Lognormal Distribution Examples: Difference between revisions

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<noinclude>{{Banner Weibull Examples}}{{Navigation box}}
<noinclude>{{Banner Weibull Examples}}{{Navigation box}}
''These examples also appear in the [[The_Weibull_Distribution#Weibull Distribution Examples|Life Data Analysis Reference book]].''
''These examples also appear in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference].''
</noinclude>
</noinclude>
===Complete Data Example===
===Complete Data Example===
Line 37: Line 37:
   & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\  
   & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\  
  & {\hat{\sigma '}}= & 1.10   
  & {\hat{\sigma '}}= & 1.10   
\end{align}</math>
\end{align}\,\!</math>


For rank regression on <math>X</math>
For rank regression on <math>X\,\!</math>  


::<math>\begin{align}
::<math>\begin{align}
   & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\  
   & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\  
  & {{{\hat{\sigma' }}}}= & 1.24   
  & {{{\hat{\sigma' }}}}= & 1.24   
\end{align}</math>
\end{align}\,\!</math>


For rank regression on <math>Y:</math>
For rank regression on <math>Y:\,\!</math>  


::<math>\begin{align}
::<math>\begin{align}
   & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\  
   & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\  
  & {{{\hat{\sigma' }}}}= & 1.36   
  & {{{\hat{\sigma' }}}}= & 1.36   
\end{align}</math>
\end{align}\,\!</math>


===Complete Data RRX Example===
===Complete Data RRX Example===


From [[Appendix: Weibull References|Kececioglu [20, p. 347]]]. Fifteen identical units were tested to failure and following is a table of their times-to-failure:
From Kececioglu [[Appendix:_Life_Data_Analysis_References|[20, p. 347]]]. 15 identical units were tested to failure and following is a table of their failure times:


<center>Times-to-Failure Data</center>
<center>'''Times-to-Failure Data'''</center>


<center><math>\begin{matrix}
<center><math>\begin{matrix}
   \text{Data Point Index} & \text{Time-to-Failure, hr}  \\
   \text{Data Point Index} & \text{Failure Times (Hr)}  \\
   \text{1} & \text{62}\text{.5}  \\
   \text{1} & \text{62}\text{.5}  \\
   \text{2} & \text{91}\text{.9}  \\
   \text{2} & \text{91}\text{.9}  \\
Line 76: Line 76:
   \text{14} & \text{410}\text{.6}  \\
   \text{14} & \text{410}\text{.6}  \\
   \text{15} & \text{550}\text{.5}  \\
   \text{15} & \text{550}\text{.5}  \\
\end{matrix}</math></center>
\end{matrix}\,\!</math></center>


'''Solution'''
'''Solution'''
Line 85: Line 85:
   {{\widehat{\mu }}^{\prime }}=5.22575  \\
   {{\widehat{\mu }}^{\prime }}=5.22575  \\
   {{\widehat{\sigma' }}}=0.62048.  \\
   {{\widehat{\sigma' }}}=0.62048.  \\
\end{matrix}</math>
\end{matrix}\,\!</math>
 


Weibull++ computed parameters for rank regression on X are:
Weibull++ computed parameters for rank regression on X are:
Line 92: Line 93:
   {{\widehat{\mu }}^{\prime }}=5.2303  \\
   {{\widehat{\mu }}^{\prime }}=5.2303  \\
   {{\widehat{\sigma'}}}=0.6283.  \\
   {{\widehat{\sigma'}}}=0.6283.  \\
\end{matrix}</math>
\end{matrix}\,\!</math>
 


The small differences are due to the precision errors when fitting a line manually, whereas in Weibull++ the line was fitted mathematically.
The small differences are due to the precision errors when fitting a line manually, whereas in Weibull++ the line was fitted mathematically.


===Complete Data Unbiased MLE Example===
===Complete Data Unbiased MLE Example===


From [[Appendix: Weibull References|Kececioglu [19, p. 406]]]. Nine identical units are tested continuously to failure and their times-to-failure were recorded at 30.4, 36.7, 53.3, 58.5, 74.0, 99.3, 114.3, 140.1, and 257.9 hours.
From Kececioglu [[Appendix:_Life_Data_Analysis_References|[19, p. 406]]]. 9 identical units are tested continuously to failure and failure times were recorded at 30.4, 36.7, 53.3, 58.5, 74.0, 99.3, 114.3, 140.1 and 257.9 hours.


