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{{Template:RGA_BOOK_SUB|Appendix B|Hypothesis Tests}} | {{Template:RGA_BOOK_SUB|Appendix B|Hypothesis Tests}} | ||
=Common Beta Hypothesis Test= | ==Common Beta Hypothesis Test== | ||
The common beta hypothesis (CBH) test is applicable to the following data types: multiple systems-concurrent operating times, repairable and fleet. As shown by Crow [[RGA_References|[17]]], suppose that <math>K\,\!</math> number of systems are under test. Each system has an intensity function given by: | The common beta hypothesis (CBH) test is applicable to the following data types: multiple systems-concurrent operating times, repairable and fleet. As shown by Crow [[RGA_References|[17]]], suppose that <math>K\,\!</math> number of systems are under test. Each system has an intensity function given by: | ||
:<math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}\,\!</math> | |||
where <math>q=1,\ldots ,K\,\!</math>. You can compare the intensity functions of each of the systems by comparing the <math>{{\beta }_{q}}\,\!</math> of each system. When conducting an analysis of data consisting of multiple systems, you expect that each of the systems performed in a similar manner. In particular, you would expect the interarrival rate of the failures across the systems to be fairly consistent. Therefore, the CBH test evaluates the hypothesis, <math>{{H}_{o}}\,\!</math>, such that <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\!</math>. Let <math>{{\tilde{\beta }}_{q}}\,\!</math> denote the conditional maximum likelihood estimate of <math>{{\beta }_{q}}\,\!</math>, which is given by: | where <math>q=1,\ldots ,K\,\!</math>. You can compare the intensity functions of each of the systems by comparing the <math>{{\beta }_{q}}\,\!</math> of each system. When conducting an analysis of data consisting of multiple systems, you expect that each of the systems performed in a similar manner. In particular, you would expect the interarrival rate of the failures across the systems to be fairly consistent. Therefore, the CBH test evaluates the hypothesis, <math>{{H}_{o}}\,\!</math>, such that <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\!</math>. Let <math>{{\tilde{\beta }}_{q}}\,\!</math> denote the conditional maximum likelihood estimate of <math>{{\beta }_{q}}\,\!</math>, which is given by: | ||
:<math>{{\tilde{\beta }}_{q}}=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{M}_{q}}}{\mathop{\sum }}}\,\ln \left( \tfrac{{{T}_{q}}}{{{X}_{iq}}} \right)}\,\!</math> | |||
where: | where: | ||
*<math>K=1.\,\!</math> | |||
*<math>{{M}_{q}}={{N}_{q}}\,\!</math> if data on the <math>{{q}^{th}}\,\!</math> system is time terminated or <math>{{M}_{q}}=({{N}_{q}}-1)\,\!</math> if data on the <math>{{q}^{th}}\,\!</math> system is failure terminated ( <math>{{N}_{q}}\,\!</math> is the number of failures on the <math>{{q}^{th}}\,\!</math> system). | |||
*<math>{{X}_{iq}}\,\!</math> is the <math>{{i}^{th}}\,\!</math> time-to-failure on the <math>{{q}^{th}}\,\!</math> system. | |||
Then for each system, assume that: | Then for each system, assume that: | ||
:<math>\chi _{q}^{2}=\frac{2{{M}_{q}}{{\beta }_{q}}}{{{{\tilde{\beta }}}_{q}}}\,\!</math> | |||
are conditionally distributed as independent chi-squared random variables with <math>2{{M}_{q}}\,\!</math> degrees of freedom. When <math>K=2\,\!</math>, you can test the null hypothesis, <math>{{H}_{o}}\,\!</math>, using the following statistic: | are conditionally distributed as independent chi-squared random variables with <math>2{{M}_{q}}\,\!</math> degrees of freedom. When <math>K=2\,\!</math>, you can test the null hypothesis, <math>{{H}_{o}}\,\!</math>, using the following statistic: | ||
:<math>F=\frac{\tfrac{\chi _{1}^{2}}{2{{M}_{1}}}}{\tfrac{\chi _{2}^{2}}{2{{M}_{2}}}}\,\!</math> | |||
If <math>{{H}_{o}}\,\!</math> is true, then <math>F\,\!</math> equals <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}\,\!</math> and conditionally has an F-distribution with <math>(2{{M}_{1}},2{{M}_{2}})\,\!</math> degrees of freedom. The critical value, <math>F\,\!</math>, can then be determined by referring to the chi-squared tables. Now, if <math>K\ge 2\,\!</math>, then the likelihood ratio procedure can be used to test the hypothesis <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\!</math>, as discussed in Crow [[RGA_References|[17]]]. Consider the following statistic: | If <math>{{H}_{o}}\,\!</math> is true, then <math>F\,\!</math> equals <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}\,\!</math> and conditionally has an F-distribution with <math>(2{{M}_{1}},2{{M}_{2}})\,\!</math> degrees of freedom. The critical value, <math>F\,\!</math>, can then be determined by referring to the chi-squared tables. Now, if <math>K\ge 2\,\!</math>, then the likelihood ratio procedure can be used to test the hypothesis <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\!</math>, as discussed in Crow [[RGA_References|[17]]]. Consider the following statistic: | ||
