, are estimated from: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 the Confidence Bounds chapter.
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.
This section includes the following subsections:
The lower and upper bounds on the mean, , are estimated from:
Since the standard deviation, 
, must be positive, 
is treated as normally
distributed, and the bounds are estimated from:
where Kα is defined by:
If δ is the
confidence level, then 
for the two-sided bounds and
α
= 1 - δ
for the one-sided bounds.
The variances and covariances of and 
are estimated from the Fisher
matrix, as follows:
Λ is the log-likelihood function of the normal distribution, described in the Statistical Background chapter and Appendix C.
The reliability of the normal distribution is:
Let 
then 
For t = T,
, and for t = 
, 
= . The above equation then becomes:
The bounds on z are estimated from:
where:
or:
The upper and lower bounds on reliability are:
The bounds around time for a given normal percentile (unreliability) are estimated by first solving the reliability equation with respect to time, as follows:
where:
and:
The next step is to calculate the variance of 
or:
The upper and lower bounds are then found by:
Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method.
In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, Λ, are given by:
(The derivations of these equations are presented in Appendix
C.) Substituting the values of Ti and solving the above
system simultaneously, we get 
= 29.58 hours, = 45 hours.
The Fisher matrix is:
Using Weibull++, the MLE method can be selected from the Set Analysis page.
The plot of the solution for this example is shown next.
As covered in the Confidence Bounds chapter, the likelihood confidence bounds are calculated by finding values for θ1 and θ2 that satisfy:
(18)
This equation can be rewritten as:
(19)
For complete data, the likelihood formula for the normal distribution is given by:
(20)
where the xi values represent the original time to failure data. (Note: Note that the procedures outlined here are for complete data only. For data with suspensions or intervals, one merely needs to expand the likelihood function to included these other data types. See the Data & Data Types chapter for more information.) For a given value of α, values for μ and σ can be found which represent the maximum and minimum values that satisfy Eqn. (16). These represent the confidence bounds for the parameters at a confidence level δ, where α = δ for two-sided bounds and α = 2δ - 1 for one-sided.
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 = 28.8 and = 11.2143. Calculate the two-sided 80% confidence bounds on
these parameters using the likelihood ratio method.
The first step is to calculate the likelihood function for the parameter estimates:
(21)
where xi are the original time-to-failure data points. We can now rearrange Eqn. (16) to the form:
Since our specified confidence level, δ,
is 80%, we can calculate the value of the chi-squared statistic, 
We can now substitute this
information into the equation:
It now remains to find the values of μ and σ which satisfy this equation. This is an iterative process that requires setting the value of μ and finding the appropriate values of σ, and vice versa.
The following table gives the values of σ based on given values of μ.
This data set is represented graphically in the following contour plot:
(Note that this plot is generated with degrees of freedom k = 1, as we are only determining bounds on one parameter. The contour plots generated in Weibull++ are done with degrees of freedom k = 2, for use in comparing both parameters simultaneously.) As can be determined from the table, the lowest calculated value for σ 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 μ 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 μ, we can perform the same procedure as before, but finding the two values of μ that correspond with a given value of σ. Using this method, we find that the two-sided 80% confidence limits on μ are 21.807 and 35.793, which are close to the initial estimates of 22 and 35.5.
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:
This can be rearranged to the form:
where Φ-1 is the inverse standard normal. This equation can now be substituted into Eqn. (20), to produce a likelihood equation in terms of σ, t and R:
(22)
The "unknown" parameter t/R depends on what type of bounds are being determined. If one is trying to determine the bounds on time for a given reliability, then R is a known constant and t is the unknown parameter. Conversely, if one is trying to determine the bounds on reliability for a given time, then t is a known constant and R is the unknown parameter. Either way, Eqn. (22) can be used to solve Eqn. (16) for the values of interest.
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 R(t) = 40% is 31.637.
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 R = 0.40 and α = 0.8 into Eqn. (22), and varying σ until the maximum and minimum values of t are found. The following table gives the values of t based on given values of σ.
This data set is represented graphically in the following contour plot:
As can be determined from the table, the lowest calculated value for t 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%.
For the data given in Example 5, determine the two-sided 80% confidence bounds on the reliability estimate for t = 30. The ML estimate for the reliability at t = 30 is 45.739%.
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 t = 30 and α = 0.8 into Eqn. (22), and varying σ until the maximum and minimum values of R are found. The following table gives the values of R based on given values of σ.
This data set is represented graphically in the following contour plot:
As can be determined from the table, the lowest calculated value for R is 24.776%, while the highest is 68.000%. These represent the 80% two-sided confidence limits on the reliability at t = 30.
From the Confidence Bounds chapter, we know that the marginal posterior distribution of μ can be written as:
where:
φ(σ) = 
is the non-informative prior
of σ.
φ(μ)
is a uniform distribution from - to +, the
non-informative prior of μ.
Using the above prior distributions, f(μ|Data) can be rewritten as:
The one-sided upper bound of μ is:
The one-sided lower bound of μ is:
The two-sided bounds of μ are:
The same method can be used to obtained the bounds of σ.
The reliable life for the normal distribution is:
The one-sided upper bound on time is:
(23)
Eqn. (22) can be rewritten in terms of μ as:
From the posterior distribution of μ:
The same method can be applied for one-sided lower bounds and two-sided bounds on time.
The one-sided upper bound on reliability is:
From the posterior distribution of μ:
The same method can be used to calculate the one-sided lower bounds and the two-sided bounds on reliability.
See Also:
The Normal Distribution
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