1P-Exponential MLE Solution for Interval Data: Difference between revisions

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* Based on lognormal approximation, the confidence interval is [496, 660]. The calculation is:
* Based on lognormal approximation, the confidence interval is [496, 660]. The calculation is:


<math>\,\!</math>
::<math>\begin{alignat}{2}
[\theta_{L},\theta_{U}]&= \hat{\theta}exp(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}})\\
&=\left [572.3\times exp(-1.96\times\tfrac{41.72}{572.3}),572.3\times exp(1.96\times\tfrac{41.72}{572.3})\right]\\
&= [496,660]\\
\end{alignat}</math>
 
 
{{Reference_Example_Heading4}}
{{Reference_Example_Heading4}}
The ML estimator for <math>\theta\,\!</math> and its variance are 572.27 and 1637.489, respectively. They are given below.
[[Image:1PE_interval_data.png|center]]
The ML estimator for <math>\theta\,\!</math> and the variance are the same as the values given in the book. The following calculation provides the step by step analytical results for calculating the variance of <math>\theta\,\!</math>.
The 95% 2-sided confidence interval for <math>\theta\,\!</math> are:
* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is

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1P-Exponential MLE Solution for Interval Data

Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.


Reference Case

Example 7.1 on page 154 in book Statistical Methods for Reliability Data by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998. The sample size of 200 data is used here.


Data

Number in State Last Inspected State F/S State End Time
41 0 F 100
44 100 F 300
24 300 F 500
32 500 F 700
29 700 F 1000
21 1000 F 2000
9 2000 F 4000


Result The cumulative distribution function for an exponential distribution is:

[math]\displaystyle{ F(t)=1-e^{-\left( \frac{t}{\theta }\right )}\,\! }[/math]


The ML estimate [math]\displaystyle{ \hat{\theta}\,\! }[/math] = 572.3, and the standard deviation is [math]\displaystyle{ se_{\hat\theta}\,\! }[/math] = 41.72. Therefore the variance is 1740.56. The 95% 2-sided confidence interval for [math]\displaystyle{ {\theta}\,\! }[/math] are:


  • Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on


[math]\displaystyle{ -2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = x^{2}_{(0.90,1)}\,\! }[/math]


The two solutions of [math]\displaystyle{ \theta\,\! }[/math] in the above equation will be the confidence bounds for [math]\displaystyle{ \theta\,\! }[/math].


  • Based on lognormal approximation, the confidence interval is [496, 660]. The calculation is:
[math]\displaystyle{ \begin{alignat}{2} [\theta_{L},\theta_{U}]&= \hat{\theta}exp(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}})\\ &=\left [572.3\times exp(-1.96\times\tfrac{41.72}{572.3}),572.3\times exp(1.96\times\tfrac{41.72}{572.3})\right]\\ &= [496,660]\\ \end{alignat} }[/math]


Results in Weibull++


The ML estimator for [math]\displaystyle{ \theta\,\! }[/math] and its variance are 572.27 and 1637.489, respectively. They are given below.

1PE interval data.png

The ML estimator for [math]\displaystyle{ \theta\,\! }[/math] and the variance are the same as the values given in the book. The following calculation provides the step by step analytical results for calculating the variance of [math]\displaystyle{ \theta\,\! }[/math].


The 95% 2-sided confidence interval for [math]\displaystyle{ \theta\,\! }[/math] are:

  • Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is
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