Survival analysis

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  • Example of right left interval
  • Journal articles exampleexpected time-to-event = 1/incidence rate
  • numerical
  • Breslau, a city in Silesia which is now the Polish city Wroclaw.)
  • The actuarial method is not computationally overwhelming and, at one time, was the predominant method used in medicine.
  • Steps
  • The actuarial method assumes that patients withdraw randomly throughout the interval; therefore, on the average, they withdraw halfway through the time represented by the interval. In a sense, this method gives patients who withdraw credit for being in the study for half of the period.
  • The results from an actuarial analysis can help answer questions that may help clinicians counsel patients or their families. For example, we might ask, If X is the length of time survived by a patient selected at random from the population represented by these patients, what is the probability that X is 6 months or greater? From Table 5, the probability is 0.80, or 4 out of 5, that a patient will live for at least 6 months.
  • In actuarial science, a life table (also called a mortality table or actuarial table) is a table which shows, for a person at each age, what the probability is that they die before their next birthday.
  • 2.0
  • Wilcoxon rank sum test ????
  • In words: the probability that if you survive to t, you will succumb to the event in the next instant.
  • Survival analysis

    1. 1. Survival analysis Dr HAR ASHISH JINDAL JR
    2. 2. Contents • • • • • • • • • Survival Need for survival analysis Survival analysis Life table/ Actuarial Kaplan Meier product limit method Log rank test Mantel Hanzel method Cox proportional hazard model Take home message
    3. 3. Survival • It is the probability of remaining alive for a specific length of time. • point of interest : prognosis of disease e.g. – 5 year survival e.g. 5 year survival for AML is 0.19, indicate 19% of patients with AML will survive for 5 years after diagnosis
    4. 4. Survival • In simple terms survival (S) is mathematically given by the formula; S = A-D/A A = number of newly diagnosed patients under observation D= number of deaths observed in a specified period
    5. 5. e.g For 2 year survival: S= A-D/A= 6-1/6 =5/6 = .83=83%
    6. 6. e.g For 5 year survival: S= A-D/A
    7. 7. Censoring • Subjects are said to be censored – if they are lost to follow up – drop out of the study, – if the study ends before they die or have an outcome of interest. • They are counted as alive or disease-free for the time they were enrolled in the study. • In simple words, some important information required to make a calculation is not available to us. i.e. censored.
    8. 8. Types of censoring Three Types of Censoring Right censoring Left censoring Interval censoring
    9. 9. Right Censoring • Right censoring is the most common of concern. • It means that we are not certain what happened to people after some point in time. • This happens when some people cannot be followed the entire time because they died or were lost to follow-up or withdrew from the study.
    10. 10. Left Censoring • Left censoring is when we are not certain what happened to people before some point in time. • Commonest example is when people already have the disease of interest when the study starts.
    11. 11. Interval Censoring • Interval censoring is when we know that something happened in an interval (i.e. not before starting time and not after ending time of the study ), but do not know exactly when in the interval it happened. • For example, we know that the patient was well at time of start of the study and was diagnosed with disease at time of end of the study, so when did the disease actually begin? • All we know is the interval.
    12. 12. 2 possibilities B AND E Survived 5 years S=6- 2/6=4/6=0.67=67% FOR 5 YEAR SURVIVAL B and E did not survive for full 5 years . S=6-4/6= 2/6= 0.33=33% Conclusion: since the observations are censored , it is not possible to know how long will subject survive . Hence the need for Special techniques to account such censored observations
    13. 13. Need for survival analysis • Investigators frequently must analyze data before all patients have died; otherwise, it may be many years before they know which treatment is better. • Survival analysis gives patients credit for how long they have been in the study, even if the outcome has not yet occurred. • The Kaplan–Meier procedure is the most commonly used method to illustrate survival curves. • Life table or actuarial methods were developed to show survival curves; although surpassed by Kaplan–Meier curves.
