For the hypothesis testing on group one, your null hypothesis is that
theta_1 = 0.10, and the alternative hypothesis is that theta_1 is less than
0.10. In practice, you would have already selected a critical region that
determines when you reject the null hypothesis (i.e., reject if test
statistic is less than 0.05, 0.10).
Next, you determine the probability of seeing the observations if the null
hypothesis is indeed true. So, for group 1, what is the probability of
seeing 4 or fewer claims, given that theta_1 = 0.10? The "4 or fewer"
condition is determined by the alternative hypothesis. If the alternative
had been theta_1 > 0.10, you would have calculated the probability that 4 or
more claims were observed, given theta_1 = 0.10.
Keep in mind that theta_1 is the mean of the poisson frequency distribution
for one exposure. The mean of the aggregate frequency distribution is the
sum of the means of the individual claim distributions. Therefore, the mean
of the group 1 aggregate frequency distribution is the sum of 100 0.10's, or
10. This is the distribution you use to test the null.
Hope this helps
Chris Coleianne
Senior Actuarial Analyst
The Harleysville Insurance Companies
215-513-8131
-----Original Message-----
From: Jennifer Ligon [SMTP:Jennifer.Ligon@usaa.com]
Sent: Tuesday, August 24, 1999 8:16 AM
To: 'studygroup4b@lists.casact.org'
Subject: freq distr question
I was looking back over some frequency distr problems in Mahler's
notes and
I'm having trouble with May 1997 #29. I don't really understand
Mahler's
solution - why does he let theta_1 equal .1 ? Can anybody explain it
differently? Thanks......
Jennifer Ligon
Actuarial Analyst, Auto Pricing
(210) 498-4360
jennifer.ligon@usaa.com