Statistical Methods for Reliability Data
Chapter 10 - Planning Life Tests
W. Q. Meeker, L. A. Escobar, and J. K. Freels
07 May 2017
CHAPTER OVERVIEW
This chapter explains
Basic ideas for planning a life test or field tracking study
The use of simulation to indicate how the results of a life test might look, to see how the data might be analyzed, and to get an idea of the expected precision for a proposed solution
The use of large-sample approximate methods to assess the precision of the results obtained from a future reliability study
How to determine an approximate sample size that provides a specified precision
How to assess the trade-offs involving sample size and study length
The use of simulation to check and "calibrate" large-sample approximate methods
Methods for assessing sensitivity of test planning conclusions to unknown inputs that must be provided
10.1 - Introduction
10.1.1 - Basic Ideas
Every test event must have clearly defined goals
Additional Considerations
Life testing can be expensive and is nearly always time-constrained
Simulation is an important tool for ensuring the test results support the conclusion and address any test planning concerns
- How many prototypes should be tested?
- Should we test the full system or just some critical components
- How precise do the results need to be?
- How long should the test be run?
It's intuitively understood that
Increasing the number of test samples will generate more information
Increasing the amount of test time will generate more information
More information allows for more precise estimates
But, it's also known that
Generating more information always requires more investment
At some point, the investment required for greater precision becomes disproportionate
Often, information exists about the system to be tested or the data to be gathered
- Design specifications
- Prior experience with similar systems or materials
- Expert knowledge regarding the expected time to failure
- Explicitly defined total test time or sample size
In the text, these prior data are referred to as "planning values" and are denoted by a \(\Box\), e.g. \(\mu^{\Box},\sigma^{\Box}\)
Simulation uses "planning values" to ensure that a proposed life-test can achieve results to support the desired conclusion.
Example 10.1 - Insulation Life Test Plan
Desired conclusion
- Determine if the \(t_{0.1}\) lifetime of a new electrical insulation at use-stress conditions is sufficient to meet a reliability requirement
Test constraints
- Total test time \(= 1000\) hours
- Elevated stress applied to produce failures within the \(1000\) hour limit
Planning Values
\[
\begin{aligned}
t_{0.12}^{\Box}&=500\;\;\text{hours}\\\\
t_{0.20}^{\Box}&=1000\;\text{hours}\\\\
p_{c}^{\Box}&=0.2
\end{aligned}
\]
For the initial evaluation, the test planners assume that the insulation lifetimes can be modeled with either a Weibull or lognormal distribution
Using the given planning values and appropriate probability paper, the test planners can graphically estimate the distribution parameters (Figure 10.1)
Weibull distribution
- \(\mu^{\Box}=8.774\;\rightarrow\;\eta^{\Box}=6464\;\text{hours}\)
- \(\sigma^{\Box}=1.244\;\rightarrow\;\beta^{\Box}=.8037\)
Lognormal distribution
- \(\mu^{\Box}=8.658\)
- \(\sigma^{\Box}=2.079\)
10.1.2 - Simulation of a Proposed Test Plan
Steps often used to evaluate a reliability test plan
- Use the chosen model & planning values to simulate data from the proposed life test
- Analyze the data, perhaps fitting more than one distribution
