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I have data from a questionnaire that uses likert scales only. The response options on the scale were regarding frequency, e.g. 'never' 'sometimes' and 'always'.

These responses were coded, with 'never' being given a 1, 'sometimes' being given a 2, and 'always' being given a 3. I understand that this is ordinal data as I used a likert scale.

However, I then summed all of the data up for the questionnaire by adding all of these coded values up for each question. So for example, if the questionnaire had 5 likert scales and someone answered all of them with the response of 'sometimes', then their answers would be coded as 2 and they therefore would have a total score of 10 for the questionnaire. Would this then become interval data or is it still somehow ordinal?

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    $\begingroup$ For "likert" read "Likert" throughout. I think you can defend almost any answer. For example, as differences between ranks are no more than that, there are no grounds for regarding 2 + 2 = 1 + 3, etc. Hence there is no rigorous basis for any arithmetic manipulation and you can't even justify the result as ordinal. In practice, people do this because they hope that the ordinal grades are approximately numerical, so therefore a sum or mean is expected to be a good way of combining information. $\endgroup$ – Nick Cox Nov 2 '15 at 16:49
  • $\begingroup$ Most university systems I've ever heard of do something equivalent with students' grades;meanwhile someone in a psychology department within probably explains at length every year to a small group of students that it is totally indefensible. I vote both ways. It's highly dubious in principle; in practice I would often try it and defend it as pragmatic if the results seemed to make sense. $\endgroup$ – Nick Cox Nov 2 '15 at 16:51
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    $\begingroup$ There is a separate small terminology issue. Calling few-points scales Likert is regarded as unforgivably loose by those who know what he actually said. It's probably too late to reverse that except within certain fields. $\endgroup$ – Nick Cox Nov 2 '15 at 16:55
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Agree with previous posts.

From my perspective, the problem with analyzing Likert scale data is that everybody uses the scale differently, i.e., some people are hard graders and only use 1 to 3, some people are easy graders and only use 3 to 5, and some people are fair graders and use 1 to 5. So a 3 from a hard grader is closer to a 5 from an easy grader than it is to a 3 from an easy grader, etc.

One thing that helps is to "row normalize" the data. What I mean by this is transform the Likert scale into a standard deviation for each respondent. If the standard deviation is zero then remove that respondent from the analysis as they are adding no information.

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    $\begingroup$ Welcome to the site, @BrianGriner. Please don't sign your posts. Your identicon, w/ a link to your userpage, is automatically added to every post for you. $\endgroup$ – gung Nov 2 '15 at 17:10
  • $\begingroup$ Ok. New to submitting answers to stack exchange so I didn't know. Thank you. $\endgroup$ – Brian Griner Nov 2 '15 at 17:12
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    $\begingroup$ No problem. Please use comments to reply, though, not as a new 'answer'. In addition, for me to be notified, you need to use the '\@username' format, as I did above, @BrianGriner. Since you're new here, why not take our tour, which has information for new users? $\endgroup$ – gung Nov 2 '15 at 17:44
  • $\begingroup$ People who give the same grade regardless do not contribute to the variation, but they contribute to the data; and they may differ one from another. I wouldn't want to see an analysis leaving them out without commentary on how different the results are leaving them in. $\endgroup$ – Nick Cox Nov 2 '15 at 21:04
  • $\begingroup$ Agreed. Row standardization should be documented as part of the methodology which includes the sample size before and after applying row standardization. If more than 5% of the sample drop out from bro standardization been one should seriously consider recruiting more respondents to fill the gap. Term of art in market research for these people are "speeders." $\endgroup$ – Brian Griner Nov 3 '15 at 17:22

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