[R] ESTIMATES SURVIVAL ANALYSIS WEIBULL

[Bert Gunter]

Consult a local statistician. You need to understand the concept of
contrasts for categorical variables, which is a basic statistical
issue, not an R issue. How R handles contrasts is explained (tersely)
in ?contrast.

Cheers,
Bert

Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll




On Fri, Apr 10, 2015 at 10:23 AM, CHIRIBOGA Xavier
<xavier.chiriboga at unine.ch> wrote:
> Dear members,
>
>
>
> I did a survival analysis, Weibul fits better for my data: this is the "head" of my data
>
>
>
>  start stop state soil volatile hours replicate
> 1     0   48     1   ca       nc    48         3
> 2     0   48     1   ca       nc    48         4
> 3     0   48     1   ca       nc    48         5
> 4     0   48     1   ca       nc    48         8
> 5     0   60     1   ca        c    60         1
> 6     0   60     1   ca        c    60         1
>
>
>
>
>
> I have 4 soils: ca, cs, cl and sand
>
>
>
> and I have 2 volatiles: nc and c
>
>
>
> However, when I get the stimates my soil "ca" does not appear and the volatile "c" neither
>
>
>
> Call:
> survreg(formula = Surv(hours, state) ~ soil * volatile, data = surgal,
>     dist = "weibull")
>                       Value Std. Error      z         p
> (Intercept)          4.3605     0.0417 104.51  0.00e+00
> soilcl              -0.0784     0.0590  -1.33  1.84e-01
> soilcs              -0.3865     0.0567  -6.81  9.59e-12
> soilsand            -0.1913     0.0579  -3.30  9.61e-04
> volatilenc          -0.1600     0.0579  -2.76  5.74e-03
> soilcl:volatilenc    0.1320     0.0823   1.60  1.09e-01
> soilcs:volatilenc    0.2679     0.0812   3.30  9.64e-04
> soilsand:volatilenc  0.0821     0.0809   1.01  3.10e-01
> Log(scale)          -1.4788     0.0442 -33.46 1.73e-245
>
> Scale= 0.228
>
> Weibull distribution
> Loglik(model)= -1036.3   Loglik(intercept only)= -1065
>         Chisq= 57.41 on 7 degrees of freedom, p= 5e-10
> Number of Newton-Raphson Iterations: 5
> n= 257
>
>
>
> Anyone knows why? or what does it mean? how should I interpret the values in the table?
>
>
>
> Thanks for your help,
>
>
>
> Xavier
>
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