Statistics 262: Intermediate Biostatistics

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Statistics 262: Intermediate Biostatistics

Kaplan-Meier methods and Parametric Regression methods

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More on Kaplan-Meier estimator of S(t)(“product-limit estimator” or “KM estimator”) When there are no censored data, the KM

estimator is simple and intuitive: Estimated S(t)= proportion of observations with failure times

> t. For example, if you are following 10 patients, and 3 of them

die by the end of the first year, then your best estimate of S(1 year) = 70%.

When there are censored data, KM provides estimate of S(t) that takes censoring into account (see last week’s lecture).

If the censored observation had actually been a failure: S(1 year)=4/5*3/4*2/3=2/5=40%

KM estimator is defined only at times when events occur! (empirically defined)

3

KM (product-limit) estimator, formally

]1[)(

at timeevent thehave number who theis

risk-at sindividual are there, event timeeach at

sevent timedistinct k

:

1

ttj j

j

jj

jj

kj

jn

dtS

td

nt

t...t t

4

KM (product-limit) estimator, formally

S(t) represents estimated survival probability at time t: P(T>t)

Observed event times

Typically dj= 1 person, unless data are grouped in time intervals (e.g., everyone who had the event in the 3rd month).

The risk set nj at time tj consists of the original sample minus all those who have been censored or had the event before tj

This formula gives the product-limit estimate of survival at each time an event happens.

dj/nj=proportion that failed at the event time tj

1- dj/nj=proportion surviving the event timeMultiply the probability of surviving

event time t with the probabilities of surviving all the previous event times.

]1[)(

at timeevent thehave number who theis

risk-at sindividual are there, event timeeach at

sevent timedistinct k

:

1

ttj j

j

jj

jj

kj

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dtS

td

nt

t...t t

5

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.

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)

1 21 31 41 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16

Conceived (event) Did not conceive (censored)

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Corresponding Kaplan-Meier Curve

S(t) is estimated at 9 event times.

(step-wise function)



1 21 31 41 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16




1 21 31 41 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16


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6 women conceived in 1st month (1st menstrual cycle). Therefore, 32/38 “survived” pregnancy-free past 1 month.

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S(t=1) = 32/38 = 84.2%

S(t) represents estimated survival probability: P(T>t)Here P(T>1).



1 2.11 31 41 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16


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.

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14


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%

Can get an estimate of the hazard rate here, h(t=2)= 5/32=15.6%. Given that you didn’t get pregnant in month 1, you have an estimated 5/32 chance of conceiving in the 2nd month.

And estimate of density (marginal probability of conceiving in month 2):f(t)=h(t)*S(t)=(.711)*(.156)=11%



1 2.11 3.11 41 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16


Risk set at 3 months includes 26 women

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17


S(t=3) = ( 84.2%)*(84.4%)*(88.5%)=62.8%

3 women conceive in the 3rd month.


23/26=88.5% “survived” event time 3 pregnancy-free.



1 21 3.11 41 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16



19


20


S(t=4) = ( 84.2%)*(84.4%)*(88.5%)*(86.4%)=54.2%

3 women conceive in the 4th month, and 1 was censored between months 3 and 4.



Hazard rates (conditional chances of conceiving, e.g. 100%-84%) look similar over time.

And estimate of density (marginal probability of conceiving in month 4):f(t)=h(t)*S(t)=(.136)* (.542)=7.4%



1 21 31 4.11 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16



22


23


S(t=6) = (54.2%)*(88.8%)=42.9%

2 women conceive in the 6th month of the study, and one was censored between months 4 and 6.



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Skipping ahead to the 9th and final event time (months=16)…

S(t=13) 22%(“eyeball” approximation)



1 21 31 41 71 71 82 82 92 92 92 113 243 243 4 4 4 6 6 9 9 9 10 13 16


2 remaining at 16 months (9th event time)

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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)!