:<math>L=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}\ln ({{\tilde{\beta }}_{q}})-M\ln ({{\beta }^{*}})\,\!</math> | |||
where: | where: | ||
*<math>M=\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}\,\!</math> | |||
*<math>{{\beta }^{*}}=\tfrac{M}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\tfrac{{{M}_{q}}}{{{{\tilde{\beta }}}_{q}}}}\,\!</math> | |||
Also, let: | Also, let: | ||
:<math>a=1+\frac{1}{6(K-1)}\left[ \underset{q=1}{\overset{K}{\mathop \sum }}\,\frac{1}{{{M}_{q}}}-\frac{1}{M} \right]\,\!</math> | |||
Calculate the statistic <math>D\,\!</math>, such that: | Calculate the statistic <math>D\,\!</math>, such that: | ||
:<math>D=\frac{2L}{a}\,\!</math> | |||
The statistic <math>D\,\!</math> is approximately distributed as a chi-squared random variable with <math>(K-1)\,\!</math> degrees of freedom. Then after calculating <math>D\,\!</math>, refer to the chi-squared tables with <math>(K-1)\,\!</math> degrees of freedom to determine the critical points. <math>{{H}_{o}}\,\!</math> is true if the statistic <math>D\,\!</math> falls between the critical points. | The statistic <math>D\,\!</math> is approximately distributed as a chi-squared random variable with <math>(K-1)\,\!</math> degrees of freedom. Then after calculating <math>D\,\!</math>, refer to the chi-squared tables with <math>(K-1)\,\!</math> degrees of freedom to determine the critical points. <math>{{H}_{o}}\,\!</math> is true if the statistic <math>D\,\!</math> falls between the critical points. | ||
==Common Beta Hypothesis Example== | ===Common Beta Hypothesis Example=== | ||
Consider the data in the following table. | Consider the data in the following table. | ||
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Given that the intensity function for the <math>{{q}^{th}}\,\!</math> system is <math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}\,\!</math>, test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}\,\!</math> while assuming a significance level equal to 0.05. Calculate the maximum likelihood estimates of <math>{{\tilde{\beta }}_{1}}\,\!</math> and <math>{{\tilde{\beta }}_{2}}\,\!</math>. Therefore: | Given that the intensity function for the <math>{{q}^{th}}\,\!</math> system is <math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}\,\!</math>, test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}\,\!</math> while assuming a significance level equal to 0.05. Calculate the maximum likelihood estimates of <math>{{\tilde{\beta }}_{1}}\,\!</math> and <math>{{\tilde{\beta }}_{2}}\,\!</math>. Therefore: | ||
:<math>\begin{align} | |||
& {{{\tilde{\beta }}}_{1}}= & 0.3753 \\ | & {{{\tilde{\beta }}}_{1}}= & 0.3753 \\ | ||
& {{{\tilde{\beta }}}_{2}}= & 0.4657 | & {{{\tilde{\beta }}}_{2}}= & 0.4657 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Then <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408\,\!</math>. Calculate the statistic <math>F\,\!</math> with a significance level of 0.05. | Then <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408\,\!</math>. Calculate the statistic <math>F\,\!</math> with a significance level of 0.05. | ||
:<math>\begin{align} | |||
F=2.0980 | F=2.0980 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Since <math>1.2408<2.0980\,\!</math> we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}\,\!</math> at the 5% significance level. | Since <math>1.2408<2.0980\,\!</math> we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}\,\!</math> at the 5% significance level. | ||
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Now suppose that we test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\!</math>. Calculate the statistic <math>D\,\!</math>. | Now suppose that we test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\!</math>. Calculate the statistic <math>D\,\!</math>. | ||
:<math>\begin{align} | |||
D=0.5260 | D=0.5260 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Using the chi-square tables with <math>K-1=2\,\!</math> degrees of freedom, the critical values at the 2.5 and 97.5 percentiles are 0.1026 and 5.9915, respectively. Since <math>0.1026<D<5.9915\,\!</math>, we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\!</math> at the 5% significance level. | Using the chi-square tables with <math>K-1=2\,\!</math> degrees of freedom, the critical values at the 2.5 and 97.5 percentiles are 0.1026 and 5.9915, respectively. Since <math>0.1026<D<5.9915\,\!</math>, we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\!</math> at the 5% significance level. | ||
=Laplace Trend Test= | ==Laplace Trend Test== | ||
