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| '''Kaplan-Meier Example'''
| | #REDIRECT [[Non-Parametric_Life_Data_Analysis]] |
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| A group of 20 units are put on a life test with the following results.
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| <center><math>\begin{matrix}
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| Number & State & State \\
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| in State & (F or S) & End Time \\
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| 3 & F & 9 \\
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| 1 & S & 9 \\
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| 1 & F & 11 \\
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| 1 & S & 12 \\
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| 1 & F & 13 \\
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| 1 & S & 13 \\
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| 1 & S & 15 \\
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| 1 & F & 17 \\
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| 1 & F & 21 \\
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| 1 & S & 22 \\
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| 1 & S & 24 \\
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| 1 & S & 26 \\
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| 1 & F & 28 \\
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| 1 & F & 30 \\
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| 1 & S & 32 \\
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| 2 & S & 35 \\
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| 1 & S & 39 \\
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| 1 & S & 41 \\
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| \end{matrix}</math></center>
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| Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.
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| '''Solution'''
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| Using the data and Eqn. (kapmeier), the following table can be constructed:
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| <center><math>\begin{matrix}
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| State & Number of & Number of & Available & {} & {} \\
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| End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} & \mathop{}_{}^{}\tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} \\
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| 9 & 3 & 1 & 20 & 0.850 & 0.850 \\
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| 11 & 1 & 0 & 16 & 0.938 & 0.797 \\
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| 12 & 0 & 1 & 15 & 1.000 & 0.797 \\
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| 13 & 1 & 1 & 14 & 0.929 & 0.740 \\
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| 15 & 0 & 1 & 12 & 1.000 & 0.740 \\
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| 17 & 1 & 0 & 11 & 0.909 & 0.673 \\
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| 21 & 1 & 0 & 10 & 0.900 & 0.605 \\
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| 22 & 0 & 1 & 9 & 1.000 & 0.605 \\
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| 24 & 0 & 1 & 8 & 1.000 & 0.605 \\
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| 26 & 0 & 1 & 7 & 1.000 & 0.605 \\
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| 28 & 1 & 0 & 6 & 0.833 & 0.505 \\
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| 30 & 1 & 0 & 5 & 0.800 & 0.404 \\
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| 32 & 0 & 1 & 4 & 1.000 & 0.404 \\
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| 35 & 0 & 1 & 3 & 1.000 & 0.404 \\
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| 39 & 0 & 1 & 2 & 1.000 & 0.404 \\
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| 41 & 0 & 1 & 1 & 1.000 & 0.404 \\
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| \end{matrix}</math></center>
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| As can be determined from the preceding table, the reliability estimates for the failure times are:
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| <center><math>\begin{matrix}
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| Failure Time & Reliability Est. \\
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| 9 & 85.0% \\
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| 11 & 79.7% \\
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| 13 & 74.0% \\
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| 17 & 67.3% \\
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| 21 & 60.5% \\
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| 28 & 50.5% \\
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| 30 & 40.4% \\
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| \end{matrix}</math></center>
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