- Assess estimate precision (typically via confidence intervals)
- Simulate and fit distributions to many samples to assess sampling variations. This assessment does not depend on large sample approximations
- Repeat this simulation-evaluation process with different sample sizes to gauge sample size and test length requirements to achieve the desired precision
- Repeat steps (1) - (5) with different planning values to gauge how sensitive the results may be to errors in our "expert knowledge"
10.1.3 - Uncertainty in Planning Values
10.2 - Approximate Variance of ML Estimators
10.2.1 - Large-Sample Approximations
Motivation
Simulation is a powerful tool for estimating test plan properties, but the results are often open to interpretation
In contrast to simulation, large sample approximations have some advantages
Can directly approximate estimation precision as a function of sample size
Estimate the required sample size for a desired precision
Variance factors allow trade-offs in test planning decisions
10.2.2 - Basic Large-Sample Approximations
Background
- For \(F(t|\mathbf{\underline{\theta}}),\;\mathbf{\underline{\theta}}=(\theta_1,...,\theta_k)\) the following results hold for large sample sizes and the standard regularity conditions (Appendix B.4)
\[
\mathbf{\widehat{\underline{\theta}}}_{_{MLE}}\sim MVN(\mathbf{\underline{\theta}}, \Sigma_{\mathbf{\widehat{\underline{\theta}}}})
\]
and
\[
\Sigma_{\mathbf{\widehat{\underline{\theta}}}}=\mathcal{I}^{-1}_{\underline{\theta}}=E\left[-\frac{\partial^2\mathcal{L}(\underline{\theta}|\underline{t})}{\partial\underline{\theta}\partial\underline{\theta}^T}\right]^{-1}=\sum_{i=1}^n E\left[-\frac{\partial^2\mathcal{L}_i(\underline{\theta}|\underline{x})}{\partial\underline{\theta}\partial\underline{\theta}^T}\right]^{-1}
\]
Results of interest in reliability tests
In many cases we're interested performing a test to gain information on some function of \(\mathbf{\underline{\widehat{\theta}}}\)
For large samples \(g(\mathbf{\underline{\widehat{\theta}}})\sim NOR\left(g(\mathbf{\underline{\theta}}),\widehat{se}_{g(\mathbf{\underline{\widehat{\theta}}})}\right)\)
where \(\widehat{se}_{g(\mathbf{\underline{\widehat{\theta}}})}=\sqrt{\widehat{Var}\left[\widehat{g}(\mathbf{\underline{\widehat{\theta}}})\right]}\)
\[
\widehat{Var}\left[\widehat{g}(\mathbf{\underline{\widehat{\theta}}})\right]=\left[\frac{\partial g(\mathbf{\underline{\theta}})}{\partial \mathbf{\underline{\theta}}}\right]^T\Sigma_{\mathbf{\underline{\widehat{\theta}}}}\left[\frac{\partial g(\mathbf{\underline{\theta}})}{\partial \mathbf{\underline{\theta}}}\right]
\]
- If \(g(\mathbf{\underline{\theta}}) \in \mathbb{R}^+,\; \forall \mathbf{\underline{\theta}}\) it is generally better to use the \(\log\) transform of the function where
\[
\log[g(\mathbf{\underline{\widehat{\theta}}})]\sim NOR\left(\log[g(\mathbf{\underline{\theta}})],\widehat{se}_{\log[\widehat{g}(\mathbf{\underline{\theta}})]}\right)
\]
and
\[
\widehat{Var}\left[\log(\widehat{g}(\mathbf{\underline{\theta}}))\right]=\left(\frac{1}{g(\mathbf{\underline{\theta}})}\right)^2\widehat{Var}[\widehat{g}(\mathbf{\underline{\theta}})]
\]
This is the same procedure presented in previous chapters for strictly positive quantities, but generalized for vector-valued functions
The approximate standard errors for \(\widehat{g}(\mathbf{\underline{\theta}})\;\text{and}\;\widehat{g}(\mathbf{\log[\underline{\theta}]})\) may be represented as
\[
