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 evaluates 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 can be used to test the hypothesis $${{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\!$$, as discussed in Crow [17]. 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.

Common Beta Hypothesis Example
Consider the data in the following table.

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 the maximum likelihood estimates of $${{\tilde{\beta }}_{1}}\,\!$$ and $${{\tilde{\beta }}_{2}}\,\!$$. Therefore:


 * $$\begin{align}

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

Then $$\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408\,\!$$. Calculate the statistic $$F\,\!$$ with a significance level of 0.05.


 * $$\begin{align}

F=2.0980 \end{align}\,\!$$

Since $$1.2408<2.0980\,\!$$ we fail to reject the null hypothesis that $${{\beta }_{1}}={{\beta }_{2}}\,\!$$ at the 5% significance level.

Now suppose that we test the hypothesis that $${{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\!$$. Calculate the statistic $$D\,\!$$.


 * $$\begin{align}

D=0.5260 \end{align}\,\!$$

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 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, $$U\,\!$$, 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, $$\alpha \,\!$$.

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 $$U\,\!$$ for System 1.


 * $$\begin{align}

U=-2.6121 \end{align}\,\!$$

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.