'''Solution'''
'''Solution'''
Line 109: Line 110:
   {{\widehat{\mu }}^{\prime }}=4.3553  \\
   {{\widehat{\mu }}^{\prime }}=4.3553  \\
   {{\widehat{\sigma' }}}=0.67677  \\
   {{\widehat{\sigma' }}}=0.67677  \\
\end{matrix}</math>
\end{matrix}\,\!</math>
 


This same data set can be entered into Weibull++ by creating a data sheet capable of handling non-grouped time-to-failure data. Since the results shown above are unbiased, the Use Unbiased Std on Normal Data option in the User Setup must be selected in order to duplicate these results.
This same data set can be entered into Weibull++ by creating a data sheet capable of handling non-grouped time-to-failure data. Since the results shown above are unbiased, the Use Unbiased Std on Normal Data option in the User Setup must be selected in order to duplicate these results.
Line 117: Line 119:
   {{\widehat{\mu }}^{\prime }}=4.3553  \\
   {{\widehat{\mu }}^{\prime }}=4.3553  \\
   {{\widehat{\sigma' }}}=0.6768  \\
   {{\widehat{\sigma' }}}=0.6768  \\
\end{matrix}</math>
\end{matrix}\,\!</math>
 


===Suspension Data Example===
===Suspension Data Example===


From [[Appendix: Weibull References|Nelson [30, p. 324]]]. Ninety-six locomotive controls were tested, 37 failed and 59 were suspended after running for 135,000 miles. Table 9.6 below shows their times-to-failure.
From Nelson [[Appendix:_Life_Data_Analysis_References|[30, p. 324]]]. 96 locomotive controls were tested, 37 failed and 59 were suspended after running for 135,000 miles. The table below shows the failure and suspension times.
 
'''Solution'''
 
The distribution used in the publication was the base-10 lognormal.
Published results (using MLE):
 
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=2.2223  \\
  {{\widehat{\sigma' }}}=0.3064  \\
\end{matrix}</math>
 
Published 95% confidence limits on the parameters:
 
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=\left\{ 2.1336,2.3109 \right\}  \\
  {{\widehat{\sigma'}}}=\left\{ 0.2365,0.3970 \right\}  \\
\end{matrix}</math>
 
Published variance/covariance matrix:
 
::<math>\left[ \begin{matrix}
  \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0020 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001  \\
  {} & {} & {}  \\
  \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001 & {} & \widehat{Var}\left( {{{\hat{\sigma '}}}} \right)=0.0016  \\
\end{matrix} \right]</math>
 
To replicate the published results (since Weibull++ uses a lognormal to the base  <math>e</math> ), take the base-10 logarithm of the data and estimate the parameters using the Normal distribution and MLE.
 
*Weibull++ computed parameters for maximum likelihood are:
 
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=2.2223  \\
  {{\widehat{\sigma' }}}=0.3064  \\
\end{matrix}</math>
 
*Weibull++ computed 95% confidence limits on the parameters:
 
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=\left\{ 2.1364,2.3081 \right\}  \\
  {{\widehat{\sigma'}}}=\left\{ 0.2395,0.3920 \right\}  \\
\end{matrix}</math>
 
*Weibull++ computed/variance covariance matrix:
 
::<math>\left[ \begin{matrix}
  \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0019 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.0009  \\
  {} & {} & {}  \\
  \widehat{Cov}({\mu }',{{{\hat{\sigma' }}}})=0.0009 & {} & \widehat{Var}\left( {{{\hat{\sigma' }}}} \right)=0.0015  \\
\end{matrix} \right]</math>


{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|-
|colspan="4" style="text-align:center"|Table - Nelson's Locomotive Data
|colspan="4" style="text-align:center"|'''Nelson's Locomotive Data'''
|-
|-
!
!
Line 259: Line 211:
|}
|}


'''Solution'''
The distribution used in the publication was the base-10 lognormal.
Published results (using MLE):
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=2.2223  \\
  {{\widehat{\sigma' }}}=0.3064  \\
\end{matrix}\,\!</math>
Published 95% confidence limits on the parameters:
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=\left\{ 2.1336,2.3109 \right\}  \\
  {{\widehat{\sigma'}}}=\left\{ 0.2365,0.3970 \right\}  \\
\end{matrix}\,\!</math>
Published variance/covariance matrix:
::<math>\left[ \begin{matrix}
  \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0020 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001  \\
  {} & {} & {}  \\
  \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001 & {} & \widehat{Var}\left( {{{\hat{\sigma '}}}} \right)=0.0016  \\
\end{matrix} \right]\,\!</math>
To replicate the published results (since Weibull++ uses a lognormal to the base <math>e\,\!</math> ), take the base-10 logarithm of the data and estimate the parameters using the normal distribution and MLE.
*Weibull++ computed parameters for maximum likelihood are:
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=2.2223  \\
  {{\widehat{\sigma' }}}=0.3064  \\
\end{matrix}\,\!</math>
*Weibull++ computed 95% confidence limits on the parameters:
::<math>\begin{matrix}
  {{\widehat{\mu }}^{\prime }}=\left\{ 2.1364,2.3081 \right\}  \\
  {{\widehat{\sigma'}}}=\left\{ 0.2395,0.3920 \right\}  \\
\end{matrix}\,\!</math>
*Weibull++ computed/variance covariance matrix:
::<math>\left[ \begin{matrix}
  \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0019 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.0009  \\
  {} & {} & {}  \\
  \widehat{Cov}({\mu }',{{{\hat{\sigma' }}}})=0.0009 & {} & \widehat{Var}\left( {{{\hat{\sigma' }}}} \right)=0.0015  \\
\end{matrix} \right]\,\!</math>