\widehat{se}_{\widehat{g}(\mathbf{\underline{\theta}})}=\frac{1}{\sqrt{n}}\sqrt{V_{\widehat{g}}}\;\;\text{and}\;\;\widehat{se}_{\widehat{g}(\mathbf{\log[\underline{\theta}]})}=\frac{1}{\sqrt{n}}\sqrt{V_{\log(\widehat{g})}}
\]
- Where the following variance factors depend on the value of \(\mathbf{\underline{\theta}}\) but do depend on the sample size \(n\)
\[
V_{\widehat{g}}=n\widehat{Var}_{\widehat{g}(\mathbf{\underline{\theta}})}\;\;\text{and}\;\;V_{\log(\widehat{g})}=n\widehat{Var}_{\widehat{g}(\mathbf{\log[\underline{\theta}]})}
\]
- Thus, substituting \(\mathbf{\underline{\theta}^{\Box}}\) allows us to choose a sample size \(n\) corresponding to an appropriate \(\widehat{se}_{(\mathbf{\cdot)}}\)
10.3 - Sample Size for Unrestricted Functions
Sample Size for Unrestricted Functions
If the goal for a test event is to make conclusions regarding the value of a function of the parameters \(g(\mathbf{\underline{\theta}})\in \mathbb{R}\)
An approximate \(100(1-\alpha)\%\) CI for \(g(\mathbf{\underline{\theta}})\) based on the large-sample approximation would be expressed as Equation 10.3
\[
\begin{aligned}
\left[\underline{g},\overline{g}\right]&=\widehat{g}\pm z_{(1-\alpha/2)}(1/\sqrt{n})\sqrt{\widehat{V}_{\widehat{g}}}\\\\
&=\widehat{g}\pm D
\end{aligned}
\]
\[
n=\frac{z^2_{(1-\alpha/2)}V^{\Box}_{\widehat{g}}}{D^2_T}
\]
where
- \(V^{\Box}_{\widehat{g}}\equiv V_{\widehat{g}}\) evaluated at \(\mathbf{\underline{\theta}^{\Box}}\)
- \(D_{_{T}}\equiv\) the targeted half-width of the confidence interval
- \(n\equiv\) the sample size required to compute a \(100(1-\alpha)\%\) confidence interval for \(\widehat{g}\)
Example 10.3
Background
From elementary statistics
\(\widehat{\mu}_{_{MLE}}=\bar{t}\)
\(Var[\bar{t}]=\sigma^2/n \rightarrow V_{\widehat{\mu}}=nVar[\bar{t}]=\sigma\)
\(V^{\Box}_{\widehat{\mu}}=(\sigma^{\Box})^2=200^2\)
Substituting these values into Equation 10.4 gives an estimate of the required sample size
\[
n=\frac{z^2_{(1-\alpha/2)}V^{\Box}_{\widehat{\mu}}}{D_T^2}=\frac{(1.96)^2(200)^2}{30^2}\approx 171 \mbox{ samples}
\]
10.4 - Sample Size for Positive Functions
Sample Size for Positive Functions
If the goal for a test event is to make conclusions regarding the value of a function of the parameters \(g(\mathbf{\underline{\theta}})\in \mathbb{R}^+\)
An approximate \(100(1-\alpha)\%\) CI for \(\log[g(\mathbf{\underline{\theta}})]\) based on the large-sample approximation would be expressed as Equation 10.5
\[
\begin{aligned}
\left[\underline{\log[g]},\overline{\log[g]}\right]&=\log[\widehat{g}]\pm (1/\sqrt{n})z_{(1-\alpha/2)}(1/\sqrt{n})\sqrt{\widehat{V}_{\log[\widehat{g}]}}\\\\
&=\log[\widehat{g}]\pm \log[R]
\end{aligned}
\]
\[
n=\frac{z^2_{(1-\alpha/2)}V^{\Box}_{\log[\widehat{g}]}}{(\log[R_T])^2}
\]
Where
- \(V^{\Box}_{\log[\widehat{g}]}\equiv V_{\log[\widehat{g}]}\) evaluated at \(\mathbf{\underline{\theta}^{\Box}}\)
- \(\log[R_T]\equiv\) the targeted half-width of the confidence interval
- \(n\equiv\) the sample size required to compute a \(100(1-\alpha)\%\) confidence interval for \(\log[\widehat{g}]\)
Example 10.4
Background
The purpose of a test is to compute a \(95\%\) CI for the mean life of a new electrical insulation
Engineers provide the following planning information
Assume that the lifetime of the insulation \(T\sim EXP(\theta)\)
Since \(\theta \in [0, \infty)\), Equation 10.6 is used to compute \(n\)
\(\theta^{\Box}=1000\;\text{hours}\)
\(R_T=1.5\) hours
From Section 7.6.3
\(\widehat{\theta}_{_{MLE}}=\frac{TTT}{r}\)
