Template:Example: Lognormal Distribution RRY

Lognormal Distribution RRY Example

Fourteen units were reliability tested and the following life test data were obtained:

Assuming the data follow a lognormal distribution, estimate the parameters and the correlation coefficient, $$\rho $$, using rank regression on Y.

Solution

Construct Table 9.2, as shown next.

$$\overset – {\mathop{\text{Table 9}\text{.2 - Least Squares Analysis}}}\,$$

$$\begin{matrix} N & t_{i} & F(t_{i}) & {t_{i}}'& y_{i} & {{t_{i}}'}^{2} & y_{i}^{2} & t_{i} y_{i} \\ \text{1} & \text{5} & \text{0}\text{.0483} & \text{1}\text{.6094}& \text{-1}\text{.6619} & \text{2}\text{.5903} & \text{2}\text{.7619} & \text{-2}\text{.6747} \\ \text{2} & \text{10} & \text{0}\text{.1170} & \text{2.3026}& \text{-1.1901} & \text{5.3019} & \text{1.4163} & \text{-2.7403} \\ \text{3} & \text{15} & \text{0}\text{.1865} & \text{2.7080}&\text{-0.8908} & \text{7.3335} & \text{0.7935} & \text{-2.4123} \\ \text{4} & \text{20} & \text{0}\text{.2561} & \text{2.9957} &\text{-0.6552} & \text{8.9744} & \text{0.4292} & \text{-1.9627} \\ \text{5} & \text{25} & \text{0}\text{.3258} & \text{3.2189}& \text{-0.4512} & \text{10.3612} & \text{0.2036} & \text{-1.4524} \\ \text{6} & \text{30} & \text{0}\text{.3954} & \text{3.4012}& \text{-0.2647} & \text{11.5681} & \text{0.0701} & \text{-0.9004} \\ \text{7} & \text{35} & \text{0}\text{.4651} & \text{3.5553} & \text{-0.0873} & \text{12.6405} & \text{-0.0076}& \text{-0.3102} \\ \text{8} & \text{40} & \text{0}\text{.5349} & \text{3.6889}& \text{0.0873} & \text{13.6078} & \text{0.0076} & \text{0.3219} \\ \text{9} & \text{50} & \text{0}\text{.6046} & \text{3.912} & \text{0.2647} & \text{15.3039} & \text{0.0701} &\text{1.0357} \\ \text{10} & \text{60} & \text{0}\text{.6742} & \text{4.0943} & \text{0.4512} & \text{16.7637} & \text{0.2036}&\text{1.8474} \\ \text{11} & \text{70} & \text{0}\text{.7439} & \text{4.2485} & \text{0.6552} & \text{18.0497}& \text{0.4292} & \text{2.7834} \\ \text{12} & \text{80} & \text{0}\text{.8135} & \text{4.382} & \text{0.8908} & \text{19.2022} & \text{0.7935} & \text{3.9035} \\ \text{13} & \text{90} & \text{0}\text{.8830} & \text{4.4998} & \text{1.1901} & \text{20.2483}&\text{1.4163} & \text{5.3552} \\ \text{14} & \text{100}& \text{1.9517} & \text{4.6052} & \text{1.6619} & \text{21.2076} &\text{2.7619} & \text{7.6533} \\ \sum_{}^{} & \text{ } & \text{ } & \text{49.222} & \text{0} & \text{183.1531} & \text{11.3646} & \text{10.4473} \\

\end{matrix}$$

The median rank values ( $$F({{t}_{i}})$$ ) can be found in rank tables or by using the Quick Statistical Reference in Weibull++.

The $${{y}_{i}}$$  values were obtained from the standardized normal distribution's area tables by entering for  $$F(z)$$  and getting the corresponding  $$z$$  value ( $${{y}_{i}}$$ ).

Given the values in the table above, calculate $$\widehat{a}$$  and  $$\widehat{b}$$:


 * $$\begin{align}

& \widehat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime }{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime })(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime 2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime })}^{2}}/14} \\ & &  \\  & \widehat{b}= & \frac{10.4473-(49.2220)(0)/14}{183.1530-{{(49.2220)}^{2}}/14} \end{align}$$

or:


 * $$\widehat{b}=1.0349$$

and:


 * $$\widehat{a}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\widehat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,t_{i}^{\prime }}{N}$$

or:


 * $$\widehat{a}=\frac{0}{14}-(1.0349)\frac{49.2220}{14}=-3.6386$$


 * Therefore:


 * $${\sigma'}=\frac{1}{\widehat{b}}=\frac{1}{1.0349}=0.9663$$

and:


 * $${\mu }'=-\widehat{a}\cdot {\sigma'}=-(-3.6386)\cdot 0.9663$$

or:


 * $${\mu }'=3.516$$

The mean and the standard deviation of the lognormal distribution are obtained using equations in section Lognormal Statistical Properties:


 * $$\overline{T}=\mu ={{e}^{3.516+\tfrac{1}{2}{{0.9663}^{2}}}}=53.6707\text{ hours}$$

and:


 * $${\sigma}=\sqrt{({{e}^{2\cdot 3.516+{{0.9663}^{2}}}})({{e}^}-1)}=66.69\text{ hours}$$

The correlation coefficient can be estimated as:


 * $$\widehat{\rho }=0.9754$$

The above example can be repeated using Weibull++, using RRY.



The mean can be obtained from the QCP and both the mean and the standard deviation can be obtained from the Function Wizard.