Template:Statistical tests for effectiveness of corrective actions crow rga

Statistical Tests for Effectiveness of Corrective Actions

Introduction

The purpose of the statistical tests is to explore the effectiveness of corrective actions during or at the end of a phase. Say we have two phases, Phase 1 and Phase 2. Suppose that corrective actions are incorporated during or at the end of Phase 1. The system is then operated during Phase 2. The general question is whether or not the corrective actions have been effective.
There are two questions that can be addressed regarding the effectiveness of the corrective actions:

Question 1. Is the average failure intensity for Phase 2 statistically less than the average failure intensity for Phase1?
Question 2. Is the average failure intensity for Phase 2 statistically less than the Crow-AMSAA (NHPP) instantaneous failure intensity at the end of Phase 1?

Average Failure Intensities Test

The purpose of this test is to compare the average failure intensity during Phase 2 with the average failure intensity during Phase 1. The average failure intensity for Phase 1 is:

[math]\displaystyle{ {{\overline{r}}_{1}}=\frac{{{N}_{1}}}{{{T}_{1}}} }[/math]

where [math]\displaystyle{ {{T}_{1}} }[/math] is the Phase 1 test time and [math]\displaystyle{ {{N}_{1}} }[/math] is the number of failures during Phase 1.

Similarly, the average failure intensity for Phase 2 is:

[math]\displaystyle{ {{\overline{r}}_{2}}=\frac{{{N}_{2}}}{{{T}_{2}}} }[/math]

where [math]\displaystyle{ {{T}_{2}} }[/math] is the Phase 2 test time and [math]\displaystyle{ {{N}_{2}} }[/math] is the number of failures during Phase 2. The overall test time, [math]\displaystyle{ T }[/math] , is:

[math]\displaystyle{ T={{T}_{1}}+{{T}_{2}} }[/math]

The overall number of failures, [math]\displaystyle{ N }[/math] , is:

[math]\displaystyle{ N={{N}_{1}}+{{N}_{2}} }[/math]

Define [math]\displaystyle{ P }[/math] as:

[math]\displaystyle{ P=\frac{{{T}_{2}}}{T} }[/math]

If the cumulative binomial probability [math]\displaystyle{ B(k;P,N) }[/math] of up to [math]\displaystyle{ {{N}_{2}} }[/math] failures is less than or equal to the statistical significance [math]\displaystyle{ \alpha }[/math] , then the average failure intensity for Phase 2 is statistically less than the average failure intensity for Phase 1 at the specific significance level. The cumulative binomial distribution probability is given by:

[math]\displaystyle{ B(k;P,N)=\underset{f=0}{\overset{k}{\mathop \sum }}\,\left( \begin{matrix} N \\ f \\ \end{matrix} \right){{P}^{f}}{{\left( 1-P \right)}^{N-f}} }[/math]

which gives the probability that the test failures, [math]\displaystyle{ f }[/math] , are less than or equal to the number of allowable failures, or acceptance number [math]\displaystyle{ k }[/math] in [math]\displaystyle{ N }[/math] trials, when each trial has a probability of succeeding of [math]\displaystyle{ P }[/math] .

Example 4

Suppose a test is being conducted and is divided into two phases. The test time for the first Phase is [math]\displaystyle{ {{T}_{1}}=27 }[/math] days and the test time for the second phase is [math]\displaystyle{ {{T}_{2}}=18 }[/math] days. The number of failures during Phase 1 is [math]\displaystyle{ {{N}_{1}}=11 }[/math] and the number of failures during Phase 2 is [math]\displaystyle{ {{N}_{2}}=2. }[/math] Using Eqn. (AFI1), the average failure intensity for Phase 1 is:

[math]\displaystyle{ {{\overline{r}}_{1}}=\frac{{{N}_{1}}}{{{T}_{1}}}=\frac{11}{27}=0.4074. }[/math]

Similarly, using Eqn. (AFI2), the average failure intensity for Phase 2 is:

[math]\displaystyle{ {{\overline{r}}_{2}}=\frac{{{N}_{2}}}{T2}=\frac{2}{18}=0.1111 }[/math]

Although the average failure intensities are different, we want to see, if at the 10% statistical significance level, the average failure intensity for Phase 2 is statistically less than the average failure intensity for Phase 1.

