The Exponential Distribution
The exponential distribution is a commonly used distribution in reliability engineering. Mathematically, it is a fairly simple distribution, which many times leads to its use in inappropriate situations. It is, in fact, a special case of the Weibull distribution where
. The exponential distribution is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out).
Exponential Probability Density Function
The 2-Parameter Exponential Distribution
The 2-parameter exponential pdf is given by:
where 
is the location parameter.
Some of the characteristics of the 2-parameter exponential distribution are discussed in Kececioglu
[19]:
- The location parameter,
, if positive, shifts the beginning of the distribution by a distance of
to the right of the origin, signifying that the chance failures start to occur only after
hours of operation, and cannot occur before. - The scale parameter is
. - The exponential pdf has no shape parameter, as it has only one shape.
- The distribution starts at
at the level of
and decreases thereafter exponentially and monotonically as
increases beyond
and is convex. - As
,
.
The 1-Parameter Exponential Distribution
The 1-parameter exponential pdf is obtained by setting
, and is given by:
where:
-
= constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.)
-
= mean time between failures, or to failure
= operating time, life, or age, in hours, cycles, miles, actuations, etc.
This distribution requires the knowledge of only one parameter,
, for its application. Some of the characteristics of the 1-parameter exponential distribution are discussed in Kececioglu
[19]:
- The location parameter,
, is zero. - The scale parameter is
. - As
is decreased in value, the distribution is stretched out to the right, and as
is increased, the distribution is pushed toward the origin. - This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate,
). - The distribution starts at
at the level of
and decreases thereafter exponentially and monotonically as
increases, and is convex. - As
,
. - The pdf can be thought of as a special case of the Weibull
pdf with
and
.
Exponential Distribution Functions
The Mean or MTTF
The mean, 
or mean time to failure (MTTF) is given by:
Note that when 
, the MTTF is the inverse of the exponential distribution's constant failure rate. This is only true for the exponential distribution. Most other distributions do not have a constant failure rate. Consequently, the inverse relationship between failure rate and MTTF does not hold for these other distributions.
The Median
The median, 
is:
The Mode
The mode, 
is:
The Standard Deviation
The standard deviation, 
, is:
The Exponential Reliability Function
The equation for the 2-parameter exponential cumulative density function, or cdf, is given by:
Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function of the 2-parameter exponential distribution is given by:
The 1-parameter exponential reliability function is given by:
The Exponential Conditional Reliability Function
The exponential conditional reliability equation gives the reliability for a mission of
duration, having already successfully accumulated
hours of operation up to the start of this new mission. The exponential conditional reliability function is:
which says that the reliability for a mission of
duration undertaken after the component or equipment has already accumulated
hours of operation from age zero is only a function of the mission duration, and not a function of the age at the beginning of the mission. This is referred to as the
memoryless property.
The Exponential Reliable Life Function
The reliable life, or the mission duration for a desired reliability goal,
, for the 1-parameter exponential distribution is:
![{\displaystyle {\begin{aligned}\ln[R({{t}_{R}})]=-\lambda ({{t}_{R}}-\gamma )\end{aligned}}\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/1e60b14069d62d296d47969da5facb817a0c5c29)
or:
![{\displaystyle {{t}_{R}}=\gamma -{\frac {\ln[R({{t}_{R}})]}{\lambda }}\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a1c1610110bf07d5e0f3bbaeebd99b04a0799301)
The Exponential Failure Rate Function
The exponential failure rate function is:
Once again, note that the constant failure rate is a characteristic of the exponential distribution, and special cases of other distributions only. Most other distributions have failure rates that are functions of time.
Characteristics of the Exponential Distribution
The primary trait of the exponential distribution is that it is used for modeling the behavior of items with a constant failure rate. It has a fairly simple mathematical form, which makes it fairly easy to manipulate. Unfortunately, this fact also leads to the use of this model in situations where it is not appropriate. For example, it would not be appropriate to use the exponential distribution to model the reliability of an automobile. The constant failure rate of the exponential distribution would require the assumption that the automobile would be just as likely to experience a breakdown during the first mile as it would during the one-hundred-thousandth mile. Clearly, this is not a valid assumption. However, some inexperienced practitioners of reliability engineering and life data analysis will overlook this fact, lured by the siren-call of the exponential distribution's relatively simple mathematical models.
