R help - Feb 2007 - Time of failure, Arrhenius and Weibull distribution

Hi,

I'm currently doing some analyses on time of failure of a product.
I found on the internet some article about the Arrhenius equation, and I can
calculate the results with R.
Equation:
k=A*exp(-Ea/R*T)

I can fit a model with lm for this purpose, so far no problem.

But for the confidence interval, how can I use the Weibull distribution?
Or should I use the weibull distribution at all? Or can I safely use the
predict.lm method to predict my time of failure at a certain temperature?
Is there anyone who has experience with this kind of calculations?

Kind regards


Bart
	[[alternative HTML version deleted]]

On 15-Feb-07 Bart Joosen wrote:
> Hi,
> 
> I'm currently doing some analyses on time of failure of a product.
> I found on the internet some article about the Arrhenius equation,
> and I can calculate the results with R.
> Equation:
> k=A*exp(-Ea/R*T)
> 
> I can fit a model with lm for this purpose, so far no problem.
> 
> But for the confidence interval, how can I use the Weibull
> distribution?
> Or should I use the weibull distribution at all? Or can I safely use
> the predict.lm method to predict my time of failure at a certain
> temperature?
> Is there anyone who has experience with this kind of calculations?
Can you clarify your query?

Since you are using lm, would you write what your model is,
in terms like

  lm( Y ~ X1 + X2 + ... )

explaining how Y relates to failure time, and what sort of
variables X1, X2, ... are, and how Arrhenius (normally used
for evaluating the rate coefficient k of a chemical reaction,
where the only quantity that might be subject to statistical
estimation is the coefficient A) comes into it?

The standard output of summary(lm(...)) will give you the
estimates, and their standard errors, of the coefficients
of X1, X2, ... in the linear model, from which confidence
intervals for these coefficients can readily be computed.

You can certainly use predict.lm to produce

a) A confidence interval for the estimated expected ("predicted")
   value of Y, taking account of the uncertainties in the
   estimates of a1, a2, ... in Y = a1*X1 + a2*X2 + ...

b) A "prediction interval" (also called "tolerance
interval")
   for the value of Y that would be observed at given values
   of X1, X2, ... , taking account both the uncertainties
   in the estimated coefficients (i.e. in the expected value
   of Y) and the estimated random scatter of Y about its
   expected value.

How these would relate to the failure time (tF say) depends
on the relationship between Y and tF.

Your query does not describe how any Weibull considerations
come into your linear model!

If you can spell it all out for us in sufficient detail, I'm
sure people will be able to contribute suggestions and explanations.

Best wishes,
Ted.

--------------------------------------------------------------------
E-Mail: (Ted Harding) <ted.harding at nessie.mcc.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 15-Feb-07                                       Time: 22:15:00
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My model is as follows:

mod <- lm(log(Degrad/Time) ~ I(1e+05/(8.617*(Temp + 273.16))), data)

Where Degrad/Time = k, and because of taking logs, the intercept = log(A)
and the coefficient of the term = -Ea.

This is how I used the Arrhenius formule in a linear model.
I know I can use predict to estimate my failure times, with confidence 
intervals, but I'm not sure wether I should use least square regression 
(assuming a normal distribution) or a MLE with a Weibull distrbution.
And if I should use MLE with Weibull, I dont know how to implement it in R.


Thanks for your reply

Bart




Ted Harding
Thu, 15 Feb 2007 14:18:16 -0800
On 15-Feb-07 Bart Joosen wrote:
>Hi,
>
>I'm currently doing some analyses on time of failure of a product.
>I found on the internet some article about the Arrhenius equation,
>and I can calculate the results with R.
>Equation:
>k=A*exp(-Ea/R*T)
>
>I can fit a model with lm for this purpose, so far no problem.
>
>But for the confidence interval, how can I use the Weibull
>distribution?
>Or should I use the weibull distribution at all? Or can I safely use
>the predict.lm method to predict my time of failure at a certain
>temperature?
>Is there anyone who has experience with this kind of calculations?
Can you clarify your query?

Since you are using lm, would you write what your model is,
in terms like

  lm( Y ~ X1 + X2 + ... )

explaining how Y relates to failure time, and what sort of
variables X1, X2, ... are, and how Arrhenius (normally used
for evaluating the rate coefficient k of a chemical reaction,
where the only quantity that might be subject to statistical
estimation is the coefficient A) comes into it?

The standard output of summary(lm(...)) will give you the
estimates, and their standard errors, of the coefficients
of X1, X2, ... in the linear model, from which confidence
intervals for these coefficients can readily be computed.

You can certainly use predict.lm to produce

a) A confidence interval for the estimated expected ("predicted")
   value of Y, taking account of the uncertainties in the
   estimates of a1, a2, ... in Y = a1*X1 + a2*X2 + ...

b) A "prediction interval" (also called "tolerance
interval")
   for the value of Y that would be observed at given values
   of X1, X2, ... , taking account both the uncertainties
   in the estimated coefficients (i.e. in the expected value
   of Y) and the estimated random scatter of Y about its
   expected value.

How these would relate to the failure time (tF say) depends
on the relationship between Y and tF.

Your query does not describe how any Weibull considerations
come into your linear model!

If you can spell it all out for us in sufficient detail, I'm
sure people will be able to contribute suggestions and explanations.

Best wishes