    14. 14. What is survival analysis? • Statistical methods for analyzing longitudinal data on the occurrence of events. • Events may include death, injury, onset of illness, recovery from illness (binary variables) or transition above or below the clinical threshold of a meaningful continuous variable (e.g. CD4 counts). • Accommodates data from randomized clinical trial or cohort study design. 15
    15. 15. Randomized Clinical Trial (RCT) Disease Random assignment Target population Intervention Diseasefree, at-risk cohort Disease-free Disease Control Disease-free Timeline
    16. 16. Randomized Clinical Trial (RCT) Cured Random assignment Target population Treatment Patient population Not cured Cured Control Not cured Timeline TIME
    17. 17. Randomized Clinical Trial (RCT) Dead Random assignment Target population Treatment Patient population Alive Dead Control Alive Timeline TIME
    18. 18. Cohort study (prospective/retrospective) Disease Exposed Target population Disease-free cohort Disease-free Disease Unexposed Disease-free Timeline TIME
    19. 19. Objectives of survival analysis  Estimate time-to-event for a group of individuals, such as time until second heart-attack for a group of MI patients.  To compare time-to-event between two or more groups, such as treated vs. placebo MI patients in a randomized controlled trial.  To assess the relationship of co-variables to time-toevent, such as: does weight, insulin resistance, or cholesterol influence survival time of MI patients? 20
    20. 20. nominal Censored observations Kaplan- meier or Actuarial Scale of measurement of dependent variable numerical Censored observations Cox Proportional Hazard Model
    21. 21. Life table
    22. 22. History of life table • John Graunt developed a life table in 1662 based on London‟s bills of mortality, but he engaged in a great deal of guess work because age at death was unrecorded and because London‟s population was growing in an unquantified manner due to migration.
    23. 23. HISTORY OF THE LIFE TABLE Edmund Halley (1656 – 1742) - ‘An estimate of the Degree of the Mortality of Mankind drawn from the curious Table of the Births and Funerals at the city of Breslau’.
    24. 24. Life table/ Actuarial methods Actuary means “someone collection and interpretation of numerical data (especially someone who uses statistics to calculate insurance premiums)” Known as the Cutler–Ederer method (1958) in the medical literature Widely used for descriptive and analytical purposes in demography, public health, epidemiology, population geography, biology and many other branches of sci ence. Describe the extent to which a generation of people dies off with age.
    25. 25. Life table A special type of analysis which takes into account the life history of a hypothetical group or cohort of people that decreases gradually by death till all members of the group died. A special measure not only for mortality but also for other vital events like reproduction, chances of survival etc.
    26. 26. Uses & Applications The probability of surviving any particular year of age Remaining life expectancy for people at different ages Moreover, can be used to assess: At the age of 5, to find number of children likely to enter primary school. At the age of 15, to find number of women entering fertile period. At age of 18, to find number of persons become eligible for voting.
    27. 27. Uses & Applications Computation of net reproduction rates. Helps to project population estimates by age & sex. To estimate the number likely to die after joining service till retirement, helping in budgeting for payment towards risk or pension.
    28. 28. If we want to construct a life table showing survival & death in a cohort of 150000 babies STEPS
    29. 29. Steps 1. These 1,50,000 babies born at same time were subjected to those mortality influences at various ages that influence population at certain period of time. 2. On the basis of mortality rates operating, we can estimate what number would be alive at first birthday by applying mortality rates during first year on them. 3. By applying mortality rates of second year on numbers of babies surviving at the end of the first year, we estimate number who would survive at the end of second year. 4. Similarly for other ages by applying mortality rates of selected year follow them till all members of cohort die. 5. These number of survivors at various ages form the basic data set out in a life table. 6. From these numbers we can calculate the average life time a person can expect to live after any age.