The Effect of lambda and gamma on the Exponential pdf
-
- The exponential pdf has no shape parameter, as it has only one shape.
- The exponential pdf is always convex and is stretched to the right as
decreases in value. - The value of the pdf function is always equal to the value of
at
(or
). - The location parameter,
, if positive, shifts the beginning of the distribution by a distance of
to the right of the origin, signifying that the chance failures start to occur only after
hours of operation, and cannot occur before this time. - The scale parameter is
. - As
,
.
The Effect of lambda and gamma on the Exponential Reliability Function
-
- The 1-parameter exponential reliability function starts at the value of 100% at
, decreases thereafter monotonically and is convex. - The 2-parameter exponential reliability function remains at the value of 100% for
up to
, and decreases thereafter monotonically and is convex. - As
,
. - The reliability for a mission duration of
, or of one MTTF duration, is always equal to
or 36.79%. This means that the reliability for a mission which is as long as one MTTF is relatively low and is not recommended because only 36.8% of the missions will be completed successfully. In other words, of the equipment undertaking such a mission, only 36.8% will survive their mission.
- The 1-parameter exponential reliability function starts at the value of 100% at
The Effect of lambda and gamma on the Failure Rate Function
-
- The 1-parameter exponential failure rate function is constant and starts at
. - The 2-parameter exponential failure rate function remains at the value of 0 for
up to
, and then keeps at the constant value of
.
- The 1-parameter exponential failure rate function is constant and starts at
Estimation of the Exponential Parameters
Probability Plotting
Estimation of the parameters for the exponential distribution via probability plotting is very similar to the process used when dealing with the Weibull distribution. Recall, however, that the appearance of the probability plotting paper and the methods by which the parameters are estimated vary from distribution to distribution, so there will be some noticeable differences. In fact, due to the nature of the exponential cdf, the exponential probability plot is the only one with a negative slope. This is because the y-axis of the exponential probability plotting paper represents the reliability, whereas the y-axis for most of the other life distributions represents the unreliability.
This is illustrated in the process of linearizing the cdf, which is necessary to construct the exponential probability plotting paper. For the two-parameter exponential distribution the cumulative density function is given by:
Taking the natural logarithm of both sides of the above equation yields:
![{\displaystyle \ln \left[1-F(t)\right]=-\lambda (t-\gamma )\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9d105abb8f6c0fe7aa32a35570af5c6fc4cc062f)
or:
![{\displaystyle {\begin{aligned}\ln[1-F(t)]=\lambda \gamma -\lambda t\end{aligned}}\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9f8307e60179122f435932455d3e751be4d3507b)
Now, let:
![{\displaystyle {\begin{aligned}y=\ln[1-F(t)]\end{aligned}}\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/29c5de2a411e2b412923f2d2bfb4f30045c76fb6)
and:
which results in the linear equation of:
Note that with the exponential probability plotting paper, the y-axis scale is logarithmic and the x-axis scale is linear. This means that the zero value is present only on the x-axis. For
,
and
. So if we were to use
for the y-axis, we would have to plot the point
. However, since the y-axis is logarithmic, there is no place to plot this on the exponential paper. Also, the failure rate,
, is the negative of the slope of the line, but there is an easier way to determine the value of
from the probability plot, as will be illustrated in the following example.
Plotting Example
1-Parameter Exponential Probability Plot Example
6 units are put on a life test and tested to failure. The failure times are 7, 12, 19, 29, 41, and 67 hours. Estimate the failure rate for a 1-parameter exponential distribution using the probability plotting method.
In order to plot the points for the probability plot, the appropriate reliability estimate values must be obtained. These will be equivalent to
since the y-axis represents the reliability and the
values represent unreliability estimates.
