Hi, and welcome to the tutorial for Exam C, problem 22. This problem covers the learning objective "Construction & Selection of Parametric Models". Here's a view of the problem. Please complete the question and select one of the five answers. If incorrect, you can click the gray X to try again.
Here is a solution to the problem. In this question we are instructed to use a likelihood ratio test to determine whether the hypothesized parameters could be changed to better represent the data. In short, our null hypothesis is that alpha equals 1.5 and theta equals 7.8. The alternate hypothesis is that the maximum likelihood estimates of 1.4 and 7.6 should be used instead. The likelihood ratio test relies on that fact that twice the ratio of the alternate likelihood and null likelihood follows a Chi-square distribution. Although the test is called likelihood ratio, ratios are rarely ever used. Typically twice the difference in loglikelihood will be used instead, as shown. For this test to be permissible, the alternative case must be a special case of the null. For instance, an exponential distribution is a special case of a Gamma, where the Gamma parameter alpha is equal to 1. Because the hypotheses we are testing are both Pareto, we are OK to do this test. Notice that in this problem, the alternative loglikelihood is given; we must only find the null loglikelihood.
Let's start by writing out the likelihood function for the Pareto distribution with a sample size of 200. This simplifies to alpha to the 200 times theta to the 200alpha, divided by the multiplication of all 200 of the identity on the bottom. Taking the natural log makes this look a lot less intimidating. We could continue to maximize this equation to find the maximum likelihood estimates if we wanted to, but remember that we were given those values already, so there's no need to do that. Instead we now plug in the hypothesized values to find the null hypothesis loglikelihood. Remember that we were given the sum of the observations plus 7.8, which is why that equals 607.64. After completing the algebra we find that the null loglikelihood is -821.77.
Now we can find the test statistic. Twice the difference in loglikelihood of the alternative and null is equal to 7.7. For this test, the degrees of freedom for the Chi-square distribution are equal to the number of free parameters in the alternative minus the number of free parameters in the null. This can be tough to understand. The null hypothesis has no free parameters because those parameters were taken as given, and were not estimated. The MLE hypothesis has two "free parameters" because an estimation technique was used to find both; no parameters in that test were initially taken as given (even though the result of the MLE test was given).
We're ready to reach a conclusion now. Checking the given Chi-square table with two degrees of freedom, we see that the test statistic is greater than the critical value for 2.5% significance, but not 1% significance. That means we reject the null hypothesis with 97.5% certainty to use the alternative-hypothesized parameters instead, provided we are comfortable with that level of certainty. Rejection at the 2.5% level corresponds with answer C.
That concludes this tutorial. Thanks for watching!