Template:Normal distribution confidence bounds

Confidence Bounds=

The method used by the application in estimating the different types of confidence bounds for normally distributed data is presented in this section. The complete derivations were presented in detail (for a general function) in Chapter 5.

Exact Confidence Bounds

There are closed-form solutions for exact confidence bounds for both the normal and lognormal distributions. However these closed-forms solutions only apply to complete data. To achieve consistent application across all possible data types, Weibull++ always uses the Fisher matrix method or likelihood ratio method in computing confidence intervals.

Fisher Matrix Confidence Bounds

Bounds on the Parameters

The lower and upper bounds on the mean, [math]\displaystyle{ \widehat{\mu } }[/math] , are estimated from:

[math]\displaystyle{ \begin{align} & {{\mu }_{U}}= & \widehat{\mu }+{{K}_{\alpha }}\sqrt{Var(\widehat{\mu })}\text{ (upper bound),} \\ & {{\mu }_{L}}= & \widehat{\mu }-{{K}_{\alpha }}\sqrt{Var(\widehat{\mu })}\text{ (lower bound)}\text{.} \end{align} }[/math]

Since the standard deviation, [math]\displaystyle{ {{\widehat{\sigma }}_{T}} }[/math] , must be positive, [math]\displaystyle{ \ln ({{\widehat{\sigma }}_{T}}) }[/math] is treated as normally distributed, and the bounds are estimated from:

[math]\displaystyle{ \begin{align} & {{\sigma }_{U}}= & {{\widehat{\sigma }}_{T}}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{T}})}}{{{\widehat{\sigma }}_{T}}}}}\text{ (upper bound),} \\ & {{\sigma }_{L}}= & \frac{{{\widehat{\sigma }}_{T}}}{{{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{T}})}}{{{\widehat{\sigma }}_{T}}}}}}\text{ (lower bound),} \end{align} }[/math]

where [math]\displaystyle{ {{K}_{\alpha }} }[/math] is defined by:

[math]\displaystyle{ \alpha =\frac{1}{\sqrt{2\pi }}\int_{{{K}_{\alpha }}}^{\infty }{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt=1-\Phi ({{K}_{\alpha }}) }[/math]

If [math]\displaystyle{ \delta }[/math] is the confidence level, then [math]\displaystyle{ \alpha =\tfrac{1-\delta }{2} }[/math] for the two-sided bounds and [math]\displaystyle{ \alpha =1-\delta }[/math] for the one-sided bounds. The variances and covariances of [math]\displaystyle{ \widehat{\mu } }[/math] and [math]\displaystyle{ {{\widehat{\sigma }}_{T}} }[/math] are estimated from the Fisher matrix, as follows:

[math]\displaystyle{ \left( \begin{matrix} \widehat{Var}\left( \widehat{\mu } \right) & \widehat{Cov}\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \\ \widehat{Cov}\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) & \widehat{Var}\left( {{\widehat{\sigma }}_{T}} \right) \\ \end{matrix} \right)=\left( \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\mu }^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \mu \partial {{\sigma }_{T}}} \\ {} & {} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial \mu \partial {{\sigma }_{T}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \sigma _{T}^{2}} \\ \end{matrix} \right)_{\mu =\widehat{\mu },\sigma =\widehat{\sigma }}^{-1} }[/math]

[math]\displaystyle{ \Lambda }[/math] is the log-likelihood function of the normal distribution, described in Chapter 3 and Appendix C.

Bounds on Reliability

The reliability of the normal distribution is:

[math]\displaystyle{ \widehat{R}(T;\hat{\mu },{{\hat{\sigma }}_{T}})=\int_{T}^{\infty }\frac{1}{{{\widehat{\sigma }}_{T}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\widehat{\mu }}{{{\widehat{\sigma }}_{T}}} \right)}^{2}}}}dt }[/math]

