Question

To investigate the performance of a missile component, a laboratory puts 85 of the components on life test without replacement under specified environmental conditions. The first 12 failures are observed after 20, 34, 40, 55, 80, 97, 110, 124, 158, 172, 190 and 205 minutes.

1. Using the Weibull model, estimate the mean life of the component.
2. Using the exponential model, estimate the mean life of the component.
3. Which estimate would you prefer? Why?

Answer 1

Solution:

a.

We fit the Weibull Model to the data in R and obtain the following output:

> failure_time<-c(20,34,40,55,80,97,110,124,158,172,190,205)
> Weibull_Fit <- fitdist(failure_time, "weibull")
> summary(Weibull_Fit)
Fitting of the distribution ' weibull ' by maximum likelihood
Parameters :
estimate Std. Error
shape 1.806601 0.4303062
scale 120.463125 20.2405018
Loglikelihood: -65.64328 AIC: 135.2866 BIC: 136.2564
Correlation matrix:
shape scale
shape 1.0000000 0.3095397
scale 0.3095397 1.0000000

So, the shape parameter is and the scale parameter is .

The mean life of the component is,

b.

We fit the Exponential Model to the data in R and obtain the following output:

> failure_time<-c(20,34,40,55,80,97,110,124,158,172,190,205)
> Exponential_Fit <- fitdist(failure_time, "exp")
> summary(Exponential_Fit)
Fitting of the distribution ' exp ' by maximum likelihood
Parameters :
estimate Std. Error
rate 0.009338521 0.002664463
Loglikelihood: -68.08329 AIC: 138.1666 BIC: 138.6515

The estimated rate parameter is = 0.009338521.

The mean is given by,

The mean is = 107.0833.

c.

We obtain the plots for the Weibull distribution:

We have the following plots for exponential fit:

Observing the plots we will prefer the estimate using Weibull Model.