The Laplace trend test evaluates the hypothesis that a trend does not exist within the data. The Laplace trend test is applicable to the following data types: multiple systems-concurrent operating times, repairable and fleet. The Laplace trend test can determine whether the system is deteriorating, improving, or if there is no trend at all. Calculate the test statistic, <math>U\,\!</math>, using the following equation: | The Laplace trend test evaluates the hypothesis that a trend does not exist within the data. The Laplace trend test is applicable to the following data types: multiple systems-concurrent operating times, repairable and fleet. The Laplace trend test can determine whether the system is deteriorating, improving, or if there is no trend at all. Calculate the test statistic, <math>U\,\!</math>, using the following equation: | ||
:<math>U=\frac{\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{X}_{i}}}{N}-\tfrac{T}{2}}{T\sqrt{\tfrac{1}{12N}}}\,\!</math> | |||
where: | where: | ||
*<math>T\,\!</math> = total operating time (termination time) | |||
*<math>{{X}_{i}}\,\!</math> = age of the system at the <math>{{i}^{th}}\,\!</math> successive failure | |||
*<math>N\,\!</math> = total number of failures | |||
The test statistic <math>U\,\!</math> is approximately a standard normal random variable. The critical value is read from the standard normal tables with a given significance level, <math>\alpha \,\!</math>. | The test statistic <math>U\,\!</math> is approximately a standard normal random variable. The critical value is read from the standard normal tables with a given significance level, <math>\alpha \,\!</math>. | ||
==Laplace Trend Test Example== | ===Laplace Trend Test Example=== | ||
Consider once again the data given in the table above. Check for a trend within System 1 assuming a significance level of 0.10. Calculate the test statistic <math>U\,\!</math> for System 1. | Consider once again the data given in the table above. Check for a trend within System 1 assuming a significance level of 0.10. Calculate the test statistic <math>U\,\!</math> for System 1. | ||
:<math>\begin{align} | |||
U=-2.6121 | U=-2.6121 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
From the standard normal tables with a significance level of 0.10, the critical value is equal to 1.645. If <math>-1.645<U<1.645\,\!</math> then we would fail to reject the hypothesis of no trend. However, since <math>U<-1.645\,\!</math> then an improving trend exists within System 1. If <math>U>1.645\,\!</math> then a deteriorating trend would exist. | From the standard normal tables with a significance level of 0.10, the critical value is equal to 1.645. If <math>-1.645<U<1.645\,\!</math> then we would fail to reject the hypothesis of no trend. However, since <math>U<-1.645\,\!</math> then an improving trend exists within System 1. If <math>U>1.645\,\!</math> then a deteriorating trend would exist. | ||
Revision as of 21:35, 2 June 2014
Common Beta Hypothesis Test
The common beta hypothesis (CBH) test is applicable to the following data types: multiple systems-concurrent operating times, repairable and fleet. As shown by Crow [17], suppose that [math]\displaystyle{ K\,\! }[/math] number of systems are under test. Each system has an intensity function given by:
- [math]\displaystyle{ {{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}\,\! }[/math]
where [math]\displaystyle{ q=1,\ldots ,K\,\! }[/math]. You can compare the intensity functions of each of the systems by comparing the [math]\displaystyle{ {{\beta }_{q}}\,\! }[/math] of each system. When conducting an analysis of data consisting of multiple systems, you expect that each of the systems performed in a similar manner. In particular, you would expect the interarrival rate of the failures across the systems to be fairly consistent. Therefore, the CBH test evaluates the hypothesis, [math]\displaystyle{ {{H}_{o}}\,\! }[/math], such that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\! }[/math]. Let [math]\displaystyle{ {{\tilde{\beta }}_{q}}\,\! }[/math] denote the conditional maximum likelihood estimate of [math]\displaystyle{ {{\beta }_{q}}\,\! }[/math], which is given by:
- [math]\displaystyle{ {{\tilde{\beta }}_{q}}=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{M}_{q}}}{\mathop{\sum }}}\,\ln \left( \tfrac{{{T}_{q}}}{{{X}_{iq}}} \right)}\,\! }[/math]
where:
- [math]\displaystyle{ K=1.\,\! }[/math]
- [math]\displaystyle{ {{M}_{q}}={{N}_{q}}\,\! }[/math] if data on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system is time terminated or [math]\displaystyle{ {{M}_{q}}=({{N}_{q}}-1)\,\! }[/math] if data on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system is failure terminated ( [math]\displaystyle{ {{N}_{q}}\,\! }[/math] is the number of failures on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system).
