Transcript Tests for time-to-event outcomes (survival analysis) Time-to-event outcome (survival data) Are the observation groups independent or correlated? Outcome Variable Time-toevent (e.g., time to fracture) independent correlated Kaplan-Meier statistics: estimates survival functions.

Tests for time-to-event
outcomes (survival analysis)
1
Time-to-event outcome
(survival data)
Are the observation groups independent or correlated?
Outcome
Variable
Time-toevent (e.g.,
time to
fracture)
independent
correlated
Kaplan-Meier statistics: estimates survival functions for
n/a (already over
time)
each group (usually displayed graphically); compares survival
functions with log-rank test
Modifications to
Cox regression
if proportionalhazards is
violated:
Time-dependent
predictors or timedependent hazard
ratios (tricky!)
Cox regression: Multivariate technique for time-to-event data;
gives multivariate-adjusted hazard ratios
2
Early example of survival
analysis, 1669
Christiaan Huygens' 1669 curve
showing how many out of 100 people
survive until 86 years.
From: Howard Wainer STATISTICAL GRAPHICS: Mapping the
Pathways of Science. Annual Review of Psychology. Vol. 52: 305-335
3
Early example of survival
analysis
Roughly, what shape
is this function?
What was a person’s
chance of surviving
This36?
is survival analysis!
past 20? Past
We are trying to estimate
this curve—only the
outcome can be any
binary event, not just
death.
4
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.
5
Randomized Clinical Trial (RCT)
Disease
Intervention
Random
assignment
Target
population
Disease-free,
at-risk cohort
Disease-free
Disease
Control
Disease-free
TIME
Randomized Clinical Trial (RCT)
Cured
Treatment
Random
assignment
Target
population
Patient
population
Not cured
Cured
Control
Not cured
TIME
Randomized Clinical Trial (RCT)
Dead
Treatment
Random
assignment
Target
population
Patient
population
Alive
Dead
Control
Alive
TIME
Cohort study
(prospective/retrospective)
Disease
Exposed
Target
population
Disease-free
cohort
Disease-free
Disease
Unexposed
Disease-free
TIME
Examples of survival analysis
in medicine
10
RCT: Women’s Health
Initiative (JAMA, 2001)
On hormones
Cumulative
incidence
On placebo
11
WHI and low-fat diet…
Prentice, R. L. et al. JAMA 2006;295:629-642.
Control
Low-fat diet
12
Retrospective cohort study:
From December 2003 BMJ:
Aspirin, ibuprofen, and mortality after myocardial
infarction: retrospective cohort study
13
Why use survival analysis?
1. Why not compare mean time-to-event
between your groups using a t-test or
linear regression?
-- ignores censoring
2. Why not compare proportion of events
in your groups using risk/odds ratios or
logistic regression?
--ignores time
14
Survival Analysis: Terms


Time-to-event: The time from entry into a
study until a subject has a particular outcome
Censoring: Subjects are said to be censored
if they are lost to follow up or drop out of the
study, or 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.
15
Review Question 1

a.
b.
c.
d.
Which of the following data sets is likely to
lend itself to survival analysis?
A case-control study of caffeine intake and breast
cancer.
A randomized controlled trial where the outcome
was whether or not women developed breast
cancer in the study period.
A cohort study where the outcome was the time it
took women to develop breast cancer.
A cross-sectional study which identified both
whether or not women have ever had breast
16
cancer and their date of diagnosis.
Introduction to Kaplan-Meier
Non-parametric estimate of the survival
function:
Simply, the empirical probability of
surviving past certain times in the
sample (taking into account censoring).
17
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.
18
Survival Data (right-censored)
Subject A
Subject B
Subject C
Subject D
Subject E
X 1. subject E dies at 4
months
Beginning of study
 Time in months 
End of study
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
 Time in months 
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
Corresponding Kaplan-Meier
Curve
100%
subject C dies at
7 months
 Time in months 
Fraction
surviving this
death = 2/3
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
4. Subjects B
and D survive
for the whole
year-long
study period
Subject E
X 1. subject E dies at 4
months
Beginning of study
 Time in months 
End of study
Corresponding Kaplan-Meier
Curve
100%
Rule from probability
theory: limit
Product
estimate of survival =
P(A&B)=P(A)*P(B)
if A and B independent
P(surviving
interval 1/at-risk up to failure 1) *
In survival analysis:
intervals are
defined by
failures (2
leading
P(surviving
interval
2/at-risk
upintervals
to failure
2) to failures here).
P(surviving intervals
2)=P(surviving
interval 1)*P(surviving interval 2)
= 4/51 *and
2/3=
.5333
 Time in months 
24
The product limit estimate