\(V_{[\widehat{\theta}]}=n\widehat{Var}[\widehat{\theta}]=\frac{n}{E\left[-\frac{\partial^2 \mathcal{L}(\theta)}{\partial\theta^2}\right]}=\frac{\theta^2}{1-\exp\left(-\frac{t_c}{\theta}\right)}\)
\(V^{\Box}_{\log[\widehat{\theta}]}=\frac{V^{\Box}_\widehat{\theta}}{(\theta^{\Box})^2}=\frac{1}{1-\exp\left(-\frac{t_c}{\theta}\right)}=\frac{1}{1-\exp\left(-\frac{500}{1000}\right)}=2.5415\)
Substituting these values into Equation 10.4 gives an estimate of the required sample size
\[
n=\frac{z^2_{(1-\alpha/2)}V^{\Box}_{\log[\widehat{\theta}]}}{(\log[R_T])^2}=\frac{(1.96)^2(2.5415)}{(\log[1.5])^2}\approx 60 \mbox{ samples}
\]
Proof of \(E\left[-\frac{\partial^2\mathscr{L}(\theta)}{\partial\theta^2}\right]\)
- Example 10.4 is based upon the following equivalence
\[E\left[-\frac{\partial^2\mathscr{L}(\theta)}{\partial\theta^2}\right] = \frac{1-\exp\left(-\frac{t_c}{\theta}\right)}{n\theta^2}\]
- To prove this, recall the likelihood function for right censored data
\[
\mathscr{L}(\theta)=\sum_{i=1}^n f(t_i|\theta)^{\delta_i}\times S(t_i|\theta)^{1-\delta_i} \quad i = 1,\cdots,n
\]
- The indicator variable \(\delta_i\) assumes the following values
\[
\delta_i=
\begin{cases}
1 \mbox{ if $t_i$ is a failure time}\\
0 \mbox{ if $t_i$ is a right-censored observation}
\end{cases}
\]
Further, Example 10.4 states that the assumed underlying distribution is \(EXP(\theta)\), therefore we can make the following substitutions
- \(f(t_i|\theta) = \frac{1}{\theta}\exp\left[-\frac{t_i}{\theta}\right]\)
- \(S(t_i|\theta) = \exp\left[-\frac{t_i}{\theta}\right]\)
Including these substitutions results in this updated likelihood function
\[
\mathscr{L}(\theta)=\sum_{i=1}^n \left(\frac{1}{\theta}\exp\left[-\frac{t_i}{\theta}\right]\right)^{\delta_i}\times\left(\exp\left[-\frac{t_i}{\theta}\right]\right)^{1-\delta_i} \quad i = 1,\cdots,n
\]
- Taking the log of this expression gives the log-likelihood function which is almost always easier to work with
\[
\begin{aligned}
\mathcal{L}(\theta)=&\sum_{i=1}^n {\delta_i}\left(\log\left[\frac{1}{\theta}\right]-\frac{t_i}{\theta}\right) + (1-\delta_i)\left(-\frac{t_i}{\theta}\right) \quad i = 1,\cdots,n\\\\
=&\sum_{i=1}^n -\delta_i\log[\theta]-\delta_i\frac{t_i}{\theta} -\frac{t_i}{\theta} +\delta_i\frac{t_i}{\theta} \quad i = 1,\cdots,n\\\\
=&\sum_{i=1}^n -\delta_i\log[\theta]-\frac{t_i}{\theta} \quad i = 1,\cdots,n\\\\
=& -\log[\theta]\sum_{i=1}^n \delta_i- \frac{1}{\theta}\sum_{i=1}^nt_i \quad i = 1,\cdots,n\\\\
=& -r\log[\theta] - \frac{TTT}{\theta}
\end{aligned}
\]
Observing the last expression
Recall that \(\delta_i = 1\) is the event at time \(t_i\) is a failure and is \(0\) otherwise
Thus, \(\sum_{i=1}^n \delta_i\) is just the number of failures \(r\)
Further, \(\sum_{i=1}^n t_i\) can easily be seen as the total time on test \(TTT\)
Our goal is to find \(E\left[\frac{\partial^2\mathscr{L}}{\partial\theta^2}\right]\), therefore our next steps are to:
The first and second derivatives are expressed as
\[
\begin{aligned}
\mathscr{L} &=-r\log[\theta] - \frac{TTT}{\theta}\\\\
\frac{\partial\mathscr{L}}{\partial\theta} &=-\frac{r}{\theta}+\frac{TTT}{\theta^2}\\\\ \frac{\partial^2\mathscr{L}}{\partial\theta^2} &= \frac{r}{\theta^2}-\frac{TTT}{\theta^3}\\\\
-\frac{\partial^2\mathscr{L}}{\partial\theta^2} &= -\frac{r}{\theta^2}+\frac{TTT}{\theta^3}
\end{aligned}
\]
Finally, we want to find the expected value of this expression
Note that the expectation operator is distributive
\[
\begin{aligned}
E\left[-\frac{\partial^2\mathscr{L}}{\partial\theta^2}\right] =\;& E\left[-\frac{r}{\theta^2}+\frac{TTT}{\theta^3}\right]\\\\
