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st: RE: Mean test in a Likert Scale


From   "David Radwin" <dradwin@mprinc.com>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: RE: Mean test in a Likert Scale
Date   Thu, 30 Aug 2012 16:13:22 -0700 (PDT)

Leonor,

No, you can't correctly calculate the mean of an ordinal-level measure
like the Likert scale you describe, although plenty of people do it
anyway. 

But you can use -ttest- with these data if you first collapse each
variable to a dichotomous (dummy) variable, because the mean of a
dichotomous variable is identical to the proportion where the value is 1.
As a guess, you might set the highest two values to 1, the lowest two
values to 0, and the middle value to missing to calculate the proportion
agreeing or somewhat agreeing.

David
--
David Radwin
Senior Research Associate
MPR Associates, Inc.
2150 Shattuck Ave., Suite 800
Berkeley, CA 94704
Phone: 510-849-4942
Fax: 510-849-0794

www.mprinc.com


> -----Original Message-----
> From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-
> statalist@hsphsun2.harvard.edu] On Behalf Of Leonor Saravia
> Sent: Thursday, August 30, 2012 3:23 PM
> To: statalist@hsphsun2.harvard.edu
> Subject: st: Mean test in a Likert Scale
> 
> Hello,
> 
> I'm working with a survey that presents 26 questions and each of them
> has as possible answer a 5 point Likert scale from Desagree (1) to
> Agree (5). This survey was applyed for a treatment and a control
> group.
> 
> As far as I know, it is possible to analyze the information given only
> by the proportions of each answer; for instance, 25% agrees, 50%
> desagree, or so.
> 
> I have two questions that maybe one of you have had before:
> 
> a) Is it possible to calculate the mean score of a sample (treatment
> or control group) - adding the individual answers - when one is
> working with a Likert scale?
> 
> b) If it is possible to calculate a mean score of a sample when using
> a Likert scale, to compare the answers of the treatment versus the
> control group, is it well done if I use the 'ttest' command?
> 
> 
> Thank you in advance.
> 
> Best regards,
> 
> Leonor
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