Hypothesis Tests
Hypothesis Tests
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 tests 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 [17] can be used to test the hypothesis [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}} }[/math] . 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.
Example
Consider the data in Table B.1.
| Table B.1 - 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 [math]\displaystyle{ {{\tilde{\beta }}_{1}} }[/math] and [math]\displaystyle{ {{\tilde{\beta }}_{2}} }[/math] using Eqn. (CondBeta). 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] . Using Eqn. (ftatistic) calculate the statistic [math]\displaystyle{ F }[/math] with a significance level of 0.05.
- [math]\displaystyle{ F=2.0980 }[/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 instead it is desired to test the hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}} }[/math] . Calculate the statistic [math]\displaystyle{ D }[/math] using Eqn. (Dtatistic).
- [math]\displaystyle{ D=0.5260 }[/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 tests 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] .
Example
Consider once again the data in Table B.1. 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 using Eqn. (Utatistic).
- [math]\displaystyle{ U=-2.6121 }[/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.
Critical Values for Cramér-von Mises Test
Table B.2 displays the critical values for the Cramér-von Mises goodness-of-fit test given the sample size, [math]\displaystyle{ M }[/math] , and the significance level, [math]\displaystyle{ \alpha }[/math] .
| Table B.2 - Critical values for Cramér-von Mises test | |||||
| [math]\displaystyle{ \alpha }[/math] | |||||
| [math]\displaystyle{ M }[/math] | 0.20 | 0.15 | 0.10 | 0.05 | 0.01 |
| 2 | 0.138 | 0.149 | 0.162 | 0.175 | 0.186 |
| 3 | 0.121 | 0.135 | 0.154 | 0.184 | 0.23 |
| 4 | 0.121 | 0.134 | 0.155 | 0.191 | 0.28 |
| 5 | 0.121 | 0.137 | 0.160 | 0.199 | 0.30 |
| 6 | 0.123 | 0.139 | 0.162 | 0.204 | 0.31 |
| 7 | 0.124 | 0.140 | 0.165 | 0.208 | 0.32 |
| 8 | 0.124 | 0.141 | 0.165 | 0.210 | 0.32 |
| 9 | 0.125 | 0.142 | 0.167 | 0.212 | 0.32 |
| 10 | 0.125 | 0.142 | 0.167 | 0.212 | 0.32 |
| 11 | 0.126 | 0.143 | 0.169 | 0.214 | 0.32 |
| 12 | 0.126 | 0.144 | 0.169 | 0.214 | 0.32 |
| 13 | 0.126 | 0.144 | 0.169 | 0.214 | 0.33 |
| 14 | 0.126 | 0.144 | 0.169 | 0.214 | 0.33 |
| 15 | 0.126 | 0.144 | 0.169 | 0.215 | 0.33 |
| 16 | 0.127 | 0.145 | 0.171 | 0.216 | 0.33 |
| 17 | 0.127 | 0.145 | 0.171 | 0.217 | 0.33 |
| 18 | 0.127 | 0.146 | 0.171 | 0.217 | 0.33 |
| 19 | 0.127 | 0.146 | 0.171 | 0.217 | 0.33 |
| 20 | 0.128 | 0.146 | 0.172 | 0.217 | 0.33 |
| 30 | 0.128 | 0.146 | 0.172 | 0.218 | 0.33 |
| 60 | 0.128 | 0.147 | 0.173 | 0.220 | 0.33 |
| 100 | 0.129 | 0.147 | 0.173 | 0.220 | 0.34 |
For application of the Cramér-von Mises critical values, refer to Sections 5.5.1 and 10.1.6.1.