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Kaplan-Meier: SAS output The LIFETEST Procedure

Product-Limit Survival Estimates

Survival Standard Number Number time Survival Failure Error Failed Left

0.0000 1.0000 0 0 0 38 1.0000 . . . 1 37 1.0000 . . . 2 36 1.0000 . . . 3 35 1.0000 . . . 4 34 1.0000 . . . 5 33 1.0000 0.8421 0.1579 0.0592 6 32 2.0000 . . . 7 31 2.0000 . . . 8 30 2.0000 . . . 9 29 2.0000 . . . 10 28 2.0000 0.7105 0.2895 0.0736 11 27 2.0000* . . . 11 26 3.0000 . . . 12 25 3.0000 . . . 13 24 3.0000 0.6285 0.3715 0.0789 14 23 3.0000* . . . 14 22 4.0000 . . . 15 21 4.0000 . . . 16 20 4.0000 0.5428 0.4572 0.0822 17 19 4.0000* . . . 17 18

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Kaplan-Meier: SAS output Survival

Standard Number Number time Survival Failure Error Failed Left

6.0000 . . . 18 17 6.0000 0.4825 0.5175 0.0834 19 16 7.0000* . . . 19 15 7.0000* . . . 19 14 8.0000* . . . 19 13 8.0000* . . . 19 12 9.0000 . . . 20 11 9.0000 . . . 21 10 9.0000 0.3619 0.6381 0.0869 22 9 9.0000* . . . 22 8 9.0000* . . . 22 7 9.0000* . . . 22 6 10.0000 0.3016 0.6984 0.0910 23 5 11.0000* . . . 23 4 13.0000 0.2262 0.7738 0.0944 24 3 16.0000 0.1508 0.8492 0.0880 25 2 24.0000* . . . 25 1 24.0000* . . . 25 0

NOTE: The marked survival times are censored observations.

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Monday “Gut Check” Problem… Calculate the product-limit estimate of

survival for the following data (n=9):Time-to-event (months) Survival

(1=died/0=censored)

10 0

2 1

4 0

8 1

12 0

14 0

10 1

1 0

3 0

Not so easy to get a plot of the actual hazard function!

In SAS, need a complicated MACRO, and depends on assumptions…here’s what I get from Paul Allison’s macro for these data…

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At best, you can get the cumulative hazard function…

t

duuh

duuhtS

etS

t

0

)(

)()(log

)( 0

See lecture 1 if you want more math!

Linear cumulative hazard function indicates a constant hazard.

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Cumulative Hazard Function

If the hazard function is increasing with time, e.g. h(t)=kt, then the cumulative hazard function will be curved up, for example h(t)=kt gives a quadratic:

ktkdut

0

If the hazard function is constant, e.g. h(t)=k, then the cumulative hazard function will be linear (and higher hazards will have steeper slopes):

2

2

0

ktktdu

t

If the hazard function is decreasing over time, e.g. h(t)=k/t, then the cumulative hazard function should be curved down, for example:

)log(0

tkdut

kt

t

duuhtS0

)()(log

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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 )

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

Survival times (months) of 44 patients with chronic active hepatitis randomised to receive prednisolone or no treatment.

http://bmj.bmjjournals.com/cgi/content/full/317/7156/468/

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Kaplan-Meier: example 2

Are these two curves different?

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.

Big drops at the end of the curve indicate few patients left. E.g., only 2/3 (66%) survived this drop.


0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0

Control group:

6 controls made it past 100 months.


0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

treated group:

5/6 of 54% rapidly drops the curve to 45%.

2/3 of 45% rapidly drops the curve to 30%.

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Point-wise confidence intervals

We will not worry about mathematical formula for confidence bands. The important point is that there is a confidence interval for each estimate of S(t). (SAS uses Greenwood’s formula.)

