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A Numerical Example for Time-Terminated Tests

Assume we now place only ten devices on test, which are replaced by similar devices as soon as they fail. Assume that we test these m = 10 devices for a time T0 = 20, and observe (as represented in Figure 3) n = 3 failures, but we do not know the exact failure times.

Even without knowing the exact times of these failures, we can still use the Chi-Square distribution with DF = 2n + 2 = 2 x 3 + 2 = 8. We obtain a "conservative" CI for the mean (or failure rate) of the underlying Exponential Distribution. For this, we again use the total test time T (= mT0 = 10 x 20 = 200) statistic and the Chi-Square percentile (now with DF = 8) and obtain a CI for the device true mean life θ (or reliability or rate = 1/α). We obtain a conservative CI, for confidence level 100(1 - α)% (say, of 80%, 90%, 95%). The statistic of the CI for the mean is given by:

Figure 3. Representation of Type I Censoring; 10 Devices Continuously on Test
Figure 3. Representation of Type I Censoring; 10 Devices Continuously on Test (Click to Zoom)

We calculate the corresponding upper and lower Chi-Square percentiles. As before, confidence coefficient (1 - α) depends on the required confidence. A 95% confidence yields = 0.05, α/2 = 0.025 and 1 - α/2 = 0.975. The Chi-Square table values are:

Equation

The corresponding CI for the true mean life θ, with a 95% confidence is ((2 x 200)/17.54; (2 x 200)/2.18) = (22.81; 183.49).

Since rate ρ = 1/θ, a CI for the true failure rate ρ, with confidence level of 95%, can be obtained by using the reciprocal values of the corresponding CI for the mean: (1/183.49, 1/22.81) = (0.00545; 0.0438). Because the Exponential is a one-parameter distribution, the reliability at any time T is given by: R(T) = P{X ≥ T} = Exp{-T/θ) = Exp{-λT}.

Then, a 95% CI for the reliability at any mission time T can be obtained by using either the mean or the failure rate CI upper and lower limits. For example we obtain, using the upper/lower limits of the CI for Rate , and for a mission time T = 10:

R1(T) = P{X ≥ T} = Exp{-T/θ1} = Exp{ρ1T} = Exp{-0.00545 x 10} = 0.947

R2(T) = P{X ≥ T} = Exp{-T/θ2) = Exp{-ρ2T} = Exp{-0.0438 x 10} = 0.645

Hence, a 95% CI for the true reliability, when mission time T = 10, is: (0.65, 0.95).

Finally, reliability bounds in this time terminated case, are resolved in the same manner as shown earlier, with the only change being that now Chi-Square has DF = 2n + 2 instead of DF = 2n. For example, a lower 97.5% bound for the reliability, from the above data is obtained by dropping the upper limit of the 95% CI, or by applying the upper bound of the corresponding CI for failure rate (-0.0438):

R2(T) = P{X ≥ T} = Exp{-T/θ2} = Exp{λ2T} = Exp{-0.0438 x 10} = 0.645

An upper 97.5% bound for device reliability is analogously obtained by:

R1(T) = P{X ≥ T} = Exp{-T/θ1} = Exp{-λ1T} = Exp{-0.00545 x 10} = 0.947

Thus, the true device reliability is equal to or less than 0.947, and greater than or equal to 0.645, 97.5% of the times.