# Parametric Binomial Example - Demonstrate MTTF

This example appears in the Life Data Analysis Reference book.

In this example, we will use the parametric binomial method to design a test that will demonstrate $MTTF=75\,\!$ hours with a 95% confidence if no failure occur during the test $f=0\,\!$ . We will assume a Weibull distribution with a shape parameter $\beta =1.5\,\!$ . We want to determine the number of units to test for ${{t}_{TEST}}=60\,\!$ hours to demonstrate this goal.

The first step in this case involves determining the value of the scale parameter $\eta \,\!$ from the $MTTF\,\!$ equation. The equation for the $MTTF\,\!$ for the Weibull distribution is:

$MTTF=\eta \cdot \Gamma (1+{\frac {1}{\beta }})\,\!$ where $\Gamma (x)\,\!$ is the gamma function of $x\,\!$ . This can be rearranged in terms of $\eta \,\!$ :

$\eta ={\frac {MTTF}{\Gamma (1+{\tfrac {1}{\beta }})}}\,\!$ Since $MTTF\,\!$ and $\beta \,\!$ have been specified, it is a relatively simple matter to calculate $\eta =83.1\,\!$ . From this point on, the procedure is the same as the reliability demonstration example. Next, the value of ${{R}_{TEST}}\,\!$ is calculated as:

${{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1\%\,\!$ The last step is to substitute the appropriate values into the cumulative binomial equation. The values of $CL\,\!$ , ${{t}_{TEST}}\,\!$ , $\beta \,\!$ , $f\,\!$ and $\eta \,\!$ have already been calculated or specified, so it merely remains to solve the binomial equation for $n\,\!$ . The value is calculated as $n=4.8811,\,\!$ or $n=5\,\!$ units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.

The procedure for determining the required test time proceeds in the same manner, determining $\eta \,\!$ from the $MTTF\,\!$ equation, and following the previously described methodology to determine ${{t}_{TEST}}\,\!$ from the binomial equation with Weibull distribution.