I think you're referring to a likelihood ratio test.
Could it happen that there were differences between the categories of an IV, but the overall test was not significant?
The null hypothesis of the LRT is that all coefficients for the categorical variable are 0, with the alternative being that at least one coefficient is not 0.
I suppose it could be the case that you could fail to reject the null of the LRT and yet find differences between categories. Those two things aren't mutually exclusive.
What additional information does the second approach really provide? If there is an overall significance, would not be there differences between some of the categories?
Evaluating the statistical significance of the categorical variables via looking at their p-values does not tell us about the categorical variable as a whole, only about the single coefficient's statistical significance.
Here is an example in R
N = 100
cat = factor(sample(1:5, N, replace = T))
x = rnorm(N)
eta = model.matrix(~x+cat)%*%c(1,2,0,0,0,0)
p = 1/(1+exp(-eta))
y = rbinom(length(p),1,p)
model = glm(y~x+cat, family = binomial())
glm(formula = y ~ x + cat, family = binomial())
Min 1Q Median 3Q Max
-2.0345 -0.7243 0.2921 0.6635 1.8355
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.6631 0.5648 1.174 0.2404
x 1.9981 0.4647 4.299 1.71e-05 ***
cat2 0.8766 0.8555 1.025 0.3056
cat3 0.3210 0.8327 0.386 0.6998
cat4 1.1713 0.8468 1.383 0.1666
cat5 1.8251 0.8722 2.093 0.0364 *
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 122.173 on 99 degrees of freedom
Residual deviance: 86.275 on 94 degrees of freedom
Number of Fisher Scoring iterations: 5
A priori, we know that the categories have no effect on the outcome, and yet
cat5 comes out as significant. So if we did not have access to the true data generating mechanism, we may be tempted to say that the
cat variable has an impact on the outcome.
But, that would be erroneous, since we are basing our decision on only one category of the variable. To determine if a model with the
cat variable does better than a model without the
cat variable, we can do a likelihood ratio test.
model0 = glm(y~x, family = binomial())
anova(model0,model, test = 'LRT')
Analysis of Deviance Table
Model 1: y ~ x
Model 2: y ~ x + cat
Resid. Df Resid. Dev Df Deviance Pr(>Chi)
1 98 92.167
2 94 86.275 4 5.8923 0.2073
We fail to reject the null from this test. That means that from our data, we can not say that at least one of the coefficients from the
cat variable is 0. And that would be correct. That the
cat5 variable is significant is just an artifact of sampling and random error.