What is a Chi Square Test?

How to do a Chi-square test for independence in SPSS?

1. Using Chi-Square Statistic in Research
This easy tutorial will show you how to run the Chi-Square test in SPSS, and how to interpret the result.

The chi-square test of independence uses to investigate the relationship between two categorical variables that have two or more categories. In addition, the test compares the proportions, that is, the frequency of cases observed in each of the categories with values that would be expected to have if there is no relationship between the variables.

First of all, the chi-square test based on a contingency table that shows the intersection of each category of one variable with each group of the other variable.

1. How to do a Chi-square test for independence in SPSS?

What is a Chi-square test for independence?
The chi-square test of independence is used to investigate the relationship between two
categorical variables that have two or more categories. The chi-square test compares the
proportions, that is, the frequency of cases observed in each of the categories with values that
would be expected to have if there is no relationship between the variables.
The chi-square test is based on a contingency table that shows the intersection of each
category of one variable with each group of the other variable.
We will test whether education (primary school, secondary school, BA, Master, PhD)
is independent of gender (male, female).
Null hypothesis:
The education level is not independent of gender.
Alternative hypothesis:
The education level is independent of gender.
How to do a Chi-square test for independence in SPSS?
1. From SPSS menu, choose Analyze – Descriptive Statistics – Crosstabs.


2. A new window will open. Transfer variables from the left box to the right boxes. One
variable transfer into box Row(s) and other into box Column(s).
Checkbox Display clustered bar charts.

2. 3. Click on Statistics tab. A new window will appear. Choose Chi-square and Phi and
Cramer's V from Nominal box. Click Continue.

4. Click on Cells… tab and a new window will appear. Check Observed in the Counts
box. Check Row, Column, Total in the Percentages box. Click Continue and then OK.

3. 5. The results of the Chi-square test for independence will appear in the output box.

4. How to report a Chi-square test for independence results?

1. The first table is Case processing Summary. It shows the number of valid, missing, and
total cases.

2. The second table is Crosstabulation. It shows the number of cases in each group.
For example, there is 11 male finished primary school, 11 male finished secondary
school, 13 male finished BA, 8 male finished Master, and 7 male finished PhD. There
is a female finished primary school, 15 females finished secondary school, 21 female
finished BA, 10 female completed Master, and 10 female finished PhD. There are 50
males and 66 females in the sample. 21 respondents finished primary school, 26
finished secondary school, 34 finished BA, 18 completed Master, and 17 finished PhD.
The total number of respondents is 116.

3. The third table shows the Chi-square test statistic and its significance.
4.
If p-value (Asymp. Sig. (2-sided)) for Pearson Chi-square statistic is higher than .05 (p
> .05), the education is not dependent of gender (we fail to reject the null hypothesis).

If p-value (Asymp. Sig. (2-sided)) for Pearson Chi-square statistic is lower than .05 (p
< .05), the education is dependent on gender (we reject the null hypothesis).

5. Since the results show that p = .892 > .05, we fail to reject the null hypothesis.

5. Following table shows the results of Cramer's V and Phi measures. If Cramer's V/Phi
is 0.10, there is a small impact, 0.30 medium, and 0.50 high.

6. In the end, there are clustered bar chart.

6. We may report the results of the Chi-square test for independence in the following way:

A chi-square test for independence was computed to determine whether
education (primary school, secondary school, BA, Master, PhD) is independent of
gender (male, female). The results are not significant, χ
2
(4) = 1.111, p = .892, Cramer's
V/phi = .892. We fail to reject the null hypothesis that education is the same across
gender (male, female) and conclude that education is not related to gender.