Logistic Regression: A Self-learning Text, Third Edition (Statistics in the Health Sciences)

To carry out thelikelihood ratio test, we would

need to compare two models. The full model is

Model A as described by the first set of results

discussed here. The reduced model is a differ-

ent model that contains the five covariables

without the CAT variable.

The null hypothesis here is that the coefficient

of the CAT variable is zero in the full model.

Under this null hypothesis, the model will

reduce to a model without the CAT variable in

it. Because we have provided neither a printout

for this reduced model nor the corresponding

log likelihood statistic, we cannot carry out the

likelihood ratio test here.

To carry out theWald testfor the significance

of the CAT variable, we must use the informa-

tion in the row of results provided for the CAT

variable. The Wald statistic is given by the esti-

mated coefficient divided by its standard error;

from the results, the estimated coefficient is

0.5978 and the standard error is 0.3520.

Dividing the first by the second gives us the

value of the Wald statistic, which is aZ, equal

to 1.70. Squaring this statistic, we get the chi-

square statistic equal to 2.88, as shown in the

table of results.

TheP-value of 0.0894 provided next to this chi

square is somewhat misleading. ThisP-value

considers a two-tailed alternative hypothesis,

whereas most epidemiologists are interested

in one-tailed hypotheses when testing for the

significance of an exposure variable. That is,

the usual question of interest is whether the

odds ratio describing the effect of CAT

controlling for the other variables is signifi-

cantlyhigherthan the null value of 1.

To obtain a one-tailed P-value from a two-

tailedP-value, we simply take half of the two-

tailed P-value. Thus, for our example, the

one-tailedP-value is given by 0.0894 divided

by 2, which is 0.0447. Because thisP-value is

less than 0.05, we can conclude, assuming this

model is appropriate, that there is a significant

effect of the CAT variable at the 5% level of

significance.

EXAMPLE (continued)

LR test:

Full model Reduced model

Model A Model A w=o CAT

H 0 : b¼ 0

whereb¼coefficient of CAT in model A

Reduced model (w/o CAT) printout

not provided here

WALD TEST:

Variable Coefficient S.E. Chi sq P

Intercept –6.7747 1.1402 35.30 0.0000

CAT 0.5978 0.3520 2.88 0.0894

AGE 0.0322 0.0152 4.51 0.0337

CHL 0.0088 0.0033 7.19 0.0073

ECG 0.3695 0.2936 1.58 0.2082

SMK 0.8348 0.3052 7.48 0.0062

HPT 0.4392 0.2908 2.28 0.1310

Z¼

0 : 5978

0 : 3520

¼ 1 : 70

Z^2 =CHISQ=2.88

P¼0.0896 misleading

(Assumes two-tailed test)

usual question: OR>1? (one-tailed)

One-tailedP¼

Two-tailedP

2

= = 0.0447

0.0894

2

P<0.05)significant at 5% level

148 5. Statistical Inferences Using Maximum Likelihood Techniques