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.
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Rick Marcantonio
Quality Assurance
IBM
IL
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Original Message:
Sent: Wed November 18, 2020 06:28 PM
From: Tom Dillon
Subject: Chi-squared test of independence?
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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#SPSSStatistics