Template:Confidence bounds logistic rga - Revision history

Richard House: Redirected page to Logistic#Confidence Bounds

2012-08-24T03:04:01Z

Redirected page to Logistic#Confidence Bounds

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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.~~]]

Nicolette Young: Created page with '==Confidence Bounds== Least squares is used to estimate the parameters of the following Logistic model. :: $\ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}}$ Th…'

2012-01-10T20:39:52Z

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…'

New page

==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'''
 
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:
 
 
:1) The parameters <math>b</math> and <math>k</math> .
 
:2) Reliability at month 5.

 
'''Solution'''
 
: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.]]

Template:Confidence bounds logistic rga - Revision history

Richard House: Redirected page to Logistic#Confidence Bounds

Nicolette Young: Created page with '==Confidence Bounds== Least squares is used to estimate the parameters of the following Logistic model. ::\ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}} Th…'

Nicolette Young: Created page with '==Confidence Bounds== Least squares is used to estimate the parameters of the following Logistic model. :: $\ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}}$ Th…'