Discuss your response to this survival analysis study.
Please read the following study:
Bonomi, L., Jiang, X., & Ohno-Machado, L. (2020). Protecting patient privacy in survival
analyses. Journal of the American Medical Informatics Association, 27(3), 366–
375. https://doi.org/10.1093/jamia/ocz195
Discuss your response to this survival analysis study. Do you have the same concerns as the
researchers regarding the patient privacy issues when presenting actuarial/survival analysis
tables? Do you have other suggestions regarding protecting patient privacy within a study?
Embed course material concepts, principles, and theories (which require supporting citations) in
your initial response along with at least 2 scholarly, peer-reviewed journal articles. Keep in mind
that these scholarly references can be found in the Saudi Digital Library by conducting an
advanced search specific to scholarly references. Use Saudi Electronic University academic
writing standards and APA style guidelines.
Module 13
Chapter 11
The Survival
Curve
Learning Objectives
• Evaluate the data for survival analysis if it
is normally distributed
• Evaluate the statistical tests used in
survival analysis
• Perform and interpret the log-rank test
• Evaluate the concept of censored survival
data
Types of Censoring
• Right censoring
– Type I
– Type II
• Left censoring
• Right truncation
• Left truncation
Kaplan–Meier Assumptions
• Kaplan-Meier estimate
– Assume censored patients have same survival prospects
– Early or late recruitment has same survival prospect
– Assume the event happens at the specified time
(Goel, Khanna, & Kishore, 2010)
Number of subjects – Number of subjects
St = Living at the start
died
Number of subjects living at the start
Kaplan-Meier versus Log-Rank Test
• Kaplan-Meier
–
Survival prospects the
same regardless of:
o
Censoring
o
Recruitment time
o
Event timing
• Log-Rank
– No assumptions
about subjects
– Helps in testing null
hypothesis
– No difference
between survival
curves
– Tests significance
only not estimates
Log-Rank Test
H0: Two survival curves are identical
H1: Two survival curves are not identical
Test statistic: χ =
2
(O jt − E jt )
2
E jt
Reject H0 if c2 > c2,df where df = k – 1
and k = number of comparison groups.
RCT to Compare Two Treatments for
Advanced Gastric Cancer
Example 11.3.
Log-Rank Test (1 of 2)
H0: Two survival curves are identical
H1: Two survival curves are not identical
Test statistic:
χ2 =
(O jt − E jt ) 2
E jt
(6 − 2.620) 2 (3 − 6.380) 2
=
+
= 6.151
2.620
6.380
Example 11.3.
Log-Rank Test (2 of 2)
• Reject H0 if c2 ≥ 3.84.
• Reject H0 because 6.151 > 3.84. We have
statistical evidence that two survival
curves are not identical.
Comparing Survival Curves
H0: Two survival curves are equal
c2 Test with df=1. Reject H0 if c2 > 3.84
c2 = 6.151. Reject H0.
Cox Proportional Hazards
Regression Assumptions
• Statistical significance
– Wald statistic value “z”
– Beta (β) coefficient significantly different from 0
• Regression coefficients
– Sign (±) of coefficient indicates hazard
o Positive (+) high risk
o Negative (-) low risk
• Hazards ratios
– Effect size
Cox Proportional Hazards
Regression
• Model
h(t) = h0(t) exp (b1X1 + b2X2 + … + bpXp)
• Where h(t) = hazard at time t (risk of
failure at time t),
h0(t) = baseline hazard,
Xi are predictors,
bi are regression coefficients.
Cox Proportional Hazards
Regression
• Model
ln(h(t)/h0(t)) = b1X1 + b2X2 + … + bpXp
• exp(bi) = hazard ratios
Example 11.5.
Cox Proportional Hazards Regression
(1 of 3)
• Framingham Study
–
–
–
–
Outcome = all-cause mortality
N = 5180 participants ≥ 45 years
10-year follow-up
Analysis with Cox proportional hazards
regression
Example 11.5.
Cox Proportional Hazards Regression
(2 of 3)
Age
Male sex
bi
p
0.11149 0.0001
0.67958 0.0001
HR
1.118
1.973
Example 11.5.
Cox Proportional Hazards Regression
(3 of 3)
• Multivariable model
bi
p
HR (95% CI)
Age
0.11691
0.0001
1.12 (1.11 – 1.14)
Male sex
0.40359
0.0001
1.50 (1.22 – 1.85)
SBP
0.11691
0.0001
1.02 (1.01 – 1.02)
Current
smoker
0.40359
0.0001
2.16 (1.76 – 2.64)
Total chol.
0.40359
0.0001
1.00 (0.99 – 1.00)
Diabetes
0.40359
0.0001
0.82 (0.62 – 1.08)
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