'Solution'
Concerning the total test time, using Eqn. (T) we have:

[math]\displaystyle{ T={{T}_{1}}+{{T}_{2}}=27+18=45 }[/math]

Using Eqn. (N), the total number of failures is equal to:

[math]\displaystyle{ N={{N}_{1}}+{{N}_{2}}=11+2=13 }[/math]

Then, by using Eqn. (P) we calculate [math]\displaystyle{ P }[/math] as:

[math]\displaystyle{ P=\frac{{{T}_{2}}}{T}=\frac{18}{45}=0.4 }[/math]

Using Eqn. (Cum.Binomial), we have:

[math]\displaystyle{ \begin{align} & B(k;P,N)= & B({{N}_{2}};P,N) \\ & = & \underset{f=0}{\overset{{{N}_{2}}}{\mathop \sum }}\,\left( \begin{matrix} N \\ {{N}_{2}} \\ \end{matrix} \right){{P}^{f}}{{\left( 1-P \right)}^{N-f}} \\ & = & \underset{f=0}{\overset{2}{\mathop \sum }}\,\left( \begin{matrix} 13 \\ f \\ \end{matrix} \right){{0.4}^{f}}{{\left( 1-0.4 \right)}^{13-f}} \\ & = & 0.058 \end{align} }[/math]

Since 0.058 is lower than 0.10, the conclusion is that at the 10% significance level the average failure intensity for Phase 2 is statistically less than the average failure intensity for Phase 1. The conclusion would be different for a different significance level. For example, at the 5% significance level, since 0.058 is not lower than 0.05, we fail to reject the null hypothesis. In other words, we cannot statistically prove any significant difference between the average failure intensities at the 5% level.

Average vs. Demonstrated Failure Intensities Test

The purpose of this test is to compare the average failure intensity during Phase 2 with the Crow-AMSAA (NHPP) instantaneous failure intensity at the end of Phase 1; in other words, the demonstrated failure intensity at the end of Phase 1. Once again, the average failure intensity for Phase 2 is given by Eqn. (AFI2):

[math]\displaystyle{ {{\overline{r}}_{2}}=\frac{{{N}_{2}}}{{{T}_{2}}} }[/math]

where [math]\displaystyle{ {{T}_{2}} }[/math] is the Phase 2 test time and [math]\displaystyle{ {{N}_{2}} }[/math] is the number of failures during Phase 2. The Crow-AMSAA (NHPP) model estimate of failure intensity at time is [math]\displaystyle{ \widehat{r}\left( {{T}_{1}} \right) }[/math] . In Confidence Interval Procedures for the Weibull Process with Applications to Reliability Growth [16], Dr. Larry H. Crow showed that the Crow-AMSAA (NHPP) estimate is approximately distributed as a random variable with standard deviation [math]\displaystyle{ \sqrt{\tfrac{{{N}_{1}}}{2}.} }[/math]
We therefore treat [math]\displaystyle{ \widehat{r}\left( {{T}_{1}} \right) }[/math] as an approximate Poisson random variable with number of failures:

[math]\displaystyle{ N_{1}^{*}=\frac{{{N}_{1}}}{2} }[/math]

We also set:

[math]\displaystyle{ T_{1}^{*}=\frac{N_{1}^{*}}{\widehat{r}\left( {{T}_{1}} \right)} }[/math]

Then we define:

[math]\displaystyle{ T=T_{1}^{*}+{{T}_{2}} }[/math]

and:

[math]\displaystyle{ N=N_{1}^{*}+{{N}_{2}} }[/math]

Let:

[math]\displaystyle{ P=\frac{{{T}_{2}}}{T} }[/math]

If the cumulative binomial probability [math]\displaystyle{ B(k;P,N) }[/math] of up to [math]\displaystyle{ {{N}_{2}} }[/math] failures is less than or equal to the statistical significance [math]\displaystyle{ \alpha }[/math] , then the average failure intensity for Phase 2 is statistically less than the Crow-AMSAA (NHPP) instantaneous failure intensity at the end of Phase 1, at the specific significance level. The cumulative binomial distribution probability is again given by Eqn. (Cum.Binomial).

Example 5

Suppose a test is being conducted and is divided into two phases. The test time for the first phase is [math]\displaystyle{ {{T}_{1}}=27 }[/math] days and the test time for the second phase is [math]\displaystyle{ {{T}_{2}}=18 }[/math] days. The number of failures during Phase 1 is [math]\displaystyle{ {{N}_{1}}=11 }[/math] and the number of failures during Phase 2 is [math]\displaystyle{ {{N}_{2}}=2. }[/math] The Crow-AMSAA (NHPP) parameters for the first phase are [math]\displaystyle{ \beta =0.7189 }[/math] and [math]\displaystyle{ \lambda =1.0288. }[/math]
The demonstrated failure intensity at the end of Phase 1 is calculated as follows:

[math]\displaystyle{ \begin{align} & \widehat{r}\left( {{T}_{1}} \right)= & \lambda \beta {{T}^{\beta -1}} \\ & = & 1.0288\cdot 0.7189\cdot {{27}^{0.7189-1}} \\ & = & 0.2929 \end{align} }[/math]

Using Eqn. (AFI2), the average failure intensity for Phase 2 is:

[math]\displaystyle{ {{\overline{r}}_{2}}=\frac{{{N}_{2}}}{T2}=\frac{2}{18}=0.1111 }[/math]

Determine if the average failure intensity for Phase 2 is statistically less than the demonstrated failure intensity at the end of Phase 1 at the 10% significance level.