Next, these points are plotted on an exponential probability plotting paper. A sample of this type of plotting paper is shown next, with the sample points in place. Notice how these points describe a line with a negative slope.
Once the points are plotted, draw the best possible straight line through these points. The time value at which this line intersects with a horizontal line drawn at the 36.8% reliability mark is the mean life, and the reciprocal of this is the failure rate
.
This is because at 
:
The following plot shows that the best-fit line through the data points crosses the
line at
hours. And because
hours,
failures/hour.
Rank Regression on Y
Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized. The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for rank regression on Y (RRY) were derived:
and:
In our case, the equations for 
and
are:
![{\displaystyle {\begin{aligned}{{y}_{i}}=\ln[1-F({{t}_{i}})]\end{aligned}}\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5e0b950906054db1119e3e72b2edb7c1404ad4d0)
and:
and the 
is estimated from the median ranks. Once
and
are obtained, then
and
can easily be obtained from above equations.
For the one-parameter exponential, equations for estimating a and b become:
The Correlation Coefficient
The estimator of 
is the sample correlation coefficient,
, given by:
RRY Example
2-Parameter Exponential RRY Example
14 units were being reliability tested and the following life test data were obtained. Assuming that the data follow a 2-parameter exponential distribution, estimate the parameters and determine the correlation coefficient,
, using rank regression on Y (RRY).
| Life Test Data | |
|---|---|
| Data point index | Time-to-failure |
| 1 | 5 |
| 2 | 10 |
| 3 | 15 |
| 4 | 20 |
| 5 | 25 |
| 6 | 30 |
| 7 | 35 |
| 8 | 40 |
| 9 | 50 |
| 10 | 60 |
| 11 | 70 |
| 12 | 80 |
| 13 | 90 |
| 14 | 100 |
Solution
Construct the following table, as shown next.
The median rank values ( 
) can be found in rank tables or they can be estimated using the
Quick Statistical Reference in Weibull++.
Given the values in the table above, calculate 
and
:
or:
and:
or:
Therefore:
and:
or:
Then:
The correlation coefficient can be estimated using equation for calculating the correlation coefficient:
This example can be repeated using Weibull++, choosing 2-parameter exponential and rank regression on Y (RRY), as shown next.
The estimated parameters and the correlation coefficient using Weibull++ were found to be:
Please note that the user must deselect the Reset if location parameter > T1 on Exponential option on the Calculations page of the Application Setup window.
The probability plot can be obtained simply by clicking the Plot icon.
Rank Regression on X
Similar to rank regression on Y, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the line is minimized.
Again the first task is to bring our exponential cdf function into a linear form. This step is exactly the same as in regression on Y analysis. The deviation from the previous analysis begins on the least squares fit step, since in this case we treat
as the dependent variable and
as the independent variable. The best-fitting straight line to the data, for regression on X (see
Parameter Estimation), is the straight line:
The corresponding equations for 
and
are:
and:
where:
and:
The values of 
are estimated from the median ranks. Once
and
are obtained, solve for the unknown
value, which corresponds to:
Solving for the parameters from above equations we get:
and:
For the one-parameter exponential case, equations for estimating a and b become:
The correlation coefficient is evaluated as before.
RRX Example
2-Parameter Exponential RRX Example
Using the same data set from the
RRY example above and assuming a 2-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate,
, using rank regression on X.
Solution
The table constructed for the RRY analysis applies to this example also. Using the values from this table, we get:
or:
and:
or:
Therefore:
and:
The correlation coefficient is found to be:
Note that the equation for regression on Y is not necessarily the same as that for the regression on X. The only time when the two regression methods yield identical results is when the data lie perfectly on a line. If this were the case, the correlation coefficient would be
. The negative value of the correlation coefficient is due to the fact that the slope of the exponential probability plot is negative.
This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on X (RRX) methods for analysis, as shown below. The estimated parameters and the correlation coefficient using Weibull++ were found to be:
The probability plot can be obtained simply by clicking the
Plot icon.