    30. 30. SECOND COLUMN (lx) SIXTH COLUMN (Px) (Lx) THIRD COLUMN ) FIFTH COLUMN(ex0(dx) age ‘x’ out of EIGHT aggregate • lx is thenumber of years are expected to attain exact of lx persons It is of persons who lived in COLUMN by cohort COLUMN person survive till his x (qx) • ItItmeasures number FOURTH of precise‘lx’ x willof given reaching •• number of births. thatof person ofamong aagewho dieabefore agenext the probability a persons Itisgives the average numbers years (Tx) between ages x & rate to which population x+1 •‘x+1’ the mortality SEVENTH COLUMN groups would be exposed, but it It is b’daybenumber 1,42,759 in lx COLUMN (x) ‘0’ year indicates the cannumber same • • Thus isLx=lx-1/2dx astolived by group fromrates obtained fromthem die. It is the expected theFIRST column against x until all of death age not the of years live under the prevailing mortality conditions. • number that begin their life together and a particular year year ofso Since a dx=lx-(lx+1) either live or die in are running first of life their person must age specific death The age exact years of age starting ••••It gives Tx=(Lx)+(Lx+1)+(Lx+2)+……………Ln. fromby Thus the 2, Lx=108163-1/2×3144=106591 For expectation of life at age x is obtained age OR registration records. qx+px=1 life0table corresponding to ‘x = 0’ • Thusis based =Tx/lx In for qx= on assumption that deaths are evenly distributed life. this the above table •••0,1,2,3…………………99. T0=129197+111899+106591+……..+9+5+2= Lx ex dx/lx. So px=1-qx • throughout the year. against q0=27124/142759=.19000 If x=2, in above 1,15,635 4638611z e20 life table, x=0/ 108163 40.66 •• Similarly dx=142759-115635=27124= indicates the number who have Forx=0,figure = 4397525 then 1 year • completed first year q1=7472/115635=.06462 second & so on. For Similarly for x=1, of life and running the p0=1-0.19000=0.81000 = also 0 dx=115635-108163=7472 • • •If ‘xx=95,then calculated = 1.5 = lx+(lx+1) Lx can 1’ e be = 32 / 21 as: Lx For 95 •• Similarly x=1, p1=1-0.06462=0.93538 2 Table 1. Life table of a birth cohort 1 2 3 4 5 6 7 8 Age X Living at age x (lx) Dying b/w x & x+1 (dx) Mortality rate (qx) Survival rate (px) Living b/w x & x+1 (Lx) Living above age x (Tx) Life expectancy at x (ex0) 0 142759 27124 .19000 .81000 129197 4638611 32.49 1 115635 7472 .06462 .93538 111899 4509414 39.00 2 108163 3144 .02907 .97093 106591 4397525 40.66 3 . . . . . . .
    31. 31. Age x Living Dying Mortality at age x b/w x & rate (qx) x+1 (dx) (lx) 1 2 3 Survival rate (px) 4 Living b/w x & x+1 (Lx) 5 Living > age x (Tx) 6 Life expectancy at x (ex0) 7 8 0 142759 27124 .19000 .81000 129197 4638611 32.49 1 115635 7472 .06462 .93538 111899 4509414 39.00 2 108163 3144 .02907 .97093 106591 4397525 40.66 3 105019 3254 .03098 .96902 103392 4290934 40.86 4 102006 2006 .01967 .98033 101121 4187542 41.05 5 100000 .01710 .98290 99145 4086420 40.86 . . . . . . . . . . . . . . . . . . . . . . . . 95 21 9 .40957 .59043 16 32 1.52 96 12 6 .42932 .57068 9 16 1.34 97 6 3 .44964 .55036 5 7 1.17 98 3 1 .47046 .52954 2 2 0.64 99 1 1 .49176 .50823 … … … 1710
    32. 32. Modified life table 1. For survival in different treatment regimens 2. Arrange the the 13 patients on etoposide plus cisplatin(treatment arm =1) according to length of time they had no progression of their disease. 3. Features of the intervals: 1. arbitrary 2. should be selected with minimum censored observations
    33. 33. Life table for sample of 13 patient treated with etoposide with cisplatin Life Table Survival Variable: Progression-Free Survival ni wi di No. of pts (13) began the study, so n1 is 13 Interval No. No. No. No. of Start entering withdrawn exposed terminal 2 patients are Time Interval du.referredto risk events to as Interval withdrawals (w1). qi = di/[ni- pi = 1–qi si = pipi–1pi(wi/2)] 2…p1 Propn Propn terminating surviving Cumul Propn Surv at End 0.0 13.0 2.0 12.0 1.0 1 patient's disease 0.0833 progressed, referred 0.9167 0.9167 to as a terminal event (d1) 3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021 6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021 9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021 Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–
    34. 34. Assumption: • The actuarial method assumes that patients withdraw randomly throughout the interval; therefore, on the average, they withdraw halfway through the time represented by the interval. • In a sense, this method gives patients who withdraw credit for being in the study for half of the period.
