- No. The model is not fully parameterized, since it
contains five parameters whereas the number of
covariate patterns is 186. - No. The model does not perfectly predict the case/
noncase status of each of the 289 subjects in the data. - a. The deviance value of 159.2017 is not calculated
using the deviance formula
Devðb^Þ¼ 2 lnðL^c=L^maxÞ.
In particular 2 lnL^c¼ 279 : 317 and 2 lnL^max¼ 0 ,
so Devð^bÞ¼ 279 : 317.
b. The other logistic model, i.e., Model 2, is the model
that is defined by the 186 covariate patterns that
comprise the fully parameterized model derived
from the four independent variables PREVHOSP,
AGE, GENDER, and PAMU. This model will
contain 185 predictor variables plus an intercept
term.
c. 159 : 2017 ¼ 2 lnL^Model 1 ð 2 lnL^Model 2 Þ,
where 2 lnL^Model 1 ¼ 279 : 3170
and 2 lnL^Model 2 ¼ 279 : 3170 159 : 2017 ¼ 120 : 1153.
d. G¼#of covariate patterns¼186 is large relative
ton¼289. - a. The HL test has aP‐value of 0.4553, which is highly
nonsignificant. Therefore, the HL test indicates
that the model does not have lack of fit.
b. Models 1 and 2 as described in question 5b.
c. Choose Model 1 since it provides adequate fit and is
more parsimonious than Model 2. However, a LR
test cannot be performed since the deviance for
Model 1 is based on a large number of covariate
patterns.
d. Neither model perfectly fits the data, since neither
model is a saturated model. - Consider the information shown in the output under
the heading “Partition for the Hosmer and Lemeshow
Test.”
a. The 10 groups are formed by partitioning the 289
predicted risks into 10 deciles of risk, where, for
example, the highest (i.e., 10th) decile contains
approximately the highest 10% of the predicted
risks.
b. Because subjects with the same value for the
predicted risk cannot be separated into different
deciles.
c. For group 5:
Expected number of cases¼sum of predicted risks
for all 29 subjects in group 5.
Expected number of noncases¼ 29 expected
number of cases
680 Test Answers