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Measures of Association (2) Attributable Risks and fractions

This article discusses the measures of association, such as attributable risks and fractions, in epidemiological studies. It explores examples and calculations of attributable risks and percentages in both cohort and case-control studies.

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Measures of Association (2) Attributable Risks and fractions

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  1. Measures of Association (2)Attributable Risks and fractions October 19 2004 Epidemiology 511 W. A. Kukull

  2. Total risks in Exposed and Unexposed Cases per 105 pop/yr Incidence due to Exposure Background Risk

  3. Attributable Risk (AR)for the exposed group • The exposure under consideration must be considered causal • Among exposed, how much disease is due to our exposure of interest? • how many cases can be prevented? • AR = [incidence in the exposed] - [incidence in the non exposed] • AR is the difference of incidence rates • it is also called the Risk Difference (RD)

  4. AttributableRisk in the exposedExample(1) hypothetical data • Cohort (ID) study of hip fracture incidence • Low,soft chairs = 0.41 per 105 person-days • Other chairs= 0.29 per 105 person-days • AR = 0.41 - 0.29 = 0.12 per 105 p-d • This means: 0.12 /105 (of the 0.41/105 ) hip Fx that occur among persons who use low soft chairs are due to those chairs (if causal)

  5. Attributable Risk for the exposed group Example(2) • Cohort (CI) : OC use and Bacteriuria • in OC users 56.0 per 1000 pop. per year • in non users 40.3 per 1000 • AR= 56.0 - 40.3 = 15.7 per 1000 • This means that 15.7 of the 56.0/1000 cases that occur annually in the population among OC users are due to OC (if causal) • Or, 15.7 cases/1000 pop. could be prevented

  6. Attributable Risk for the exposed group Example(3) • Cohort (CI) : CHD incidence among smokers and non smokers: • Smokers: 28.0 per 1000 pop. per year • Non Smokers 17.4 per 1000 • AR = 28.0 - 17.4 = 10.6 per 1000 pop. per yr • 10.6 cases per 1000 of the 28.0 cases per 1000 of CHD that occur among smokers annually are due to smoking (if causal)

  7. AR % or Attributable fraction ={Proportion of incidence among the exposed that is attributable to the exposure} = AR / Ie • Cohort (ID): hip fx incidence in persons using low, soft chairs vs. other chairs • LSC: 0.41 per 105 person-days • Other: 0.29 per 105 person-days • AR% = (0.41 - 0.29) / (0.41) = (0.12) / (0.41) • AR% = 0.292 or 29.2% • This means that 29.2 % of hip fx’s among persons using LSC are due to LSC (if causal)

  8. AR%Example (2) • Cohort (CI) : OC use and Bacteriuria incidence • in OC users 56.0 per 1000 pop. per year • in non users 40.3 per 1000 • AR%= (56.0 -40.3) / (56.0) = 15.7 / 56.0 ~ 28% • This means that 28% of the 56.0 cases/1000 that occur in OC users are due to OC (if causal) • Or, these 28% could be prevented by eliminating OC use

  9. AR% Example (3) • Cohort (CI) : CHD incidence among smokers and non smokers • Smokers: 28.0 per 1000 pop. per year • Non Smokers 17.4 per 1000 • AR% = (28.0 - 17.4) / (28.0) = 10.6 / 28.0= 37.8% • This means that 37.8% of the 28.0 per 1000 cases of CHD that occur among smokers are due to smoking (if causal)

  10. Population AR:Attributable risk for a total population • What will be the impact of eliminating the “exposure” upon the “disease” rates of our State (or community) ? • PAR = I(pop) - I(not exposed) • But we hardly ever know I(pop) • what will we do? • I(pop) = I(e) P(e) + I(ne) P(ne) • P(e) is the proportion “ exposed”

  11. PAR (2) • CHD and smoking (previous example) • I (e) = 28.0 per 1000; I(ne) = 17.4 per 1000 • Suppose we have community survey data that says 44% of our community are smokers : P(e) = .44 • Then: I(pop) = (.028)(.44) + (.0174)(.56) • I(pop) = .0221 or 22.1 per 1000 • PAR = I(pop) - I (ne) = ?

  12. Population Attributable Risk Percent {PAR%} • The proportion of the incidence in the total population that is attributable to exposure • PAR/ I(pop) = PAR % {also AR ( P(e))} • From previous slide: PAR = 4.7 per 1000 • PAR % = (.0047) / (.0221) = .213 or 21.3% • So how much of a reduction in total incidence of CHD could we expect?

  13. Attributable Risks:Case Control Studies • AR and PARcannot be calculated directly from a case-control study • the incidence rates in exposed and unexposed is necessary [and P(e) ] • AR% and PAR%can be calculated in a case-control study, given: • RR estimate (e.g. Odds Ratio) • Proportion exposed in population [P(e)]

  14. Case-Control Studies:AR% and PAR% (OR - 1) AR% = x 100 OR (Pe) (OR - 1) [if Pe in the control group is an estimate of population exposure] PAR% = (Pe) (OR - 1) + 1 PAR% = AR% x (proportion of exposed cases) N.B.: The association must be “causal”

  15. Useful equivalents • In a case-control study Incidence is not available directly • Incidence in the exposed = RR [I(not exposed)] • assume that the OR is an estimate of RR • Then: AR% = (OR - 1) / OR

  16. Hypothetical Case-control study Osteoporosis YES NO Alcohol Abuse Hx 113 75 YES 637 1425 NO • We observed an OR = • What percent of osteoporosis among the persons with a hx of alcohol abuse is due to a hx of alcohol abuse?

  17. Useful equivalents • If we know the proportion of the population exposed [P(e)] and the RR (or OR), then: • PAR%= {[P(e) (RR -1)] / [P(e) (RR-1) +1]} • What if exposed persons were 2.7 times more likely to get the specific disease and and 10% of the population is exposed? • What percent of the disease in the population could be prevented by removing the exposure?

  18. Another look… • If you are given incidence in the population (I(pop) ) then: • PAR% = PAR divided by [ I(pop)] • If incidence in the population is 4351.5/105 and 316 cases per 105 population are due to the exposure; what percent of the disease in the population could be prevented by eliminating the exposure?

  19. One more time… • PAR = P(e) [AR] • Suppose 20.2% of the population was exposed to HRT; and the 1566/105 cases of CHD among the exposed were due to HRT. • Assuming HRT is a cause of CHD, what is the excess risk of CHD in the population that is due to HRT?

  20. Exposures and Populations • AR among the exposed will always be greater than the corresponding PAR. • To find the impact of eliminating more than one exposure for a specific disease • You cannot simply add PAR%s for each independent exposure; you must study the incidence in the jointly exposed and calculate a new PAR%.

  21. PAR% depends on exposure prevalence and the RR estimate (roughly like this) 100 RR=10 80 PAR% RR=5 60 RR=3 RR=2 40 20 100% Exposure Prevalence

  22. Summary Table In EXPOSED In POPULATION Incidence attributable to exposure I (e) - I (ne) I (pop) - I (ne) {AR} {PAR} Proportion of incidence attributable to exposure I (e) - I (ne) I (pop) - I (ne) I (e) I (pop) {AR %} {PAR % } N.B. “Useful equivalents” also apply

  23. Comparison *Age adjusted

  24. Comparison table comments • Relative risk of death for smokers from Lung Ca and from CHD ? • Attributable risks • Greatest impact on mortality ?

  25. Terminology Table [AR ] [AR% ] [PAR ] [PAR%]

  26. Summary • Measures of risk in the exposed • Measures of risk in the population • Excess disease associated with exposure • Usefulness in Public health

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