    35. 35. Life table for sample of 13 patient treated with etoposide with cisplatin Life Table Survival Variable: Progression-Free Survival ni wi Interval No. No. Start entering withdrawn Time Interval du. Interval One-half of the number of patients di qi = di/[niwithdrawing is subtracted from the pi = 1–qi (wi/2)] number beginning the interval, so the EXPOSED TO RISK during the period, 13 – No. of2), or 12 in first (½ No. Propn Propn interval. terminal terminating surviving exposed to risk si = pipi–1pi2…p1 Cumul Propn Surv at End events 0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167 3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021 6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021 9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021 Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–
    36. 36. Life table for sample of 13 patient treated with etoposide with cisplatin Life Table Survival Variable: Progression-Free Survival ni wi Interval No. No. Start entering withdrawn Time Interval du. Interval di No. exposed to risk qi = di/[ni- pi = 1–qi si = pipi–1pi(wi/2)] 2…p1 The of No. proportion terminating (q1 Propn Propn = d1/[n1-(w1/2]) is 1/12 = terminal terminating surviving 0.0833. events Cumul Propn Surv at End 0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167 3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021 6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021 9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021 Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–
    37. 37. Life table for sample of 13 patient treated with etoposide with cisplatin Life Table Survival Variable: Progression-Free Survival ni wi Interval No. No. Start entering withdrawn Time Interval du. Interval di No. exposed to risk qi = di/[ni- pi = 1–qi si = pipi–1pi(wi/2)] 2…p1 No. of Propn Propn Cumul Propn terminal proportion surviving (p1 = 1-q1) is 1End Surv at – The terminating surviving events 0.0833 = 0.9167 0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 we are still in 0.8021 because 6.0 5.0 4.0 3.0 0.0 0.0000 9.0 1.0 1.0 0.5 0.0 0.0000 0.9167 the first period, the cumulative survival is 1.0000 0.8021 0.9167 1.0000 0.8021 Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–
    38. 38. Life table for sample of 13 patient treated with etoposide with cisplatin Life Table Survival Variable: Progression-Free Survival ni wi di Interval No. No. No. No. of the Start At entering withdrawn exposed terminal Time beginning of Interval du. to risk events the second four Interval patients withdraw w2 = 4 interval, only 0.0 10 patients 13.0 2.0 12.0 1.0 remain.n2=10 3.0 10.0 4.0 8.0 1.0 6.0 5.0 4.0 3.0 0.0 9.0 1.0 1.0 0.5 0.0 qi = di/[ni- pi = 1–qi si = pipi–1pi(wi/2)] 2…p1 Propn Propn terminating surviving 0.0833 0.9167 one's 0.1250 0.8750 disease progressed, so d2 = 1 Cumul Propn Surv at End 0.9167 0.8021 0.0000 1.0000 0.8021 0.0000 1.0000 0.8021 Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–
    39. 39. Life table for sample of 13 patient treated with etoposide with cisplatin Life Table Survival Variable: Progression-Free Survival ni wi di Interval No. No. No. No. of Start entering withdrawn exposed terminal Time Interval du. to risk events Interval the proportion terminating (q2 nd 0.0 13.0 = d2/[n2-(w2/2]) during 2 2.0 12.0 1.0 interval is 1/[10 – (4/2)] = 1/8, or 0.1250. qi = di/[ni- pi = 1–qi si = pipi–1pi(wi/2)] 2…p1 Propn Propn proportion the Cumul Propn terminating surviving no with Surv at End progression is 1 – 0.1250, or 0.8750 0.9167 0.0833 0.9167 3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021 6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021 9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021 Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–