===Interval Data Example===
===Interval Data Example===
Line 265: Line 270:
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"  
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"  
|-
|-
|colspan="3" style="text-align:center"| Non-Grouped Data Times-to-Failure with intervals (lnterval and left censored)
|colspan="3" style="text-align:center"| '''Non-Grouped Data Times-to-Failure with Intervals'''
|-  
|-  
!Data point index
!Data point index
Line 295: Line 300:
   & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\  
   & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\  
  & {{{\hat{\sigma' }}}}= & 0.18   
  & {{{\hat{\sigma' }}}}= & 0.18   
\end{align}</math>
\end{align}\,\!</math>


For rank regression on <math>X\ </math>:   
 
For rank regression on <math>X\ \,\!</math>:   


::<math>\begin{align}
::<math>\begin{align}
   & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\  
   & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\  
  & {{{\hat{\sigma' }}}}= & 0.17   
  & {{{\hat{\sigma' }}}}= & 0.17   
\end{align}</math>
\end{align}\,\!</math>
 


For rank regression on <math>Y\ </math>:   
For rank regression on <math>Y\ \,\!</math>:   


::<math>\begin{align}
::<math>\begin{align}
   & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\  
   & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\  
  & {{{\hat{\sigma' }}}}= & 0.21   
  & {{{\hat{\sigma' }}}}= & 0.21   
\end{align}</math>
\end{align}\,\!</math>

Latest revision as of 21:46, 18 September 2023

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These examples also appear in the Life data analysis reference.

Complete Data Example

Determine the lognormal parameter estimates for the data given in the following table.

Non-Grouped Times-to-Failure Data
Data point index State F or S State End Time
1 F 2
2 F 5
3 F 11
4 F 23
5 F 29
6 F 37
7 F 43
8 F 59

Solution

Using Weibull++, the computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & {\hat{\sigma '}}= & 1.10 \end{align}\,\! }[/math]

For rank regression on [math]\displaystyle{ X\,\! }[/math]

[math]\displaystyle{ \begin{align} & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & {{{\hat{\sigma' }}}}= & 1.24 \end{align}\,\! }[/math]

For rank regression on [math]\displaystyle{ Y:\,\! }[/math]

[math]\displaystyle{ \begin{align} & {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & {{{\hat{\sigma' }}}}= & 1.36 \end{align}\,\! }[/math]

Complete Data RRX Example

From Kececioglu [20, p. 347]. 15 identical units were tested to failure and following is a table of their failure times:

Times-to-Failure Data
[math]\displaystyle{ \begin{matrix} \text{Data Point Index} & \text{Failure Times (Hr)} \\ \text{1} & \text{62}\text{.5} \\ \text{2} & \text{91}\text{.9} \\ \text{3} & \text{100}\text{.3} \\ \text{4} & \text{117}\text{.4} \\ \text{5} & \text{141}\text{.1} \\ \text{6} & \text{146}\text{.8} \\ \text{7} & \text{172}\text{.7} \\ \text{8} & \text{192}\text{.5} \\ \text{9} & \text{201}\text{.6} \\ \text{10} & \text{235}\text{.8} \\ \text{11} & \text{249}\text{.2} \\ \text{12} & \text{297}\text{.5} \\ \text{13} & \text{318}\text{.3} \\ \text{14} & \text{410}\text{.6} \\ \text{15} & \text{550}\text{.5} \\ \end{matrix}\,\! }[/math]

Solution

Published results (using probability plotting):

[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=5.22575 \\ {{\widehat{\sigma' }}}=0.62048. \\ \end{matrix}\,\! }[/math]


Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=5.2303 \\ {{\widehat{\sigma'}}}=0.6283. \\ \end{matrix}\,\! }[/math]


The small differences are due to the precision errors when fitting a line manually, whereas in Weibull++ the line was fitted mathematically.

Complete Data Unbiased MLE Example

From Kececioglu [19, p. 406]. 9 identical units are tested continuously to failure and failure times were recorded at 30.4, 36.7, 53.3, 58.5, 74.0, 99.3, 114.3, 140.1 and 257.9 hours.

