Solved – quick way to convert z-scores into percentile scores

change-scoresquantilesr

Does anyone know of a function or R package that can help me turn z scores into percentile scores?
The final goal is to classify or rank a group of respondents into four categories based on the heights of their z scores (20% lowest scores, 30%, 30%, 20% highest scores).

Any ideas? Many thanks in advance!

Best Answer

pnorm(z) will do it.

> pnorm(1.96)
[1] 0.9750021
> pnorm(0)
[1] 0.5
> pnorm(-1)
[1] 0.1586553

Or if you insist on a percentile, boom. Then try

round(pnorm(1.96)*100,0)

Related Solutions

Solved – How to convert percentile rank into z score in R

You would apply the inverse cdf to the percentile ranks to convert them to quantiles, so if you want standard normals, $z=\Phi^{-1}(p)$ should do what you seem to be asking for.

However, this transformation won't of itself make an ordinal scale into an interval scale.

In R, you would do this as:

z <- qnorm(p)

Solved – Plotting error ellipsoid from 3×3 covariance matrix in R

I'm interested in this too, so I studied a bit and came up with an incomplete answer. This will produce an ellipsoid and mark its radii, but I haven't figured out how to rotate it using the eigenvectors.

Loading the rgl package and running demo(shapes3d) will give you this ellipsoid3d function:

ellipsoid3d <- function(rx=1,ry=1,rz=1,n=30,ctr=c(0,0,0),
                         qmesh=FALSE,
                         trans = par3d("userMatrix"),...) {
   if (missing(trans) && !rgl.cur()) trans <- diag(4)
   degvec <- seq(0,pi,length=n)
   ecoord2 <- function(p) {
     c(rx*cos(p[1])*sin(p[2]),ry*sin(p[1])*sin(p[2]),rz*cos(p[2])) }
   v <- apply(expand.grid(2*degvec,degvec),1,ecoord2)
   if (qmesh) v <- rbind(v,rep(1,ncol(v))) ## homogeneous
   e <- expand.grid(1:(n-1),1:n)
   i1 <- apply(e,1,function(z)z[1]+n*(z[2]-1))
       i2 <- i1+1
   i3 <- (i1+n-1) %% n^2 + 1
   i4 <- (i2+n-1) %% n^2 + 1
   i <- rbind(i1,i2,i4,i3)
   if (!qmesh)
     quads3d(v[1,i],v[2,i],v[3,i],...)
   else return(rotate3d(qmesh3d(v,i,material=list(...)),matrix=trans))
 }

For demonstration, I'll use this covariance matrix: $$\left(\begin{array} {cc}2.2&.4&.2\\.4&2.8&.3\\.2&.3&2.4\end{array}\right)$$

Covariance.Matrix=matrix(c(2.2,.4,.2,.4,2.8,.3,.2,.3,2.4),3)
Eigenvalue.Roots=eigen(Covariance.Matrix)$values^.5->X

require(rgl)
wire3d(ellipsoid3d(X[1],X[2],X[3],qmesh=T),col='lightgreen')
axes3d("x",pos=c(0,0,0),at=c(0,X[1]),labels='',lwd=2,col='maroon')
axes3d("y",pos=c(0,0,0),at=c(0,X[2]),labels='',lwd=2,col='maroon')
axes3d("z",pos=c(0,0,0),at=c(0,X[3]),labels='',lwd=2,col='maroon')

Rotating the ellipsoid3d object is possible by using rotate3d() inside wire3d(), but I'm not sure how to find the angles $\phi,\theta,\psi$ from the eigenvectors. This is my best guess so far (based on this):

Phi=with(eigen(Covariance.Matrix),atan(vectors[3,2]/vectors[3,1]))
Theta=acos(eigen(Covariance.Matrix)$vectors[3,3])
Psi=with(eigen(Covariance.Matrix),atan(vectors[2,3]/-vectors[1,3]))

If these are known, the euler function given here may help produce a matrix for rotate3d's matrix argument...but I'm not sure it works. With sample data, I seem to get a different rotation using John Fox's scatter3d function. If you have a dataset of observation trios, not just a covariance matrix, scatter3d seems to work well.

Best Answer

Related Solutions

Solved – How to convert percentile rank into z score in R

Solved – Plotting error ellipsoid from 3×3 covariance matrix in R

Related Question