Exponential Distribution for Grouped Data Example - Revision history

Lisa Hacker at 21:46, 18 September 2023

2023-09-18T21:46:26Z

← Older revision		Revision as of 21:46, 18 September 2023
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	<noinclude>		<noinclude>
	{{Banner Weibull Examples}}		{{Banner Weibull Examples}}
	''This example appears in the [~~[The_Exponential_Distribution#Exponential_Distribution_for_Grouped_Data\|~~Life ~~Data Analysis Reference book]~~]''.		''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''.
	</noinclude>		</noinclude>

Nicolette Young at 17:16, 16 December 2015

2015-12-16T17:16:39Z

← Older revision		Revision as of 17:16, 16 December 2015
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	<br>		<br>
	<br>		<br>
	[[Image:Exponential Distribution Example 8 Data.png\|center\|~~650px~~\|]]		[[Image:Exponential Distribution Example 8 Data.png\|center\|750px\|]]

Richard House at 18:13, 25 September 2012

2012-09-25T18:13:41Z

← Older revision		Revision as of 18:13, 25 September 2012
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	'''Solution'''		'''Solution'''

	1. For the 2-parameter exponential distribution and for <math>\hat{\gamma }=100</math> hours (first failure), the partial of the log-likelihood function, <math>\lambda</math>, becomes:		1. For the 2-parameter exponential distribution and for <math>\hat{\gamma }=100\,\!</math> hours (first failure), the partial of the log-likelihood function, <math>\lambda\,\!</math>, becomes:

	::<math> \begin{align}		::<math> \begin{align}
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	\end{align}		\end{align}
	</math>		\,\!</math>


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	Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, <math>\gamma </math>, at 100 hours.		Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, <math>\gamma \,\!</math>, at 100 hours.


	7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time <math>{{T}_{i}}</math> where <math>i</math> indicates the group number. In this example, the total number of groups is <math>N=6</math> and the total number of units is <math>{{N}_{T}}=20</math>. Thus, the median rank values will be estimated for 20 units and for the total failed units (<math>{{N}_{{{F}_{i}}}}</math>) up to the <math>{{i}^{th}}</math> group, for the <math>{{i}^{th}}</math> rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.		7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time <math>{{T}_{i}}\,\!</math> where <math>i\,\!</math> indicates the group number. In this example, the total number of groups is <math>N=6\,\!</math> and the total number of units is <math>{{N}_{T}}=20\,\!</math>. Thus, the median rank values will be estimated for 20 units and for the total failed units (<math>{{N}_{{{F}_{i}}}}\,\!</math>) up to the <math>{{i}^{th}}\,\!</math> group, for the <math>{{i}^{th}}\,\!</math> rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.


	For example, the median rank value of the fourth group will be the <math>{{17}^{th}}</math> rank out of a sample size of twenty units (or 81.945%).		For example, the median rank value of the fourth group will be the <math>{{17}^{th}}\,\!</math> rank out of a sample size of twenty units (or 81.945%).


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	\text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\		\text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\
	\sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\		\sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\
	\end{matrix}</math></center>		\end{matrix}\,\!</math></center>


	Given the values in the table above, calculate <math>\hat{a}</math> and <math>\hat{b}</math>:		Given the values in the table above, calculate <math>\hat{a}\,\!</math> and <math>\hat{b}\,\!</math>:


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	& & \\		& & \\
	& \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6}		& \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6}
	\end{align}</math>		\end{align}\,\!</math>

	or:		or:

	::<math>\hat{b}=-0.005392</math>		::<math>\hat{b}=-0.005392\,\!</math>

	and:		and:

	::<math>\hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}</math>		::<math>\hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\!</math>

	or:		or:

	::<math>\hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793</math>		::<math>\hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\!</math>

	Therefore:		Therefore:


	::<math>\hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}</math>		::<math>\hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\!</math>


	and:		and:

	::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}</math>		::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\!</math>

	or:		or:

	::<math>\hat{\gamma }\simeq 51.8\text{ hours}</math>		::<math>\hat{\gamma }\simeq 51.8\text{ hours}\,\!</math>

	Then:		Then:
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	\hat{\lambda }= & 0.0054\text{ failures/hour} \\		\hat{\lambda }= & 0.0054\text{ failures/hour} \\
	\hat{\gamma }= & 51.82\text{ hours}		\hat{\gamma }= & 51.82\text{ hours}
	\end{align}</math>		\end{align}\,\!</math>

	The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the <math>15^{th}</math> decimal point.		The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the <math>15^{th}\,\!</math> decimal point.

