Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models

# Exercise 2¤

This table shows the failure times for steel specimens at four stress levels:

Stress | 0.870 | 0.990 | 1.090 | 1.180
--- | --- | --- | --- | ---
Failure times | 1.670 | 0.800 | 0.012 | 0.073
2.200 | 1.000 | 0.180 | 0.089
2.510 | 1.370 | 0.200 | 0.117
2.900 | 2.250 | 0.240 | 0.135
3.000 | 2.950 | 0.260 | 0.175
4.700 | 3.700 | 0.320 | 0.262
7.530 | 6.070 | 0.320 | 0.270
14.700 | 6.650 | 0.420 | 0.350
27.900 | 7.050 | 0.440 | 0.386
37.400 | 7.350 | 0.880 | 0.456

1. Plot the product-limit (Kaplan-Meier) estimates of the ln(-ln( i(t)) against the times to failure and assess whether a proportional hazards model is appropriate.
2. Engineering background suggests that at stress level y, failure times should have approximately a Weibull distribution with a scale parameter related to stress s by a power law relationship = csp and with a shape parameter which is independent of s.
   Assess this continuous model graphically and against the discrete alternative that failure times at the ith stress level have a Weibull distribution with parameters i and , (i = 1,2,3 and 4).
3. Are you satisfied that should be considered the same for all four stresses?

Click here for the charts and the answers to the questions.