1. Lecture 11: Hypothesis Testing III
Stratified Tests
Renyi and Other Tests

2. Stratified Tests Adjust for a covariate
Allows you to control for a confounder without using a regression approach
However
Like regression, if interaction is present, it won’t be detected
Assumes the ‘treatment’ effect is the same across strata

3. Sometime Confusing “Stratified” analysis Sometimes
Subgroup analysis
Stratified “combined” test
In this case, combined test
Recall Mantel-Haenszel odds ratio

4. Notation Now three variables J = 1, 2,…., K indexes groups
Outcome (time to event)
Group variable (i.e. treatment)
Strata variable (i.e. gender, cancer grade)
J = 1, 2,…., K indexes groups
S = 1, 2,…, M indexes strata

5. Similar to the Standard Test
Formal hypothesis
Now, Zj.(t) is represented by a sum

6. From there, inference is the same
Chi-square test with K – 1 d.f. where S-1 is the inverse of the estimated variance covariance matrix
For the 2 group scenario it can be reduced to a Z-score

7. Asymptotics Just like unstratified test, requires large N
Here requires even larger- think about dividing the sample into M strata
In most cases, there probably is not sufficient N

8. Small Example 20 subjects received 1 of two treatments
9 patients on treatment 1
11 patients received treatment 2
Patients also categorized by disease type
2 strata
Question:
Does the data show a treatment effect after adjusting for disease type?

9. Time
Death
Censor
Trt
Disease 1 5 6 2 8 37 49 58 79 11 50 51 62 67 73 86 90 96 97

11. What first Data in standard format
Trt 1: 1, 5, 5+,6+, 8, 37, 49, 58, 79+
Trt 2: 11+, 50, 51, 62, 73, 86, 90, 96, 97, 97
We might first conduct a global test
What is our hypothesis

12. Constructing Statistic

13. Calculate Statistic
Z-statistic
c2 statistic

14. Now Let’s Adjust for Disease Type
Steps:
Divide the data according to strata
Calculate Zjs.(t) and
Sum Zjs(t) and across strata to get Zj.(t) &
Calculate your test statistic according to

15. Divide data By Strata Disease 1 Disease 2 Time Death Censor Trt 1 5 85 8 49 11 2 50 62 67 73 86
Time
Death
Censor
Trt 6 1 37 58 79 51 2 90 96 97

16. Calculate and sgsms

17. Calculate and sgsms

18. Calculate the Statistic
Z (or chi-square)
What is our conclusion

19. R Code
>times<-c(1,5,5,6,8,11,37,49,50,51,58,62,67,73,79,86,90,96,97,97) >trt<- c(1,1,1,1,1,2,1,1,2,2,1,2,2,2,1,2,2,2,2,2) >strat<-c(1,1,1,2,1,1,2,1,1,2,2,1,1,1,2,1,2,2,2,2) >death<-c(1,1,0,0,1,0,1,1,1,1,1,1,1,1,0,1,1,1,1,1) #Global >survdiff(st~trt) Call: survdiff(formula = st ~ trt) N Observed Expected (O-E)^2/E (O-E)^2/V trt= trt= Chisq= 6.1 on 1 degrees of freedom, p=

20. R Code
#Stratified survdiff(st~trt + strata(strat)) Call: survdiff(formula = st ~ trt + strata(strat)) N Observed Expected (O-E)^2/E (O-E)^2/V trt= trt= Chisq= 9.5 on 1 degrees of freedom, p=

21. BMT: Hodgkin’s & Non-Hodgkin’s Lymphoma
Study included 43 BMT patients
Is there a difference in hazard rates between
Allogenic transplant = HLA matched sibling donor (N=16)
Autogenic transplant = Own “cleaned” marrow (N=27)
But want to adjust for disease state
Non-Hodgkin’s lymphoma (N=23)
Hodgkin’s disease (N=20)