Lisa Hacker at 00:47, 16 August 2012

2012-08-16T00:47:56Z

← Older revision		Revision as of 00:47, 16 August 2012
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	</noinclude>		</noinclude>

	~~Twenty~~ units were reliability tested with the following results:		20 units were reliability tested with the following results:

Chris Kahn at 08:16, 13 August 2012

2012-08-13T08:16:02Z

← Older revision		Revision as of 08:16, 13 August 2012
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	Then:		Then:

	::<math>f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}</math>		::<math>f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\!</math>

Chris Kahn at 08:02, 13 August 2012

2012-08-13T08:02:55Z

← Older revision		Revision as of 08:02, 13 August 2012
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	\hat{\gamma }= & 51.82\text{ hours}		\hat{\gamma }= & 51.82\text{ hours}
	\end{align}</math>		\end{align}</math>

			The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the <math>15^{th}</math> decimal point.

Chris Kahn at 07:57, 13 August 2012

2012-08-13T07:57:20Z

← Older revision		Revision as of 07:57, 13 August 2012
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	Twenty units were reliability tested with the following results:		Twenty units were reliability tested with the following results:


	{\| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"		{\| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
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	\|2 \|\|600		\|2 \|\|600
	\|}		\|}


	1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.		1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.
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	\end{align}		\end{align}
	</math>		</math>


	2. Enter the data in a Weibull++ standard folio and calculate it as shown next.		2. Enter the data in a Weibull++ standard folio and calculate it as shown next.
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	<br>		<br>
	[[Image:Exponential Distribution Example 8 Data.png\|center\|650px\|]]		[[Image:Exponential Distribution Example 8 Data.png\|center\|650px\|]]


	3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.		3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.
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	[[Image:Exponential Distribution Example 8 Plot.png\|center\|650px\|]]		[[Image:Exponential Distribution Example 8 Plot.png\|center\|650px\|]]

	~~You can select different plots in the '''Plot Type''' drop-down list, shown next.~~

	~~[[Image:Exponential Distribution Example 8 Plot Type~~.~~png\|center]]~~		4. View the Reliability vs. Time plot.


	~~4. View the Reliability vs. Time plot~~.		[[Image:Exponential Distribution Example 8 Rel Plot.png\|center\|650px\|]]


	~~[[Image:Exponential Distribution Example 8 Rel Plot.png\|center\|650px\|]]~~

	5. View the ''pdf'' plot.		5. View the ''pdf'' plot.
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	[[Image:Exponential Distribution Example 8 Pdf Plot.png\|center\|650px\|]]		[[Image:Exponential Distribution Example 8 Pdf Plot.png\|center\|650px\|]]



	6. View the Failure Rate vs. Time plot.		6. View the Failure Rate vs. Time plot.
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	\hat{\gamma }= & 51.82\text{ hours}		\hat{\gamma }= & 51.82\text{ hours}
	\end{align}</math>		\end{align}</math>

	~~The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the <math>{{15}^{th}}</math> decimal point.~~
	~~<br>~~

Chris Kahn at 07:55, 13 August 2012

2012-08-13T07:55:38Z

← Older revision		Revision as of 07:55, 13 August 2012
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	Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, <math>\gamma </math>, at 100 hours.		Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, <math>\gamma </math>, at 100 hours.


	7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time <math>{{T}_{i}}</math> where <math>i</math> indicates the group number. In this example, the total number of groups is <math>N=6</math> and the total number of units is <math>{{N}_{T}}=20</math>. Thus, the median rank values will be estimated for 20 units and for the total failed units (<math>{{N}_{{{F}_{i}}}}</math>) up to the <math>{{i}^{th}}</math> group, for the <math>{{i}^{th}}</math> rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.		7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time <math>{{T}_{i}}</math> where <math>i</math> indicates the group number. In this example, the total number of groups is <math>N=6</math> and the total number of units is <math>{{N}_{T}}=20</math>. Thus, the median rank values will be estimated for 20 units and for the total failed units (<math>{{N}_{{{F}_{i}}}}</math>) up to the <math>{{i}^{th}}</math> group, for the <math>{{i}^{th}}</math> rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.


	For example, the median rank value of the fourth group will be the <math>{{17}^{th}}</math> rank out of a sample size of twenty units (or 81.945%).		For example, the median rank value of the fourth group will be the <math>{{17}^{th}}</math> rank out of a sample size of twenty units (or 81.945%).


	The following table is then constructed.		The following table is then constructed.