Let [math]\displaystyle{ \widehat{z}(t;\hat{\mu },{{\hat{\sigma }}_{T}})=\tfrac{t-\widehat{\mu }}{{{\widehat{\sigma }}_{T}}}, }[/math] then [math]\displaystyle{ \tfrac{dz}{dt}=\tfrac{1}{{{\widehat{\sigma }}_{T}}}. }[/math] For [math]\displaystyle{ t=T }[/math] , [math]\displaystyle{ \widehat{z}=\tfrac{T-\widehat{\mu }}{{{\widehat{\sigma }}_{T}}} }[/math] , and for [math]\displaystyle{ t=\infty , }[/math] [math]\displaystyle{ \widehat{z}=\infty . }[/math] The above equation then becomes:

[math]\displaystyle{ \hat{R}(\widehat{z})=\int_{\widehat{z}(T)}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz }[/math]

The bounds on [math]\displaystyle{ z }[/math] are estimated from:

[math]\displaystyle{ \begin{align} & {{z}_{U}}= & \widehat{z}+{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \\ & {{z}_{L}}= & \widehat{z}-{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \end{align} }[/math]

where:

[math]\displaystyle{ Var(\widehat{z})={{\left( \frac{\partial z}{\partial \mu } \right)}^{2}}Var(\widehat{\mu })+{{\left( \frac{\partial z}{\partial {{\sigma }_{T}}} \right)}^{2}}Var({{\widehat{\sigma }}_{T}})+2\left( \frac{\partial z}{\partial \mu } \right)\left( \frac{\partial z}{\partial {{\sigma }_{T}}} \right)Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) }[/math]

or:

[math]\displaystyle{ Var(\widehat{z})=\frac{1}{\widehat{\sigma }_{T}^{2}}\left[ Var(\widehat{\mu })+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{T}})+2\cdot \widehat{z}\cdot Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \right] }[/math]

The upper and lower bounds on reliability are:

[math]\displaystyle{ \begin{align} & {{R}_{U}}= & \int_{{{z}_{L}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (upper bound)} \\ & {{R}_{L}}= & \int_{{{z}_{U}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (lower bound)} \end{align} }[/math]

Bounds on Time

The bounds around time for a given normal percentile (unreliability) are estimated by first solving the reliability equation with respect to time, as follows:

[math]\displaystyle{ \hat{T}(\widehat{\mu },{{\widehat{\sigma }}_{T}})=\widehat{\mu }+z\cdot {{\widehat{\sigma }}_{T}} }[/math]

where:

[math]\displaystyle{ z={{\Phi }^{-1}}\left[ F(T) \right] }[/math]

and:

[math]\displaystyle{ \Phi (z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z(T)}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz }[/math]

The next step is to calculate the variance of [math]\displaystyle{ \hat{T}(\widehat{\mu },{{\widehat{\sigma }}_{T}}) }[/math] or:

[math]\displaystyle{ \begin{align} Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \mu } \right)}^{2}}Var(\widehat{\mu })+{{\left( \frac{\partial T}{\partial {{\sigma }_{T}}} \right)}^{2}}Var({{\widehat{\sigma }}_{T}}) \\ & +2\left( \frac{\partial T}{\partial \mu } \right)\left( \frac{\partial T}{\partial {{\sigma }_{T}}} \right)Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \\ Var(\hat{T})= & Var(\widehat{\mu })+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{T}})+2\cdot z\cdot Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \end{align} }[/math]

The upper and lower bounds are then found by:

[math]\displaystyle{ \begin{align} & {{T}_{U}}= & \hat{T}+{{K}_{\alpha }}\sqrt{Var(\hat{T})}\text{ (upper bound)} \\ & {{T}_{L}}= & \hat{T}-{{K}_{\alpha }}\sqrt{Var(\hat{T})}\text{ (lower bound)} \end{align} }[/math]

Example 4

Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method.