23. Global Test 2 1 43 16
0.628
0.234 4 42 15
0.643
0.230 28 41 14
0.659
0.225 30 40 13
-0.325
0.219 32 39
0.667
0.222 …
132 22 7
-0.318
0.217
140 21
-0.333
252 18
-0.389
0.238
357
0.563
0.246
Sum
0.886
5.841

24. Global Results Global Test Results
> dat<-read.csv("C:\\BJW\\AutoAllo.csv")
> d<-dat$death; t<-dat$time
> dis<-dat$disease; type<-dat$graft
> nostrat<-survdiff(Surv(t, d)~type)
> nostrat
Call:
survdiff(formula = Surv(t, d) ~ type)
N Observed Expected (O-E)^2/E (O-E)^2/V
type=
type=
Chisq= 0.1 on 1 degrees of freedom, p= 0.714

25. Stratified by Disease Type
Non-Hodgkin’s Lymphoma subjects 28 1 23 11
0.522
0.250 32 22 10
0.545
0.248 42 21 9
-0.429
0.245 49 20
0.550 53 19 8
-0.421
0.244 57 18
-0.444
0.247 63 17
-0.471
0.249 81 2 16
-1.000
0.467 84 14
0.429
140 13 7
-0.538
252
-0.636
0.231
357
0.300
0.210
524 6
-0.750
0.188
Sum
-2.344
3.319

26. Stratified by Disease Type
Hodgkin’s Disease subjects 2 1 20 5
0.750
0.188 4 19
0.789
0.166 30 18 3
-0.167
0.139 36 17
-0.176
0.145 41 16
-0.188
0.152 52 15
-0.200
0.160 62 14
-0.214
0.168 72 13
0.769
0.178 77 12
0.833 79 11
0.909
0.083
108 10
0.000
132 9
sum
3.106
1.518

27. Stratified Results Stratified Test Results
> strat<-survdiff(Surv(t, d)~type + strata(dis))
> strat
Call:
survdiff(formula = Surv(t, d) ~ type + strata(dis))
N Observed Expected (O-E)^2/E (O-E)^2/V
type=
type=
Chisq= 0.1 on 1 degrees of freedom, p= 0.729

28. Stratified Results Stratified Test Results Again we fail to reject
This seems in error (recall (our survival curves looked VERY different)

29. Problem? The treatment effect is not the same in the 2 disease states
They are in different directions
ZAllo =
ZAuto = 3.106
Stratified approach is NOT appropriate

30. Alternative to Stratified Analysis
Alternatives
Define 4 groups and conduct a K-sample log rank test
Allogenic and NHL
Allogenic and Hodgkin’s
Autogenic and NHL
Autogenic and Hodgkin’s
Subgroup analysis (by disease) should be performed
Allo|Hodgkins
Allo|Non-Hodgkins

31. R Code- K sample test
> allgrp<-ifelse(dis==1 & type==1, 1, 0) > allgrp<-ifelse(dis==1 & type==2, 2, allgrp) > allgrp<-ifelse(dis==2 & type==1, 3, allgrp) > allgrp<-ifelse(dis==2 & type==2, 4, allgrp) > grp4<-survdiff(Surv(t, d)~allgrp) > grp4 Call: survdiff(formula = Surv(t, d) ~ allgrp) N Observed Expected (O-E)^2/E (O-E)^2/V allgrp= allgrp= allgrp= allgrp= Chisq= 11.1 on 3 degrees of freedom, p=

32. R Code- Subgroup analysis
> ### Subgroup (NHL) > subNHL<-survdiff(Surv(t,d)[which(dis==1)]~type[which(dis==1)]) > subNHL Call: survdiff(formula = Surv(t, d)[which(dis == 1)] ~ type[which(dis ==1)]) N Observed Expected (O-E)^2/E (O-E)^2/V type[which(dis == 1)]= type[which(dis == 1)]= Chisq= 1.7 on 1 degrees of freedom, p= > ### Subgroup (Hodgkins) > subHD<-survdiff(Surv(t,d)[which(dis==2)]~type[which(dis==2)]) > subHD survdiff(formula = Surv(t, d)[which(dis == 2)] ~ type[which(dis ==2)]) N Observed Expected (O-E)^2/E (O -E)^2/V type[which(dis == 2)]= type[which(dis == 2)]= Chisq= 6.4 on 1 degrees of freedom, p=

33. Summary: Stratified Testing
Alternative to a regression approach to control for a 2nd covariate when examining treatment effect.
Sample size needs to be larger that in the case of testing K-groups for test results to be valid.
One needs to be cautious about misinterpreting null results when interactions exist.
We can use a subgroup approach if this fails.