The probability of surviving in the entire year,
taking into account censoring
= (4/5) (2/3) = 53%
NOTE:  40% (2/5) because the one drop-out
survived at least a portion of the year.
AND <60% (3/5) because we don’t know if
the one drop-out would have survived until
the end of the year.
25
Example 1: time-to-conception
for subfertile women
“Failure” here is a good thing.
38 women (in 1982) were treated for infertility with
laparoscopy and hydrotubation.
All women were followed for up to 2-years to describe
time-to-conception.
The event is conception, and women "survived" until
they conceived.
Example from: BMJ, Dec 1998; 317: 1572 - 1580.
26
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
S(t) is estimated at 9 event times.
(step-wise function)
28
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
6 women conceived in 1st month
(1st menstrual cycle). Therefore,
32/38 “survived” pregnancy-free
past 1 month.
31
Corresponding Kaplan-Meier
Curve
S(t=1) = 32/38 = 84.2%
S(t) represents estimated survival probability: P(T>t)
Here P(T>1).
32
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2.1
3
4
7
7
8
8
9
9
9
11
24
24
Important detail of how the data were coded:
Censoring at t=2 indicates survival PAST the 2nd cycle (i.e., we
know the woman “survived” her 2nd cycle pregnancy-free).
Thus, for calculating KM estimator at 2 months, this person
should still be included in the risk set.
Think of it as 2+ months, e.g., 2.1 months.
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
34
Corresponding Kaplan-Meier
Curve
5 women conceive in 2nd month.
The risk set at event time 2 included 32
women.
Therefore, 27/32=84.4% “survived” event
time 2 pregnancy-free.
S(t=2) = ( 84.2%)*(84.4%)=71.1%
35
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2.1
3.1
4
7
7
8
8
9
9
9
11
24
24
Risk set at 3
months includes
26 women
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
37
Corresponding Kaplan-Meier
Curve
3 women conceive in the 3rd month.
The risk set at event time 3 included 26
women.
23/26=88.5% “survived” event time 3
pregnancy-free.
S(t=3) = ( 84.2%)*(84.4%)*(88.5%)=62.8%
38
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3.1
4
7
7
8
8
9
9
9
11
24
24
Risk set at 4
months includes
22 women
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
40
Corresponding Kaplan-Meier
Curve
3 women conceive in the 4th month, and 1
was censored between months 3 and 4.
The risk set at event time 4 included 22
women.
19/22=86.4% “survived” event time 4
pregnancy-free.
S(t=4) = ( 84.2%)*(84.4%)*(88.5%)*(86.4%)=54.2%
41
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4.1
7
7
8
8
9
9
9
11
24
24
Risk set at 6
months includes
18 women
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
43
Corresponding Kaplan-Meier
Curve
2 women conceive in the 6th month of the
study, and one was censored between
months 4 and 6.
The risk set at event time 5 included 18
women.
16/18=88.8% “survived” event time 5
pregnancy-free.
S(t=6) = (54.2%)*(88.8%)=42.9%
44
Skipping ahead to the 9th and
final event time (months=16)…
S(t=13)  22%
(“eyeball” approximation)
45
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
2 remaining at 16
months (9th event
time)
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Skipping ahead to the 9th and
final event time (months=16)…
S(t=16) =( 22%)*(2/3)=15%
Tail here just represents that the
final 2 women did not conceive
(cannot make many inferences
from the end of a KM curve)!
47
Comparing 2 groups
Use log-rank test to test the null hypothesis of no difference
between survival functions of the two groups
Kaplan-Meier: example 2
Researchers randomized 44 patients with chronic active
hepatitis were to receive prednisolone or no treatment
(control), then compared survival curves.
Example from: BMJ 1998;317:468-469 ( 15 August )
49
Survival times (months) of 44 patients with chronic active hepatitis randomised to receive
prednisolone or no treatment.
Prednisolone (n=22)
Control (n=22)
2
2
6
3
12
4
54
7
56 *
10
68
22
89
28
96
29
96
32
125*
37
128*
40
131*
41
140*
54
141*
61
143
63
145*
71
146
127*
148*
140*
162*
146*
168
158*
173*
167*
181*
182*
Data from: BMJ 1998;317:468-469 ( 15 August )
*=censored
Kaplan-Meier: example 2
Are these two curves
different?
Big drops at the end of
the curve indicate few
patients left. E.g., only
2/3 (66%) survived this
drop.
Misleading to the eye—
apparent convergence by end
of study. But this is due to 6
controls who survived fairly
long, and 3 events in the
treatment group when the
sample size was small.
51
Log-rank test
Test of Equality over Strata
Test
Log-Rank
Chi-Square
4.6599
DF
1
Pr >
Chi-Square
0.0309
Chi-square test (with 1 df) of the (overall)
difference between the two groups.
Groups appear significantly different.
52
Caveats

