Template:Confidence bounds logistic rga: Difference between revisions

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(Created page with '==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> Th…')
 
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==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.]]

Latest revision as of 03:04, 24 August 2012

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