Estimation, Evaluation, and Selection of Actuarial Models

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64 CHAPTER 4. MODEL EVALUATION AND SELECTION Definition 4.5 Ahypothesistestisuniformly most powerful if no other test exists that has the same or lower significance level and for a particular value within the alternative hypothesis has a smaller probability of making a Type II error. Example 4.6 For the ongoing example, determine the probability of making a Type II error when the alternative hypothesis is true with µ =2, 000. µ ¯X − 1, 200 Pr 3, 435/ √ 20 < 1.645|µ =2, 000 =Pr(¯X − 1, 200 < 1, 263.51|µ =2, 000) =Pr(¯X 0.1257) = 0.45, the null hypothesis is rejected when ¯X − 1, 200 3, 435/ √ 20 > 0.1257. In this example, the test statistic is 0.292 which is in the rejection region and thus the null hypothesis is rejected. Of course, few people would place confidence in the results of a test that was designed to make errors 45% of the time. Because Pr(Z >0.292) = 0.3851, the null hypothesis is rejected for those who select a significance level that is greater than 38.51% and is not rejected by those who use a significance level that is less than 38.51%. ¤ Few people are willing to make errors 38.51% of the time. Announcing this figure is more persuasive than the earlier conclusion based on a 5% significance level. When a significance level is used, readers are left to wonder what the outcome would have been with other significance levels. The value of 38.51% is called a p-value. A working definition is:
4.2. A REVIEW OF HYPOTHESIS TESTS 65 Definition 4.8 For a hypothesis test, the p-value is the probability that the test statistic takes on a value that is less in agreement with the null hypothesis than the value obtained from the sample. Tests conducted at a significance level that is greater than the p-value will lead to a rejection of the null hypothesis, while tests conducted at a significance level that is smaller than the p-value will lead to a failure to reject the null hypothesis. Also, because the p-value must be between 0 and 1, it is on a scale that carries some meaning. The closer to zero the value is, the more support the data give to the alternative hypothesis. Common practice is that values above 10% indicate that the data provide no evidence in support of the alternative hypothesis, while values below 1% indicate strong support for the alternative hypothesis. Values in between indicate uncertainty as to the appropriate conclusion and may call for more data or a more careful look at the data or the experiment that produced it. The test used in the ongoing example assumed a normal distribution with a known standard deviation. Actuarial problems are rarely so simple. That leaves two strategies. One is to develop unique tests for each problem that might arise. Generally, this means a new test is needed for each distribution that might be considered. For some, but not all, problems such tests have been determined. In this note, life will be made much easier. A single test procedure will be introduced that can be used in many situations. Definition 4.9 The likelihood ratio test is conducted as follows. First, let the likelihood function be written as L(θ). Letθ 0 be the value of the parameter(s) that maximizes the likelihood function. However, only values of the parameter(s) that are within the null hypothesis may be considered. Let L 0 = L(θ 0 ).Letˆθ be the maximum likelihood estimator (where the parameter(s) can vary over all possible values) and then let L 1 = L(ˆθ). The test statistic is T =2ln(L 1 /L 0 )=2(lnL 1 − ln L 0 ). The null hypothesis is rejected if T>cwhere c is calculated from α =Pr(T>c) where T has a chi-square distribution with degrees of freedom equal to the number of free parameters in the model less the number of free parameters if the null hypothesis is true. This test makes some sense. When the alternative hypothesis is true, forcing the parameter to be selected from the null hypothesis should produce a likelihood value that is significantly smaller. It should be noted that the continuing example must now end. Both the null and alternative hypothesis have one free parameter, though it is not as free under the null hypothesis. If the null hypothesis were changed to µ =1, 200, then it has no free parameters and the test can be carried out with one degree of freedom. Example 4.10 You want to test the hypothesis that the population that produced Data Set B has a mean that is other than 1,200. Assume that the population has a gamma distribution and conduct the likelihood ratio test at a 5% significance level. Also, determine the p-value. The hypotheses are H 0 : µ =1, 200 H 1 : µ 6= 1, 200. From earlier work the maximum likelihood estimates are ˆα = 0.55616 and ˆθ = 2, 561.1. The loglikelihood at the maximum is ln L 1 = −162.293. Next, the likelihood must be maximized,
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