- [math]\displaystyle{ {{X}_{iq}}\,\! }[/math] is the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] time-to-failure on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system.
Then for each system, assume that:
- [math]\displaystyle{ \chi _{q}^{2}=\frac{2{{M}_{q}}{{\beta }_{q}}}{{{{\tilde{\beta }}}_{q}}}\,\! }[/math]
are conditionally distributed as independent chi-squared random variables with [math]\displaystyle{ 2{{M}_{q}}\,\! }[/math] degrees of freedom. When [math]\displaystyle{ K=2\,\! }[/math], you can test the null hypothesis, [math]\displaystyle{ {{H}_{o}}\,\! }[/math], using the following statistic:
- [math]\displaystyle{ F=\frac{\tfrac{\chi _{1}^{2}}{2{{M}_{1}}}}{\tfrac{\chi _{2}^{2}}{2{{M}_{2}}}}\,\! }[/math]
If [math]\displaystyle{ {{H}_{o}}\,\! }[/math] is true, then [math]\displaystyle{ F\,\! }[/math] equals [math]\displaystyle{ \tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}\,\! }[/math] and conditionally has an F-distribution with [math]\displaystyle{ (2{{M}_{1}},2{{M}_{2}})\,\! }[/math] degrees of freedom. The critical value, [math]\displaystyle{ F\,\! }[/math], can then be determined by referring to the chi-squared tables. Now, if [math]\displaystyle{ K\ge 2\,\! }[/math], then the likelihood ratio procedure can be used to test the hypothesis [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\! }[/math], as discussed in Crow [17]. Consider the following statistic:
- [math]\displaystyle{ L=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}\ln ({{\tilde{\beta }}_{q}})-M\ln ({{\beta }^{*}})\,\! }[/math]
where:
- [math]\displaystyle{ M=\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}\,\! }[/math]
- [math]\displaystyle{ {{\beta }^{*}}=\tfrac{M}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\tfrac{{{M}_{q}}}{{{{\tilde{\beta }}}_{q}}}}\,\! }[/math]
Also, let:
- [math]\displaystyle{ a=1+\frac{1}{6(K-1)}\left[ \underset{q=1}{\overset{K}{\mathop \sum }}\,\frac{1}{{{M}_{q}}}-\frac{1}{M} \right]\,\! }[/math]
Calculate the statistic [math]\displaystyle{ D\,\! }[/math], such that:
- [math]\displaystyle{ D=\frac{2L}{a}\,\! }[/math]
The statistic [math]\displaystyle{ D\,\! }[/math] is approximately distributed as a chi-squared random variable with [math]\displaystyle{ (K-1)\,\! }[/math] degrees of freedom. Then after calculating [math]\displaystyle{ D\,\! }[/math], refer to the chi-squared tables with [math]\displaystyle{ (K-1)\,\! }[/math] degrees of freedom to determine the critical points. [math]\displaystyle{ {{H}_{o}}\,\! }[/math] is true if the statistic [math]\displaystyle{ D\,\! }[/math] falls between the critical points.
