Template:Avg failure intensities test crow rga

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.