8.
Assessing Product Reliability
8.2. Assumptions/Prerequisites 8.2.3. How can you test reliability model assumptions?


Formal Trend Tests should accompany Trend Plots and Duane Plots. Three are given in this section 
In this section we look at
formal statistical tests that can allow us to quantitatively determine
whether or not the repair times of a system show a significant trend (which
may be an improvement or a degradation trend). The section on trend
and growth plotting contained a discussion of visual tests for trends
 this section complements those visual tests as several numerical tests
are presented.
Three statistical test procedures will be described:


The Reverse Arrangement Test (RAT test) is simple and makes no assumptions about what model a trend might follow 
The Reverse Arrangement Test
Assume there are \(r\) repairs during the observation period and they occurred at system ages \(T_1, \, T_2, \, T_3, \, \ldots, \, T_r\). (We set the start of the observation period to \(T = 0\)). Let \(I_1 = T_1, \, I_2 = T_2  T_1, \, I_3 = T_3  T_2, \, \ldots, \, I_r = T_r  T_{r1}\) be the interarrival times for repairs (i.e., the sequence of waiting times between failures). Assume the observation period ends at time \(T_{end} > T_r\). Previously, we plotted this sequence of interarrival times to look for evidence of trends. Now, we calculate how many instances we have of a later interarrival time being strictly greater than an earlier interarrival time. Each time that happens, we call it a reversal. If there are a lot of reversals (more than are likely from pure chance with no trend), we have significant evidence of an improvement trend. If there are too few reversals we have significant evidence of degradation. A formal definition of the reversal count and some properties of this count are:
In the example, we saw 7 reversals (2 more than average). Is this strong evidence for an improvement trend? The following table allows us to answer that at a 90 % or 95 % or 99 % confidence level  the higher the confidence, the stronger the evidence of improvement (or the less likely that pure chance alone produced the result). 

A useful table to check whether a reliability test has demonstrated significant improvement 
Onesided test means before looking at the data we expected improvement trends, or, at worst, a constant repair rate. This would be the case if we know of actions taken to improve reliability (such as occur during reliability improvement tests). For the \(r\) = 5 repair times example above where we had \(R\) = 7, the table shows we do not (yet) have enough evidence to demonstrate a significant improvement trend. That does not mean that an improvement model is incorrect  it just means it is not yet "proved" statistically. With small numbers of repairs, it is not easy to obtain significant results. For numbers of repairs beyond 12, there is a good approximation formula that can be used to determine whether \(R\) is large enough to be significant. Calculate 

Use this formula when there are more than 12 repairs in the data set 
$$ z = \frac{R  \frac{r(r1)}{4} + 0.5}{\sqrt{\frac{(2r+5)(r1)r}{72}}} \, , $$
and if \(z\) > 1.282,
we have at least 90 % significance. If \(z\)
> 1.645, we have 95 % significance, and a \(z\)
> 2.33 indicates 99 % significance
since \(z\)
has an approximate standard normal distribution.
That covers the (onesided) test for significant improvement trends. If, on the other hand, we believe there may be a degradation trend (the system is wearing out or being over stressed, for example) and we want to know if the data confirms this, then we expect a low value for \(R\) and we need a table to determine when the value is low enough to be significant. The table below gives these critical values for \(R\).
For numbers of repairs \(r\) > 12, use the approximation formula above, with \(R\) replaced by \([r(r1)/2  R]\). 

Because of the success of the Duane model with industrial improvement test data, this Trend Test is recommended  The Military
Handbook Test
This test is better at finding significance when the choice is between no trend and a NHPP Power Law (Duane) model. In other words, if the data come from a system following the Power Law, this test will generally do better than any other test in terms of finding significance. As before, we have \(r\) times of repair \(T_1, \, T_2, \, \ldots, \, T_r\) with the observation period ending at time \(T_{end} > T_r\). Calculate $$ \chi_{2r}^2 = 2 \sum_{i=1}^r \mbox{ ln } \frac{T_{end}}{T_i} \, , $$ and compare this to percentiles of the ChiSquare distribution with \(2r\) degrees of freedom. For a onesided improvement test, reject no trend (or HPP) in favor of an improvement trend if the chi square value is beyond the 90 (or 95, or 99) percentile. For a onesided degradation test, reject no trend if the chisquare value is less than the 10 (or 5, or 1) percentile. Applying this test to the 5 repair times example, the test statistic has value 13.28 with 10 degrees of freedom, and the chisquare percentile is 79 %. This test is better at finding significance when the choice is between no trend and a NHPP Exponential model. In other words, if the data come from a system following the Exponential Law, this test will generally do better than any test in terms of finding significance. As before, we have \(r\) times of repair \(T_1, \, T_2, \, \ldots, \, T_r\) with the observation period ending at time \(T_{end} > T_r\). Calculate $$ z = \frac{\sqrt{12r} \sum_{i=1}^r \left( T_i  \frac{T_{end}}{2}\right)}{r T_{end}} \, , $$ and compare this to high (for improvement) or low (for degradation) percentiles of the standard normal distribution. 

Formal tests generally confirm the subjective information conveyed by trend plots 
Case Study 1: Reliability Test Improvement Data (Continued from earlier work)
The failure data and Trend plots and Duane plot were shown earlier. The observed failure times were: 5, 40, 43, 175, 389, 712, 747, 795, 1299 and 1478 hours, with the test ending at 1500 hours. Reverse Arrangement Test: The interarrival times are: 5, 35, 3, 132, 214, 323, 35, 48, 504 and 179. The number of reversals is 33, which, according to the table above, is just significant at the 95 % level. The Military Handbook Test: The ChiSquare test statistic, using the formula given above, is 37.23 with 20 degrees of freedom and has significance level 98.9 %. Since the Duane Plot looked very reasonable, this test probably gives the most precise significance assessment of how unlikely it is that sheer chance produced such an apparent improvement trend (only about 1.1 % probability). 