BookmarkSubscribeRSS Feed
Kodzo
Calcite | Level 5

I have the R program bellow to generate a funnel plot.  I can use dummy_facid N OE ration to generate the plot.

My question is how can I change the program for variance and dispersion to have the normal approximation instead of negative binomial?

Can someone help???

Thanks

Emmanuel

#############################################################

## Funnel Plots

## -------------------------------

## First is the funnel.plot function

## Last is the code putting the funnel plots in a PDF

funnel.plot = function(nvar, response,id=numeric(1),xlab="",ylab="",main="",pch=1)

{

ns = nvar        ## This sets the x-axis

mu = 0.04385      ## mu value doesn't really matter if we adjust for

            ## dispersion in variance calculations.

#######

## Calculate the dispersion of the data where

## we expect responses to be centered around 1

vars = 1/((ns*mu)^2) * ns*mu*(1-mu)

dispersion = sum((response-1)^2/vars,na.rm=T)/(length(vars))

gensd = function(n)    ## This function is called when we generate the confidence bands

{

    sqrt(dispersion*(1/(n*mu))^2 * n * mu * (1-mu))

}

plot(ns,response,

    xlab=xlab,

    ylab=ylab, ylim=c(-1.5,5.5),main=main,pch=pch,cex=0.8)             ## Plot the data

xgrid.min = min(ns,na.rm=T)

xgrid.max = max(ns,na.rm=T)

xgrid.spc = (xgrid.max-xgrid.min)/100

xgrid = seq(max(c(xgrid.min-xgrid.spc*10,1),na.rm=T),

        xgrid.max+xgrid.spc*10,

        xgrid.spc)                ## Generate an appropriate x grid

                            ## for confidence lines

yu = 1+qnorm(0.95)*gensd(xgrid)        ## 90% CI lines centered at 1

yl = 1-qnorm(0.95)*gensd(xgrid)

lines(xgrid,yu,lty=5)

lines(xgrid,yl,lty=5)

yu = 1+qnorm(0.975)*gensd(xgrid)        ## 95% CI lines centered at 1

yl = 1-qnorm(0.975)*gensd(xgrid)

lines(xgrid,yu,lty=3)

lines(xgrid,yl,lty=3)

yu = 1+qnorm(0.995)*gensd(xgrid)        ## 99% CI lines centered at 1

yl = 1-qnorm(0.995)*gensd(xgrid)

lines(xgrid,yu,lty=2)

lines(xgrid,yl,lty=2)

abline(h=1.0,lty=2)

legend("topright", inset=.005, title="Confidence Lines",

     c("90%","95%","99%"), lty=c(5,3,2), horiz=TRUE)  ## Add legend

#rect(10, -0.15, 25, 0.15, border="green",lwd=1.5,lty="dashed") # coloured zoom-in box

####################### Label Outliers

#offset = (xgrid.max - xgrid.min)/45

offset = 0.0

pos.outlier = (response > (1+qnorm(0.95)*gensd(ns)))

text(ns[pos.outlier]+offset,response[pos.outlier],label=id[pos.outlier],font=2,cex=0.8,adj=c(-0.3,0.2),col=rgb(.8,0,0))

neg.outlier = (response < (1-qnorm(0.95)*gensd(ns)))

text(ns[neg.outlier]+offset,response[neg.outlier],label=id[neg.outlier],font=2,cex=0.8,adj=c(-0.3,0.2),col="blue")

}

setwd ("C:/2012 Analyses/Peds Pilot/Validation")

dta <- read.table("Allmortality.csv",sep=",",header=T)

attach(dta)

pdf("TQIP Ped FunnelPlot.pdf",width=10,height=8)

## N_TQIP_Pt , Overall_OE (TQIP cases and O/E for Overall Mortality)

funnel.plot(N,OE,dummy_facid,

    xlab = "Total Number of TQIP TBI Patients",

    ylab = "Mortality  O/E Ratios",

    main = "Risk Adjusted Mortality: ALL Pediatric Patients")

dev.off()

1 REPLY 1
Rick_SAS
SAS Super FREQ

Why not use SAS?  I wrote a series of four articles on various aspects of computing funnel plots in SAS

I discuss the normal approximation and provide SAS code.

hackathon24-white-horiz.png

The 2025 SAS Hackathon has begun!

It's finally time to hack! Remember to visit the SAS Hacker's Hub regularly for news and updates.

Latest Updates

How to Concatenate Values

Learn how use the CAT functions in SAS to join values from multiple variables into a single value.

Find more tutorials on the SAS Users YouTube channel.

SAS Training: Just a Click Away

 Ready to level-up your skills? Choose your own adventure.

Browse our catalog!

Discussion stats
  • 1 reply
  • 638 views
  • 0 likes
  • 2 in conversation