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.
The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.
Normal Probability Density Function
The pdf of the normal distribution is given by:
- [math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}}\,\! }[/math]
where:
- [math]\displaystyle{ \mu\,\! }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T}\,\! }[/math],
- [math]\displaystyle{ \theta\,\! }[/math] = standard deviation of the times-to-failure
It is a 2-parameter distribution with parameters [math]\displaystyle{ \mu \,\! }[/math] (or [math]\displaystyle{ \bar{T}\,\! }[/math] ) and [math]\displaystyle{ {{\sigma }}\,\! }[/math] (i.e., the mean and the standard deviation, respectively).
Normal Statistical Properties
The Normal Mean, Median and Mode
The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu .\,\! }[/math] Because the normal distribution is symmetrical, the median and the mode are always equal to the mean:
- [math]\displaystyle{ \mu =\tilde{T}=\breve{T}\,\! }[/math]
The Normal Standard Deviation
As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}\,\! }[/math].
The Normal Reliability Function
The reliability for a mission of time [math]\displaystyle{ T\,\! }[/math] for the normal distribution is determined by:
- [math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx\,\! }[/math]
There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.
The Normal Conditional Reliability Function
The normal conditional reliability function is given by:
- [math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}\,\! }[/math]
Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.
The Normal Reliable Life
Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:
- [math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt\,\! }[/math]
for [math]\displaystyle{ T\,\! }[/math].
The Normal Failure Rate Function
The instantaneous normal failure rate is given by:
- [math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}\,\! }[/math]
Characteristics of the Normal Distribution
Some of the specific characteristics of the normal distribution are the following:
- The normal pdf has a mean, [math]\displaystyle{ \bar{T}\,\! }[/math], which is equal to the median, [math]\displaystyle{ \breve{T}\,\! }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T}\,\! }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T}\,\! }[/math]. This is because the normal distribution is symmetrical about its mean.
- The mean, [math]\displaystyle{ \mu \,\! }[/math], or the mean life or the [math]\displaystyle{ MTTF\,\! }[/math], is also the location parameter of the normal pdf, as it locates the pdf along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty \,\! }[/math].
- The normal pdf has no shape parameter. This means that the normal pdf has only one shape, the bell shape, and this shape does not change.
- The standard deviation, [math]\displaystyle{ {{\sigma }}\,\! }[/math], is the scale parameter of the normal pdf.
- As [math]\displaystyle{ {{\sigma }}\,\! }[/math] decreases, the pdf gets pushed toward the mean, or it becomes narrower and taller.
- As [math]\displaystyle{ {{\sigma }}\,\! }[/math] increases, the pdf spreads out away from the mean, or it becomes broader and shallower.
- The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty \,\! }[/math].
- The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }}\,\! }[/math], and vice versa.
- The standard deviation is also the distance between the mean and the point of inflection of the pdf, on each side of the mean. The point of inflection is that point of the pdf where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the pdf has a value of zero.
- The normal pdf starts at [math]\displaystyle{ t=-\infty \,\! }[/math] with an [math]\displaystyle{ f(t)=0\,\! }[/math]. As [math]\displaystyle{ t\,\! }[/math] increases, [math]\displaystyle{ f(t)\,\! }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T}\,\! }[/math]. Thereafter, [math]\displaystyle{ f(t)\,\! }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0\,\! }[/math] at [math]\displaystyle{ t=+\infty \,\! }[/math].
Weibull++ Notes on Negative Time Values
One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).
Normal Distribution Examples
The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.
Complete Data Example
6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:
Objectives
- 1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method
- 2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.
- 3. Obtain the pdf plot for the data.
- 4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.
- 5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.
- 6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.
Solution
The following figure shows the data as entered in Weibull++, as well as the calculated parameters.
The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.
Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.
To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.
Suspension Data Example
19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.
| Non-Grouped Data Times-to-Failure Data with Suspensions | ||
| Data point index | Last Inspected | State End Time |
|---|---|---|
| 1 | F | 2 |
| 2 | S | 3 |
| 3 | F | 5 |
| 4 | S | 7 |
| 5 | F | 11 |
| 6 | S | 13 |
| 7 | S | 17 |
| 8 | S | 19 |
| 9 | F | 23 |
| 10 | F | 29 |
| 11 | S | 31 |
| 12 | F | 37 |
| 13 | S | 41 |
| 14 | F | 43 |
| 15 | S | 47 |
| 16 | S | 53 |
| 17 | F | 59 |
| 18 | S | 61 |
| 19 | S | 67 |
Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]
If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]
For the rank regression on y (RRY) method, the parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]
Interval Censored Data Example
8 units are being reliability tested, and the following is a table of their failure times:
| Non-Grouped Interval Data | ||
| Data point index | Last Inspected | State End Time |
|---|---|---|
| 1 | 30 | 32 |
| 2 | 32 | 35 |
| 3 | 35 | 37 |
| 4 | 37 | 40 |
| 5 | 42 | 42 |
| 6 | 45 | 45 |
| 7 | 50 | 50 |
| 8 | 55 | 55 |
This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]
For rank regression on x:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]
If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]
The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.
Mixed Data Types Example
Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.
| Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored) | ||||
|---|---|---|---|---|
| Data point index | Number in State | Last Inspection | State (S or F) | State End Time |
| 1 | 1 | 10 | F | 10 |
| 2 | 1 | 20 | S | 20 |
| 3 | 2 | 0 | F | 30 |
| 4 | 2 | 40 | F | 40 |
| 5 | 1 | 50 | F | 50 |
| 6 | 1 | 60 | S | 60 |
| 7 | 1 | 70 | F | 70 |
| 8 | 2 | 20 | F | 80 |
| 9 | 1 | 10 | F | 85 |
| 10 | 1 | 100 | F | 100 |
Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]
If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]
For the rank regression on y (RRY) method, the parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]
Comparison of Analysis Methods
8 units are being reliability tested, and the following is a table of their failure times:
| Non-Grouped Times-to-Failure Data | ||
| Data point index | State F or S | State End Time |
|---|---|---|
| 1 | F | 2 |
| 2 | F | 5 |
| 3 | F | 11 |
| 4 | F | 23 |
| 5 | F | 29 |
| 6 | F | 37 |
| 7 | F | 43 |
| 8 | F | 59 |
Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]
If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]
For the rank regression on y (RRY) method, the parameters are:
- [math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]
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]
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] .