Survival estimates can be unreliable
toward the end of a study when there
are small numbers of subjects at risk of
having an event.
53
WHI and breast cancer
Small
numbers
left
Limitations of Kaplan-Meier
•
•
•
•
Mainly descriptive
Doesn’t control for covariates
Requires categorical predictors
Can’t accommodate predictor variables
that change over time
55
Review question 2

a.
b.
c.
d.
e.
Investigators studied a cohort of individuals who joined a
weight-loss program by tracking their weight loss over 1
year. Which of the following statistical test is likely the
most appropriate test for evaluating the effectiveness of
the weight loss program?
A two-sample t-test.
ANOVA
Repeated-measures ANOVA
Chi-square
Kaplan-Meier methods
56
Review question 3
Investigators compared mean cholesterol level between
cases with heart disease and controls without heart
disease. Which of the following is likely the most
appropriate statistical test for this comparison?
a.
b.
c.
d.
e.
A two-sample t-test.
ANOVA.
Repeated-measures ANOVA
Chi-square
Kaplan-Meier methods
57
Review question 4
What is another way to analyze the data from
review question 3?
a.
b.
c.
d.
e.
Logistic regression
Cox regression
Linear regression
Kaplan-Meier methods
There is no other way.
58
Review question 5
Which statement about this K-M curve
is correct?
a.
b.
c.
d.
The mortality rate was higher in the
control group than the treated group.
The probability of surviving past 100
days was about 50% in the treated
group.
The probability of surviving past 100
days was about 70% in the control
group.
Treatment should be recommended.
59
Introduction to Cox Regression



Also called proportional hazards
regression
Multivariate regression technique where
time-to-event (taking into account
censoring) is the dependent variable.
Estimates adjusted hazard ratios.