=\;& E\left[-\frac{r}{\theta^2}\right]+E\left[\frac{TTT}{\theta^3}\right]\\\\
=\;& -\frac{E[r]}{\theta^2}+\frac{E[TTT]}{\theta^3}
\end{aligned}
\]
\[
\begin{aligned}
E[r] &= n \times F(t_c)\\\\
&= n\left(1-\exp\left[-\frac{t_c}{\theta}\right]\right)
\end{aligned}
\]
Now, lets look at \(E[TTT]\) the expected total time on test
We know that \(TTT = \sum_{i=1}^r t_i + (n-r)t_c\), applying the expectation operator gives
\[
E[TTT] = E\left[\sum_{i=1}^r t_i|r \right] + (n-r)t_c
\]
10.5 - Sample Sizes for Log-Location-Scale Distributions with Censoring
10.5.1 - Large-Sample Approximation for \(\Sigma_{(\mu,\sigma)}\)
In this section
\[
\Sigma_{(\widehat{\mu},\widehat{\sigma})}=\left[\begin{array}{cc} \widehat{Var}[\widehat{\mu}] & \widehat{Cov}[\widehat{\mu},\widehat{\sigma}]\\ \widehat{Cov}[\widehat{\mu},\widehat{\sigma}] & \widehat{Var}[\widehat{\mu}]\end{array}\right]=\frac{1}{n}\left[\begin{array}{cc} V_{\widehat{\mu}} & V_{(\widehat{\mu},\widehat{\sigma})}\\ V_{(\widehat{\mu},\widehat{\sigma})} & V_{\widehat{\sigma}} \end{array}\right]=\mathcal{I}^{-1}_{(\widehat{\mu},\widehat{\sigma})}
\]
Table C.20 lists the large-sample approximate values for the variance-covariance matrix elements assuming a normal distribution
The values in the table are defined with respect to the standardized censoring time (the number of standard deviations \(t_c\) is from \(\mu\))
\[
\zeta_c=\frac{t_c-\mu}{\sigma}
\]
- Note that the values in Table C.20 may also be applied to when the lognormal distribution is assumed, in which case
\[
\zeta_c=\frac{\log[t_c]-\mu}{\sigma}
\]
So, how do we use Table C.20?
First, we use existing prior information or "expert" knowledge to compute the planning values \(\mu^{\Box}, \sigma^{\Box}\)
The censoring time, \(t_c\), will be determined before testing begins
Thus, we can compute \(\zeta_c^{\Box}\) - assuming a lognormal distribution
\[
\zeta_c^{\Box}=\frac{\log[t_c]-\mu^{\Box}}{\sigma^{\Box}}
\]
- Once \(\zeta_c^{\Box}\) is known, we can find the appropriate line on Table C.20 and read off the values
So, what are the values displayed in Table C.20?
\(100\Phi(\zeta_c)\equiv\) the population fraction failing up to time \(t_c\) (given \(\mu^{\Box}\) and \(\sigma^{\Box}\))
\((1/\sigma^2)V_{\widehat{\mu}}\equiv\) the scaled large-sample variance factor for \(\mu\)
\((1/\sigma^2)V_{\widehat{\sigma}}\equiv\) the scaled large-sample variance factor for \(\sigma\)
\((1/\sigma^2)V_{(\widehat{\mu},\widehat{\sigma})}\equiv\) the scaled large-sample covariance factor for \(\sigma\) and \(\sigma\)
\(\rho(\widehat{\mu},\widehat{\sigma})\equiv\) the large sample correlation between \(\widehat{\mu}\) and \(\widehat{\sigma}\)
\(f_{11}, f_{22}, f_{12}\equiv\) the scaled Fisher information matrix for a single observation from the corresponding location scale distribution
Example 10.5
Background
In Example 10.1 we used the available planning information to find \(\mu^{\Box}=8.774\) and \(\sigma^{\Box}=1.244\)
Now, we're interested in computing a \(95\%\) CI for \(\beta=1/\sigma, s.t.\) the endpoints are \(50\%\) away from \(\beta_{_{MLE}}\)
Since \(\beta\) is a strictly positive quantity, our confidence interval precision factor is expressed as
\[
R_{_{T}}=1.5=\frac{\overset{\sim}{g}=1.5\widehat{g}}{\widehat{g}}=\frac{\widehat{g}=1.5\underset{\sim}{g}}{\underset{\sim}{g}}=\sqrt{\frac{1.5\widehat{g}}{\frac{\widehat{g}}{1.5}}}=\sqrt{(1.5)^2}
\]
Under these assumptions, what is the required sample size?
