Statistik for MPH: september Analyse af kohortestudier, rater, standardisering (Silva: 70-80,

Transcription

1 Statistik for MPH: september Analyse af kohortestudier, rater, standardisering (Silva: 70-80, ) Per Kragh Andersen 1

2 Fra den. 2. uges statistikundervisning: skulle jeg gerne 1. forstå at den relative risiko er et fornuftigt mål for association mellem determinant og udfald i kohorteundersøgelser 2. kunne beregne relativ risiko med tilhørende sikkerhedsgrænser ud fra en given tabel 3. kunne beregne chi-square teststørrelsen for hypotesen: relativ risiko=1 ud fra en given tabel 4. så småt begynde at vænne mig til begreberne: nulhypotese, teststørrelse, fordeling af teststørrelse, P -værdi, signifikansniveau, accept/forkastelse af hypotese 5. forstå at der er en tæt sammenhæng mellem sikkerhedsgrænser og hypotesetest 2

3 Fra den. 2. uges statistikundervisning: behøver jeg derimod ikke nødvendigvis 1. at have forstået, hvordan formlen for SD(ln R) fremkommer 2. at have forstået, hvordan standard afvigelsen i nævneren af teststørrelsen fremkommer 3. at have forstået, hvorfor teststørrelsens værdi netop skal slås op i χ 2 -tabellen med 1. frihedsgrad 4. at have forstået alle begreberne nævnt under 4. ovenfor til bunds 3

4 Cohort studies Classical example, The Evans County Study: 609 white men, aged 40-76, followed for 7 years from CHD CHL>260mg/100mL Yes No Total Yes No Total Summarizing data in this way, it has been assumed that everyone has been followed for 7 years - this is more an exception than the rule: People leave the study due to loss of follow-up People may enter the study at different time points We wish to take the varying times at risk into account. 4

5 The 7-year CHD-risk ( cumulative incidence ) expresses the proportion of the cohort that has a CHD-event during the 7 years of follow-up. It is, therefore, the 5-year risk and the 10-year risk Alternative frequency measure: the rate, r(t) expressing risk per time unit at time t ( speed at t). For h > 0 (small): r(t) Prob(Disease before time t + h given no disease time t)/h. We estimate a constant rate as (Silva, p.63): r = number of (CHD-) events obs. total person-time at risk = a y. Alternative names: incidence rate, incidence density, hazard function, hazard rate. 5

6 Calculation of person-years Calculation of person-years of observation for six individuals (see also Silva, p.62). Date of birth Date observation Date observation Death began ended (0=no, 1=yes) Jan. 4, 1897 July 11, 1954 Dec. 31, Sept. 5, 1884 Aug. 3, 1954 Nov. 25, Dec. 16, 1904 Oct. 25, 1954 Dec 31, Jan. 16, 1899 Nov. 1, 1954 Dec. 31, Apr. 9, 1912 Feb. 19, 1957 Dec. 31, Feb. 22, 1910 Dec. 8, 1957 Aug. 18, Years Months Days Years

7 r = years = per year = 51 per 1000 years Note that while the risk is between 0 and 1 (it is a probability), the rate has no upper limit (the value depends on the time unit used). Interpretation of a rate of r = per year: If we observe a cohort where the rate has this value and somehow collect 1000 person years of observation, then we expect to see 51 events. Some times: approximations are used for person-years when dates are unavailable (see, e.g. Silva p.71). What does the fraction 2 6 tell us? Not much! If it is to be thought of as a risk estimate then it is quite unclear to which follow-up period it corresponds. 7

8 Confidence interval for a rate Better (and more complicated) method: 1. Calculate l = ln(r) 2. Calculate L 2 = l a Calculate L 1 = l a 3. The desired 95% confidence limits are: from exp(l 1 ) to exp(l 2 ) Example: l = ln(0.051) = 2.98 L 2 = = 1.59 L 1 = = % confidence limits from exp( 4.36) = per year to exp( 1.59) = per year. 8

9 Relation between rate and risk If the rate is (constant) = r (per year) what is then the T -year risk (if there are no other competing events)? h = T N h = T N h = T N 0 T Probability of no event before T = (1 r h) (1 r h) (1 r h) = (1 r h) N = ( 1 r T N ) N exp( r T ). That is: T -year risk = 1 exp( rt ). If the cumulative rate rt is small then T -year risk rt (Silva, p.77). 9

10 Example Assume that r = per year and consider a follow-up period of T = 5 years. The T -year risk is then 1 exp( rt ) = 1 exp( ) = 1 exp( 0.255) = The corresponding cumulative rate is r T = = Note that: 1 exp( rt ) = < rt = Note also that it is quite smart that, given a (constant) rate, it is possible to estimate the risk for any chosen follow-up period, T. 10

11 Corresponding measure of association RATE RATIO, some times called Relative risk which is bad terminology, but not totally silly if the disease is rare, because then: Risk 1 Rate 1 T Risk 2 Rate 2 T = Rate 1. Rate 2 95% confidence limits for rate ratio, RR: 1. Calculate ln(rr) 2. Calculate L 2 = ln(rr) L 1 = ln(rr) The 95% confidence limits are: from exp(l 1 ) to exp(l 2 ) 1 a a 2 1 a a 2 11

12 Example: Nurses health study (p. 179) BC cases Person-years Age Group OC Non-OC OC Non-OC a 1 = 204 a 2 = 240 y 1 = y 2 = Table 8.12: Cohort study of breast cancer (BC) and oral contraceptive (OC) use. OC users Non-OC users r O = r N = = per year l O =ln( ) l N =ln( ) = = L 2 = L 2 = = L 1 = L 1 = exp(l 1 )= per year exp(l 2 )= per year exp(l 1 )= per year exp(l 2 )= per year 12