    40. 40. Life table for sample of 13 patient treated with etoposide with cisplatin Life Table Survival Variable: Progression-Free Survival ni wi Interval No. No. Start entering withdrawn Time Interval du. Interval 0.0 3.0 13.0 10.0 2.0 di No. exposed to risk No. of terminal events qi = di/[ni- pi = 1–qi si = pipi–1pi(wi/2)] 2…p1 Propn Propn terminating surviving the 12.0 cumulative proportion of surv 1.0 0.0833 0.9167 4.0 8.0 6.0 from probability theory: 5.0 4.0 Rule 3.0 =p1*p2= 0.0.9167 × 0.8750= 0.8021 Cumul Propn Surv at End 0.9167 1.0 0.1250 0.8750 0.8021 0.0 0.0000 1.0000 0.8021 0.0 0.0000 1.0000 0.8021 P(A&B)=P(A)*P(B) if A and B independent 9.0 1.0 1.0 0.5 Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–9
    41. 41. Life table • Like Cancer treatment, life table of survivorship after any treatment as treatment of cancer by irradiation or drugs or after operation, such as of cancer cervix or breast can be prepared & made use in probabilities of survival at beginning or at any point of time. • More recently, survival can be enquired after– Heart operation like bypass, angioplasty, ballooning, stenting, heart transplantation. – Kidney, lung, liver & other organ transplantation.
    42. 42. Life Table • This computation procedure continues until the table is completed. • pi = the probability of surviving interval i only; to survive interval i, a patient must have survived all previous intervals as well. • The probability of survival at one period is treated as though it is independent of the probability of survival at others • Thus, pi is an example of a conditional probability because the probability of surviving interval i is dependent, or conditional, on surviving until that point. • This is called survival function.
    43. 43. Limitation • The assumption that all withdrawals during a given interval occur, on average, at the midpoint of the interval. • This assumption is of less consequence when short time intervals are analyzed; however, considerable bias can occur : • if the intervals are large, • if many withdrawals occur, & • if withdrawals do not occur midway in the interval. • The Kaplan–Meier method overcomes this problem.
    44. 44. Kaplan Meier product limit method
    45. 45. Kaplan-Meier Product limit method • Similar to actuarial analysis except time since entry in the study is not divided into intervals for analysis. • Survival is estimated each time a patient has an event. • Withdrawals are ignored • It gives exact survival times in comparison to actuarial because it does not group survival time into intervals 46
    46. 46. Introduction to Kaplan-Meier • Non-parametric estimate of the survival function. • Commonly used to describe survivorship of study population/s. • Commonly used to compare two study populations. • Intuitive graphical presentation. 47
    47. 47. Survival Data (right-censored) Subject A Subject B Subject C Subject D Subject E X 1. subject E dies at 4 months 0 Beginning of study 12  Time in months  End of study
    48. 48. Corresponding Kaplan-Meier Curve 100% Probability of surviving to 4 months is 100% = 5/5 Fraction surviving this death = 4/5 Subject E dies at 4 months 4  Time in months 
    49. 49. Survival Data Subject A Subject B 2. subject A drops out after 6 months Subject C 3. subject C dies X at 7 months Subject D Subject E X 1. subject E dies at 4 months Beginning of study  Time in months  End of study
    50. 50. Corresponding Kaplan-Meier Curve 100% Fraction surviving this death = 2/3 subject C dies at 7 months 4 7  Time in months 
    51. 51. Survival Data Subject A 2. subject A drops out after 6 months Subject B 3. subject C dies X at 7 months Subject C Subject D Subject E 4. Subjects B and D survive for the whole year-long study period X 1. subject E dies at 4 months Beginning of study  Time in months  End of study
    52. 52. Corresponding Kaplan-Meier Curve Rule from probability theory: 100% P(A&B)=P(A)*P(B) if A and B independent In kaplan meier : intervals are defined by failures(2 intervals leading to failures here). P(surviving intervals 1 and 2)=P(surviving interval 1)*P(surviving interval 2) Product limit estimate of survival = P(surviving interval 1/at-risk up to failure 1) * P(surviving interval 2/at-risk up to failure 2) = 4/5 * 2/3= .5333 The probability of surviving in the entire year, taking into account censoring = (4/5) (2/3) = 53% 0  Time in months  12
