Template:Confidence bounds logistic rga

Confidence Bounds
Least squares is used to estimate the parameters of the following Logistic model.


 * $$\ln (\frac{1}-1)=\ln (b)-k{{T}_{i}}$$

Thus the confidence bounds on the parameters are given by:


 * $$b=\hat{b}{{e}^{{{t}_{n-2,\alpha /2}}SE(\ln \hat{b})}}$$


 * where:


 * $$SE(\ln \hat{b})=\sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{({{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}},\ \ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}$$


 * $$\sigma =\sqrt{SSE/(n-2)}$$


 * and:


 * $$k=\hat{k}\pm {{t}_{n-2,\alpha /2}}SE(\hat{k})$$


 * where:


 * $$SE(\hat{k})=\frac{\sigma }{\sqrt},\ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}$$

Since the reliability is always between 0 and 1, the logit transformation is used to obtain the confidence bounds on reliability.


 * $$CB=\frac{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}$$

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 $$b$$  and  $$k$$.


 * 2)	Reliability at month 5.

Solution
 * 1)	The values of $$\widehat{b}$$  and  $$\widehat{k}$$  estimated from the least squares analysis in Example 1:


 * $$\begin{align}

& \widehat{b}= & 3.3991 \\ & \widehat{\alpha }= & 0.7398 \end{align}$$

Thus the 2-sided 90% confidence bounds on parameter $$b$$  using Eqn. (LogCBb) are:


 * $$\begin{align}

& {{b}_{lower}}= & 2.5547 \\ & {{b}_{upper}}= & 4.5225 \end{align}$$

The 2-sided 90% confidence bounds on parameter $$k$$  using Eqn. (logCBk) are:


 * $$\begin{align}

& {{k}_{lower}}= & 0.6798 \\ & {{k}_{upper}}= & 0.7997 \end{align}$$


 * 2)	First calculate the reliability estimation at month 5:


 * $$\begin{align}

& {{R}_{5}}= & \frac{1}{1+b{{e}^{-5k}}} \\ & = & 0.9224 	\end{align}$$ Thus the 2-sided 90% confidence bounds on reliability at month 5 using Eqn. (LogCR) are:


 * $$\begin{align}

& {{[{{R}_{5}}]}_{lower}}= & 0.8493 \\ & {{[{{R}_{5}}]}_{upper}}= & 0.9955 \end{align}$$

Figure Logig86 shows a graph of the reliability plotted with 2-sided 90% confidence bounds.