Since \(\beta=1/\sigma\) is strictly positive, Equation 10.6 should be used
\[
n=\frac{z^2_{(1-\alpha/2)}V^{\Box}_{\log[\widehat{\beta}]}}{(\log[R_T])^2}=\frac{(1.96)^2 V^{\Box}_{\log[\widehat{\beta}]}}{(\log[1.5])^2}
\]
- Thus, the only quantity yet to be computed is
\[
V^{\Box}_{\log[\widehat{\beta}]}=V^{\Box}_{\log[\widehat{\sigma}]}=\frac{1}{(\sigma^{\Box})^2}V^{\Box}_{\widehat{\sigma}}
\]
Note, this example assumes that the test data are best modeled using a Weibull distribution - therefore Table C.20 should not be used.
The SMRD package contains the table.lines or lsinf functions to compute these quantities for other location-scale distributions
The lsinf function has three required arguments
z - the value of \(\zeta^{\Box}_c\)censor.type - the type of censoringdistribution - the assumed underlying distribution
For this example, the value of \(\zeta^{\Box}_c\) is
\[
\zeta^{\Box}_c = \frac{t_c - \mu^{\Box}}{\sigma^{\Box}} =\frac{1000-8.774}{1.244}
= -1.5
\]
- Using the
lsinf function, assuming a weibull distribution, with right censored observation and \(\zeta^{\Box}_c=-1.5\) gives
$f11
[1] 0.1999893
$f12
[1] -0.3112703
$f22
[1] 0.6954854
$matrix
f_i1 f_i2
f_1j 0.1999893 -0.3112703
f_2j -0.3112703 0.6954854
However, note that lsinf returns the scaled Fisher information matrix, while we want \(V_{\log[\widehat{\sigma}]} =\frac{1}{\sigma^2}V_{\widehat{\sigma}}\)
We can easily get the desired result by transforming the matrix returned by lsinf
To find \(V_{\widehat{\sigma}}\), note that
\[
\begin{aligned}
\frac{\sigma^2}{n}\mathcal{I}_{(\mu,\sigma)}&=\left[\begin{array}{cc}f_{11}&f_{12}\\f_{12}&f_{22}\end{array}\right]\\\\
\mathcal{I}_{(\mu,\sigma)}&=\frac{n}{\sigma^2}\left[\begin{array}{cc}f_{11}&f_{12}\\f_{12}&f_{22}\end{array}\right]\\\\
\mathcal{I}^{-1}_{(\mu,\sigma)}&=\frac{\sigma^2}{n}\left[\begin{array}{cc}f_{11}&f_{12}\\f_{12}&f_{22}\end{array}\right]^{-1}
\end{aligned}
\]
- From this result, the following equivalence can be made using the equations given in section 10.5.1
\[
\begin{aligned}
\mathcal{I}^{-1}_{(\widehat{\mu},\widehat{\sigma})}=\frac{1}{n}\left[\begin{array}{cc}V_{\widehat{\mu}}&V_{\widehat{\mu},\widehat{\sigma}}\\V_{\widehat{\mu},\widehat{\sigma}}&V_{\widehat{\sigma}}\end{array}\right]&=\frac{\sigma^2}{n}\left[\begin{array}{cc}f_{11}&f_{12}\\f_{12}&f_{22}\end{array}\right]^{-1}\\\\
\frac{1}{\sigma^2}\left[\begin{array}{cc}V_{\widehat{\mu}}&V_{\widehat{\mu},\widehat{\sigma}}\\V_{\widehat{\mu},\widehat{\sigma}}&V_{\widehat{\sigma}}\end{array}\right]&=\left[\begin{array}{cc}f_{11}&f_{12}\\f_{12}&f_{22}\end{array}\right]^{-1}
\end{aligned}
\]
- Inverting the matrix from
lsinf gives
f_1j f_2j
f_i1 16.480523 7.375995
f_i2 7.375995 4.739033
We then find \(1/(\sigma^{\Box})^2 V^{\Box}_{\widehat{\sigma}}=\) f_inv[2,2] \(\approx 4.74\)
And can finally compute the number of samples that should be tested
\[
n=\frac{(1.96)^2(4.74)}{[\log(1.5)]^2}\approx 111\;\text{samples}
\]
- Note that an alternative to
lsinf is the function table.lines that returns the columns in Table C.20 for a give value of \(\zeta_c\)
$z [1] -1.5
$phib [1] 0.0668072
$f11 [1] 0.2790593
$f22 [1] 0.805458
$f12 [1] -0.4478958
$v11 [1] 33.33856
$v22 [1] 11.55049
$v12 [1] 18.53877
$rho [1] 0.944729
$vmugsigma [1] 3.583468
$vsigmagmu [1] 1.24153
$z [1] -1.5
$phib [1] 0.1999893
$f11 [1] 0.1999893
$f22 [1] 0.6954854