34. Renyi Tests
Previous tests we discussed all use weighted integral of estimated difference in cumulative hazard rates
Doesn’t address situation where early differences favor one group, and later differences favor another group
Solution: Renyi tests
i.e. addresses issue of crossing hazard rates

35. Renyi Test
Censored data analogs of Kolmogrov-Smirnov statistic when comparing to uncensored samples
Recall KS is a test of equality of one-dimensional probability distributions used to compare two samples

36. Komolgrov-Smirnov Test
Recall empirical distribution function
Hypothesis
The KS statistic is

37. Example of a KS test Two groups observed for a continuous outcome:
1: -0.2, 3.7, 4.3, 5.0, 7.7, 8.6
2: -0.9, 0.4, 0.5, 2.6, 3.0, 12.1
We want to determine if the distribution of the outcomes are different (without assuming any distributional form…)

38. Constructing KS statisticx
P(X1 < x)
P(X2 < x)
|P(X1 < x)-P(X2 < x)|
-0.9
1/6
-0.2
0.4
1/3
0.5
1/2
2.6
2/3
3.0
5/6
3.7
4.3
5.0
7.7
8.6 1
12.1

40. K-S Test

41. Renyi Test Approach Find the value of Z(ti) for each failure time
Note different from Z(t) which sums over all ti < t
Calculate series of Z(ti) :
Estimate the standard error of Z(t) (all times)

42. Renyi Statistic
When hazard rates cross, the absolute value of Z(t) will have max value at some value t < t
Hypothesis test:
Note that multiple tests are made, because we are taking the max over Z(t)

43. Test Statistic Q
Use the same variance estimate for test statistic as in standard two-sample approach
Test statistic
Q is approximated by distribution of
sup{|B(x)|, 0 < x < 1}
where B is a standard Brownian motion process
Use table C.5 to find associated p-value

44. Small Example Given the following data Group 1: (7, 8+, 9, 15, 17)

45. Constructing the statisticdk
dk1 Yk
Yk1
Var

46. Calculating Q First we can calculate Q
Once we have Q we compare to table C.5

47. Example 2: Kidney Infection
Data on 119 kidney dialysis patients
Comparing time to kidney infection between two groups
Catheters placed percutaneously (n = 76)
Catheters placed surgically (n = 43)

48. Example: Kidney Infection

49. R Code: Kidney Infection
> kidney<-read.csv("H:\\public_html\\BMRTY722_Summer2015\\Data\\Kidney.csv") > time<-kidney$Time > infect<-kidney$d > percut<-kidney$cath > st<-Surv(time, infect) > LRtest<-survdiff(st~percut) > LRtest Call: survdiff(formula = st ~ percut) N Observed Expected (O-E)^2/E (O-E)^2/V percut= percut= Chisq= 2.5 on 1 degrees of freedom, p= 0.112

50. How to Test This in R?
We could write our own R function to conduct the Renyi test…
BUT, it turns out there was a package released in April that has the Renyi test (and all weight functions from K & M included )

51. R Code: Kidney Infection
> library(survMisc) > RYtest<-comp(survfit(st~percut)) > RYtest $tne t n e n_percut=1 e_percut=1 n_percut=2 e_percut=2 1: : … 16: $tests$lrTests ChiSq df p Log-rank Gehan-Breslow (mod~ Wilcoxon) Tarone-Ware Peto-Peto Mod~ Peto-Peto (Andersen) Flem~-Harr~ with p=1, q= $tests$supTests Q p Log-rank Gehan-Breslow (mod~ Wilcoxon) Tarone-Ware Peto-Peto Mod~ Peto-Peto (Andersen) Renyi Flem~-Harr~ with p=1, q=