Maximum Likelihood Estimation
As outlined in Parameter Estimation, maximum likelihood estimation works by developing a likelihood function based on the available data and finding the values of the parameter estimates that maximize the likelihood function. This can be achieved by using iterative methods to determine the parameter estimate values that maximize the likelihood function. This can be rather difficult and time-consuming, particularly when dealing with the three-parameter distribution. Another method of finding the parameter estimates involves taking the partial derivatives of the likelihood equation with respect to the parameters, setting the resulting equations equal to zero, and solving simultaneously to determine the values of the parameter estimates. The log-likelihood functions and associated partial derivatives used to determine maximum likelihood estimates for the exponential distribution are covered in Appendix D.
MLE Example
MLE for the Exponential Distribution
Using the same data set from the RRY and RRX examples above and assuming a 2-parameter exponential distribution, estimate the parameters using the MLE method.
Solution
In this example, we have complete data only. The partial derivative of the log-likelihood function,
is given by:
![{\displaystyle {\frac {\partial \Lambda }{\partial \lambda }}={\underset {i=1}{\overset {{F}_{e}}{\mathop {\sum } }}}\,\left[{\frac {1}{\lambda }}-\left({{T}_{i}}-\gamma \right)\right]={\underset {i=1}{\overset {14}{\mathop {\sum } }}}\,\left[{\frac {1}{\lambda }}-\left({{T}_{i}}-\gamma \right)\right]=0\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/df9882227bb5be7d2369e9b5d19e621cf020d2ea)
Complete descriptions of the partial derivatives can be found in
Appendix D. Recall that when using the MLE method for the exponential distribution, the value of
is equal to that of the first failure time. The first failure occurred at 5 hours, thus
hours
Substituting the values for
and
we get:
or:
Using Weibull++:
The probability plot is:
Confidence Bounds
In this section, we present the methods used in the application to estimate the different types of confidence bounds for exponentially distributed data. The complete derivations were presented in detail (for a general function) in the chapter for
Confidence Bounds.
At this time we should point out that exact confidence bounds for the exponential distribution have been derived, and exist in a closed form, utilizing the
distribution. These are described in detail in Kececioglu
[20], and are covered in the section in the
test design chapter. For most exponential data analyses, Weibull++ will use the approximate confidence bounds, provided from the Fisher information matrix or the likelihood ratio, in order to stay consistent with all of the other available distributions in the application. The
confidence bounds for the exponential distribution are discussed in more detail in the
test design chapter.
Fisher Matrix Bounds
Bounds on the Parameters
For the failure rate 
the upper (
) and lower (
) bounds are estimated by Nelson
[30]:
![{\displaystyle {\begin{aligned}&{{\lambda }_{U}}=&{\hat {\lambda }}\cdot {{e}^{\left[{\tfrac {{{K}_{\alpha }}{\sqrt {Var({\hat {\lambda }})}}}{\hat {\lambda }}}\right]}}\\&&\\&{{\lambda }_{L}}=&{\frac {\hat {\lambda }}{{e}^{\left[{\tfrac {{{K}_{\alpha }}{\sqrt {Var({\hat {\lambda }})}}}{\hat {\lambda }}}\right]}}}\end{aligned}}\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d91cb74d66d344a2c5ab6f51eaf1958f0006a21c)
where 
is defined by:
If 
is the confidence level, then
for the two-sided bounds, and
for the one-sided bounds.
The variance of 
is estimated from the Fisher matrix, as follows:
where 
is the log-likelihood function of the exponential distribution, described in
Appendix D.
Note that, for fixed 
, the log-likelihood function is increasing in
. This means that the MLE solution for
cannot be found by setting
to zero. (Since
is increasing in
,
can never be zero.) Instead, the MLE solution for
is simply its largest possible value allowed by the sample; namely,
, the first failure time. The MLE solution for
is found in the usual fashion by setting
to zero and solving. Weibull++ treats
as a constant when computing bounds; i.e.,
. (See the discussion in Appendix D for more information.)