A hazard ratio is a ratio of rates (hazard
rates)
60
History


“Regression Models and Life-Tables” by
D.R. Cox, published in 1972, is one of
the most frequently cited journal
articles in statistics and medicine
Introduced “maximum partial likelihood”
61
Introduction to Cox Regression
Distinction between hazard/rate ratio and
odds ratio/risk ratio:
 Hazard ratio: ratio of rates
 Odds/risk ratio: ratio of proportions
 All are measures of relative risk!
By
takingregression
into account
you are the
taking
into
account
Logistic
aimstime,
to estimate
odds
ratio;
Cox
more information
just binary
yes/no.
regression
aims tothan
estimate
the hazard
ratio
Gain power/precision.
62
Example 1:
Study of publication bias
By
KaplanMeier
methods
From: Publication bias: evidence of delayed publication in a cohort study of clinical research projects BMJ 1997;315:640-645 (13 September) 63
Univariate Cox regression
Table 4 Risk factors for time to publication using univariate Cox regression analysis
Characteristic
# not published
# published
Hazard ratio (95% CI)
Null
29
23
1.00
Non-significant
trend
16
4
0.39 (0.13 to 1.12)
Significant
47
99
2.32 (1.47 to 3.66)
From: Publication bias: evidence of delayed publication in a cohort study of clinical research projects BMJ 1997;315:640-645 (13 September)
Interpretation: Significant results have a 2-fold higher
incidence of publication compared to null results.
64
Example 2:
Study of mortality in
academy award winners for screenwriting
KaplanMeier
methods
From: Longevity of screenwriters who win an academy award: longitudinal study BMJ 2001;323:1491-1496 ( 22-29 December ) 65
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)
Characteristics of Cox
Regression


Can accommodate both discrete and
continuous measures of event times
Easy to incorporate time-dependent
covariates—covariates that may change in
value over the course of the observation
period
67
Characteristics of Cox
Regression, continued



Cox models the effect of covariates on the
hazard rate but leaves the baseline
hazard rate unspecified.
Does NOT assume knowledge of absolute
risk.
Estimates relative rather than absolute
risk.
68
Assumptions of Cox Regression


Proportional hazards assumption: the
hazard for any individual is a fixed
proportion of the hazard for any other
individual
Multiplicative risk
69
The Hazard function
P(t  T  t  t / T  t )
h(t )  lim
t 
 0
t
In words: the probability that if you survive to t,
you will succumb to the event in the next instant.
70
The model
Components:
•A baseline hazard function that is left unspecified but must be
positive (=the hazard when all covariates are 0)
•A linear function of a set of k fixed covariates that is exponentiated.
(=the relative risk)
hi (t )  0 (t )e
1xi1 ...  k xik
Can take on any form!
log hi (t )  log 0 (t )  1 xi1  ...  k xik
71
The model
Proportional hazards:
Hazard for person i (eg a smoker)
Hazard
ratio
hi (t ) 0 (t )e 1xi1 ...  k xik
1 ( xi1  x j 1 ) ... 1 ( xik  x jk )
HRi , j 


e
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!
72
The model: binary predictor

HRlung cancer / smoking
(1)  
sage
hi ( t )  0 ( t )e smoking
 smoking (1 0 )



e
h j ( t )  0 ( t )e  smoking ( 0 )  sage ( 60 )
HRlung cancer / smoking  e
( 60 )
 smoking
This is the hazard ratio for smoking adjusted for age.
73
The model:continuous
predictor

HRlung cancer / 10  years increase in age
( 0 ) 
sage
hi ( t )  0 ( t )e smoking
 age ( 70  60 )



e
h j ( t )  0 ( t )e  smoking ( 0 )  sage ( 60 )
HRlung cancer / 10  years increase in age  e
( 70 )
 age (10 )
This is the hazard ratio for a 10-year increase in age,
adjusted for smoking.
Exponentiating a continuous predictor gives you the
hazard ratio for a 1-unit increase in the predictor.
74
Review Question 6

a.
b.
c.
d.
Exponentiating a beta-coefficient from
linear regression gives you what?
Odds ratios
Risk ratios
Hazard ratios
None of the above
75
Review Question 7

a.
b.
c.
d.
Exponentiating a beta-coefficient from
logistic regression gives you what?
Odds ratios
Risk ratios
Hazard ratios
None of the above
76
Review Question 8

a.
b.
c.
d.
Exponentiating a beta-coefficient from
Cox regression gives you what?
Odds ratios
Risk ratios
Hazard ratios
None of the above
77
Intention-to-Treat Analysis in
Randomized Trials
Intention-to-treat analysis: compare
outcomes according to the groups to
which subjects were initially
randomized, regardless of which
intervention (if any) they actually
followed.
78
Intention to treat

