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Chi-squared test of independence?

  • 1.  Chi-squared test of independence?

    Posted Thu November 19, 2020 03:24 PM

    Hello,

    I am running an experiment on aggression. I have a control and an experimental group. I have administered a 5 point Likert scale to each group. I am interested in their answers, not individual responses, just higher values indicating greater aggression. How can I enter this in SPSS and do the chi-square test of independence based on individuals' overall value within each group? 



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    Tom Dillon
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  • 2.  RE: Chi-squared test of independence?

    Posted Thu November 19, 2020 04:45 PM
    Edited by Rick Marcantonio Thu November 19, 2020 05:00 PM
    Hi, Tom. If I understand correctly, you want to compare the responses for each category of the scale to the control group. SPSS doesn't have a preset way to do that through the menus, but I think this might do what you want.

    *I have created some mock data, assuming 10 subjects in each group.

    data list free /likert_score group.
    begin data.
    2 0
    3 0
    4 0
    3 0
    4 0
    5 0
    4 0
    4 0
    3 0
    2 0
    5 1
    4 1
    4 1
    3 1
    5 1
    5 1
    5 1
    4 1
    3 1
    5 1
    end data.

    * Then I aggregate it to get the counts for each possible value of the scale (here I assume the scale goes 1 through 5).

    AGGREGATE /OUTFILE=* MODE=replace /BREAK=group
     /likert_score_1=CIN(likert_score 1 1)
     /likert_score_2=CIN(likert_score 2 2)
     /likert_score_3=CIN(likert_score 3 3)
     /likert_score_4=CIN(likert_score 4 4)
     /likert_score_5=CIN(likert_score 5 5).

    * We have to transpose that to get back to columns for the test.

    flip variables=likert_score_1 to likert_score_5.

    * Rename the variables for clarity.

    rename variables (var001 var002=control experimental).

    * Now compute chi-square using the control group as the expected counts. If you'd rather,
    * you can go into the menu here and do a chi-square specifying equal cell counts for the
    * "experimental" variable.

    do if (control ne experimental).
    compute ev=((experimental-control)**2)/control.
    else.
    compute ev=0.
    end if.

    * We need the sum of the expected values...

    frequencies variables=ev /format notable /statistics sum.

    * Finally, enter the sum as the first argument below.

    compute chi_prob=sig.chisq(18.58,4).
    execute.

    * The value in the last column is the p-value for the chi-square.

    ------------------------------
    Rick Marcantonio
    Quality Assurance
    IBM
    IL
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  • 3.  RE: Chi-squared test of independence?

    Posted Thu November 19, 2020 07:30 PM
    Wow, that will be helpful in the future so thank you! However, I really just want to look at the overall of the Likert scale. That being the case would I still need to break it into categories, say, 1 strongly disagree 2 disagree, and so forth? Or could I do a variable for scores with the total for each participant and run the chi-square between groups and scores? I apologize if I am not making much sense still in my undergrad. Thanks for any more information or advice you can provide. 



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    Tom Dillon
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