52. R Code: Kidney Infection
> library(survMisc) > RYtest<-comp(survfit(st~percut), FHp=0, FHq=0) > RYtest $tne t n e n_percut=1 e_percut=1 n_percut=2 e_percut=2 1: : … 16: $tests$lrTests ChiSq df p Log-rank Gehan-Breslow (mod~ Wilcoxon) Tarone-Ware Peto-Peto Mod~ Peto-Peto (Andersen) Flem~-Harr~ with p=0, q= $tests$supTests Q p Log-rank Gehan-Breslow (mod~ Wilcoxon) Tarone-Ware Peto-Peto Mod~ Peto-Peto (Andersen) Renyi Flem~-Harr~ with p=0, q=

53. Example 3: Gastric Cancer
Clinical trial of chemotherapy vs. chemotherapy combined with radiotherapy
45 Patients randomized to each of two arms
Followed for up to 8 years

55. R Code: Gastric Cancer
> RYtest<-comp(survfit(Surv(tm, dth)~x, data=dat)) > RYtest $tne t n_x=1 e_x=1 n_x=2 e_x=2 n e 1: … 80: $tests$lrTests ChiSq df p Log-rank Gehan-Breslow (mod~ Wilcoxon) Tarone-Ware Peto-Peto Mod~ Peto-Peto (Andersen) Flem~-Harr~ with p=1, q= $tests$supTests Q p Log-rank Gehan-Breslow (mod~ Wilcoxon) Tarone-Ware Peto-Peto Mod~ Peto-Peto (Andersen) Renyi Flem~-Harr~ with p=1, q=

56. Compare To Log Rank Renyi test 0.05< p <0.06
What would you expect to see from the log rank test? More or less significant?

57. LR Results
> LRtest<-survdiff(Surv(tm, dth)~x) > LRtest Call: survdiff(formula = Surv(tm, dth) ~ x) N Observed Expected (O-E)^2/E (O-E)^2/V x= x= Chisq= 0.2 on 1 degrees of freedom, p= 0.63

58. Final Comments on the Renyi Test
Simulations comparing the Renyi vs. log-rank
Hazards cross  Renyi test performs better
Renyi test has little loss of power if proportional hazard assumption holds (with limited censoring)
However, with large amounts of censoring, advantages of the Renyi test decline
So this tests provides a good alternative when hazard rates cross.
But caution still needs to be taken when there is a large amount of censoring.

59. Other Tests for Crossing Hazards
Cramer-von Mises test(s):
Based on the integrated squared difference between two curves
T-test analog:
Requires estimation of the mean
Compared area under S1(t) and S2(t)
Brookmeyer-Crowley
Censored version of two-sample median test

60. Cramer-Von Mises Test
Based on the Nelson-Aalen estimator for the hazard rate and it’s associated variance
Ideally we integrate over time 0 to t but this integral is estimated by summing over distinct death times

61. 2-Sample T-test analog
Again this test is based on the difference in the area under the survival curve between two groups
Components of the test include:
Order all observed times (event and censored)
Calculate dij, cij, and Yij or both groups
Calculate the KM estimator for survival and censoring
Calculate the pooled KM estimate of survival

62. 2-Sample T-test analog Once these estimates are obtained:
Construct weight function
Construct the test statistic
Construct the variance of the test statistic
Calculate a Z-score according to

63. Summary of Other 2-Sample Tests
When the hazard rates cross, both the Cramer-Von Mises and the 2-sample t-test analog have greater power than log-rank.
When hazard rates are proportional, both show power loss relative to log-rank.
Performance is similar to the Renyi test when hazards cross but Renyi has better power for proportional hazards.

64. Test Based on Fixed Points in Time
Complicated description in K&M (chapter 7.8)
However, pretty simple idea when you are comparing two groups:

65. Next Time
We will begin our discussion of semi-parametric regression modeling in survival analysis.