	<center><math>\begin{matrix}		<center><math>\begin{matrix}
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	::<math>f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}</math>		::<math>f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}</math>


	Using Weibull++, the estimated parameters are:		Using Weibull++, the estimated parameters are:

Chris Kahn at 07:49, 13 August 2012

2012-08-13T07:49:00Z

← Older revision		Revision as of 07:49, 13 August 2012
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	[[Image:Exponential Distribution Example 8 Data.png\|center\|650px\|]]		[[Image:Exponential Distribution Example 8 Data.png\|center\|650px\|]]

	3. ~~The~~ exponential ~~probability~~ plot is shown next		3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

	[[Image:Exponential Distribution Example 8 Plot.png\|center\|650px\|]]		[[Image:Exponential Distribution Example 8 Plot.png\|center\|650px\|]]

	~~The~~ '''Plot Type''' drop-down ~~box allows you to select different plot types~~, shown next.		You can select different plots in the '''Plot Type''' drop-down list, shown next.

	[[Image:Exponential Distribution Example 8 Plot Type.png\|center]]		[[Image:Exponential Distribution Example 8 Plot Type.png\|center]]


	4. ~~Select '''~~Reliability vs. Time~~'''~~.		4. View the Reliability vs. Time plot.


	[[Image:Exponential Distribution Example 8 Rel Plot.png\|center\|~~550px~~\|]]		[[Image:Exponential Distribution Example 8 Rel Plot.png\|center\|650px\|]]

	5. ~~The exponential <math>~~pdf~~</math>~~ plot ~~is shown next~~.		5. View the ''pdf'' plot.


	[[Image:Exponential Distribution Example 8 Pdf Plot.png\|center\|~~550px~~\|]]		[[Image:Exponential Distribution Example 8 Pdf Plot.png\|center\|650px\|]]

	6. ~~The exponential failure rate~~ plot ~~is shown next~~.		6. View the Failure Rate vs. Time plot.


	[[Image:Exponential Distribution Example 8 Failure Rate Plot.png\|center\|~~550px~~\|]]		[[Image:Exponential Distribution Example 8 Failure Rate Plot.png\|center\|650px\|]]


	Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the ~~failure rate~~ plot does ~~not exist~~ for times before the location parameter, <math>\gamma </math>, at 100 hours.		Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, <math>\gamma </math>, at 100 hours.

	7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. ~~That~~ is because the median rank values are determined from the total number of failures observed by time <math>{{T}_{i}}</math> where <math>i</math> indicates the group number. In this example the total number of groups is <math>N=6</math> and the total number of units is <math>{{N}_{T}}=20</math>. Thus, the median rank values will be estimated for ~~twenty~~ units and for the total failed units (<math>{{N}_{{{F}_{i}}}}</math>) up to the <math>{{i}^{th}}</math> group, for the <math>{{i}^{th}}</math> rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference.		7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time <math>{{T}_{i}}</math> where <math>i</math> indicates the group number. In this example, the total number of groups is <math>N=6</math> and the total number of units is <math>{{N}_{T}}=20</math>. Thus, the median rank values will be estimated for 20 units and for the total failed units (<math>{{N}_{{{F}_{i}}}}</math>) up to the <math>{{i}^{th}}</math> group, for the <math>{{i}^{th}}</math> rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

	For example, the median rank value of the fourth group will be the <math>{{17}^{th}}</math> rank out of a sample size of twenty units (or 81.945%).		For example, the median rank value of the fourth group will be the <math>{{17}^{th}}</math> rank out of a sample size of twenty units (or 81.945%).
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	::<math>f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}</math>		::<math>f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}</math>

	Using Weibull++ , the estimated parameters are:		Using Weibull++, the estimated parameters are:

	::<math>\begin{align}		::<math>\begin{align}
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	\end{align}</math>		\end{align}</math>

	The small difference in the values from Weibull++ is due to rounding. In ~~Weibull++~~ the calculations and the rank values are carried out up to the <math>{{15}^{th}}</math> decimal point.		The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the <math>{{15}^{th}}</math> decimal point.
	<br>		<br>

Chris Kahn at 07:45, 13 August 2012

2012-08-13T07:45:00Z

← Older revision		Revision as of 07:45, 13 August 2012
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	4. Show the Reliability vs. Time plot for the results.		4. Show the Reliability vs. Time plot for the results.

	5. Show the '''pdf''' plot for the results.		5. Show the ''pdf'' plot for the results.

	6. Show the Failure Rate vs. Time plot for the results.		6. Show the Failure Rate vs. Time plot for the results.