    53. 53. Example :kaplan–Meier survival curve in detail for patients on etoposide plus cisplatin Event Time (T) Number at Risk ni Number of Events di Mortality qi = di/ni Survival pi = 1 - qi Cumulative Survival S = pip(i-1)…p2p1 1.0 13 1 0.076 0.9231 0.9231 2.4 12 2.8 11 3.1 10 3.7 9 4.4 8 4.6 7 4.7 6 6.5 5 7.1 4 8.0 3 8.1 2 12.0 1 • In this method first step is to list the times when a death or drop out occurs, as in the column “Event Time”. 1 0.1250 0.8750 0.8077 • One patient's disease progressed at 1 month and another at 4.4 months, and they are listed under the column “Number of Events.” • Then, each time an event or outcome occurs, the mortality, survival, and cumulative survival are calculated in the same manner as with the life table method.
    54. 54. Contd… • If the table is published in an article, it is often formatted in an abbreviated form, such as in Table 5. Kaplan–Meier survival curve in abbreviated form for patients on etoposide plus cisplatin Event Time (T) Number at Risk ni Number of Events di Mortality qi = di/ni Survival pi = 1 - qi Cumulative Survival S = pip(i-1)…p2p1 1.0 13 1 0.076 0.9231 0.9231 4.4 8 1 0.1250 0.8750 0.8077 .. .. .. 12.0
    55. 55. 2.0 4.0 6.0 8.0 10.0 12.0 Kaplan meir survival curve for patients on etoposide & cisplatin (Source: Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–91.)
    56. 56. Limitations of Kaplan-Meier • Requires nominal predictors only • Doesn‟t control for covariates Cox progressive hazard model solves these problems 57
    57. 57. Kaplan meir survival curve with 95 % confidence limits for patients on irinotecan & cisplatin (Source: Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–91.)
    58. 58. Comparison between 2 survival curve • Don’t make judgments simply on the basis of the amount of separation between two lines
    59. 59. Comparison between 2 survival curve • For comparison if no censored observations occur, the Wilcoxon rank sum test introduced, is appropriate for comparing the ranks of survival time. • If some observations are censored, methods may be used to compare survival curves. – the Logrank statistic – the Mantel–Haenszel chi-square statistic.
    60. 60. Logrank test • The log rank statistic is one of the most commonly used methods to learn if two curves are significantly different. • This method also known as Mantel-logrank statistics or Cox-Mantel-logrank statistics • The logrank test compares the number of observed deaths in each group with the number of deaths that would be expected based on the number of deaths in the combined groups that is, if group membership did not matter.
    61. 61. Hazard ratio • The logrank statistic calculates the hazard ratio • It is estimated by O1st group/E1st group divided by O2nd group/E2group • The hazard ratio is interpreted in a similar manner as the odds ratio • Using the hazard ratio assumes that the hazard or risk of death is the same throughout the time of the study.
    62. 62. Mantel– Haenszel chi test • Another method for comparing survival distributions is an estimate of the odds ratio developed by Mantel and Haenszel that follows (approximately) a chi-square distribution with 1 degree of freedom. • The Mantel– Haenszel test combines a series of 2 2 tables formed at different survival times into an overall test of significance of the survival curves. • The Mantel–Haenszel statistic is very useful because it can be used to compare any distributions, not simply survival curves
    63. 63. Cox progressive hazard model
    64. 64. Why is called cox proportional hazard model • Cox =scientist‟s name(Sir David Roxbee Cox) – British statistician – In 1972 developed it. • Uses hazard function • covariates have a multiplicative or a proportional , effect on the probability of event
    65. 65. What does cox model do> • It examines two pieces of information: – The amount of time since the event first happened to a person – The person‟s observations on the independent variables.