13 Example (contd.) RR = = ln(rr) = L 2 = = L 1 = = exp(l 1 ) = 0.964, exp(l 2 ) =

14 Alternative measure of association: rate difference The rate difference is r 2 r 1 and is estimated by a 2 y 2 a 1 y 1 with SD(r 2 r 1 ) = a1 y a 2 y2 2. In the example this is r O r N = per year (or 30 per years) with 95% confidence interval from 204 r O r N = = (per year) to 204 r O r N = = (per year). 14

15 Exercise: 1. Calculate the rates for stomach cancer (SC) in Cali and Birmingham (with 95% confidence limits) based on the table in Silva, p Calculate the rate ratio with 95% confidence limits 3. Calculate the chi-square test (to be defined in the following) for comparing the two rates. 15

16 Solution (1) SC cases Person-years Cali Birmingham Cali Birmingham r C = = per year r B = ln(r C ) = ln(r B ) = L 2 = L = = L 1 = L = = exp(l 1 )= per year exp(l 2 )= per year exp(l 1 )= per year exp(l 2 )= per year = per year 16

17 Solution (2) RR = = 0.59, ln(rr) = L 2 = = 0.447, 1 L 1 = = 0.618, exp(l 1 ) = 0.54, exp(l 2 ) =

18 Test for comparing two rates Nurses health study: Events (BC) Person-years OC-users 204 (a 1 ) (y 1 ) Non-OC-users 240 (a 2 ) (y 2 ) 444 (a) (y) Here: a = a 1 + a 2, y = y 1 + y 2. 18

19 Mantel-Haenszel idea Compare: OBServed = 204 = a 1 with : EXPected = = y 1 a y = 187.6, SD(a 1 ) = = a y1 y 2 y y = Test statistic = ( OBS EXP SD ) 2 ( = ) = The chi-square table (1 d.f.) gives 0.10 < P <

20 Solution to exercise (3) OBS = EXP = = SD = = 27.0 ( ) 2 OBS EXP = chi-square (1 d.f.) SD P <

21 And now for something completely different 21

22 How can we adjust/standardize for age when comparing two groups? Two types of standardization exist: DIRECT cf. Silva INDIRECT p Direct : Compare weighted averages of the age-specific rates in the two groups using some standard age-distribution as weights Indirect : In each group, compare the observed number of events with what one would have expected, had the group been exposed to some standard age-specific rates. Other ways of adjustment (also for other factors than age): STRATIFIED REGRESSION ANALYSIS 22

23 Example: 1970 US mortality data California (a) Maine (b) Age Pop.in No.of Rate per Pop.in No.of Rate per 1000 deaths 1000 ys deaths 1000 ys. < Total (Kahn & Sempos, 1989) The total rates are often denoted crude rates, and the values are: (a) California: = 8.3 per 1000 ys. (b) Maine: = 11.1 per 1000 ys. leading to RR =

24 The age distributions in California and Maine differ so the apparent difference may be (partly) ascribed to this. California Maine Age Pop. in 1000 % Pop. in 1000 % < Total

25 Example: 1970 US mortality data United States Age Pop.in Weight No.of Rate per 1000 deaths 1000 ys. < Total Directly standardized rates: DSR a = = 8.8 per 1000 ys. DSR b = = 9.9 per 1000 ys. leading to RR =

26 Example from Silva, p.71 Male stomach cancer cases: Cali Birmingham Age Popu- Person- No.of Rate per Popu- Person- No.of Rate per lation years cancers ys. lation years cancers ys Total Age distributions (%): Age Cali Birmingham Standard Total

27 The age distributions in Cali and Birhingham differ so the difference between the crude rates may be (partly) ascribed to this. Directly standardized rates: DSR C = = DSR B = = (both per ys.) 27

28 Indirect standardization (a) California: Obs. number of deaths = Expected number of deaths = = Standardized mortality ratio: SMR a = = = (b) Maine: Obs. = Exp. = SMR b = = =

29 If one wants to express the results as rates then indirectly standardized rates may be computed: ISR a = SMR a = 8.8 per 1000 ys., ISR b = SMR b = 9.9 per 1000 ys., where per year=9.4 per 1000 ys. is the overall crude rate in the standard population. Silva (p. 74) illustrates indirect standardization by computing the number of cancer cases expected in Cali if the age-specific rates from Birmingham apply. 29

30 Discussion Standardization is a classical technique for age-adjustment. For comparing groups statistically, stratified (and regression) analysis (to be discussed later) is superior. The average calculated in direct standardization frequently provides an over-simplification for analysis (but it may be OK for graphical displays etc.). Indirect standardization (SMR) may be used for comparing a sample with standard rates (for this, we need a confidence interval, see below) but for comparing two (or more) groups (samples) stratified (and regression) analysis is better. 30

31 Confidence interval for SMR Best method (using ln!): 1. Calculate ln(sm R) 2. Calculate ln(smr) OBS = L 2 ln(smr) OBS = L 1 3. The desired 95% confidence limits are from exp(l 1 ) to exp(l 2 ) Example: Maine compared to US: 1. ln(1.050) = = L 2 = 0.068, = L 1 = exp(l 1 ) = 1.030, exp(l 2 ) = Exercise: California! 31

32 Solution SM R for California: 1. ln(smr) = = L 1 = = L 2 = = exp(l 1 ) = exp(l 2 ) =