I have just tried two ways to perform a linear discriminant analysis.
This is the data, divided in two tables (printed from screen using R commander):
They belong respectively to the two already known groups.
I ran the next code to get the coefficients and critic value of the function:
m1=cov.wt(Craneos1)$center m1=data.matrix(m1) m2=cov.wt(Craneos2)$center m2=data.matrix(m2) D=((nrow(Craneos1)-1)*var(Craneos1)+(nrow(Craneos2)-1)*var(Craneos2))/30 d=solve(D)%*%(m1-m2) d # Estos son los coeficientes de la función 0.5*t(m1+m2)%*%d # Este es el punto crítico
I get the next results:
> d # Coefficients of the discriminant analysis [,1] x1 -0.089306662 x2 0.155774683 x3 0.005231617 x4 -0.177194601 x5 -0.177408670 > 0.5*t(m1+m2)%*%d # Este es el punto crítico [,1] [1,] -30.46349
These results are exactly the same as those ones that appeared in a book I am studying.
I reordered the last two tables into only one with the first column as a factor:
In this case, I used the function lda (from the package MASS), since it has been widely used in examples on internet.
I ran the next command:
Which displayed the next results:
> lda(group~.,data=Datos) Call: lda(group ~ ., data = Datos) Prior probabilities of groups: CRANEOS1 CRANEOS2 0.53125 0.46875 Group means: X1 X2 X3 X4 X5 CRANEOS1 174.8235 139.3529 132.0000 69.82353 130.3529 CRANEOS2 185.7333 138.7333 134.7667 76.46667 137.5000 Coefficients of linear discriminants: LD1 X1 0.047726591 X2 -0.083247929 X3 -0.002795841 X4 0.094695000 X5 0.094809401
My questions are:
Is there anything wrong in the code displayed above?
Why are the both coefficients different depending on the mode used?
Which of them must be more reliable?
How can I get the critic value using the Mode 2?