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{ \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 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{ \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 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{ \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.

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.