Exam MAS-I Practice Question 96
Study guide: Exam MAS-I
An insurer's five territories carry the codes 1 to 5, assigned in the order the territories were added to its system. Exposure and claims by territory:
| Territory code | Exposure | Claims | Claim frequency |
|---|---|---|---|
| 1 | 4,000 | 240 | 0.0600 |
| 2 | 4,000 | 380 | 0.0950 |
| 3 | 4,000 | 220 | 0.0550 |
| 4 | 4,000 | 360 | 0.0900 |
| 5 | 4,000 | 290 | 0.0725 |
An analyst fits a Poisson GLM with a log link and an offset of \ln(\text{exposure}) to these five rows, using the territory code as the only predictor:
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.67824 0.06167 -43.428 <2e-16 ***
territory 0.02685 0.01833 1.465 0.143
Null deviance: 67.620 on 4 degrees of freedom
Residual deviance: 65.472 on 3 degrees of freedom
AIC: 107.03
Which conclusion is best supported?
- A) The code was fitted as one slope although its order means nothing; refit territory as categorical
- B) The territory coefficient is not significant at the 5% level, so territory can be dropped from the model
- C) Each step up in territory code raises frequency by about 2.7%, so territory 5 has the highest frequency
- D) The residual deviance of 65.5 on 3 degrees of freedom shows overdispersion; refit the same model as quasi-Poisson
- E) The poor fit shows exposure should be entered as a predictor with an estimated coefficient instead of an offset
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