R help - Nov 2015 - post a message - R help. Surivival analysis and goodness of fit

Hello there,

I have registered for both through nabble and R-help-request with this email.

I want to post the following question:

I want to implement a Hosmer Lemeshow Goodness of Fit test to my survival
analysis.
In R we can use the hoslem.test() function.
The x values are our observations, and y are our fitted probabilities.

So we can take the following data:

s <- Surv(ovarian$futime, ovarian$fustat)
sWei <- survreg(s ~ age,dist='weibull',data=ovarian)

so here we assume a weibull distribution. I want to do a HL GOF test to see, if
this is an invalid assumption.
Now how do we get our predicted probabilities. I assume it is with the predict()
function.
I have tried the code, but it is not correct

predict(sWei, newdata=list(ovarian$age), type = 'response')
 		 	   		  
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> On Nov 5, 2015, at 6:34 AM, Luke Gaylor <luke.gaylor at live.com.au>
wrote:
> 
> Hello there,
> 
> I have registered for both through nabble and R-help-request with this
email.
> 
> I want to post the following question:
> 
> I want to implement a Hosmer Lemeshow Goodness of Fit test to my survival
analysis.
> In R we can use the hoslem.test() function.
> The x values are our observations, and y are our fitted probabilities.
> 
> So we can take the following data:
> 
> s <- Surv(ovarian$futime, ovarian$fustat)
> sWei <- survreg(s ~ age,dist='weibull',data=ovarian)
> 
> so here we assume a weibull distribution. I want to do a HL GOF test to
see, if this is an invalid assumption.
> Now how do we get our predicted probabilities. I assume it is with the
predict() function.
> I have tried the code, but it is not correct
(It would be better to post the reason for your conclusion of ?incorrectitude?.
Error? Unexpected result? )

> predict(sWei, newdata=list(ovarian$age), type = 'response?)
This code had an unmatched single quote because the flanking character was
a?smart-quote. I?m side-stepping the wisdom of using an HL GOF test of a
distrubutional assumption. I suspect that the argument to newdata will not have
a name. Try:

 predict(sWei, newdata=list(age = ovarian$age), type = ?response')

Or, since you jsut want the original data to be the basis:

 predict(sWei,  type = ?response')

? 
David.
> 		 	   		  
> 	[[alternative HTML version deleted]]
> 
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David Winsemius
Alameda, CA, USA