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Unformatted text preview: g0 (“LN Exam I for Stat 4930  Fall 2003 October 10, 2003 1. (15 points) Let S'U) be the KaplanMeier estimator of the survivor function and recall logmx) = log[$/(1 — Derive Vm:(logit(5'(t))) and use it to construct a 95% conﬁdence interval for
3(t). Hint: Var(S(t)) = [S (10]2 2le “Had”, where k is the number of unique death times. ULSQ w‘kjlu’ 50‘”! among) m 3(9); 103.1(93= loje —oj(Ie‘) i i."  ..
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(a) [10 points) Show that h(tX) = h0(t) exp(ﬁ:z:) implies S(tX) = SMQWPUW),
(b) (5 points) Show that the Breslow estimator of the baseline cumulative hazard function in the Cox model, EUR) = 233:1 W simpliﬁes to the N elsonAalen estimator
(3') if there are no covariates. a) Mm): tome?”
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5 mo 2 r e w ~ 3. (20 points) Use the below data to answer the following questions. Note: 6 = 1 corresponds to
failure, 6 = 0 corresponds to censoring. Subject Time 6 ‘l‘ M g 3; 5.1
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5 3 0 s 1 l 2 a l u l (a) Compute the KaplanMeier and NelsonAalen estimator for S(t). Compare. (b) Estimate the median survival time using the two estimators and S (50). Compare. (c) Does patient 6 contribute to the estimator? Why or why not? ((1) Can you estimate the mean? Explain (Do NOT estimate it!). N  a‘jt‘j
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{ma 5(E\ 3m9s(l_;_[): O “Scﬂr'qu .]: g 4. (50 points) The data is from the Mayo Clinic trial in primary biliary cirrhosis (PBC) of the
liver conducted between 1974 and 1984. A total of 424 PBC patients, referred to Mayo Clinic during that ten—year interval, met eligibility criteria for the randomized placebo controlled
trial of the drug D—penicillamine (drug). Age is measured in years. (3) Estimate the hazard ratio and give a 95% conﬁdence interval for a 10 year change in age. > summary . coxph (pbc . age)
Call: coxph(formula = Surv(time, status) " age, data = pbc) n= 312
coef expCcoaf) se(coef) z 13
age 0.04 1.04 0.0088 4.54 5.7e06 ‘03 (not; {1r ,,\ 1 a,” (bub/3" p:‘oq inf a lo ([«anjg ﬁlo : ICE): “#0 féio: (:5. 34.01. 130007593 _. ( log?) L773) (b) The investigator wanted to ﬁt the stratiﬁed log rank test to examine the impact of drug.
stratifying on sex. He ﬁt the below cox model. Give a test statistic and p—value that ’roughly’ corresponds to the stratiﬁed log rank test described (Be Carefull). What do
you conclude? Justify your choice. Call:
coxph(formu1a = Surv(time, status) " sex + drug, data = pbc) n= 312
coef exp(coef) se (coef) z p
sex 0.4824 0.617 0.237 2.039 0.041
drug 0.0503 0.951 0.179 0.281 0.780 exp(coef) exp(coef) lower .95 upper .95
sex 0.517 1.62 0.388 0.981
drug 0.951 1.05 0.669 1.351 quuare= 0.012 (max possible= 0.983 ) Likelihood ratio test: 3.85 on 2 df, p=0.146 Wald test = 4.27 on 2 df, p=0.119
Score (logrank) test = 4.35 on 2 df, p=0.114 Sign “wk LU? WU‘IJ (UN 23"“) +0 545b,.) .1’ Q's/“j; QJJ‘HLLJ £4” 39X.
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at. in no Jo J m. .J Ara; we ((2) Test the hypothesis that Bl z 82 z 0. i.e., age and bilrubin are not needed in the model.
State your conclusions. coxph(formula = Surv(time, status) " age + bili + drug, data = pbc) n= 312
coef expCcoef) seCcoef) z p age 0.0397 1.041 0.00913 4.355 1.3e—05 5111 0.1479 1.159 0.01308 11.307 0.0e+00 ( I) drug 0.0566 0.945 0.18724 0.302 7.6901 expCcoef) exp(coef) lower .95 upper .95 age 1.041 0.961 1.022 1.06
bili 1.159 0.863 1.130 1.19
drug 0.945 1.058 0.655 1.36 quuare= 0.285 (max possible: 0.983 )
Likelihood ratio test= 105 on 3 df,
Wald test = 150 on 3 df,
Score (logrank) test = 214 on 3 df, "U'lﬂ‘d
000 ******************t** Call:
coxph(formula = Surv(time, status) “ drug, data = pbc) n= 312 coef exp(coef) seCcoef) 2 p ( )
drug 0.0572 0.944 0.179 —0 319 0.75 a exp(coef) exp(—coef) lower .95 upper .95
drug 0.944 1.06 0.665 1.34 quuare= 0 (max possible= 0.983 ) Likelihood ratio test= 0.1 on 1 df, p=0.75
Wald test = 0.1 on 1 df, p=0.75
Score (logrank) test = 0.1 on 1 df, p=0.75 LRT~; Q[L(nL(&\]= O§~.l:.loqﬂ
“.01., 0.} 30”.) 7.. 7.11.3 U... (d) The investigator was interested in whether the effect of drug varied by sex. Test this hypothesis and state your conclusions. Estimate the hazard ratio for drug when sex=0
and when sex=1. Call:
coxph(formula = Surv(time, status) ” sex + drug + sex * drug,
data = pbc)
n= 312
coef exp(coef) se(coef) z p sex 1.159 0.314 0.714 1.622 0.10
drug 0.444 0.641 0.444 1.000 0.32
sex:drug 0.475 1.607 0.486 0.977 0.33 exp(coef) exp(—coef) lower .95 upper .95
sex 0.314 3.185 0.0774 1.27
drug 0.641 1.559 0.2686 1.53
sex:drug 1.607 0.622 0.6204 4.16 quuare= 0.015 (max possible= 0.983 )
Likelihood ratio test= 4.83 on 3 df, p=0.185 Wald test = 5.63 on 3 df, p=0.131
Score (logrank) test = 5.82 on 3 df, p=0.121 uh" 65101 :0
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141.15“ I’ll/k3 O) ejllgc‘ (e) If the Breslow method had been used to handle ties (the Efron method was used in the problems above), describe how you would expect the estimated coefficients to look
relative to those given above in part b). go LA LA) Exam I for Stat 6934  Fall 2003 October 10, 2003 1. ( 15 points) Discuss the use of weights in the (weighted) log rank test, including when and why we should use particular sets of weights . Why does the (weighted) 10g rank test have
low power when the hazards cross? Explain. Sue Mk 2. (15 points) Show that in the Cox model with t continuous covariate X2, i.e., h(t]X1, X2) 2 he
ratio for X = 1 vs X we covariates, a binary covariate X1 and a (t) expwlzl + 6232), that [31 is the log hazard
= 0 holding the continuous covariate X2 constant. +63%;
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