4.4 – 1) / 4.4] * 100 = 77%
B.
Among cocaine users, 77% of the low birth weight births are attributable to cocaine use over the study period. In other words, if we assume causality and cocaine users quit using cocaine, then 77% of the low birth weight cases in the (previously) exposed population (i.e. 77% of the cases among cocaine users) would be eliminated.
C.
APe = [(4.4 – 1) / 4.4] * 100 = 77% (previously calculated)
Pe|d = 110 / 1183 = 0.093
APt = APe * Pe|d = 77% * 0.093 = 7.18%
Assuming a causal relationship between cocaine use and low birth weight, eliminating cocaine use in this population would eliminate 7.18% of cases of low birth weight.
D.
APe * Pe|d = 77% * 0.02 = 1.54%
Assuming a causal relationship between cocaine use and low birth weight, eliminating cocaine use in California would eliminate 1.54 % of cases of low birth weight. This proportion is smaller than the proportion from the previous question because in California, less of the risk of low birth weight is attributable to cocaine use (the prevalence of the exposure among the cases) than in the other population, so eliminating cocaine use is not as effective in reducing incidence of low birth weight.
E.
APe * Pe|d = 77% * 0.5 = 38.5%
Since such a high proportion of cases are cocaine users, eliminating that exposure would eliminate a great deal of low birth weight cases.
Question 11
APe = ((Re-Ru)/Re) x 100 = ([(27/482)-(77/1908)]/(27/482)) x 100 = 28%
Question 12
APt = ((Rt-Ru)/Rt) x 100 = ([(104/2390)-(77/1908)]/(104/2390)) x 100 = 7%
Question 13
Yes
Yes
Yes
No (this measure requires information about the proportion of the population that is exposed)
Question 14
a)
Deaths
Person-time
Women
15
728
Men
31
1988
b)
Incidence density ratio (IDR)
c)
IDR = IDwomen = 15/728 = 1.32
IDmen 31/1988
The rate of mortality among women was 1.32 times the rate of mortality among men in homeless shelters in New York City between 1987 and 1994.
d)
IDD = IDwomen – IDmen = 15/728 – 31/1988 = 0.0050 deaths per person-year
Between 1987 and 1994, there was an excess of 0.0050 deaths per person-year (5.0 deaths per 1000 person-years) among women living in homeless shelters compared to men in New York City.
Question 15
A.
Low Birth Weight
Normal Birth Weight
Total
Cocaine Users
110
289
399
Cocaine Nonusers
1073
15896
16969
Total
1183
16185
17368
B.
Risk = New Cases / Population at Risk = 110/399
= 0.276 or 27.6 per 100 cocaine users or 27.6% (either answer is acceptable)
C.
Relative Risk = Risk Users / Risk Nonusers
RR = 0.276 / 0.063 = 4.38
D.
The risk of low birth weight among infants born to mothers who used cocaine is 4.38 times
the risk of low birth weight among infants born to mothers who did not use cocaine, over the study period.
Question 16
They are equally associated with diet, because 1 / 0.60 = 1.67. Remember that the RR is on the multiplicative scale and the equal strength of association in the opposite direction is the inverse (1/RR). Also remember that a ratio of any kind is bounded by 0 at the lower end (in epidemiology, as negative cases cannot occur) and infinity at the higher end.
Question 17
A study examined mortality among homeless shelter residents in New York City from 1987 to 1994. There were 15 deaths observed among women aged 25-34, with 728 person-years of observation. Among men aged 25-34, 31 deaths were observed, with 1988 person-years of observation.
a.
IRR = IRwomen = 15/728 = 1.32
IRmen 31/1988
The rate of mortality among women was 1.32 times the rate of mortality among men in homeless shelters in New York City between 1987 and 1994.
b.
IRD = IRwomen – IRmen = 15/728 – 31/1988 = 0.0050 deaths per person-year
Between 1987 and 1994, there was an excess of 0.0050 deaths per person-year (5.0 deaths per 1000 person-years) among women living in homeless shelters compared to men in New York City.
Question 18
a.
|
|
Intervention
|
Control
|
Totals
|
|
Death
|
9
|
16
|
25
|
|
No Death
|
91
|
84
|
175
|
|
Totals
|
100
|
100
|
200
|
b.