Solution to Example 4

In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, [math]\displaystyle{ \Lambda , }[/math] are given by:

[math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{T}_{i}}-\mu )=0 \\ \frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{T}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0 \end{align} }[/math]

(The derivations of these equations are presented in Appendix C.) Substituting the values of [math]\displaystyle{ {{T}_{i}} }[/math] and solving the above system simultaneously, we get [math]\displaystyle{ \widehat{\sigma }=29.58 }[/math] hours [math]\displaystyle{ , }[/math] [math]\displaystyle{ \widehat{\mu }=45 }[/math] hours [math]\displaystyle{ . }[/math]

The Fisher matrix is:

[math]\displaystyle{ \left[ \begin{matrix} \widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 \\ {} & {} & {} \\ \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500 \\ \end{matrix} \right] }[/math]

  Using Weibull++ , the MLE method can be selected from the Set Analysis page.

The plot of the solution for this example is shown next.

Likelihood Ratio Confidence Bounds

Bounds on Parameters

As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for [math]\displaystyle{ {{\theta }_{1}} }[/math] and [math]\displaystyle{ {{\theta }_{2}} }[/math] that satisfy:

[math]\displaystyle{ -2\cdot \text{ln}\left( \frac{L({{\theta }_{1}},{{\theta }_{2}})}{L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})} \right)=\chi _{\alpha ;1}^{2} }[/math]

This equation can be rewritten as:

[math]\displaystyle{ L({{\theta }_{1}},{{\theta }_{2}})=L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}} }[/math]

For complete data, the likelihood formula for the normal distribution is given by:

[math]\displaystyle{ L(\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-\mu }{\sigma } \right)}^{2}}}} }[/math]

where the [math]\displaystyle{ {{x}_{i}} }[/math] values represent the original time to failure data. For a given value of [math]\displaystyle{ \alpha }[/math] , values for [math]\displaystyle{ \mu }[/math] and [math]\displaystyle{ \sigma }[/math] can be found which represent the maximum and minimum values that satisfy Eqn. (lratio3). These represent the confidence bounds for the parameters at a confidence level [math]\displaystyle{ \delta , }[/math] where [math]\displaystyle{ \alpha =\delta }[/math] for two-sided bounds and [math]\displaystyle{ \alpha =2\delta -1 }[/math] for one-sided.

Example 5

Five units are put on a reliability test and experience failures at 12, 24, 28, 34, and 46 hours. Assuming a normal distribution, the MLE parameter estimates are calculated to be [math]\displaystyle{ \widehat{\mu }=28.8 }[/math] and [math]\displaystyle{ \widehat{\sigma }=11.2143. }[/math] Calculate the two-sided 80% confidence bounds on these parameters using the likelihood ratio method.

Solution to Example 5

The first step is to calculate the likelihood function for the parameter estimates:

[math]\displaystyle{ \begin{align} L(\widehat{\mu },\widehat{\sigma })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\widehat{\mu },\widehat{\sigma })=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{\widehat{\sigma }\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-\widehat{\mu }}{\widehat{\sigma }} \right)}^{2}}}} \\ L(\widehat{\mu },\widehat{\sigma })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{11.2143\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-28.8}{11.2143} \right)}^{2}}}} \\ L(\widehat{\mu },\widehat{\sigma })= & 4.676897\times {{10}^{-9}} \end{align} }[/math]

where [math]\displaystyle{ {{x}_{i}} }[/math] are the original time-to-failure data points. We can now rearrange Eqn. (lratio3) to the form:

[math]\displaystyle{ L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0 }[/math]

Since our specified confidence level, [math]\displaystyle{ \delta }[/math] , is 80%, we can calculate the value of the chi-squared statistic, [math]\displaystyle{ \chi _{0.8;1}^{2}=1.642374. }[/math] We can now substitute this information into the equation:

[math]\displaystyle{ \begin{align} L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\ \\ L(\mu ,\sigma )-4.676897\times {{10}^{-9}}\cdot {{e}^{\tfrac{-1.642374}{2}}}= & 0, \\ \\ L(\mu ,\sigma )-2.057410\times {{10}^{-9}}= & 0. \end{align} }[/math]

It now remains to find the values of [math]\displaystyle{ \mu }[/math] and [math]\displaystyle{ \sigma }[/math] which satisfy this equation. This is an iterative process that requires setting the value of [math]\displaystyle{ \mu }[/math] and finding the appropriate values of [math]\displaystyle{ \sigma }[/math] , and vice versa.

