Template:Example: Normal Distribution RRY

Normal Distribution RRY Example

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

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

Solution

Construct a table like the one shown next.

$$\overset – {\mathop{\text{Table}\text{ - Least Squares Analysis}}}\,$$ $$\begin{matrix} \text{N} & \text{T}_{i} & \text{F(T}_{i}\text{)} & \text{y}_{i} & \text{T}_{i}^{2} & \text{y}_{i}^{2} & \text{T}_{i}\text{ y}_{i} \\ \text{1} & \text{5} & \text{0}\text{.0483} & \text{-1}\text{.6619} & \text{25} & \text{2}\text{.7619} & \text{-8}\text{.3095} \\ \text{2} & \text{10} & \text{0}\text{.1170} & \text{-1}\text{.1901} & \text{100} & \text{1}\text{.4163} & \text{-11}\text{.9010} \\ \text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.8908} & \text{225} & \text{0}\text{.7935} & \text{-13}\text{.3620} \\ \text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.6552} & \text{400} & \text{0}\text{.4292} & \text{-13}\text{.1030} \\ \text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.4512} & \text{625} & \text{0}\text{.2036} & \text{-11}\text{.2800} \\ \text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.2647} & \text{900} & \text{0}\text{.0701} & \text{-7}\text{.9422} \\ \text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.0873} & \text{1225} & \text{0}\text{.0076} & \text{-3}\text{.0542} \\ \text{8} & \text{40} & \text{0}\text{.5349} & \text{0}\text{.0873} & \text{1600} & \text{0}\text{.0076} & \text{3}\text{.4905} \\ \text{9} & \text{50} & \text{0}\text{.6046} & \text{0}\text{.2647} & \text{2500} & \text{0}\text{.0701} & \text{13}\text{.2370} \\ \text{10} & \text{60} & \text{0}\text{.6742} & \text{0}\text{.4512} & \text{3600} & \text{0}\text{.2036} & \text{27}\text{.0720} \\ \text{11} & \text{70} & \text{0}\text{.7439} & \text{0}\text{.6552} & \text{4900} & \text{0}\text{.4292} & \text{45}\text{.8605} \\ \text{12} & \text{80} & \text{0}\text{.8135} & \text{0}\text{.8908} & \text{6400} & \text{0}\text{.7935} & \text{71}\text{.2640} \\ \text{13} & \text{90} & \text{0}\text{.8830} & \text{1}\text{.1901} & \text{8100} & \text{1}\text{.4163} & \text{107}\text{.1090} \\ \text{14} & \text{100} & \text{0}\text{.9517} & \text{1}\text{.6619} & \text{10000} & \text{2}\text{.7619} & \text{166}\text{.1900} \\ \mathop{}_{}^{} & \text{630} & {} & \text{0} & \text{40600} & \text{11}\text{.3646} & \text{365}\text{.2711} \\ \end{matrix}$$


 * The median rank values ( $$F({{t}_{i}})$$ ) can be found in rank tables, available in many statistical texts, or they can be estimated by using the Quick Statistical Reference in Weibull++.
 * The $${{y}_{i}}$$  values were obtained from standardized normal distribution's area tables by entering for  $$F(z)$$  and getting the corresponding  $$z$$  value ( $${{y}_{i}}$$ ).  As with the median rank values, these standard normal values can be obtained with the Quick Statistical Reference.

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


 * $$\begin{align}

& \widehat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}})(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}})}^{2}}/14} \\ & &  \\  & \widehat{b}= & \frac{365.2711-(630)(0)/14}{40,600-{{(630)}^{2}}/14}=0.02982 \end{align}$$

and:


 * $$\widehat{a}=\overline{y}-\widehat{b}\overline{T}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\widehat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}$$

or:


 * $$\widehat{a}=\frac{0}{14}-(0.02982)\frac{630}{14}=-1.3419$$

Therefore:


 * $$\widehat{\sigma}=\frac{1}{\hat{b}}=\frac{1}{0.02982}=33.5367$$

and:


 * $$\widehat{\mu }=-\widehat{a}\cdot \widehat{\sigma }=-(-1.3419)\cdot 33.5367\simeq 45$$

or $$\widehat{\mu }=45$$  hours $$.$$

The correlation coefficient can be estimated using:


 * $$\widehat{\rho }=0.979$$

The preceding example can be repeated using Weibull++.


 * Create a new folio for Times-to-Failure data, and enter the data given in this example.
 * Choose Normal from the Distributions list.
 * Go to the Analysis page and select Rank Regression on Y (RRY).
 * Click the Calculate icon located on the Main page.



The probability plot is shown next.