epidemiology for MPH, Lecture notes of Epidemiology

Download epidemiology for MPH and more Lecture notes Epidemiology in PDF only on Docsity! Comparing Disease Occurrence By Tilaye Workineh (BSc, MPH/Epid, Asst. Prof. of Epidemiology) Lecture 3 Outline = Introduction " Absolute comparison of disease occurrence " Relative comparison of disease occurrence - Epidemiologic measures of association b/n exposure and outcome 2 comparing disease occurence Aug 2021 Absolute comparison methods A. Attributable risk /AR/ " AR indicate how much of the risk is due to (attributed to) the exposure. " Quantify the excess risk among the exposed that can be attributable to the exposure " This is simply the risk difference b/n the Incidences for the two groups: AR = Risk in exposed - Risk in non-exposed = Cl Exposed — Cl Unexposed 5 comparing disease occurence Aug 2021 Eg. Use the following 2x2 table to calculate risk difference in IMR among illiterate and educated mothers MATERNAL NO. OF NO. IMR/ RISK EDUCATION | INFANTS DEATH | 1000 RATIO ILLITERATE =| 5021 579 115.3 |1.59 JUNIOR+ 1368 99 72.4 1.00 TOTAL 6389 678 106.1 |- = U ucated mothers - IMR among educated mothers = 115.3/1000-72.4/1000 = 42.9/1000 Conclusion: 42.9/1000 live birth infants die because of mothers being uneducated/excess death due to illiteracy/ 6 comparing disease occurence Aug 2021 B. Attributable Risk Percent (AR%) "AR% among exposed: the proportion of the disease that could be prevented by eliminating the exposure. " This number refers to the proportion of the risk among the exposed pop? which could be attributed to the exposure AR% = Risk in the exposed - Risk in unexposee¢—*100 Risk in exposed AR% = Cl, - Cl) x100 7 Cl comparing disease occurence Aug 2021 1 “Estimate the impact of HRT exposure on: i) users and ii) the entire community (where fAaneaihkial\ Exposure Number of Person-years of yeh cases of breast follow-up cancer None 923 344,942 Any estrogen 393 120,356 or progestin use Total 1,316 465,298 “wnat is the question asking for? p> 10 comparing disease occurence Aug 2021 Answer 1 AR = Cl, - Clo = 393/120356 - 923/344,942 = 3.265/1000 PYs - 2.676/1000 PYs = 58.9 cases/100,000 person-years Exposure Number of Person-years hr par} cases of breast | of follow-up cancer None 923 344,942 Any estrogen 393 120,356 or progestin use Total 1,316 465,298 Answer 2 AR% = Cl, - Clo Cl, = 3.265 / 1000 PYs - 2.676 / 1000 PYs 3.265 / 1000 PYs = 18.0% Exposure Number of Person-years yeh cases of of follow-up breast cancer None 923 344,942 Any estrogen 393 120,356 or progestin USC T. |= | 31 3243120 AckKE 700 Association b/n exposure & Disease " The basic question in identifying association is: Is there excess risk associated with a given exposure? (is exposure and outcome related?) "Methods: Use one of the epidemiologic study designs Cohort 15 Case-control occurence Aug 2021 Relative Risk/Risk ratio/ " Assess the incidence Rate (IR,_.) of disease among exposed group (absolute Risk) " Assess the incidence (IR;.) of disease among unexposed group (absolute Risk) e.g. Coronary Heart Disease (CHD) Risk among Smokers vs non- smokers 1-year risk of CHD among smokers (Cl_;,) vs Non Smoker (CI;.) CHD Yes NoTotal Smokers 84 2916 3000 Non-smokers 87 4913 5000 IRE, = 84/3000 = 28/1000/yr (1-risk of CHD among smokers) IR_.= 87/5000=17.4/1000/yr (1-yr risk of CHD among non-smokers) 16 comparing disease occurence Aug 2021 Cont. Assessment of Excess Risk (Two methods in cohort Design) a.Ratio...RR RR (Ratio of two risks; Risk Ratio; Relative Risk) Cl,, /Cl,. = 28/17.4 =1.6 Interpretation of RR Smokers were 1.6 times as likely to develop CHD as were non-smokers b.Difference ...AR/RD Difference of two risks (Risk Difference)* Cl,.,- Cl. = 28.0 - 17.4 = 10.6 17 comparing disease occurence Aug 2021 Interpretation of RD OR in case-control Study Probability of case being exposed = P.... Probability of case being non-exposed =1-P... Odds of case being exposed = Probability of control being exposed = P control Probability of control being non-exposed =1-P comparing diSease occurence’ Aug 2021 control Derivation of OR in case-control Study Probability of being exposed among | |; J{- e a b a+b a /(a + c) Probability of being non-exposed an| -\& |! |&4 C I(a + c) atc [bed = | Atcebed Odds of being exposed among cases = a/c Probability of being exposed among controls = b/(b + d) Probability of being unexposed among controls = d/(b + d) Odds of being exposed among controls b/d comparing disease occurence Aug 2021 Example OR in case-control Study Past surgery HCV status HCV+ HCV- Yes 59 168 No 54 48 113 216 p> 22 comparing disease occurence Aug 2021 Standardization... * Standardization is process by which you derive a summary figure to compare health