The following table gives the values of [math]\displaystyle{ \sigma }[/math] based on given values of [math]\displaystyle{ \mu }[/math] .

[math]\displaystyle{ }[/math]

[math]\displaystyle{ \begin{matrix} \text{ }\!\!\mu\!\!\text{ } & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{1}}} & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{2}}} & \text{ }\!\!\mu\!\!\text{ } & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{1}}} & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{2}}} \\ \text{22}\text{.0} & \text{12}\text{.045} & \text{14}\text{.354} & \text{29}\text{.0} & \text{7.849}& \text{19.909} \\ \text{22}\text{.5} & \text{11}\text{.004} & \text{15}\text{.310} & \text{29}\text{.5} & \text{7}\text{.876} & \text{17}\text{.889} \\ \text{23}\text{.0} & \text{10}\text{.341} & \text{15}\text{.894} & \text{30}\text{.0} & \text{7}\text{.935} & \text{17}\text{.844} \\ \text{23}\text{.5} & \text{9}\text{.832} & \text{16}\text{.328} & \text{30}\text{.5} & \text{8}\text{.025} & \text{17}\text{.776} \\ \text{24}\text{.0} & \text{9}\text{.418} & \text{16}\text{.673} & \text{31}\text{.0} & \text{8}\text{.147} & \text{17}\text{.683} \\ \text{24}\text{.5} & \text{9}\text{.074} & \text{16}\text{.954} & \text{31}\text{.5} & \text{8}\text{.304} & \text{17}\text{.562} \\ \text{25}\text{.0} & \text{8}\text{.784} & \text{17}\text{.186} & \text{32}\text{.0} & \text{8}\text{.498} & \text{17}\text{.411} \\ \text{25}\text{.5} & \text{8}\text{.542} & \text{17}\text{.377} & \text{32}\text{.5} & \text{8}\text{.732} & \text{17}\text{.227} \\ \text{26}\text{.0} & \text{8}\text{.340} & \text{17}\text{.534} & \text{33}\text{.0} & \text{9}\text{.012} & \text{17}\text{.004} \\ \text{26}\text{.5} & \text{8}\text{.176} & \text{17}\text{.661} & \text{33}\text{.5} & \text{9}\text{.344} & \text{16}\text{.734} \\ \text{27}\text{.0} & \text{8}\text{.047} & \text{17}\text{.760} & \text{34}\text{.0} & \text{9}\text{.742} & \text{16}\text{.403} \\ \text{27}\text{.5} & \text{7}\text{.950} & \text{17}\text{.833} & \text{34}\text{.5} & \text{10}\text{.229} & \text{15}\text{.990} \\ \text{28}\text{.0} & \text{7}\text{.885} & \text{17}\text{.882} & \text{35}\text{.0} & \text{10}\text{.854} & \text{15}\text{.444} \\ \text{28}\text{.5} & \text{7}\text{.852} & \text{17}\text{.907} & \text{35}\text{.5} & \text{11}\text{.772} & \text{14}\text{.609} \\ \end{matrix} }[/math]

This data set is represented graphically in the following contour plot:

(Note that this plot is generated with degrees of freedom [math]\displaystyle{ k=1 }[/math] , as we are only determining bounds on one parameter. The contour plots generated in Weibull++ are done with degrees of freedom [math]\displaystyle{ k=2 }[/math] , for use in comparing both parameters simultaneously.) As can be determined from the table, the lowest calculated value for [math]\displaystyle{ \sigma }[/math] is 7.849, while the highest is 17.909. These represent the two-sided 80% confidence limits on this parameter. Since solutions for the equation do not exist for values of [math]\displaystyle{ \mu }[/math] below 22 or above 35.5, these can be considered the two-sided 80% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on [math]\displaystyle{ \mu }[/math] , we can perform the same procedure as before, but finding the two values of [math]\displaystyle{ \mu }[/math] that correspond with a given value of [math]\displaystyle{ \sigma . }[/math] Using this method, we find that the two-sided 80% confidence limits on [math]\displaystyle{ \mu }[/math] are 21.807 and 35.793, which are close to the initial estimates of 22 and 35.5.

