You are right Jamie - a response of 1 was given for most imporant. I will try using your v1=10-v1 and then the method Jens suggests.

thank you very much to both you and Jens for your help

Christine

Christine Roseveare Analyst Regional Public Health 570 9194 027 495 9671

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epidata-list@lists.umanitoba.ca 24/08/2007 1:33 p.m. >>>

An important question is whether a response of 1 most important or least important? If 1 is most important, then make the default value 11 and not zero.

I would just do:

describe v1-v25

and scan the medians to find those with the lowest median (if 1 is most

important and default is 11) or highest median (10 is most important and default is 0). Items that are infrequently ranked will have a median equal to the default. Items most frequently ranked will have a median that is based mostly on ranks.

I haven't thought about the formal statistical analysis here, but you'll get a pretty good assessment.

Jens approach works very well when the most important response is 10. If most important is 1, then you can also do this:

v1 = 10-v1 v2 = 10- v2 etc and use aggregate.

jamie

Christine wrote: I have been asked to set up a database for a questionnaire that has already been written. I was going to set it up in epidata but my problem is how to deal witha question that asks people to rank options. The question gives a list of topics and asks the person filling in the questionnaire to rank them from 1-10 in order of importance.

Jens wrote:

I suggest to do the following: For each question (say 25 different themes) enter the priority the person gave to that issue. setup the qes: v1 item1 ..... ## v2 item2 ..... ## etc v25 text for item 25 ##

in chk make the defaultvalue 0: before file DEFAULTVALUE v1-v25 0 end

Then in analysis after entering the data: (say you had the 25

variables)

read response aggregate /sum="v1,v2,v3,v4,v5" aggregate /sum="v6,v7,v8,v9,v10" etc

• the tables will show the sum of the variables, that is the overall

ranking sum.