Mann Reverse Arrangement Test

In the analysis of repairable equipment, one may need to make the assumption of constant failure rate. In my opinion, anytime one wants to make the assumption of constant failure rate, it should be validated.

For a piece of equipment (not a group of equipment lumped together) with a record of times to repair we can check if the data shows a trend either improving or degrading over time. The basic idea is the times between failure will either be getting further apart of closer together over time in a convincing manner.

One such test is the Kendall-Mann Reverse Arrangement Test. A key element is the concept of reversals. A reversal in time to failure (inter-arrival times) {X₁, X₂, …, X_n is an occasion when X_i < X_j for i < j.

Let’s consider an example and work through the steps to test if the failure rate is trending up or down, or not.

Example: A piece of equipment received repairs which tool a very small amount of time to accomplish at 35, 60, 98, 138, 177 and 219 days. The question our maintenance manager asked concerns the equipment and minor improvement implemented during the repairs. Is there convincing evidence in the data that the equipment is getting more reliable over time?

We have been told to use 95% confidence. This is the same as a 0.05 significance level, given 1-confidence = α.

Setting up the solution first we state the null and alternative hypothesis statements.

The null hypothesis, H₀: constant (stationary) inter-arrival times and we also assume the observations of repair times are independent and identically distributed (iid). This means they are just random and not distinguishable from a constant failure rate. For example, we cannot assume iid if we perform the repairs on a fixed monthly schedule separate from when the equipment fails.

The alternative hypothesis is one of three.

• H₁₁: decreasing inter-arrival times (degradation)
• H₁₂: increasing inter-arrival times (improvement)
• H₁₃: non-constant inter-arrival times (either degradation or improvement)

In this case, we need H₁₂ as we are interested in showing (or not) that the equipment is improving over time.

The next step is to determine the sequence of inter-arrival times. First, list the failure times in chronological order. In this example, we were given the failures as a list on which day they failed, which is this first step.

35, 60, 98, 138, 177, 219

The inter-arrival times is the time between successive repairs. We disregard the negligible repair time, and the repair times have to be negligible for this test to work. The time between the start of tracking this equipment and the first repair is 35 – 0 = 35. The time between the repair at 60 days and 35 days is 25. Following this process, we determine the ordered list of inter-arrival times.

35, 25, 38, 40, 39, 42

Next, we determine the number of reversals in this sequence of inter-arrival times. 35 is X₁ and we then count how many of the later in time inter-arrival times are larger than 35. In this case, 38, 40, 39, and 42 are larger for a count of 4. Then consider X2 = 25, and count the number of times that are larger. Here all the remaining times are larger than 25 for a count of 4. Skipping ahead to X4 = 40, the count is only one as 39 is less than 40, and 42 is larger.

The set of reversals then is 4, 4, 3, 1, and 1. The total number of reversals, R, is the sum of these counts or R = 13. The next set of steps determine the critical value for the test and permit us to compare this R = 13 to a value and determine if we should accept or reject the alternative hypothesis.

The R random variable has some interesting properties. Related to work by Kendall (1938) the mean and standard deviation of R reversals quickly converges to a normal distribution with

$$ \large\displaystyle \text{Mean }{{\mu }_{R}}=\frac{n\left( n-1 \right)}{4}$$

where, n is the number of inter-arrival observations. And,

$$ \large\displaystyle \text{Standard Deviation }{{\sigma }_{R}}=\frac{\sqrt{n\left( n-1 \right)\left( n+2.5 \right)}}{6}$$

For this example, we find a mean of 7.5 and standard deviation of 2.66145.

The next step is to determine the critical value of the test. In 1991 De La Mare determined corrected values for the critical values for specific significance levels. These provide corrections to the z_α values for the cases when n ≤ 30. The following table lists a subset of the corrections, c_α.

α	0.100	0.050	0.025	0.010
zα	1.28155	1.64485	1.95996	2.32635
cα	1.62	1.48	1.34	1.09

The c_α is the offset constant used to determine the critical value for the hypothesis test. It is used in the following formulas for each of the three possible alternative hypotheses.

• For H₁₁, conclude H₁₁ if $$ \large\displaystyle R\le {{r}_{1-\alpha }}\approx \text{Ceiling }\left[ {{\mu }_{r}}-{{z}_{\alpha }}{{\sigma }_{R}}-{{c}_{\alpha }} \right]={n\left( n-1 \right)}/{2-}\;{{r}_{\alpha }}$$
• For H₁₂, conclude H₁₂ if $$ \large\displaystyle R\ge {{r}_{\alpha }}\approx \text{Floor }\left[ {{\mu }_{r}}+{{z}_{\alpha }}{{\sigma }_{R}}+{{c}_{\alpha }} \right]={n\left( n-1 \right)}/{2-}\;{{r}_{1-\alpha }}$$
• For H₁₃, conclude H₁₃ if $$ \large\displaystyle {{r}_{\alpha /2}}\le R\le {{r}_{1-\alpha /2}}$$

Given this examples information, we should calculate the H₁₂ value and find the critical value z_α as 13.3577.

Since, R is smaller than z_α we accept H₀ and conclude that the reliability is not improving with 95% confidence.

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Related:

Laplace’s Trend Test (article)

Weakest Link (article)

Data to collect to optimize maintenance (article)