$f12 [1] -0.3112703
$v11 [1] 16.48052
$v22 [1] 4.739033
$v12 [1] 7.375995
$rho [1] 0.8346229
$vmugsigma [1] 5.000268
$vsigmagmu [1] 1.437845
10.5.3 - Large-Sample Estimate of \(\widehat{Var}[g(\widehat{\mu},\widehat{\sigma})]\)
Remember the Delta Method
- For functions of \(\widehat{\mu}, \widehat{\sigma}\) the large sample variance factors are
\[
\begin{aligned}
V_{\widehat{g}}&=\left(\frac{\partial g}{\partial\mu}\right)^2 V_{\widehat{\mu}}+\left(\frac{\partial g} {\partial\sigma}\right)^2 V_{\widehat{\sigma}}+\left(\frac{\partial g}{\partial\mu}\right)\left(\frac{\partial g}{\partial\sigma}\right)V_{(\widehat{\mu},\widehat{\sigma})}\\\\
V_{\log[\widehat{g}]}&=\left(\frac{1}{g}\right)^2 V_{\widehat{g}}, \;\;\;g>0\\\\
V_{\exp[\widehat{g}]}&=g^2V_{\widehat{g}}
\end{aligned}
\]
10.5.4 - Sample size to Estimate \(\log[t_p;\mu,\sigma]\)
- Recall, for log-location scale distributions
\[
\log(t_p|\mu,\sigma)=\mu+\Phi^{-1}_{(\cdot)}(p)\sigma
\]
- After taking derivatives of our function \(g(\mu,\sigma)= \log(t_p|\mu,\sigma)\), we have
\[
V_{\log[t_p]}=V_{\widehat{\mu}}+\left[\Phi^{-1}_{(\cdot)}(p)\right]^2 V_{\widehat{\sigma}}+2\Phi^{-1}_{(\cdot)}(p) V_{(\widehat{\mu},\widehat{\sigma})}
\]
\[
n=\frac{z_{(1-\alpha/2)}V^{\Box}_{\log[t_p]}}{(\log[R_T])^2}
\]
10.5.5 - Sample Size to Estimate \(h(\log[t_e\;\mu,\sigma])\)
For log-location scale distributions
\[
h(t_e|\mu,\sigma)=\frac{\phi(\zeta_e)}{t_e\sigma[1-\Phi(\zeta_e)]}
\]
After taking derivatives, we have
\[
V_{\log[\widehat{h}]}=\frac{1}{h^2}V_{\widehat{h}}=\left[\left(\frac{\partial h}{\partial\mu} \right)^2 V_{\widehat{\mu}}+\left(\frac{\partial h}{\partial\sigma}\right)^2 V_{\widehat{\sigma}} +2\left(\frac{\partial h}{\partial\mu}\right)\left(\frac{\partial h}{\partial\sigma}\right) V_{(\widehat{\mu},\widehat{\sigma})}\right]
\]
and
\[
n=\frac{z_{(1-\alpha/2)}V^{\Box}_{\log[\widehat{h}]}}{(\log[R_T])^2}
\]
Example 10.10
Recall the electrical insulation test discussed in Examples 10.1, 10.5, and 10.7
Suppose that we now wish to plan a life test to compute a \(95\%\) CI for \(h(1000)\) wherein the endpoints are approximately \(50\%\) away from \(\widehat{h}(1000)_{_{MLE}}\)
In Example 10.1 we noted
- \(p^{\Box}_c=0.2\) the fraction of units expected to fail when the test is ended
- \(p_e=0.2\)
Since \(t_e=t_c=1000\;\text{hours}\rightarrow p_e=p^{\Box}_c=0.2\)
Observing where the \(p_e=0.2\) and \(p^{\Box}_c=0.2\) lines intersect in Figure 10.9, we see that
\[
V_{\log[\widehat{h}(1000)]}\approx 8.2
\]
and
\[
n=\frac{z_{(1-\alpha/2)}V^{\Box}_{\log[\widehat{h}(1000)]}}{(\log[R_T])^2}=\frac{(1.96)^2(8.2)}{(\log[1.5])^2}\approx 191
\]
10.6 - Test Plans to Demonstrate Conformance with a Reliability Standard
10.6.1 - Reliability Demonstration Plans
It's often necessary to plan tests to demonstrate the reliability performance of a product
Objective: Given a performance specification and desired confidence level, want to specify
- The required test length
- The required sample size
To demonstrate if the measure of interest exceeds the specification with the \(100(1-\alpha)\%\) confidence
Example: For \(t_e=8760\;\text{hours (1 year)}\) want to demonstrate with \(100(1-\alpha)\%\) confidence that
\[
\underline{t}_{.01}>t_e\iff \overline{F}(t_e)<.01
\]
Often a reliability test will conclude without observing any failures
What does this mean?
