Fit and compare hierarchical loglinear models for complex survey data.
survey design object
loglinear model from
p-value approximation: see Details
Compute the exact asymptotic p-value (slow)?
Report the intercept?
The loglinear model is fitted to a multiway table with probabilities
svymean and with the sample size equal to the
observed sample size, treating the resulting table as if it came from iid
multinomial sampling, as described by Rao and Scott. The
variance-covariance matrix does not include the intercept term, and so
by default neither does the
coef method. A Newton-Raphson
algorithm is used, rather than iterative proportional fitting, so
starting values are not needed.
anova method computes the quantities that would be the score
(Pearson) and likelihood ratio chi-squared statistics if the data were
an iid sample. It computes four p-values for each of these, based on the
exact asymptotic distribution (see
saddlepoint approximateion to this distribution, a scaled
chi-squared distribution, and a scaled F-distribution. When testing the
two-way interaction model against the main-effects model in a two-way
table the score statistic and p-values match the Rao-Scott tests
anova method can only compare two models if they are for
exactly the same multiway table (same variables and same order). The
update method will help with this. It is also much faster to use
svyloglin for a large data set: its time
complexity depends only on the size of the model, not on the size of the
It is not possible to fit a model using a variable created inline, eg
I(x<10), since the multiway table is based on all variables used
in the formula.
Object of class
Rao, JNK, Scott, AJ (1984) "On Chi-squared Tests For Multiway Contingency Tables with Proportions Estimated From Survey Data" Annals of Statistics 12:46-60.
data(api) dclus1<-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) a<-svyloglin(~stype+comp.imp,dclus1) b<-update(a,~.^2) an<-anova(a,b) an #> Analysis of Deviance Table #> Model 1: y ~ stype + comp.imp #> Model 2: y ~ stype + comp.imp + stype:comp.imp #> Deviance= 8.376767 p= 0.0676402 #> Score= 9.013874 p= 0.05738682 print(an, pval="saddlepoint") #> Analysis of Deviance Table #> Model 1: y ~ stype + comp.imp #> Model 2: y ~ stype + comp.imp + stype:comp.imp #> Deviance= 8.376767 p= 0.05376131 #> Score= 9.013874 p= 0.05547346 ## Wald test regTermTest(b, ~stype:comp.imp) #> Wald test for stype:comp.imp #> in update(a, ~.^2) #> F = 6.382223 on 2 and 13 df: p= 0.011722 ## linear-by-linear association d<-update(a,~.+as.numeric(stype):as.numeric(comp.imp)) an1<-anova(a,d) an1 #> Analysis of Deviance Table #> Model 1: y ~ stype + comp.imp #> Model 2: y ~ stype + comp.imp + as.numeric(stype):as.numeric(comp.imp) #> Deviance= 6.739685 p= 0.03853491 #> Score= 7.293848 p= 0.03089338