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Log-rank test

Test of Equality over Strata

Pr > Test Chi-Square DF Chi-Square

Log-Rank 4.6599 1 0.0309Wilcoxon 6.5435 1 0.0105-2Log(LR) 5.4096 1 0.0200

Chi-square test (with 1 df) of the (overall) difference between the two groups.

Groups appear significantly different.

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Log-rank test

Log-rank test is just a Cochran-Mantel-Haenszel chi-square test!

Anyone remember (know) what this is?

CMH test of conditional independence

Group 1

Group 2

Event No Event

a b

c d

K Strata = unique event times

21

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Group 2

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How do you know that this is a chi-square with 1 df? Why is this the

expected value in each stratum?

Variance is the variance of a hypergeometric distribution

Group 1

Group 2

Event No Event

a b

c d


0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0

Event time 1 (2 months), control group:At risk=22

1st event at month 2.


0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

Event time 1 (2 months), treated group: At risk=22

1st event at month 2.

Stratum 1= event time 1

treated

control

Event No Event

1 21

1 21

Event time 1:

1 died from each group. (22 at risk in each group)

44

244.)43(44

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0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0

Event time 2 (3 months), control group: At risk=21

Next event at month 3.


0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

Event time 2 (3 months), treated group:At risk=21

No events at 3 months

Stratum 2= event time 2

treated

control

Event No Event

0 21

1 20

Event time 2:

At 3 months, 1 died in the control group.

At that time 21 from each group were at risk

42

25.)41(42

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0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0

Event time 3 (4 months), control group:At risk=20

1 event at month 4.


0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

Event time 3 (4 months), treated group:At risk=21

Stratum 3= event time 3 (4 months)

treated

control

Event No Event

0 21

1 19

Event time 3:

At 4 months, 1 died in the control group.

At that time 21 from the treated group and 20 from the control group were at-risk. 41

25.)40(41

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54

Log-rank test, et al.

Test of Equality over Strata

Pr > Test Chi-Square DF Chi-Square

Log-Rank 4.6599 1 0.0309Wilcoxon 6.5435 1 0.0105-2Log(LR) 5.4096 1 0.0200

Likelihood Ratio test is not ideal here because it assumes exponential distribution (constant hazard).

Wilcoxon is just a version of the log-rank test that weights strata by their size (giving more weight to earlier time points).

More sensitive to differences at earlier time points.

Log-rank test has most power to test differences that fit the proportional hazards model—so works well as a set-up for subsequent Cox regression.

55

Estimated –log(S(t))

Maybe hazard function decreases a little then increases a little? Hard to say exactly…

56

Approximated h(t)

57

One more graph from SAS…

log(-log(S(t))=

log(cumulative hazard)

If group plots are parallel, this indicates that the proportional hazards assumption is valid.

Necessary assumption for calculation of Hazard Ratios…

58

Uses of Kaplan-Meier Commonly used to describe

survivorship of study population/s. Commonly used to compare two

study populations. Intuitive graphical presentation.

59

Limitations of Kaplan-Meier

• Mainly descriptive• Doesn’t control for covariates• Requires categorical predictors

• SAS does let you easily discretize continuous variables for KM methods, for exploratory purposes.

• Can’t accommodate time-dependent variables

60

Parametric Models for the hazard/survival function The class of regression models

estimated by PROC LIFEREG is known as the accelerated failure time models.

61

Parameters of the Weibull distribution

Shape parameter (inverse of the scale parameter):

<1: hazard rate is decreasing

>1 hazard rate is increasing

62

Constant hazard rate (special case of Weibull where shape parameter =1.0)

63

Recall: two parametric models

Components:

•A baseline hazard function (that may change over time).

•A linear function of a set of k fixed covariates that when exponentiated (and a few other things) gives the relative risk.

ikkii xxth ...)(log 11

Exponential model assumes fixed baseline hazard that we can estimate.

ikkii xxtth ...log)(log 11

Weibull model models the baseline hazard as a function of time. Two parameters (baseline hazard and scale) must be estimated to describe the underlying hazard function over time.

