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| Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data. | | Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data. |
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| {{Reference_Example_Heading1}} | | {{Reference_Example_Heading1}} |
| 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. | | Example 7.1 on page 154 in the 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. |
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| The ML estimator for <math>\theta\,\!</math> and its variance are 572.27 and 1637.489, respectively. They are given below. | | The ML estimator for <math>\theta\,\!</math> and its variance are 572.27 and 1740.52, respectively. They are given below. |
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| [[Image:1PE_interval_data.png|center]] | | [[Image:1PE_interval_data.png|center]] |
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| 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 ML estimator for <math>\theta\,\!</math> and the variance are the same as the values given in the book. |
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| * Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is | | * Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is |
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| | ::<math>F(t)= 1-e^{-(\frac{t}{\theta})}</math> |
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| | The ML estimate <math>\hat{\theta}\,\!</math> = 572.3, and the standard deviation is <math>se_{\hat{\theta}}\,\!</math> = 41.72. Therefore the variance is 1740.56. |
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| | The 95% 2-sided confidence interval for <math>\theta\,\!</math> are: |
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| | * Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on |
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| | ::<math>-2ln \frac{L(\theta)}{L(\hat{\theta})} = X^{2}_{(0.90,1)}</math> |
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| | The two solutions of <math>\theta\,\!</math> in the above equation will be the confidence bounds for <math>\theta\,\!</math>. |
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1P-Exponential MLE Solution for Interval Data
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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 the 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 1740.52, respectively. They are given below.
The ML estimator for [math]\displaystyle{ \theta\,\! }[/math] and the variance are the same as the values given in the book.
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
- [math]\displaystyle{ F(t)= 1-e^{-(\frac{t}{\theta})} }[/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 \frac{L(\theta)}{L(\hat{\theta})} = 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].