Pintervention = 9/100 = 0.09
c.
Oddsintervention= pintervention/1-pintervention= .09/0.91 = 0.099
d.
Odds ratio = (pintervention/1-pintervention)/(pcontrol/1-pcontrol) = (.09/.91)/(.16/.84) = 0.519 Alternatively, OR=(9/91)/(16/84)=(9*84)/(91*16)=0.519
e.
Cumulative incidence ratio = CIR = CIintervention/CIcontrol = (9/100)/(16/100) = 0.563 over unknown study duration
f.
The OR is always farther from the null than the RR (except in instances of special sampling designs, to be covered later in the course). (0.519 is farther from 1.0 than 0.563)
g.
Var[ln(OR)] = (1/a) + (1/b) + (1/c) + (1/d) = (1/9)+(1/91)+(1/16)+(1/84) = 0.197
Var[ln(CIR)] = (b / a (a+b)) + (d / c(c+d)) = [91/(9*(9+91))]+ [84/(16*(16+84))] = 0.154
Question 19
The odds of laboratory workers in company X developing skin cancer is 3.2 times the odds of accountants in company X developing skin cancer, over the study period.
Question 20
a.
Age-standardized. The percent of people over 65 has increased from 1940 to 1980, so examining the crude rates would make it appear that the cancer mortality rate has increased when the increase in crude rates is really largely attributable to the change in the age distribution in the population. Age-standardized rates would adjust for the difference between the age distributions, making the rates comparable. (Hennekens pg 68)
b.
Total
1,267
308,000
1,267/308,000 =
0.00411 =
4.11/1000
3,290
940,000
3,290/940,000 =
0.00350 =
3.50/1000
The crude mortality rate for A is 4.11/1,000 and for B is 3.50/1,000. Population B appears healthier, or at least less likely to die, at least on a crude level.
c.
Population A
Population B
Age
Deaths
Population
Rate (A)
Deaths
Population
Rate (B)
0‑4
80
22,000
80/22,000=0.004
400
100,000
400/100,000=0.004
5‑19
97
85,000
0.001
340
290,000
0.001
20‑44
140
100,000
0.001
300
300,000
0.001
45‑64
350
70,000
0.005
850
170,000
0.005
65+
600
31,000
0.019
1,400
80,000
0.018
Total
1,267
308,000
3,290
940,000
The age-specific mortality rates are nearly identical.
d.
Age (yrs)
U.S. 1970 Population size
Population A Rate
Expected Deaths
Population B Rate
Expected Deaths
0‑4
9,500
0.004
0.004*9,500=38
0.004
0.004*9,500 =38
5‑19
22,000
0.001
22
0.001
22
20‑44
41,000
0.001
41
0.001
41
45‑64
30,000
0.005
150
0.005
150
65+
10,000
0.019
190
0.018
180
Total
198,000
441
431
The adjusted numbers of deaths are nearly the same.
e.
Population A Age adjusted total rate = 441/198,000 = 0.00223 Population B Age adjusted total rate = 431/198,000 = 0.00218
They have nearly the same adjusted total mortality rates.
*Remember, that Direct Adjustment uses the observed rates, and a standard population.
Question 21
C.
Population A (crude mortality risk: 503.3/100,000) appears to be healthier than Population B (crude mortality risk: 701.3/100,000) when we examine the crude mortality risks for the entire population.
D.
The age-specific mortality risks for population A and B are exactly the same for all age strata.
E.
The two populations might be equally healthy, but the difference in their overall crude mortality risk is due to differences in their age distribution, not differences in their actual health status. (Examine the population size columns).
G.
The populations are the same at all age-specific strata for mortality risks (except for the overall population mortality risk, which we saw from part C is an artifact due to different age distributions within the population),nso the populations are equally healthy.
So although the total expected deaths are different when you calculate using the crude mortality risk from part B, if you were to sum up the number of expected deaths as calculated here in table 2, we’d see that these populations are equally healthy when adjusted for age
Question 22
Group A: 60/600 = 0.10
Group B: 63/600 = 0.10
The age-standardized rates are very similar.
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