Thank you very much Peter - very enlightening to do it from first principles!
Regards,
Paul
On 4 October 2016 at 10:23:52 am, peter dalgaard (pda...@gmail.com) wrote:
> On 04 Oct 2016, at 00:30 , Paul Sanfilippo <prs...@gmail.com> wrote:
>
> Hi,
>
> I am trying to replicate a test in the Hosmer - Applied Logistic regression
> text (pp 289, 3rd ed) that uses a Multivariable Wald test to test the
> equality of coefficients across the 2 logits of a 3 category response
> multinomial model. I’d like to see whether (from a statistical standpoint) it
> is acceptable to collapse the 2 response categories and then simply use a
> binary logistic regression. The idea is that if the coefficients across the 2
> logits are similar (non-significant p value with Wald test), then it is
> reasonable to pool the categories.
>
> There does not seem to be a built in way to do this in R?
>
> Using the mtcars dataset as an example (for the sake of the example, using
> cyl as a 3-factor response), does anyone have any ideas how to do this
>
> library(nnet)
> data(mtcars)
> mtcars$cyl <- as.factor(mtcars$cyl)
> mtcars$am <- as.factor(mtcars$am)
> mod <- multinom(cyl ~ am + hp, data=mtcars)
> summary(mod)
>
>> summary(mod)
> Call:
> multinom(formula = cyl ~ am + hp, data = mtcars)
>
> Coefficients:
> (Intercept) am1 hp
> 6 -42.03847 -3.77398 0.4147498
> 8 -92.30944 -26.27554 0.7836576
>
> So, I want to simultaneously test whether the 3 coefficients across the 2
> logits are similar.
R and R-packages do not always produce every single test that someone have
thought up. Sometimes, you just get the tools to roll your own, and are
expected to know enough theory to do so.
The generic technique for a Wald test would be to
(a) extract the variance-covariance matrix (V) of the coefficients (beta)
(b) write the linear relation that you wish to test in matrix form A beta = 0
(c) the variance-covariance matrix of A beta will be A V A'
(d) the Wald test is (A beta)' (A V A')^{-1} (A beta)
For the case of multinom(), a little extra care is needed because it presents
the coefficients as a matrix, and the variance covariance matrix is ordered as
if the coefficients were organized as a vector _by row_:
> vcov(mod)
6:(Intercept) 6:am1 6:hp 8:(Intercept) 8:am1
6:(Intercept) 771.682250 69.0782649 -7.61945359 168.212743 -463.676388
6:am1 69.078265 10.6015542 -0.71221686 13.069175 -38.850954
6:hp -7.619454 -0.7122169 0.07550636 -1.647338 4.560144
8:(Intercept) 168.212743 13.0691754 -1.64733776 1019.860307 -1473.837691
8:am1 -463.676388 -38.8509537 4.56014436 -1473.837691 2195.306673
8:hp -3.169803 -0.2992327 0.03147124 -7.719801 11.719345
8:hp
6:(Intercept) -3.16980299
6:am1 -0.29923273
6:hp 0.03147124
8:(Intercept) -7.71980097
8:am1 11.71934471
8:hp 0.06548745
> coef(mod)
(Intercept) am1 hp
6 -42.03847 -3.77398 0.4147498
8 -92.30944 -26.27554 0.7836576
So the thing to do would be roughly like this (the code can surely be
improved):
> beta <- as.vector(t(coef(mod)))
> A <- rbind(c(1,0,0,-1,0,0), c(0,1,0,0,-1,0), c(0,0,1,0,0,-1))
> A
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 0 0 -1 0 0
[2,] 0 1 0 0 -1 0
[3,] 0 0 1 0 0 -1
> A %*% beta
[,1]
[1,] 50.2709610
[2,] 22.5015565
[3,] -0.3689079
> t(A %*% beta) %*% solve(A %*% vcov(mod) %*% t(A), A %*% beta)
[,1]
[1,] 3.592326
And then get the p-value from the asymptotic chi-square on 3-df
> pchisq(3.59, 3, lower=FALSE)
[1] 0.3092756
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd....@cbs.dk Priv: pda...@gmail.com
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