# exam1s - STAT 526 MIDTERM KEY Here are the results 1 | 0 2...

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Unformatted text preview: STAT 526 MIDTERM KEY Here are the results: 1 | 0 2 | 0 3 | 0166 4 | 559 5 | 78 6 | 5 7 | 4 8 | 0 9 | 2 10 | 0 1. (45 points) The data frame Insurance in the MASS package concerns the number of car insurance claims made by the policyholders of an insurance company in the third quarter of 1973. The data frame consists of the following variables: District District of policyholder (1 to 4): 4 is major cities. Group Group of car: < 1 litre, 1 − 1 . 5 litre, 1 . 5 − 2 litre, > 2 litre. Age Age of driver in 4 ordered groups: < 25, 25 − 29, 30 − 35, > 35. Holders Number of policyholders . Claims Number of claims. A model is fitted to the data using the following program. library(MASS); data(Insurance) options(contrasts=c("contr.treatment","contr.treatment")) fit <- glm(Claims~District+Group+Age+offset(log(Holders)), data=Insurance,family=poisson) Part of the results are summarized below ( summary(fit) ). Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.82174 0.07679 -23.724 < 2e-16 *** District2 0.02587 0.04302 0.601 0.547597 District3 0.03852 0.05051 0.763 0.445657 District4 0.23421 0.06167 3.798 0.000146 *** Group1-1.5l 0.16134 0.05053 3.193 0.001409 ** Group1.5-2l 0.39281 0.05500 7.142 9.18e-13 *** Group>2l 0.56341 0.07232 7.791 6.65e-15 *** Age25-29-0.19101 0.08286-2.305 0.021149 * Age30-35-0.34495 0.08137-4.239 2.24e-05 *** Age>35-0.53667 0.06996-7.672 1.70e-14 ***--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 October 19, 2009 STAT 526 MIDTERM KEY (Dispersion parameter for poisson family taken to be 1) Null deviance: 236.26 on 63 degrees of freedom...
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## This note was uploaded on 12/14/2009 for the course CHEM CHEM 140B taught by Professor Burkhart during the Spring '08 term at UCSD.

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exam1s - STAT 526 MIDTERM KEY Here are the results 1 | 0 2...

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