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Date:   Wed, 10 Sep 2008 10:11:18 +0200
Reply-To:   Spousta Jan <JSpousta@CSAS.CZ>
Sender:   "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
From:   Spousta Jan <JSpousta@CSAS.CZ>
Subject:   Re: Selecting the top 3% of Service Users
Comments:   To: "Secrist, Kevin" <KSecrist@buttecounty.net>
In-Reply-To:   A<04B618C2F0D4574098D8B3BDDC1B3B2F020C8477@bc-mail-03.bci.buttecounty.net>
Content-Type:   text/plain; charset="iso-8859-2"

Hi Kevin,

Try this (the first part is only for testing purposes).

* create a fake dataset. INPUT PROGRAM. - loop ident = 1 to 1000. - compute service_minutes = uniform(100). - end case. - end loop. - END FILE. END INPUT PROGRAM. form ident (f4). execute.

**** the solution ****. * create percentiles. RANK service_minutes (D) /NTILES (100) into percentiles . * select top 3%. select if percentiles <= 3. execute.

Best regards,

Jan

-----Original Message----- From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of Secrist, Kevin Sent: Wednesday, September 10, 2008 12:18 AM To: SPSSX-L@LISTSERV.UGA.EDU Subject: Selecting the top 3% of Service Users

Hi SPSS Guru's,

I have a question that is probably very simple, but here it goes. I need to select the top 3% of service users based on service minutes in a mental health department. The Population is all clients. What would the code for this be?

Thanks

Kevin Secrist, Administrative Analyst, Associate

Butte County Department of Behavioral Health

Administrative Services Division

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