Estimation, Evaluation, and Selection of Actuarial Models

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76 CHAPTER 4. MODEL EVALUATION AND SELECTION j y j F ∗ (x) F n (x) summand 0 50 0.0000 0.0000 0.0399 1 82 0.0391 0.0526 0.0388 2 115 0.0778 0.1053 0.0126 3 126 0.0904 0.1579 0.0332 4 155 0.1227 0.2105 0.0070 5 161 0.1292 0.2632 0.0904 6 243 0.2138 0.3158 0.0501 7 294 0.2622 0.3684 0.0426 8 340 0.3033 0.4211 0.0389 9 384 0.3405 0.4737 0.0601 10 457 0.3979 0.5263 0.1490 11 680 0.5440 0.5789 0.0897 12 855 0.6333 0.6316 0.0099 13 877 0.6433 0.6842 0.0407 14 974 0.6839 0.7368 0.0758 15 1,193 0.7594 0.7895 0.0403 16 1,340 0.7997 0.8421 0.0994 17 1,884 0.8983 0.8947 0.0592 18 2,558 0.9561 0.9474 0.0308 19 3,476 0.9860 1.0000 0.0141 20 ∞ 1.0000 1.0000 The total is 1.0226 and the test statistic is −19(1) + 19(1.0226) = 0.4292. Because the test statistic is less than the critical value of 2.492, the exponential model is viewed as plausible. For Data Set B censored at 1,000, the results are in the following table. j y j F ∗ (x) Fn(x) ∗ summand 0 0 0.0000 0.00 0.0376 1 27 0.0369 0.05 0.0718 2 82 0.1079 0.10 0.0404 3 115 0.1480 0.15 0.0130 4 126 0.1610 0.20 0.0334 5 155 0.1942 0.25 0.0068 6 161 0.2009 0.30 0.0881 7 243 0.2871 0.35 0.0493 8 294 0.3360 0.40 0.0416 9 340 0.3772 0.45 0.0375 10 384 0.4142 0.50 0.0575 11 457 0.4709 0.55 0.1423 12 680 0.6121 0.60 0.0852 13 855 0.6960 0.65 0.0093 14 877 0.7052 0.70 0.0374 15 974 0.7425 0.75 0.0092 16 1000 0.7516 0.75
4.5. HYPOTHESIS TESTS 77 The total is 0.7602 and the test statistic is −20(0.7516) + 20(0.7602) = 0.1713. Because the test statistic does not exceed the critical value of 2.492, the exponential model is viewed as plausible. ¤ Exercise 81 Perform the Anderson-Darling test of the Weibull distribution for the continuing example. 4.5.3 Chi-square goodness-of-fit test Unlike the previous two tests, this test allows for some discretion. It begins with the selection of k − 1 arbitrary values, t = c 0
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