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.

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.
Once this is done, simply go to Data > Weight
Cases… Select Weight cases by, and use the new “Frequency” variable.
Next, to actually perform the chi square, simply go to Analyze > Descriptive
statistics > Crosstabs. Put the Fruit in Row(s): and the
Group in Column(s):. Then in the Statistics… dialogue box, check
the Chisquare box. Hit Continue and then OK. The results of the chi square show the same
crosstabulation box as we had above as well as the actual chi square
statistic. By doing it in SPSS instead of by hand or using Excel, we can get
the test statistic information we’re looking for much faster.
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.
