leading to w^2 ¼ 8 :918 with 5 dfðp¼: 0636 Þ, indicating that at least one
covariate must be moderately related significantly to the number of complaints.
Note: For model 1, the SAS program would include this instruction:
MODEL CASES = GENDER RESIDENCY REVENUE HOURS/
DIST = POISSON LINK = LOG OFFSET = LN;
and for model 2,
MODEL CASES = /DIST = POISSON LINK = LOG
OFFSET = LN;
(See the note after Example 10.6 for other details of the program.)
Test for a Single Variable Let us assume that we now wish to test whether the
addition of one particular independent variable of interest adds significantly to
the prediction of the response over and above that achieved by other factors
already present in the model. The null hypothesis for this test may be stated as:
‘‘FactorXidoes not have any value added to the prediction of the response
given that other factors are already included in the model.’’ In other words,
H 0 :bi¼ 0To test such a null hypothesis, one can use
zi¼bb^i
SEðbb^iÞwherebb^iis the corresponding estimated regression coe‰cient and SEðbb^iÞis the
estimate of the standard error ofbb^i, both of which are printed by standard
computer-packaged programs such as SAS. In performing this test, we refer the
value of thezscore to percentiles of the standard normal distribution; for
example, we compare the absolute value ofzto 1.96 for a two-sided test at the
5% level.
Example 10.8 Refer to the data set on emergency service of Example 10.5
(Table 10.2) with all four covariates. We have the results shown in Table 10.4.
Only the e¤ect of workload (hours) is significant at the 5% level.
Note: Use the same SAS program as in Examples 10.6 and 10.7.
Given a continuous variable of interest, one can fit a polynomial model and
use this type of test to check for linearity (seetype 1 analysisin the next sec-
tion). It can also be used to check for a single product representing an e¤ect
modification.
POISSON REGRESSION MODEL 363