introduction to survival analysis
DESCRIPTION
Introduction to Survival Analysis. Consider the following clinical out-come evaluation situation:. Goal: To determine the effectiveness of a new therapeutic intervention with sexually abused children. Consider the following clinical outcome evaluation:. Procedure: - PowerPoint PPT PresentationTRANSCRIPT
Introduction to Survival Analysis
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Consider the following clinical out-come evaluation situation:
• Goal: To determine the effectiveness of a new therapeutic intervention with sexually abused children.
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Consider the following clinical outcome evaluation:
Procedure:1. Identify N=9 children
who are referred to your agency after being sexually abused.
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2. Obtain a baseline measure on each (e.g., number of behavioral or emotional symptoms).
3. Provide therapeutic intervention.4. Obtain post-treatment measurement.
ResultsID Name PRE- # of
symptoms POST- # of symptoms
Improvement
∆ in # of symptoms
1 Bob 54 42 12
2 Susan 62 46 16
3 Richard 58 44 14
4 Debbie 64 43 21
5 Todd 55 57 -2 (symptom increase)
6 Cindy 60 52 8
7 Sam 67 52 15
8 Carrie 62 48 14
9 Jim 61 51 10
How would we summarize the data?
Analysis: Change in # of symptoms
N Mean St.Dev. Min Max
9 12.00 6.42 -2.00 21.00
Summary: Pre and Post Data
Group N Mean St. Dev. St. Error Mean
PRE 9 60 4.15 1.38
POST 9 48 5.42 1.81
Outcome Analysis
• Are the children showing significant improvement in number of symptoms?– What type of statistical test should I do?– Answer: A paired t-test (pre/post data)
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Pre/Post Findings
Analysis: Paired Sample T-test
t value df P-value (2 tail)
Pair One 5.444 8 0.001
Pre/Post Findings
• Interpretation: There was a significant improvement in number of symptoms between the beginning of treatment and the end of treatment.
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Analysis: Paired Sample T-test
t value df P-value (2 tail)
Pair One 5.444 8 0.001
Pre/Post Findings
Concern: But would the children have improved the same amount without our treatment?
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Pre/Post Findings
Concern: But would the children have improved the same amount without our treatment?– We need a control group!
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How Do We Compare Pre/Post Data Between Groups?
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How Do We Compare Pre/Post Data Between Groups?
• Answer: Independent t-test on change in number of symptoms.
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AnalysisRx Versus Control N Mean St.Dev
Symptom Reduction
Cntrl9 12.00 6.42
RX 9 8.78 5.59
AnalysisRx Versus Control
N Mean St.Dev
Symptom Reduction
Cntrl9 12.00 6.42
RX 9 8.78 5.59
Independent t-test (Ctrl versus Rx Groups)
Symptom Reduction
t-value dfP-value
(2 tailed)
1.136 16 .273
Findings
• Interpretation: No significant difference in symptom reduction between control and treatment groups
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New Type of Problem
• How would you analyze the following problem?
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TIME TO FAILUREDid our client fail over time?
Yes/No
• Recidivism• Rehospitalization• Acting Out• Relapse• Dropping Out• Death
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Time is a Variable
• The question is not simply, “Did our client experience a failure?”
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• Rather, “Did our client last longer before failure?”
– e.g. “Did our clients go longer before being rehospitalized?”
Chi-Square??
• Problem: What time period are you talking about?
Failure
New
Treatm
ent
Yes No
Yes 60% 40%
No 80% 20%
Particularly Difficult In Agencies
• Unlike controlled experiments, it is difficult to study cohort groups.
• Clients come and go in unpredictable manner.– How do you randomize
over time?
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Path Of Failure Over Time Is Important
Time to Rehospitalization
% of clients NOT
rehospitalized
100%
6 months 1 year
Rx as Usual
Additionally, Path Of Failure Is Important
Time to Rehospitalization
% of clients NOT
rehospitalized
100%
6 months 1 year
Rx as Usual
New Rx
Survival Analysis
• Compares number of failures over TIME – Recorded as binary yes/no
• Path of that Failure
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Instead Of Looking at Symptom Reduction, Let’s Look at the Symptom Avoidance?
EXAMPLE:Sexually Reactive Children
Sexually Reactive" children are pre-pubescent boys and girls who have been exposed to, or had contact with, inappropriate sexual activities. The sexually reactive child may engage in a variety of age-inappropriate sexual behaviors as a result of his or her own exposure to sexual experiences, and may begin to act out, or engage in, sexual behaviors or relationships that include excessive sexual play, inappropriate sexual comments or gestures, mutual sexual activity with other children, or sexual molestation and abuse of other children.