Solution

The results published were obtained by using the unbiased model. Published Results (using MLE):

[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=4.3553 \\ {{\widehat{\sigma' }}}=0.67677 \\ \end{matrix}\,\! }[/math]


This same data set can be entered into Weibull++ by creating a data sheet capable of handling non-grouped time-to-failure data. Since the results shown above are unbiased, the Use Unbiased Std on Normal Data option in the User Setup must be selected in order to duplicate these results. Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=4.3553 \\ {{\widehat{\sigma' }}}=0.6768 \\ \end{matrix}\,\! }[/math]

Suspension Data Example

From Nelson [30, p. 324]. 96 locomotive controls were tested, 37 failed and 59 were suspended after running for 135,000 miles. The table below shows the failure and suspension times.

Nelson's Locomotive Data
Number in State F or S Time
1 1 F 22.5
2 1 F 37.5
3 1 F 46
4 1 F 48.5
5 1 F 51.5
6 1 F 53
7 1 F 54.5
8 1 F 57.5
9 1 F 66.5
10 1 F 68
11 1 F 69.5
12 1 F 76.5
13 1 F 77
14 1 F 78.5
15 1 F 80
16 1 F 81.5
17 1 F 82
18 1 F 83
19 1 F 84
20 1 F 91.5
21 1 F 93.5
22 1 F 102.5
23 1 F 107
24 1 F 108.5
25 1 F 112.5
26 1 F 113.5
27 1 F 116
28 1 F 117
29 1 F 118.5
30 1 F 119
31 1 F 120
32 1 F 122.5
33 1 F 123
34 1 F 127.5
35 1 F 131
36 1 F 132.5
37 1 F 134
38 59 S 135

Solution

The distribution used in the publication was the base-10 lognormal. Published results (using MLE):

[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=2.2223 \\ {{\widehat{\sigma' }}}=0.3064 \\ \end{matrix}\,\! }[/math]


Published 95% confidence limits on the parameters:

[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=\left\{ 2.1336,2.3109 \right\} \\ {{\widehat{\sigma'}}}=\left\{ 0.2365,0.3970 \right\} \\ \end{matrix}\,\! }[/math]


Published variance/covariance matrix:

[math]\displaystyle{ \left[ \begin{matrix} \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0020 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001 \\ {} & {} & {} \\ \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001 & {} & \widehat{Var}\left( {{{\hat{\sigma '}}}} \right)=0.0016 \\ \end{matrix} \right]\,\! }[/math]


To replicate the published results (since Weibull++ uses a lognormal to the base [math]\displaystyle{ e\,\! }[/math] ), take the base-10 logarithm of the data and estimate the parameters using the normal distribution and MLE.

  • Weibull++ computed parameters for maximum likelihood are:
[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=2.2223 \\ {{\widehat{\sigma' }}}=0.3064 \\ \end{matrix}\,\! }[/math]


  • Weibull++ computed 95% confidence limits on the parameters:
[math]\displaystyle{ \begin{matrix} {{\widehat{\mu }}^{\prime }}=\left\{ 2.1364,2.3081 \right\} \\ {{\widehat{\sigma'}}}=\left\{ 0.2395,0.3920 \right\} \\ \end{matrix}\,\! }[/math]


  • Weibull++ computed/variance covariance matrix:
[math]\displaystyle{ \left[ \begin{matrix} \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0019 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.0009 \\ {} & {} & {} \\ \widehat{Cov}({\mu }',{{{\hat{\sigma' }}}})=0.0009 & {} & \widehat{Var}\left( {{{\hat{\sigma' }}}} \right)=0.0015 \\ \end{matrix} \right]\,\! }[/math]

Interval Data Example

Determine the lognormal parameter estimates for the data given in the table below.

Non-Grouped Data Times-to-Failure with Intervals
Data point index Last Inspected State End Time
1 30 32
2 32 35
3 35 37
4 37 40
5 42 42
6 45 45
7 50 50
8 55 55

Solution

This is a sequence of interval times-to-failure where the intervals vary substantially in length. Using Weibull++, the computed parameters for maximum likelihood are calculated to be:

[math]\displaystyle{ \begin{align} & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & {{{\hat{\sigma' }}}}= & 0.18 \end{align}\,\! }[/math]


For rank regression on [math]\displaystyle{ X\ \,\! }[/math]:

[math]\displaystyle{ \begin{align} & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & {{{\hat{\sigma' }}}}= & 0.17 \end{align}\,\! }[/math]


For rank regression on [math]\displaystyle{ Y\ \,\! }[/math]:

[math]\displaystyle{ \begin{align} & {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & {{{\hat{\sigma' }}}}= & 0.21 \end{align}\,\! }[/math]