Answers to Practice Exercises
Practice
Exercises
- The data listing is in events trials (ET) format. There
are eight lines of data corresponding to the distinct
covariate patterns defined by the model; each line con-
tains the number of cases (i.e., events) and the number
of subjects (i.e., trials) for each covariate pattern. - There are eight covariate patterns:
Pattern 1:X¼(CAT¼0, AGE¼0, ECG¼0)
Pattern 2:X¼(CAT¼0, AGE¼1, ECG¼0)
Pattern 3:X¼(CAT¼0, AGE¼0, ECG¼1)
Pattern 4:X¼(CAT¼0, AGE¼1, ECG¼1)
Pattern 5:X¼(CAT¼1, AGE¼0, ECG¼0)
Pattern 6:X¼(CAT¼1, AGE¼1, ECG¼0)
Pattern 7:X¼(CAT¼1, AGE¼0, ECG¼1)
Pattern 8:X¼(CAT¼1, AGE¼1, ECG¼1) - No. The model contains four parameters, whereas
there are eight covariate patterns. - No. The model does not perfectly predict the case/
noncase status of each of the 609 subjects in the data. - a. No. The deviance value of 0.9544 is not calculated
using the deviance formula
Devðb^Þ¼ 2 lnðL^c=L^maxÞ:
In particular 2 lnL^c¼ 418 : 181 and 2 lnL^max¼ 0 ,
so Devð^bÞ¼ 418 : 181.
b. Model 1 :Logit PðXÞ¼aþbCATþg 1 AGEþg 2 ECG
Model 2 :Logit PðXÞ¼aþbCATþg 1 AGEþg 2 ECG
þg 3 AGEECG
þd 1 CATAGE
þd 2 CATECG
þd 3 CATAGEECG
c. 0 : 9544 ¼ 2 lnL^Model 1 ð 2 lnL^Model 2 Þ,
where 2 lnL^Model 1 ¼ 418 : 1810 and
2 lnL^Model 2 ¼ 418 : 1810 0 : 9544 ¼ 417 : 2266 :
d. H 0 :d 1 ¼d 2 ¼d 3 ¼0, i.e., the deviance is used to
test for whether the coefficients of all the product
terms in Model 2 are collectively nonsignificant.
e. G¼no. of covariate patterns¼ 8 <<n¼609. - a. The HL test has aP-value of 0.9177, which is highly
nonsignificant. Therefore, the HL test indicates
that the model does not have lack of fit.
b. The model contains only eight covariate patterns,
so it is not possible to obtain more than eight
distinct predicted risk values from the data. The
degrees of freedom is 4 because it is calculated as
the number of groups (i.e., 6) minus 2.
c. Models 1 and Models 2 as stated in the answer to
question 5b.
342 9. Assessing Goodness of Fit for Logistic Regression