outcomes of groups " The process can be used for mortality, natality, or morbidity data "Two methods of standardization: Direct standardization Indirect standardization, 25 comparing disease occurence Aug 2021 1. Direct standardization of Rates " Based On specific mortality/morbidity rates of study pop to compare with standard popn e.g. comparing Age specific mortality rates of two countries " We need | A standard/reference pop» with age specific rates distribution Distribution of age specific rates of study pop» p> 26 comparing disease occurence Aug 2021 Standard Population is representative of the study populations: e.g. US 1940 population, US 2000 population, European standard population, WHO world population, .... Choice of SP is arbitrary since we do not interested in the absolute value of the adjusted rates but interested in the comparison Note: If the mortality rates of a specific community are compared to the national pop", then the national pop» is considered as a “standard” pop». p> 27 comparing disease occurence Aug 2021 table 2: Calculation of tne number or expected deaths for countries A and B applied to a Standard population 0-29 100,000 30-59)65,000 b+ 20,000 —s A a B Expected deaths Expected deaths 29 .0012 x 100,000 = 120 .0042 x 100,000 = 420 30-59 0036 x 65,000 = 234 .0055 x 65,000 = 357.5 }o0+ .048 x 20,000 = 960 .05 x 20,000 = 1,000 ae 1,777.5 {ieaths lage adjusted rate 1,314/ 185,000 = 7.1 per 1,000 |1,777.5/185,000 =9.6 per 1,000 pyrs pyrs [ Age standard rate ratio (B:A)= 9.6/7.1 = 1.35 “tnterpretation:—after—controltting Total |185,000 for the confounding affects of age, the mortality in Country B is 35% higher than in country A "The reason for the d/ce b/n the crude mortality rates b/n country A and 8B is that these two pap"s have markedly.d/nt.age:structuses Example 2 " Given the age specific death rate of Sweden & Ethiopia as follows, how do we compare death rate of the two countries based on the given data set? 31 comparing disease occurence Aug 2021 Example ... Age-specific mortality rates (per 1,000 person years) Sweden __ Ethiopia Standard (N) 0-29 1.1 5.3 56,000 30-59 3.6 5.2 “33,000 60 + 45.7 50.1 11,000 Age-standardiz@l¥ #aefust@@pArtality rate (Sweden) = (1.1 x 56,000) + (3.6 x 33,000) + (45.7 x 11,000) 100,000 =6.8 per 1,000 PErSQAPYBAdSease occurence Aug 2021 2. Indirect standardization (IDS) "IDS is used when ASMER for the study pop” are not available, " Specific rates may be unstable b/c they were based on small numbers "In such cases, where ASMR for the study pop" are not available, we need to use rates of the standard pop’, = This method of standardization is called indirect standardization 35 comparing disease occurence Aug 2021 Indirect Standardization ... Indirect method requires: " Age structure of the study population * Total death in the study population, " Age-specific rates for a standard population, "Summary figure is a standardized mortality ratio (SMR), SMR = Observed x 100 Expected 36 comparing disease occurence Aug 2021 Procedure for indirect standardization 1. Choose a reference or standard population 2. Calculate the observed # of deaths in the popn (s) of interest 3. Multiply # of people in each age group of the pops) of interest by age-specific mortality rate in the comparable age group of reference pop» 4.Sum the total # of expected deaths for each popn of interest 5. Divide the total number of observed deaths of the population(s) of interest by the expected deaths 37 comparing disease occurence Aug 2021 Example 1... Age Standard Ethiopian Expected Rate population deaths 0-29 0.0011 741,000 815.1 30-59 0.0036 275,000 990.0 60+ 0.0457 59,000 2696.3 4501.4 Total expected deaths = 4501.4 Total deaths Ethiopia = 8281.0 (given above) 40 comparing disease occurence Aug 2021 Example 1... » Standardised mortality ratio (SMR) = Observed deaths = O/E Expected deaths SMR (%) = 8281 *100% = 184 % 4501.4 Interpretation : Mortality rate in Ethiopia was 84% higher than that of Sweden based on the data during the given period 41 comparing disease occurence Aug 2021 Example 2 "In 1951, census reported popn of male farmers by age with total observed death of 1,464 people = From the census, standard death rate of the country’s pop was also available " Compare the death rates of the male farmers with that of the general pop of the country using the following data comparing disease occurence Aug 2021 Interpretation of SMR " SMR = 100% Rates are similar to the standard population, " SMR < 100% Fewer deaths occurred than expected (rates are lower than the standard) " SMR > 100% More deaths occurred than expected (rates are higher than the standard) p 45 comparing disease occurence Aug 2021 p> 46 comparing disease occurence Aug 2021