Common Beta Hypothesis Example
Consider the data in the following table.
| Repairable system data | |||
| System 1 | System 2 | System 3 | |
| Start | 0 | 0 | 0 |
| End | 2000 | 2000 | 2000 |
| Failures | 1.2 | 1.4 | 0.3 |
| 55.6 | 35 | 32.6 | |
| 72.7 | 46.8 | 33.4 | |
| 111.9 | 65.9 | 241.7 | |
| 121.9 | 181.1 | 396.2 | |
| 303.6 | 712.6 | 444.4 | |
| 326.9 | 1005.7 | 480.8 | |
| 1568.4 | 1029.9 | 588.9 | |
| 1913.5 | 1675.7 | 1043.9 | |
| 1787.5 | 1136.1 | ||
| 1867 | 1288.1 | ||
| 1408.1 | |||
| 1439.4 | |||
| 1604.8 | |||
Given that the intensity function for the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system is [math]\displaystyle{ {{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}\,\! }[/math], test the hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}\,\! }[/math] while assuming a significance level equal to 0.05. Calculate the maximum likelihood estimates of [math]\displaystyle{ {{\tilde{\beta }}_{1}}\,\! }[/math] and [math]\displaystyle{ {{\tilde{\beta }}_{2}}\,\! }[/math]. Therefore:
- [math]\displaystyle{ \begin{align} & {{{\tilde{\beta }}}_{1}}= & 0.3753 \\ & {{{\tilde{\beta }}}_{2}}= & 0.4657 \end{align}\,\! }[/math]
Then [math]\displaystyle{ \tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408\,\! }[/math]. Calculate the statistic [math]\displaystyle{ F\,\! }[/math] with a significance level of 0.05.
- [math]\displaystyle{ \begin{align} F=2.0980 \end{align}\,\! }[/math]
Since [math]\displaystyle{ 1.2408\lt 2.0980\,\! }[/math] we fail to reject the null hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}\,\! }[/math] at the 5% significance level.
Now suppose that we test the hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\! }[/math]. Calculate the statistic [math]\displaystyle{ D\,\! }[/math].
- [math]\displaystyle{ \begin{align} D=0.5260 \end{align}\,\! }[/math]
Using the chi-square tables with [math]\displaystyle{ K-1=2\,\! }[/math] degrees of freedom, the critical values at the 2.5 and 97.5 percentiles are 0.1026 and 5.9915, respectively. Since [math]\displaystyle{ 0.1026\lt D\lt 5.9915\,\! }[/math], we fail to reject the null hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\! }[/math] at the 5% significance level.
Laplace Trend Test
The Laplace trend test evaluates the hypothesis that a trend does not exist within the data. The Laplace trend test is applicable to the following data types: multiple systems-concurrent operating times, repairable and fleet. The Laplace trend test can determine whether the system is deteriorating, improving, or if there is no trend at all. Calculate the test statistic, [math]\displaystyle{ U\,\! }[/math], using the following equation:
- [math]\displaystyle{ U=\frac{\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{X}_{i}}}{N}-\tfrac{T}{2}}{T\sqrt{\tfrac{1}{12N}}}\,\! }[/math]
where:
- [math]\displaystyle{ T\,\! }[/math] = total operating time (termination time)
- [math]\displaystyle{ {{X}_{i}}\,\! }[/math] = age of the system at the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] successive failure
- [math]\displaystyle{ N\,\! }[/math] = total number of failures
The test statistic [math]\displaystyle{ U\,\! }[/math] is approximately a standard normal random variable. The critical value is read from the standard normal tables with a given significance level, [math]\displaystyle{ \alpha \,\! }[/math].
Laplace Trend Test Example
Consider once again the data given in the table above. Check for a trend within System 1 assuming a significance level of 0.10. Calculate the test statistic [math]\displaystyle{ U\,\! }[/math] for System 1.
- [math]\displaystyle{ \begin{align} U=-2.6121 \end{align}\,\! }[/math]
From the standard normal tables with a significance level of 0.10, the critical value is equal to 1.645. If [math]\displaystyle{ -1.645\lt U\lt 1.645\,\! }[/math] then we would fail to reject the hypothesis of no trend. However, since [math]\displaystyle{ U\lt -1.645\,\! }[/math] then an improving trend exists within System 1. If [math]\displaystyle{ U\gt 1.645\,\! }[/math] then a deteriorating trend would exist.