64

To get Hazard Ratios (relative risk)…

•Weibull (and thus exponential) are proportional hazards models, so hazard ratio can be calculated.

•For other parametric models, you cannot calculate hazard ratio (hazards are not necessarily proportional over time).

scaleeHR

eHR

:Model Weibull

:Model lExponentia

More tricky to get confidence intervals here!

65

What’s a hazard ratio?

Distinction between hazard/rate ratio and odds ratio/risk ratio:

Hazard/rate ratio: ratio of incidence rates

Odds/risk ratio: ratio of proportions

66

Example 1

Using data from pregnancy study…

Recall: roughly, hazard rates were similar over time

(implies exponential model should be a good fit).

The LIFEREG Procedure

Analysis of Parameter Estimates

Standard 95% Confidence Chi-

Parameter DF Estimate Error Limits Square Pr > ChiSq

Intercept 1 2.2636 0.2049 1.8621 2.6651 122.08 <.0001

Scale 1 1.0217 0.1638 0.7462 1.3987

Weibull Shape 1 0.9788 0.1569 0.7149 1.3401

Scale of 1.0 makes a Weibull an exponential, so looks exponential.

68

Parametric estimates of survival function based on a Weibull model (left) and exponential (right).

Compare to KM:

69

Example 2: 2 groups

Using data from hepatitis trial, I fit exponential and Weibull models in SAS using LIFEREG (Weibull is default in LIFEREG)…

The LIFEREG Procedure

Dependent Variable Log(time)

Right Censored Values 17

Left Censored Values 0

Interval Censored Values 0

Name of Distribution Exponential

Log Likelihood -68.03461345




Intercept 1 4.4886 0.2500 3.9986 4.9786 322.37 <.0001

group 1 0.9008 0.3917 0.1332 1.6685 5.29 0.0214

Scale 0 1.0000 0.0000 1.0000 1.0000

Weibull Shape 0 1.0000 0.0000 1.0000 1.0000

P-value for group very similar to p-value from log-rank test.

Scale parameter is set to 1, because it’s exponential.

-2Log Likelihood = 2*68= 136

Hazard ratio (treated vs. control):

e-0.9008 = .406

Interpretation: median time to death was decreased 60% in treated group; or, equivalently, mortality rate is 60% lower in treated group.

Model Information

Dependent Variable Log(time)

Right Censored Values 17

Left Censored Values 0

Interval Censored Values 0

Name of Distribution Weibull

Log Likelihood -66.94904552




Intercept 1 4.4811 0.3169 3.8601 5.1022 200.00 <.0001

group 1 1.0544 0.5096 0.0556 2.0533 4.28 0.0385

Scale 1 1.2673 0.2139 0.9103 1.7643

Weibull Shape 1 0.7891 0.1332 0.5668 1.0985

Hazard ratio (treated vs. control):

e-1.05/1.267 = .43

P-value for group very similar to p-value from log-rank test and exponential model.

Scale parameter is greater than 1, indicating decreasing hazard with time.

-2Log Likelihood = 2*67= 134

Shape parameter is just 1/scale parameter!

Comparison of models using Likelihood Ratio test:

-2LogLikelihood(simpler model)—2LogLikelihood(more complex) = chi-square with 1 df (1 extra parameter estimated for weibull model).

=136-134 = 2

NS

No evidence that Weibull model is much better than exponential.

Parametric estimates of cumulative survival based on Weibull model (left) and exponential (right), by group.

Compare to KM:

73

Compare to Cox regression:

Parameter Standard Hazard 95% Hazard Ratio Variable DF Estimate Error Chi-Square Pr > ChiSq Ratio Confidence Limits

group 1 -0.83230 0.39739 4.3865 0.0362 0.435 0.200 0.948

Date post:	07-Feb-2016
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Statistics 262: Intermediate Biostatistics

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