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In an ecological seed removal experiment, we have seed removal data from 720 seed plates with 25 seeds each. For each plate, we know the number of "surviving" seeds at several times ti until the end of the experiment. This results in right-censored data and a survival object for each plate that looks like this:

> s1<-with(Site_80D1, Surv(Gone,Status))
> s1
 [1]  66  128  128  128  183  183  183  183  183  183  183  183  183  183  247  247 
[17] 247  247  367  367  367  367  367  367+ 367+

using survreg(), we estimate a parametric model assuming exponential survival distribution (and thus a constant hazard $h(t)=λ$).

> model1<-survreg(s1 ~ 1, dist="exponential")
> summary(model1)

Call:
survreg(formula = s1 ~ 1, dist = "exponential")
            Value Std. Error    z         p
(Intercept)  5.54      0.209 26.6 2.43e-155

Scale fixed at 1 

Exponential distribution
Loglik(model)= -150.3   Loglik(intercept only)= -150.3
Number of Newton-Raphson Iterations: 3 
n= 25 

assuming that $h(t) = λ = \exp( -\text{Intercept})$, we calculate $λ$ by

> exp(-(summary(model1)$table[,1]))
[1] 0.00394038

The goal is to eventually be able to gather constant hazards for all 720 plates and conduct a LME (since the plates are grouped on 2 different spatial scales) including predictors that might act on the hazard as dependent variable.

My question now is: Am I correct in assuming that $h(t) = λ = \exp( -\text{Intercept})$ returns the constant hazard rate?

Thanks a lot in advance, and bear with me since this is my first time to go to CrossValidated for help on statistics.

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2 Answers 2

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Yes that should be the estimate of the constant hazard rate.

To my understanding, the model is of the form $\log T = \alpha + W$, so $\alpha$ should represent the log of the (population) mean survival time. For an exponential model at least, 1/mean.survival will be the hazard rate, so I believe you're correct. As a result, $\exp(-\hat{\alpha})$ should be the MLE of the constant hazard rate.

Presumably those times are days, in which case that estimate would be the instantaneous hazard rate (on the per-day scale). [edit: no turns out they're minutes so substitute minute for day everywhere above]

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  • $\begingroup$ Yes, $exp(α)$ returns the mean survival time. The times are minutes, so the hazard's unit would be (removal-)events per minute. Thank you very much for taking the time to respond and helping to resolve the doubts I had! $\endgroup$
    – Jan
    Jun 26, 2015 at 17:05
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Yes, you are right that $\lambda$ is exp(-survregObj$coefficients["(Intercept)"]).

When you fit the Weibull model with survreg the output is even more confusing. Suppose you have done survregObj <- survreg(Surv(time, event) ~ 1, dist="weibull"), to fit a single Weibull model to some data with no covariates. The shape parameter of the fitted Weibull model is 1 / survregObj$scale and the scale parameter is exp(survregObj$coefficients["(Intercept)"]).

When I say "shape" and "scale" I am referring to the parameterization that is used by the R functions such as pweibull and Wikipedia (Wikipedia calls this the "standard parameterization"). Obviously survreg uses "shape" and "scale" with completely different meanings.

The parameterization that survreg uses for the Weibull distribution is explained in ?survreg (but only in the Examples section, so I didn't notice it for a long time).

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  • $\begingroup$ Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own answer. $\endgroup$
    – Silverfish
    Jun 1, 2016 at 9:04

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