Participants will be counted in the
intervention group to which they were
originally assigned, even if they:





Refused the intervention after randomization
Discontinued the intervention during the study
Followed the intervention incorrectly
Violated study protocol
Missed follow-up measurements
79
Dietary Modification Trial




<20% of diet from fat
>=5 servings of fruit and vegetables
>=6 servings of whole grains
Primary outcomes: breast and colorectal
cancer
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Participant Flow in the Dietary Modification Component of the Women's Health Initiative
Prentice, R. L. et al. JAMA 2006;295:629-642.
Copyright restrictions may apply.
81
Why intention to treat?

Preserves the benefits of randomization.


Randomization balances potential confounding factors in
the study arms. This balance will be lost if the data are
analyzed according to how participants self-selected
rather than how they were randomized.
Simulates real life, where patients often don’t
adhere perfectly to treatment or may
discontinue treatment altogether

Evaluates effectiveness, rather than efficacy
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Baseline Demographics of Participants in Women's Health Initiative Dietary Modification Trial*
Copyright restrictions may apply.
Prentice, R. L. et al. JAMA 2006;295:629-642.
Benefits of randomization…
Nutrient Consumption Estimates and Body Weight at Baseline and Year 1
Prentice, R. L. et al. JAMA 2006;295:629-642.
Copyright restrictions may apply.
86
Real-world effectiveness…
Only 31 percent of treatment participants got their
dietary fat below 20% in the first year.
Effect of intention to treat on
the statistical analysis

Intention-to-treat analyses tend to
underestimate treatment effects;
increased variability due to switching
“waters down” results.
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Example






Take the following hypothetical RCT:
Treated subjects have a 25% chance of dying during
the 2-year study vs. placebo subjects have a 50%
chance of dying.
TRUE RR= 25%/50% = .50 (treated have 50% less
chance of dying)
You do a 2-yr RCT of 100 treated and 100 placebo
subjects.
If nobody switched, you would see about 25 deaths in
the treated group and about 50 deaths in the placebo
group (give or take a few due to random chance).
Observed RR .50
Example, continued


BUT, if early in the study, 25 treated subjects
switch to placebo and 25 placebo subjects
switch to control.
You would see about
25*.25 + 75*.50 = 43-44 deaths in the placebo
group

And about
25*.50 + 75*.25 = 31 deaths in the treated group
Observed RR = 31/44  .70
Diluted effect! (but not biased)
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The researchers factored this
into their power calculation…


The study was powered to find a 14%
difference in breast cancer risk between
treatment and control.
They assumed a 50% reduction in risk
with perfect adherence, but calculated
that this would translate to only a 14%
reduction in risk with imperfect
adherence.
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Alternatives to ITT
Per-protocol analysis

Restricts analysis to only those who followed the
assigned intervention until the end.
Treatment-received analysis
 Censored analysis: Subjects are dropped from the
analysis at the time of stopping the assigned
treatment
 Transition analysis: e.g., controls who cross over
to treatment contribute to the denominator for the
control group until they cross over; then they
contribute to the denominator for the treatment
group.

But becomes an observational study…
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Review question 9
I randomized 600 people to receive treatment (n=300)
or placebo (n=300). Of these,



10 treatment and 8 placebo subjects never started their study drug.
An additional 30 dropped out of each group before the end of the trial (or were lost
to followup).
18 treated subjects and 3 placebo subjects discontinued their treatment because of
side effects.
How many subjects do I include in my primary
statistical analysis?
a.
b.
c.
d.
e.
290
272
272
242
300
treatment, 292
treatment, 289
treatment, 292
treatment, 259
treatment, 300
placebo
placebo
placebo
placebo
placebo
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Homework




Finish reading textbook (if not already
done!)
Homework 9
Read intention-to-treat article (on
Coursework)
Review for the final exam
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