Suppose we are interested in decreasing or avoiding sexual
reactivity.
• What kinds of things are going to change with respect to our methods for evaluating the data?
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What would the data look like?
ID Name Months Status
1 Bob 5 Reactive
2 Susan 10 Reactive
3 Richard 3 Reactive
4 Debbie 11 Non-reactive
5 Todd 4 Non-reactive
6 Cindy 8 Non-reactive
7 Sam 7 Non-reactive
8 Carrie 5 Reactive
9 Jim 2 Non-reactive
Summary:
Median survival = 10 months or
6-month survival = 58% +/- 19%
S = start;
0 = last measurement of observed non-reactive behavior;
x = Reactive behavior occurred
Summary: What is Survival Analysis?
• Outcome variable: Time until an event occurs• Time = Survival time• Event (Assume single event) = failure
– Death– Relapse– Sexual reactivity– Assault– Rehospitalization
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Censored Data
• Censoring: Don’t know survival time exactly
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Why censor?
1. Study ends (no event)
2. Lost to follow-up
3. Withdraws
Terminology and Notation
• t = observed survival time
• = (0,1) random variable» 1 if failure» 0 if censored
• S(t) = survivor function
• h(t) = hazard function
• t(j) = time period
δ
Probability of survival beyond a certain point in time
Failure rate
How do we use this data in a survival analysis?
ID Name Months Status
1 Bob 5 Reactive
2 Susan 10 Reactive
3 Richard 3 Reactive
4 Debbie 11 Non-reactive
5 Todd 4 Non-reactive
6 Cindy 8 Non-reactive
7 Sam 7 Non-reactive
8 Carrie 5 Reactive
9 Jim 2 Non-reactive
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•t = observed survival time• = (0,1) random variable
»1 if failure»0 if censored)
δ
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Graphical Visual of Table
t(0) t(1) t(2) t(3)
Out of the 9 children survived > 0 months:
0 failed
1 was censored
Out of the 8 children survived > 3 months:
1 failed
1 was censored
Out of the 6 children survived > 5 months):
2 failed
2 were censored
Out of the 2 children survived > 10 months:
1 failed
1 was censored
t(j) = failure time period; t = month when failure occurred; m = # of failures; q = # censored; R = risk set
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Graphical Visual of Table
t(0) t(1) t(2) t(3)
Out of the 9 children survived > 0 months:
0 failed
1 was censored
Out of the 8 children survived > 3 months:
1 failed
1 was censored
Out of the 6 children survived > 5 months):
2 failed
2 were censored
Out of the 2 children survived > 10 months:
1 failed
1 was censored
Based upon this data set, what is the probability that a child entering our program will remain non-sexually reactive for each failure time period?
S(0)= 9/9 = 1
Interpretation: Having been admitted to the program, there is 100% chance all children will start the Rx
S(3)= 7/8 = .875
Interpretation: Having started the program, there is a 87.5% chance that a given child will survive to 3 months without becoming reactive
S(5)= 4/6 = .66.7
Interpretation: Having survived 3 months, there is a 66.7% chance that a given child will survive to 5 months without becoming reactive
S(10)= 1/2 = .500
Interpretation: Having survived 5 months, there is a 50.0% chance that a given child will survive to 10 months without becoming reactive
Probability Paths
p(Heads) = .50
p(Tails) =
.50
What are your chances of flipping heads the 1st time?
Answer: 50%
Probability Paths
p(Heads) = .50
p(Tails) =
.50
Having already flipped heads once, what is the chance your next flip is heads?
.50
.50
Answer: 50%
Probability Paths
p(Heads) = .50
p(Tails) =
.50
Having already flipped heads twice, what is the chance your next flip is heads?
.50
.50
Answer: 50% .50
.50
Probability Paths
p(Heads) = .50
p(Tails) =
.50
What are your chances of flipping heads three times in a row?
.50
.50
Answer: .50 x .50 x .50 =
.125 or there is a 12.5% chance of flipping heads three times in a row.
.50
.50
Probability Paths
p(Heads) = .50
p(Tails) =
.50
What are your chances of flipping two heads and one tail in that order?
.50
.50
Answer: .50 x .50 x .50 =
.125 or there is a 12.5% chance of flipping heads three times in a row.