Bounds on Reliability
The reliability of the two-parameter exponential distribution is:
The corresponding confidence bounds are estimated from:
These equations hold true for the 1-parameter exponential distribution, with
.
Bounds on Time
The bounds around time for a given exponential percentile, or reliability value, are estimated by first solving the reliability equation with respect to time, or reliable life:
The corresponding confidence bounds are estimated from:
The same equations apply for the one-parameter exponential with
Likelihood Ratio Confidence Bounds
Bounds on Parameters
For one-parameter distributions such as the exponential, the likelihood confidence bounds are calculated by finding values for
that satisfy:
This equation can be rewritten as:
For complete data, the likelihood function for the exponential distribution is given by:
where the 
values represent the original time-to-failure data. For a given value of
, values for
can be found which represent the maximum and minimum values that satisfy the above likelihood ratio equation. These represent the confidence bounds for the parameters at a confidence level
where
for two-sided bounds and
for one-sided.
Example: LR Bounds for Lambda
Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming an exponential distribution, the MLE parameter estimate is calculated to be
. Calculate the 85% two-sided confidence bounds on these parameters using the likelihood ratio method.
Solution
The first step is to calculate the likelihood function for the parameter estimates:
where 
are the original time-to-failure data points. We can now rearrange the likelihood ratio equation to the form:
Since our specified confidence level,
, is 85%, 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
which satisfy this equation. Since there is only one parameter, there are only two values of
that will satisfy the equation. These values represent the
two-sided confidence limits of the parameter estimate
. For our problem, the confidence limits are:
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 exponential reliability equation into the likelihood function. The exponential reliability equation can be written as:
This can be rearranged to the form:
This equation can now be substituted into the likelihood ratio equation to produce a likelihood equation in terms of
and
The unknown parameter 
depends on what type of bounds are being determined. If one is trying to determine the bounds on time for the equation for the mean and the Bayes's rule equation for single parametera given reliability, then
is a known constant and
is the unknown parameter. Conversely, if one is trying to determine the bounds on reliability for a given time, then
is a known constant and
is the unknown parameter. Either way, the likelihood ratio function can be solved for the values of interest.
Example: LR Bounds on Time
For the data given above for the
LR Bounds on Lambda example (five failures at 20, 40, 60, 100 and 150 hours), determine the 85% two-sided confidence bounds on the time estimate for a reliability of 90%. The ML estimate for the time at
is
.
Solution
In this example, we are trying to determine the 85% two-sided confidence bounds on the time estimate of 7.797. This is accomplished by substituting
and
into the likelihood ratio bound equation. It now remains to find the values of
which satisfy this equation. Since there is only one parameter, there are only two values of
that will satisfy the equation. These values represent the
two-sided confidence limits of the time estimate
. For our problem, the confidence limits are:
Example: LR Bounds on Reliability
Again using the data given above for the
LR Bounds on Lambda example (five failures at 20, 40, 60, 100 and 150 hours), determine the 85% two-sided confidence bounds on the reliability estimate for a
. The ML estimate for the time at
is
.
Solution
In this example, we are trying to determine the 85% two-sided confidence bounds on the reliability estimate of 50.881%. This is accomplished by substituting
and
into the likelihood ratio bound equation. It now remains to find the values of
which satisfy this equation. Since there is only one parameter, there are only two values of
that will satisfy the equation. These values represent the
two-sided confidence limits of the reliability estimate
. For our problem, the confidence limits are:
-
.
Bayesian Confidence Bounds
Bounds on Parameters
From
Confidence Bounds, we know that the posterior distribution of
can be written as:
where 
, is the non-informative prior of
.
With the above prior distribution, 
can be rewritten as:
The one-sided upper bound of 
is:
The one-sided lower bound of 
is:
The two-sided bounds of 
are:
Bounds on Time (Type 1)
The reliable life equation is:
For the one-sided upper bound on time we have:
The above equation can be rewritten in terms of
as:
From the above posterior distribuiton equation, we have:
The above equation is solved w.r.t.
The same method is applied for one-sided lower and two-sided bounds on time.