- How many samples are required to demonstrate that the reliability of the population
- How long must the test continue
10.6.2 - Weibull \(\min(n)\) reliability demonstration plans w/ given \(\beta\)
10.6.3 - Extensions for Other Reliability Demonstration Test Plans
The \(f_{ij}\) elements in Appendix C.20 are the elements of the scaled Fisher information matrix, that is
\[
\left[\begin{array}{cc}f_{11}&f_{12}\\f_{21}&f_{22}\end{array}\right]=
\sigma^{2}
\left[\begin{array}{cc}
E\left\{-\frac{\partial^{2}\mathscr{L}(\mu,\sigma)}{\partial\mu^{2}}\right\}&
E\left\{-\frac{\partial^{2}\mathscr{L}(\mu,\sigma)}{\partial
\mu \partial \sigma}\right\}\\
E\left\{-\frac{\partial^{2}\mathscr{L}(\mu,\sigma)}{\partial
\mu \partial \sigma}\right\}&
E\left\{-\frac{\partial^{2}\mathscr{L}(\mu,\sigma)}{\partial\sigma^{2}}\right\}
\end{array}\right]=\left[\begin{array}{cc}.96841&-0.11490\\-0.11490&1.56779\end{array}\right]
\]
The variance terms in Table C.20 are then found from
\[
\begin{aligned}
\frac{1}{\sigma^{2}}\left[\begin{array}{cc}V_{\bar{\mu}}&V_{(\bar{\mu},\bar{\sigma})}\\V_{(\bar{\mu},\bar{\sigma})}&V_{\bar{\sigma}}\end{array}\right]&=\frac{n}{\sigma^{2}}\left[\begin{array}{cc}Avar(\bar{\mu})&Acov(\bar{\mu},\bar{\sigma})\\Acov(\bar{\mu},\bar{\sigma})&Avar(\bar{\sigma})\end{array}\right]=\left[\begin{array}{cc}f_{11}&f_{12}\\f_{12}&f_{22}\end{array}\right]^{-1}\\\\
&=\frac{1}{f_{11}f_{22}-f_{12}^{2}}\left[\begin{array}{cc}f_{22}&-f_{12}\\-f_{12}&f_{11}\end{array}\right]\\\\
&=\frac{1}{0.96841\times 1.56779-(-0.11490)^{2}}\left[\begin{array}{cc}0.96841&0.11490\\0.11490&1.56779\end{array}\right]\\\\
&=\left[\begin{array}{cc}1.04168&0.07634\\ 0.7634&0.64344\end{array}\right]
\end{aligned}
\]
The asymptotic correlation is then computed as
\[
\rho(\bar{\mu},\bar{\sigma})
=\frac{V_{(\bar{\mu},\bar{\sigma})}}{\sqrt{V_{\bar{\mu}}V_{\bar{\sigma}}}}
=\frac{0.07634}{\sqrt{1.04168\times 0.64344}}=0.09325
\]
Similarly, the scaled asymptotic variances for either a known \(\mu\) or a known \(\sigma\) are
\[
\begin{aligned}
\frac{n}{\sigma^{2}}Avar(\bar{\mu}|\sigma)&=\frac{1}{\sigma^{2}}V_{\bar{\mu}|\sigma}=\left[f_{11}\right]^{-1}=\left[0.96841\right]^{-1}=1.03262\\
\frac{n}{\sigma^{2}}Avar(\bar{\sigma}|\mu)&=\frac{1}{\sigma^{2}}V_{\bar{\sigma}|\mu}=\left[f_{22}\right]^{-1}=\left[1.56779\right]^{-1}=0.63784
\end{aligned}
\]
10.7 - Some Extensions
10.7.1 - Failure (Type II) Censoring
10.7.2 - Variance factors for location-scale parameters and multiple censoring
10.7.3 - Test planning for distributions that are not log-location scale