Bounds on Time and Reliability

In order to calculate the bounds on a time estimate for a given reliability, or on a reliability estimate for a given time, the likelihood function needs to be rewritten in terms of one parameter and time/reliability, so that the maximum and minimum values of the time can be observed as the parameter is varied. This can be accomplished by substituting a form of the normal reliability equation into the likelihood function. The normal reliability equation can be written as:

[math]\displaystyle{ R=1-\Phi \left( \frac{t-\mu }{\sigma } \right) }[/math]

This can be rearranged to the form:

[math]\displaystyle{ \mu =t-\sigma \cdot {{\Phi }^{-1}}(1-R) }[/math]

where [math]\displaystyle{ {{\Phi }^{-1}} }[/math] is the inverse standard normal. This equation can now be substituted into Eqn. (normlikelihood), to produce a likelihood equation in terms of [math]\displaystyle{ \sigma , }[/math] [math]\displaystyle{ t }[/math] and [math]\displaystyle{ R\ \ : }[/math]

[math]\displaystyle{ L(\sigma ,t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-\left[ t-\sigma \cdot {{\Phi }^{-1}}(1-R) \right]}{\sigma } \right)}^{2}}}} }[/math]

The unknown parameter [math]\displaystyle{ t/R }[/math] depends on what type of bounds are being determined. If one is trying to determine the bounds on time for a given reliability, then [math]\displaystyle{ R }[/math] is a known constant and [math]\displaystyle{ t }[/math] is the unknown parameter. Conversely, if one is trying to determine the bounds on reliability for a given time, then [math]\displaystyle{ t }[/math] is a known constant and [math]\displaystyle{ R }[/math] is the unknown parameter. Either way, Eqn. (normliketr) can be used to solve Eqn. (lratio3) for the values of interest.

Example 6

For the data given in Example 5, determine the two-sided 80% confidence bounds on the time estimate for a reliability of 40%. The ML estimate for the time at [math]\displaystyle{ R(t)=40% }[/math] is 31.637.

Solution to Example 6

In this example, we are trying to determine the two-sided 80% confidence bounds on the time estimate of 31.637. This is accomplished by substituting [math]\displaystyle{ R=0.40 }[/math] and [math]\displaystyle{ \alpha =0.8 }[/math] into Eqn. (normliketr), and varying [math]\displaystyle{ \sigma }[/math] until the maximum and minimum values of [math]\displaystyle{ t }[/math] are found. The following table gives the values of [math]\displaystyle{ t }[/math] based on given values of [math]\displaystyle{ \sigma }[/math] .

[math]\displaystyle{ }[/math]

This data set is represented graphically in the following contour plot:

As can be determined from the table, the lowest calculated value for [math]\displaystyle{ t }[/math] is 25.046, while the highest is 39.250. These represent the 80% confidence limits on the time at which reliability is equal to 40%.

Example 7

For the data given in Example 5, determine the two-sided 80% confidence bounds on the reliability estimate for [math]\displaystyle{ t=30 }[/math] . The ML estimate for the reliability at [math]\displaystyle{ t=30 }[/math] is 45.739%.

Solution to Example 7

In this example, we are trying to determine the two-sided 80% confidence bounds on the reliability estimate of 45.739%. This is accomplished by substituting [math]\displaystyle{ t=30 }[/math] and [math]\displaystyle{ \alpha =0.8 }[/math] into Eqn. (normliketr), and varying [math]\displaystyle{ \sigma }[/math] until the maximum and minimum values of [math]\displaystyle{ R }[/math] are found. The following table gives the values of [math]\displaystyle{ R }[/math] based on given values of [math]\displaystyle{ \sigma }[/math] .

This data set is represented graphically in the following contour plot:

As can be determined from the table, the lowest calculated value for [math]\displaystyle{ R }[/math] is 24.776%, while the highest is 68.000%. These represent the 80% two-sided confidence limits on the reliability at [math]\displaystyle{ t=30 }[/math] .