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I have a survey based on a Likert-scale (1-5) with around 20 variables and 140 cases. Since some people only cross from 1 to 3 for all questions (variables) and others cross 1-5, I want to standardize the values. So I want to calculate the average of all variables of one person to use the result for the standardization (to have a standard deviation of one and a mean of zero). Does anybody know how this standardization is called and how I can do it in SPSS? Thanks a lot!

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I think you need to spell out much more about what you think is happening.

For example, suppose you think some people were confused and thought that questions should be answered on a scale 1-3, rather than 1-5. If that's what you think then you could try recoding 2 to 3 and 3 to 5.

But, but, but:

  1. You would need to be really sure of that interpretation. If there is a report to the effect that some people thought they should use a 3-point scale, then that helps. But perhaps those people really did not want to use 4 and 5. You can see human variability everywhere: on this site, some very active users are generous with up-votes, but others much more sparing.

  2. If I were reviewing this work, I would want to see investigation of the consequences of any adjustment, i.e. analysis without adjustment and analysis with.

  3. You could try separate analyses of the 1..5 people and the 1..3 people.

  4. Yet another take is that if people really misunderstood the protocol, their answers can't be used.

  5. It may be that you can check people answering 1..3 if you have other variables. Perhaps something else singles them out.

Either way, it is not clear that this is anything to do with standardization or anything based on calculating the mean.

I think you need really good reasons not to take people's answers as given at face value.

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  • $\begingroup$ I agree. If anything, I'd put it in even stronger terms. I'd be amazed if there was strong evidence that some people really misunderstood the questions. $\endgroup$
    – Peter Flom
    Commented Oct 9, 2013 at 11:49
  • $\begingroup$ I strengthened my answer by adding more emphasis. $\endgroup$
    – Nick Cox
    Commented Oct 9, 2013 at 11:54

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