[R] Time of failure, Arrhenius and Weibull distribution

[(Ted Harding)]

On 16-Feb-07 Bart Joosen wrote:
> 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.

There is no law of the Universe which requires that failure
times follow a Weibull (or other) distribution.

The use of the Arrhenius formula suggests that the failure
mechanism in your application is the result of a progressive
chemical reaction, which ultimately reaches a certain level.

Random variations in the time taken to reach a fixed level
will depend on random fluctutations in the rate of reaction,
and then there will be the variations from item to item in
the level at which items fail. You will know more about how
these things behave than any of the rest of us!

Nevertheless, I would suggest fitting your model

  mod <- lm(Y ~ X)

where

  Y = log(Degrad/Time)
  X = I(1e+05/(8.617*(Temp + 273.16)))

and then looking at the residuals from the fit:

a) As a histogram of residuals: hist(mod$res)

b) As a plot of residuals against fitted values:

     plot(mod$fit, mod$res)

If (a) looks approximately normal, and (b) does not suggest
that there is a strong systematic trend in the size of
residuals relative to the fitted values, then I think I would
be satisfied to treat the problem as adequately represented
by a least squares regression with normally distributed
residuals.

If not, then it's back to you to consider how the underlying
mechanisms behave, anf then how to represent them in a model;
or to infer what you can from the results of (a) and (b).

Hoping this helps,
Ted.

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Date: 16-Feb-07                                       Time: 18:41:41
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