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 $$K$$  number of systems are under test. Each system has an intensity function given by:


 * $${{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}$$

where $$q=1,\ldots ,K$$. You can compare the intensity functions of each of the systems by comparing the $${{\beta }_{q}}$$  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, $${{H}_{o}}$$, such that  $${{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}$$. Let $${{\tilde{\beta }}_{q}}$$  denote the conditional maximum likelihood estimate of  $${{\beta }_{q}}$$, which is given by:


 * $${{\tilde{\beta }}_{q}}=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{\mathop{\sum }}}\,\ln \left( \tfrac \right)}$$

where:
 * •	 $$K=1.$$
 * •	 $${{M}_{q}}={{N}_{q}}$$ if data on the  $${{q}^{th}}$$  system is time terminated or  $${{M}_{q}}=({{N}_{q}}-1)$$  if data on the  $${{q}^{th}}$$  system is failure terminated ( $${{N}_{q}}$$  is the number of failures on the  $${{q}^{th}}$$  system).
 * •	 $${{X}_{iq}}$$ is the  $${{i}^{th}}$$  time-to-failure on the  $${{q}^{th}}$$  system.

Then for each system, assume that:


 * $$\chi _{q}^{2}=\frac{2{{M}_{q}}{{\beta }_{q}}}$$

are conditionally distributed as independent Chi-Squared random variables with $$2{{M}_{q}}$$  degrees of freedom. When $$K=2$$, you can test the null hypothesis,  $${{H}_{o}}$$ , using the following statistic:


 * $$F=\frac{\tfrac{\chi _{1}^{2}}{2{{M}_{1}}}}{\tfrac{\chi _{2}^{2}}{2{{M}_{2}}}}$$

If $${{H}_{o}}$$  is true, then  $$F$$  equals  $$\tfrac$$  and conditionally has an F-distribution with  $$(2{{M}_{1}},2{{M}_{2}})$$  degrees of freedom. The critical value, $$F$$, can then be determined by referring to the Chi-Squared tables. Now, if $$K\ge 2$$, then the likelihood ratio procedure [17] can be used to test the hypothesis  $${{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}$$. Consider the following statistic:


 * $$L=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}\ln ({{\tilde{\beta }}_{q}})-M\ln ({{\beta }^{*}})$$

where:
 * •	 $$M=\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}$$
 * •	 $${{\beta }^{*}}=\tfrac{M}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\tfrac}$$

Also, let:


 * $$a=1+\frac{1}{6(K-1)}\left[ \underset{q=1}{\overset{K}{\mathop \sum }}\,\frac{1}-\frac{1}{M} \right]$$

Calculate the statistic $$D$$, such that:


 * $$D=\frac{2L}{a}$$

The statistic $$D$$  is approximately distributed as a Chi-Squared random variable with  $$(K-1)$$  degrees of freedom. Then after calculating $$D$$, refer to the Chi-Squared tables with  $$(K-1)$$  degrees of freedom to determine the critical points. $${{H}_{o}}$$ is true if the statistic  $$D$$  falls between the critical points. Example Consider the data in Table B.1.

Given that the intensity function for the $${{q}^{th}}$$  system is  $${{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}$$, test the hypothesis that  $${{\beta }_{1}}={{\beta }_{2}}$$  while assuming a significance level equal to 0.05. Calculate $${{\tilde{\beta }}_{1}}$$  and  $${{\tilde{\beta }}_{2}}$$  using Eqn. (CondBeta). Therefore:


 * $$\begin{align}

& {_{1}}= & 0.3753 \\ & {_{2}}= & 0.4657  \end{align}$$

Then $$\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408$$. Using Eqn. (ftatistic) calculate the statistic $$F$$  with a significance level of 0.05.


 * $$F=2.0980$$

Since $$1.2408<2.0980$$  we fail to reject the null hypothesis that  $${{\beta }_{1}}={{\beta }_{2}}$$  at the 5% significance level. Now suppose instead it is desired to test the hypothesis that  $${{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}$$. Calculate the statistic $$D$$  using Eqn. (Dtatistic).


 * $$D=0.5260$$

Using the Chi-Square tables with $$K-1=2$$  degrees of freedom, the critical values at the 2.5 and 97.5 percentiles are 0.1026 and 5.9915, respectively. Since $$0.1026<D<5.9915$$, we fail to reject the null hypothesis that  $${{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}$$  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, $$U$$, using the following equation:


 * $$U=\frac{\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{X}_{i}}}{N}-\tfrac{T}{2}}{T\sqrt{\tfrac{1}{12N}}}$$

where:
 * •	 $$T$$ = total operating time (termination time)
 * •	 $${{X}_{i}}$$ = age of the system at the  $${{i}^{th}}$$  successive failure
 * •	 $$N$$ = total number of failures

The test statistic $$U$$  is approximately a standard normal random variable. The critical value is read from the Standard Normal tables with a given significance level, $$\alpha $$. 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 $$U$$  for System 1 using Eqn. (Utatistic).


 * $$U=-2.6121$$

From the Standard Normal tables with a significance level of 0.10, the critical value is equal to 1.645. If $$-1.6451.645$$  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, $$M$$, and the significance level,  $$\alpha $$.

For application of the Cramér-von Mises critical values, refer to Sections 5.5.1 and 10.1.6.1.