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]]
(Ted Harding)
2007-Feb-15 22:15 UTC
[R] Time of failure, Arrhenius and Weibull distribution
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 ------------------------------ XFMail ------------------------------
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