There is not a meaningful alternative way since the way you propose is not 
meaningful.  The Wald tests have some know problems even in the well defined 
cases.  Both types of tests are designed to test a predefined hypothesis, not a 
conditional hypothesis on the stepwise procedure.  It is best to use other 
approaches than stepwise selection (it has been shown to give biased results) 
such as the lasso.  If you need to use stepwise, then you should bootstrap the 
entire selection process to get better estimates/standard errors.  

Frank Harrell's book and package go into more detail on this and provide some 
tools to help (as well as the other packages that can be used).

Hope this helps,

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Peter-Heinz Fox
> Sent: Tuesday, May 05, 2009 8:02 AM
> To: r-help@r-project.org
> Subject: [R] Stepwise logistic regression with significance testing -
> stepAIC
> 
> Hello R-Users,
> 
> I have one binary dependent variable and a set of independent variables
> (glm(formula,…,family=”binomial”) ) and I am using the function stepAIC
> (“MASS”) for choosing an optimal model. However I am not sure if
> stepAIC considers significance properties like Likelihood ratio test
> and Wald test (see example below).
> 
> > y <- rbinom(30,1,0.4)
> > x1 <- rnorm(30)
> > x2 <- rnorm(30)
> > x3 <- rnorm(30)
> > xdata <- data.frame(x1,x2,x3)
> >
> > fit1 <- glm(y~ . ,family="binomial",data=xdata)
> > stepAIC(fit1,trace=FALSE)
> 
> Call:  glm(formula = y ~ x3, family = "binomial", data = xdata)
> 
> Coefficients:
> (Intercept)           x3
>     -0.3556       0.8404
> 
> Degrees of Freedom: 29 Total (i.e. Null);  28 Residual
> Null Deviance:      40.38
> Residual Deviance: 37.86        AIC: 41.86
> >
> > fit <- glm( stepAIC(fit1,trace=FALSE)$formula  ,family="binomial")
> > my.summ <- summary(fit)
> > # Wald Test
> > print(my.summ$coeff[,4])
> (Intercept)          x3
>   0.3609638   0.1395215
> >
> > my.anova <- anova(fit,test="Chisq")
> > #LR Test
> > print(my.anova$P[2])
> [1] 0.1121783
> >
> 
> Is there an alternative function or a possible way of checking if the
> added variable and the new model are significant within the regression
> steps?
> 
> Thanks in advance for your help
> 
> Regards
> 
> Peter-Heinz Fox
> 
> 
> 
> 
>       [[alternative HTML version deleted]]

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