    66. 66. Cox progressive hazard model • Used to assess the simultaneous effect of several variables on length of survival. • It allows the covariates(independent variables) in the regression equation to vary with time. • Both numerical and nominal independent variables may be used in this model.
    67. 67. COX regression coefficient • Determines relative risk or odd‟s ratio associated with each independent variable and outcome variable, adjusted for the effect of all other variables .
    68. 68. Hazard function • Opposite to survival function • Hazard function is the derivative of the survival function over time h(t)=dS(t)/dt • instantaneous risk of event at time t (conditional failure rate) • It is the probability that a person will die in the next interval of time, given that he survived until the beginning of the interval.
    69. 69. Hazard function • Hazard function given by h(t,x1,x2…x5)=ƛ0 (t)eb1x1+b2x2+….b5x5 • ƛ0 is the baseline hazard at time t i.e. ƛ0(t) • For any individual subject the hazard at time t is hi(t). • hi(t) is linked to the baseline hazard h0(t) by loge {hi(t)} = loge{ƛ0(t)} + β1X1 + β2X2 +……..+ βpXp • where X1, X2 and Xp are variables associated with the subject
    70. 70. Proportional hazards: Hazard ratio Hazard for person i (eg a smoker) hi (t ) 0 (t )e 1xi1 ...  k xik  ( x  x ) ... 1 ( xik  x jk ) HRi , j    e 1 i1 j 1 h j (t ) 0 (t )e 1x j1 ...  k x jk Hazard for person j (eg a non-smoker) Hazard functions should be strictly parallel! Produces covariate-adjusted hazard ratios! 71
    71. 71. The model: binary predictor  HRlung cancer/ smoking (1)   hi (t ) 0 (t )e smoking age  (10 )    e smoking h j (t ) 0 (t )e  smoking ( 0)  age ( 60) HRlung cancer/ smoking  e ( 60 )  smoking This is the hazard ratio for smoking adjusted for age. 72
    72. 72. Table 2. Death rates for screenwriters who have won an academy award.* Values are percentages (95% confidence intervals) and are adjusted for the factor indicated Basic analysis Adjusted analysis Demographic: Year of birth Relative increase in death rate for winners 37 (10 to 70) HR=1.37; interpretation: 37% higher incidence of death for winners compared with nominees 32 (6 to 64) Sex 36 (10 to 69) Documented education 39 (12 to 73) All three factors 33 (7 to 65) Professional: Film genre Total films Total four star films Total nominations Age at first film HR=1.35; interpretation: 35% higher incidence of death for winners compared with nominees even after adjusting for potential confounders 37 (10 to 70) 39 (12 to 73) 40 (13 to 75) 43 (14 to 79) 36 (9 to 68) Age at first nomination 32 (6 to 64) All six factors 40 (11 to 76) All nine factors 35 (7 to 70)
    73. 73. Importance • Provides the only valid method of predicting a time dependent outcome , and many health related outcomes related to time. • Can be interpreted in relative risk or odds ratio • Gives survival curves with control of confounding variables. • Can be used with multiple events for a subject.
    74. 74. Take Home Message • survival analysis deals with situations where the outcome is dichotomous and is a function of time • In survival data is transformed into censored and uncensored data • all those who achieve the outcome of interest are uncensored” data • those who do not achieve the outcome are “censored” data
    75. 75. Take Home Message • The actuarial method adopts fixed class intervals which are most often year following the end of treatment given. • The Kaplan-Meier method uses the next death, whenever it occurs, to define the end of the last class interval and the start of the new class interval. • Log-Rank test used to compare 2 survival curves but does not control for confounding. • Mantel henzel test can compare any curve not only survial curves • For control for confounding use another test called as „Cox Proportional Hazards Regression.’
    76. 76. Thank you

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