So let’s say you had some data one two nominal variables and wanted to test the independence between the two variables. However, instead of having the data on an individuallevel scale, you had it in the form like what is shown below—in a crosstabular form. You could compute the chi square test statistic by hand, but this can be cumbersome. However, there is a very simple way in SPSS to conduct the chi square test using only the numbers shown below.
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Group
 
Fruit chosen

Group 1

Group 2

Group 3

Apple

10

7

22

Orange

8

19

9

Grape

17

13

12

So with this data, we have three groups (1, 2, and 3), and three fruits (apple, orange, and grape). The first thing to do to conduct the chi square test of independence in SPSS would be to set up the two grouping variables. So in SPSS, one variable will be “Apple, Apple, Apple, Orange, Orange, Orange, Grape, Grape, Grape” while the other variable will be “1, 2, 3, 1, 2, 3, 1, 2, 3” as shown below.
By doing so, we have taken care of each combination of fruit chosen and group. Now the frequencies for each group combination are needed. So the first combination is “Apple, Group 1,” which has a frequency of 10. So simply input “10” for the first item in a new variable. Continue to do this until each fruit chosen and group combination has the correct frequency, as shown below.
Fruit * Group Crosstabulation
 
Count
 
Group

Total
 
1.00

2.00

3.00
 
Fruit

Apple

10

7

22

39

Grape

17

13

12

42
 
Orange

8

19

9

36
 
Total

35

39

43

117

ChiSquare Tests
 
Value

df

Asymp. Sig. (2sided)
 
Pearson ChiSquare

15.659^{a}

4

.004

Likelihood Ratio

15.184

4

.004

N of Valid Cases

117
 
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 10.77.