Solution
From Eqn. (N1*), we can calculate [math]\displaystyle{ N_{1}^{*}\ \ : }[/math]

[math]\displaystyle{ N_{1}^{*}=\frac{{{N}_{1}}}{2}=\frac{11}{2}=5.5 }[/math]

Using Eqn. (T1*), we have:

[math]\displaystyle{ T_{1}^{*}=\frac{N_{1}^{*}}{\widehat{r}\left( {{T}_{1}} \right)}=\frac{5.5}{0.2929}=18.78 }[/math]

Concerning the total test time, using Eqn. (T for test 2) we have:

Concerning the total number of failures, using Eqn. (N for test 2) we have:

[math]\displaystyle{ N=N_{1}^{*}+{{N}_{2}}=5.5+2=7.5 }[/math]

Then, by using Eqn.(P for test 2) we calculate [math]\displaystyle{ P }[/math] as:

[math]\displaystyle{ P=\frac{{{T}_{2}}}{T}=\frac{18}{36.78}=0.4894 }[/math]

Since the number of failures [math]\displaystyle{ N=7.5 }[/math] is not an integer, we are going to calculate the cumulative binomial probabilities for [math]\displaystyle{ N=7 }[/math] and [math]\displaystyle{ N=8 }[/math] , and then interpolate to [math]\displaystyle{ N=7.5. }[/math]
For [math]\displaystyle{ N=7 }[/math] , using Eqn. (Cum.Binomial) we have:

[math]\displaystyle{ \begin{align} & B(k;P,N)= & B({{N}_{2}};P,N) \\ & = & B(2;0.4894,7) \\ & = & \underset{f=0}{\overset{{{N}_{2}}}{\mathop \sum }}\,\left( \begin{matrix} N \\ {{N}_{2}} \\ \end{matrix} \right){{P}^{f}}{{\left( 1-P \right)}^{N-f}} \\ & = & \underset{f=0}{\overset{2}{\mathop \sum }}\,\left( \begin{matrix} 7 \\ f \\ \end{matrix} \right){{0.4894}^{f}}{{\left( 1-0.4894 \right)}^{7-f}} \\ & = & 0.244 \end{align} }[/math]

And for [math]\displaystyle{ N=8 }[/math] we have:

[math]\displaystyle{ \begin{align} & B(k;P,N)= & B({{N}_{2}};P,N) \\ & = & B(2;0.4894,8) \\ & = & \underset{f=0}{\overset{{{N}_{2}}}{\mathop \sum }}\,\left( \begin{matrix} N \\ {{N}_{2}} \\ \end{matrix} \right){{P}^{f}}{{\left( 1-P \right)}^{N-f}} \\ & = & \underset{f=0}{\overset{2}{\mathop \sum }}\,\left( \begin{matrix} 8 \\ f \\ \end{matrix} \right){{0.4894}^{f}}{{\left( 1-0.4894 \right)}^{8-f}} \\ & = & 0.158 \end{align} }[/math]

A linear interpolation between two data points [math]\displaystyle{ \left( {{x}_{a}},{{y}_{a}} \right) }[/math] and [math]\displaystyle{ \left( {{x}_{b}},{{y}_{b}} \right) }[/math] at the [math]\displaystyle{ \left( x,y \right) }[/math] interpolant is given by:

[math]\displaystyle{ y={{y}_{a}}+\left( x-{{x}_{a}} \right)\frac{\left( {{y}_{b}}-{{y}_{a}} \right)}{\left( {{x}_{b}}-{{x}_{a}} \right)} }[/math]

So for [math]\displaystyle{ N=7.5 }[/math] we would have that:

[math]\displaystyle{ \begin{align} & B({{N}_{2}};P,7.5)= & B({{N}_{2}};P,7)+\left( 7.5-7 \right)\frac{B({{N}_{2}};P,8)-B({{N}_{2}};P,7)}{8-7} \\ & = & 0.244+\left( 7.5-7 \right)\frac{0.158-0.244}{8-7} \\ & = & 0.201 \end{align} }[/math]

Since [math]\displaystyle{ 0.201 }[/math] is greater than [math]\displaystyle{ 0.10 }[/math] the conclusion is that at the 10% significance level the average failure intensity for Phase 2 is not statistically different compared to the demonstrated failure intensity at the end of Phase 1.
Figure STECA RGA7 shows the application of the two statistical tests for the Crow Extended - Continuous Evaluation model in RGA 7 for Example 2 of this chapter.

Statistical test for corrective actions in RGA 7.