Template:Non-Parametric LDA Example (Kaplan-Meier)

Problem Statement (Kaplan-Meier)
A group of 20 units are put on a life test with the following results.

$$\begin{matrix} Number & State & State \\ in State & (F or S) & End Time \\ 3 & F & 9 \\ 1 & S & 9 \\ 1 & F & 11 \\ 1 & S & 12 \\ 1 & F & 13 \\ 1 & S & 13 \\ 1 & S & 15 \\ 1 & F & 17 \\ 1 & F & 21 \\ 1 & S & 22 \\   1 & S & 24  \\ 1 & S & 26 \\ 1 & F & 28 \\ 1 & F & 30 \\ 1 & S & 32 \\ 2 & S & 35 \\ 1 & S & 39 \\ 1 & S & 41 \\ \end{matrix}$$

Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.

Solution (Kaplan-Meier)
Using the data and Eqn. (kapmeier), the following table can be constructed:

$$\begin{matrix} State & Number of & Number of & Available & {} & {}  \\ End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}} & \mathop{}_{}^{}\tfrac{{{n}_{i}}-{{r}_{i}}} \\ 9 & 3 & 1 & 20 & 0.850 & 0.850 \\   11 & 1 & 0 & 16 & 0.938 & 0.797  \\   12 & 0 & 1 & 15 & 1.000 & 0.797  \\   13 & 1 & 1 & 14 & 0.929 & 0.740  \\   15 & 0 & 1 & 12 & 1.000 & 0.740  \\   17 & 1 & 0 & 11 & 0.909 & 0.673  \\   21 & 1 & 0 & 10 & 0.900 & 0.605  \\   22 & 0 & 1 & 9 & 1.000 & 0.605  \\   24 & 0 & 1 & 8 & 1.000 & 0.605  \\   26 & 0 & 1 & 7 & 1.000 & 0.605  \\   28 & 1 & 0 & 6 & 0.833 & 0.505  \\   30 & 1 & 0 & 5 & 0.800 & 0.404  \\   32 & 0 & 1 & 4 & 1.000 & 0.404  \\   35 & 0 & 1 & 3 & 1.000 & 0.404  \\   39 & 0 & 1 & 2 & 1.000 & 0.404  \\   41 & 0 & 1 & 1 & 1.000 & 0.404  \\ \end{matrix}$$

As can be determined from the preceding table, the reliability estimates for the failure times are:

$$\begin{matrix} Failure Time & Reliability Est. \\ 9 & 85.0% \\   11 & 79.7%  \\   13 & 74.0%  \\   17 & 67.3%  \\   21 & 60.5%  \\   28 & 50.5%  \\   30 & 40.4%  \\ \end{matrix}$$