Bounds on Reliability (Type 2)
The one-sided upper bound on reliability is given by:
The above equaation can be rewritten in terms of
as:
From the equation for posterior distribution we have:
The above equation is solved w.r.t.
The same method can be used to calculate one-sided lower and two sided bounds on reliability.
Exponential Distribution Examples
Grouped Data
20 units were reliability tested with the following results:
| Table - Life Test Data | |
| Number of Units in Group | Time-to-Failure |
|---|---|
| 7 | 100 |
| 5 | 200 |
| 3 | 300 |
| 2 | 400 |
| 1 | 500 |
| 2 | 600 |
1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.
2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)
3. Show the Probability plot for the analysis results.
4. Show the Reliability vs. Time plot for the results.
5. Show the pdf plot for the results.
6. Show the Failure Rate vs. Time plot for the results.
7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).
Solution
1. For the 2-parameter exponential distribution and for
hours (first failure), the partial of the log-likelihood function,
, becomes:
![{\displaystyle {\begin{aligned}{\frac {\partial \Lambda }{\partial \lambda }}=&{\underset {i=1}{\overset {6}{\mathop {\sum } }}}\,{N_{i}}\left[{\frac {1}{\lambda }}-\left({{T}_{i}}-100\right)\right]=0\\\Rightarrow &7[{\frac {1}{\lambda }}-(100-100)]+5[{\frac {1}{\lambda }}-(200-100)]+\ldots +2[{\frac {1}{\lambda }}-(600-100)]\\=&0\\\Rightarrow &{\hat {\lambda }}={\frac {20}{3100}}=0.0065{\text{fr/hr}}\end{aligned}}\,\!}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6db2031bacd4cd430331740683344a1af04b28f6)
2. Enter the data in a Weibull++ standard folio and calculate it as shown next.
3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.
4. View the Reliability vs. Time plot.
5. View the pdf plot.
6. View the Failure Rate vs. Time plot.
Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter,
, at 100 hours.
7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time
where
indicates the group number. In this example, the total number of groups is
and the total number of units is
. Thus, the median rank values will be estimated for 20 units and for the total failed units (
) up to the
group, for the
rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.
For example, the median rank value of the fourth group will be the
rank out of a sample size of twenty units (or 81.945%).
The following table is then constructed.
Given the values in the table above, calculate
and
:
or:
and:
or:
Therefore:
and:
or:
Then:
Using Weibull++, the estimated parameters are:
The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the
decimal point.
Using Auto Batch Run
A number of leukemia patients were treated with either drug 6MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately [21, p.175].
| Table - Leukemia Treatment Results | |||
| Time (weeks) | Number of Patients | Treament | Comments |
|---|---|---|---|
| 1 | 2 | placebo | |
| 2 | 2 | placebo | |
| 3 | 1 | placebo | |
| 4 | 2 | placebo | |
| 5 | 2 | placebo | |
| 6 | 4 | 6MP | 3 patients completed |
| 7 | 1 | 6MP | |
| 8 | 4 | placebo | |
| 9 | 1 | 6MP | Not completed |
| 10 | 2 | 6MP | 1 patient completed |
| 11 | 2 | placebo | |
| 11 | 1 | 6MP | Not completed |
| 12 | 2 | placebo | |
| 13 | 1 | 6MP | |
| 15 | 1 | placebo | |
| 16 | 1 | 6MP | |
| 17 | 1 | placebo | |
| 17 | 1 | 6MP | Not completed |
| 19 | 1 | 6MP | Not completed |
| 20 | 1 | 6MP | Not completed |
| 22 | 1 | placebo | |
| 22 | 1 | 6MP | |
| 23 | 1 | placebo | |
| 23 | 1 | 6MP | |
| 25 | 1 | 6MP | Not completed |
| 32 | 2 | 6MP | Not completed |
| 34 | 1 | 6MP | Not completed |
| 35 | 1 | 6MP | Not completed |
Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.
Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.
The software will create two data sheets, one for each subset ID, as shown next.
Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.