.50
.50
Kaplin-Mier Method Using Probability Paths
1.00
0
What is the chance of a child surviving (being non-reactive) until 3 months after being assigned to the program? .875
.125
Answer: S(t(3)) = 1x .875 = .875
or there is a 87.5% chance of a child being non-reactive for 3 months after being assigned to the program.
S(0)= 9/9 = 1
Interpretation: Having been admitted to the program, there is 100% chance all children will start the Rx
S(3)= 7/8 = .875
Interpretation: Having started the program, there is a 85.5% chance that a given child will survive to 3 months without becoming reactive
S(5)= 4/6 = .66.7
Interpretation: Having survived 3 months, there is a 66.7% chance that a given child will survive to 5 months without becoming reactive
S(10)= 1/2 = .500
Interpretation: Having survived 5 months, there is a 50.0% chance that a given child will survive to 10 months without becoming reactive
Kaplin-Mier Method Using Probability Paths
1.00
0
What is the chance of a child surviving (being non-reactive) until 5 months after being assigned to the program? .875
.125
Answer: S(t(5)) = 1 x .875 x .667 = .584
or there is a 58.4% chance of a child being non-reactive for 5 months after being assigned to the program.
.667
.333
S(0)= 9/9 = 1
Interpretation: Having been admitted to the program, there is 100% chance all children will start the Rx
S(3)= 7/8 = .875
Interpretation: Having started the program, there is a 85.5% chance that a given child will survive to 3 months without becoming reactive
S(5)= 4/6 = .66.7
Interpretation: Having survived 3 months, there is a 66.7% chance that a given child will survive to 5 months without becoming reactive
S(10)= 1/2 = .500
Interpretation: Having survived 5 months, there is a 50.0% chance that a given child will survive to 10 months without becoming reactive
Kaplin-Mier Method Using Probability Paths
1.00
0
What is the chance of a child surviving (being non-reactive) until 3 months after being assigned to the program? .875
.125
Answer: S(t(10)) = 1 x .875 x .667 x .500 = .292
or there is a 29.2% chance of a child being non-reactive for 10 months after being assigned to the program.
.667
.333
S(0)= 9/9 = 1
Interpretation: Having been admitted to the program, there is 100% chance all children will start the Rx
S(3)= 7/8 = .875
Interpretation: Having started the program, there is a 85.5% chance that a given child will survive to 3 months without becoming reactive
S(5)= 4/6 = .66.7
Interpretation: Having survived 3 months, there is a 66.7% chance that a given child will survive to 5 months without becoming reactive
S(10)= 1/2 = .500
Interpretation: Having survived 5 months, there is a 50.0% chance that a given child will survive to 10 months without becoming reactive
.50
.50
Use survival data summary table to make survival curve
S(t)
Theoretical S(t)
1
0
t
8
Sur
viva
l Pro
babi
lity
1
.80
.60
.40
.20
02 4 6 8 10 12
S(t) in practice:
The Kaplin-Mier Survival Curve
t m q R S(t(j))
t(0) 0 0 1 9 1
t(1) 3 1 1 8 .875
t(2) 5 2 2 6 .584
t(3) 10 1 1 2 .292
Additionally, we know that 1 child survived > (at least 11 months)
Sur
viva
l Pro
babi
lity
1
.80
.60
.40
.20
02 4 6 8 10 12
We can compare the Kaplin-Mier survival curves of two Rx groups
Are these two curves significantly different?
Simpliest method is known as the Log-rank Test
Sur
viva
l Pro
babi
lity
1
.80
.60
.40
.20
02 4 6 8 10 12
We can compare the Kaplin-Mier survival curves of two Rx groups
Note: For one group to have a significantly different outcome from the other, the curves cannot cross
Logrank Test (aka Mantel-Cox Test)
The Logrank Test statistic compares estimates of the hazard functions of the two groups at each observed event time.
The Logrank statistic compare each O1j to its expectation E1j under the null hypothesis and is defined as:
Z =O1 j −E1 j( )j=1
J∑Vjj=1
J∑
Logrank Test (aka Mantel-Cox Test)
The Logrank Test statistic compares estimates of the hazard functions of the two groups at each observed event time.
The Logrank statistic compare each O1j to its expectation E1j under the null hypothesis and is defined as:
Z =O1 j −E1 j( )j=1
J∑Vjj=1
J∑Z =
x−μσ