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
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.