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SPSS- How to group a variable according to 1 Standard deviation from the mean?! watch

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    Hi,

    I'm trying to plot a significant interaction my regression analysis has revealed, on a graph. One of my interaction variables is continuous (scale), and so I was advised to group the scores into high and low, with high equating to more than 1 standard deviation larger than the mean, and low being less than 1 SD above the mean.


    I really can't work out how to figure that out on SPSS? Any advice would be greatly appreciated!
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    Can I use z-scores to determine that?!
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    Are you asking how to get SPSS to work out which are above/below 1 SD above the mean?
    Without looking at your data...

    Get the SD
    Create a new variable (to record if it is above/below 1SD)
    - If you have a small amount of data, manually compare each score to the SD and code 2 for above 1SD, code 1 for below 1SD and code 0 for anything below the mean
    - If you have a large amount of data, copy into Excel and use a formula to do this automatically

    That should give you a variable assigning the scores into high, low and other (those below the mean).
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    (Original post by _Sinnie_)
    Are you asking how to get SPSS to work out which are above/below 1 SD above the mean?
    Without looking at your data...

    Get the SD
    Create a new variable (to record if it is above/below 1SD)
    - If you have a small amount of data, manually compare each score to the SD and code 2 for above 1SD, code 1 for below 1SD and code 0 for anything below the mean
    - If you have a large amount of data, copy into Excel and use a formula to do this automatically

    That should give you a variable assigning the scores into high, low and other (those below the mean).
    Thanks, but I think I've found a direct way to do it in SPSS after, if you go onto analyse> descriptive statistics> descriptives, and then select the variable you're interested in, and check the "save standardised values as variables" box, the z-score for each data point then appears on the data view tab.

    I know this is really stupid, but I am right in thinking that the z-score gives a score of how many standard deviations a score is from the mean isn't it?


    Any idea as to why this is more appropriate than just doing a median split?
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    (Original post by Twinpeaks)
    Thanks, but I think I've found a direct way to do it in SPSS after, if you go onto analyse> descriptive statistics> descriptives, and then select the variable you're interested in, and check the "save standardised values as variables" box, the z-score for each data point then appears on the data view tab.

    I know this is really stupid, but I am right in thinking that the z-score gives a score of how many standard deviations a score is from the mean isn't it?


    Any idea as to why this is more appropriate than just doing a median split?
    The problem with these questions is that I remember all the names and words, but not always how they go together. So after some googling.

    Yes, the z score is the number of SD's above or below the mean. You still need a new variable to categorise those which are above 1, those below 1 and those that are negative - so that you can use that variable for the graph.

    With the median split, it depends what you're trying to say about your data. I'm no stats theory expert, but it'd probably come down to whether the categories you're looking at are pre-defined, have definite, measurable or 'known' traits, or whether you're defining the classifications (examples: an IQ of 70 is the official boundary between learning disability and not, disqualified car drivers will always have at least 12 points on their licence, performance on a arbitrary test can be categorised as you please - though you have to rationalise your decisions for each threshold). A median split will assume that your sample splits neatly in the middle, whereas, the majority of drivers have less than 12 points, so the median split will put people into the disqualified category erroneously (if I understand how it is used correctly).

    I think.
 
 
 
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