|
|
| Line 1: |
Line 1: |
| ==Confidence Bounds==
| | #REDIRECT [[Logistic#Confidence_Bounds]] |
| Least squares is used to estimate the parameters of the following Logistic model.
| |
| | |
| ::<math>\ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}}</math>
| |
| | |
| Thus the confidence bounds on the parameters are given by:
| |
| | |
| ::<math>b=\hat{b}{{e}^{{{t}_{n-2,\alpha /2}}SE(\ln \hat{b})}}</math>
| |
| | |
| :where:
| |
| | |
| ::<math>SE(\ln \hat{b})=\sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{({{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}},\ \ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}</math>
| |
| | |
| ::<math>\sigma =\sqrt{SSE/(n-2)}</math>
| |
| | |
| :and:
| |
| | |
| ::<math>k=\hat{k}\pm {{t}_{n-2,\alpha /2}}SE(\hat{k})</math>
| |
| | |
| :where:
| |
| | |
| ::<math>SE(\hat{k})=\frac{\sigma }{\sqrt{{{S}_{xx}}}},\ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}</math>
| |
| | |
| Since the reliability is always between 0 and 1, the logit transformation is used to obtain the confidence bounds on reliability.
| |
| | |
| ::<math>CB=\frac{{{{\hat{R}}}_{i}}}{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}</math>
| |
| | |
| '''Example 4'''
| |
| <br>
| |
| For the data given for Example 1 in Table 8.1, calculate the 2-sided 90% confidence bounds under the Logistic model for the following:
| |
| <br>
| |
| <br>
| |
| :1) The parameters <math>b</math> and <math>k</math> .
| |
| <br>
| |
| :2) Reliability at month 5.
| |
| | |
| <br>
| |
| '''Solution'''
| |
| <br>
| |
| :1) The values of <math>\widehat{b}</math> and <math>\widehat{k}</math> estimated from the least squares analysis in Example 1:
| |
| | |
| ::<math>\begin{align}
| |
| & \widehat{b}= & 3.3991 \\
| |
| & \widehat{\alpha }= & 0.7398
| |
| \end{align}</math>
| |
| | |
| Thus the 2-sided 90% confidence bounds on parameter <math>b</math> using Eqn. (LogCBb) are:
| |
| | |
| ::<math>\begin{align}
| |
| & {{b}_{lower}}= & 2.5547 \\
| |
| & {{b}_{upper}}= & 4.5225
| |
| \end{align}</math>
| |
| | |
| The 2-sided 90% confidence bounds on parameter <math>k</math> using Eqn. (logCBk) are:
| |
| | |
| ::<math>\begin{align}
| |
| & {{k}_{lower}}= & 0.6798 \\
| |
| & {{k}_{upper}}= & 0.7997
| |
| \end{align}</math>
| |
| | |
| | |
| :2) First calculate the reliability estimation at month 5:
| |
|
| |
| ::<math>\begin{align}
| |
| & {{R}_{5}}= & \frac{1}{1+b{{e}^{-5k}}} \\
| |
| & = & 0.9224
| |
| \end{align}</math>
| |
|
| |
|
| |
| Thus the 2-sided 90% confidence bounds on reliability at month 5 using Eqn. (LogCR) are:
| |
| | |
| ::<math>\begin{align}
| |
| & {{[{{R}_{5}}]}_{lower}}= & 0.8493 \\
| |
| & {{[{{R}_{5}}]}_{upper}}= & 0.9955
| |
| \end{align}</math>
| |
| | |
| Figure Logig86 shows a graph of the reliability plotted with 2-sided 90% confidence bounds.
| |
| | |
| [[Image:rga8.6.png|thumb|center|400px|Logistic Reliability vs. Time plot with 2-sided 90% confidence bounds.]] | |