The c^{2} test can be used to determine whether a difference between 2 categorical variables in a sample is likely to reflect a real difference between these 2 variables in the population.
Note: in the case of 2 variables being compared, the test can also be interpreted as determining if there is an association (or relationship) between the two variables.
The sample data is used to calculate a single number (or test statistic), the size of which reflects the probability (pvalue) that the observed difference between the 2 variables has occurred by chance, ie due to sampling error.
Worked example
The maternity wards of two hospitals had different preparation for childbirth schemes. A study of mothers who had participated in the schemes asked them to assess their satisfaction with the scheme with the following results:
(Observed counts) 
Hospital


A

B

Total

Very satisfied 
38

72

110

Satisfied 
33

57

90

Neutral 
42

38

80

Dissatisfied a little 
26

44

70

Dissatisfied a lot 
11

29

40

Total 
200

240

440

To answer the question 'is there any evidence of a difference in the satisfaction of the mothers between the two schemes at the two hospitals?', the chisquare test is used.
Suitable null and alternative hypotheses might be:
 H_{0}: There is no difference in satisfaction of the mothers between the two schemes, and
 H_{1}: There is a difference in satisfaction of the mothers between the two schemes.
To perform a chisquared test, the number of mothers expected in each cell of the table if the null hypothesis is true, is calculated.
Calculations
The following calculations are for demonstration and, hopefully, to aid understanding– a computer package will do the appropiate calculations.
The expected numbers (under the null hypothesis) in each cell are equal to
Thus for the very satisfied/hospital A cell the expected number is
To calculate the chisquared (?2) statistic the value of
needs to be calculated for each cell in the table. For the very satisfied/hospital A cell this is
The chisquare statistic is calculated to be total of these values
(Expected counts) 
Hospital


A

B

Total

Very satisfied 
50.0

60.0

110

Satisfied 
63.6

76.4

90

Neutral 
36.4

43.6

80

Dissatisfied a little 
31.8

38.2

70

Dissatisfied a lot 
18.2

21.8

40

Total 
200

240

440

From these expected and the observed values the chisquared teststatistic is computed, and the resulting pvalue is examined.
Computer Output
Chisquared test in Minitab
Data should be entered in 2 columns, then select
Stat > Tables > Cross Tabulation… > ChiSquare Test
Alternatively, if the values in the contingency table have already been calculated, select
Stat>Tables>ChiSquare Test
ChiSquare Test: red, yellow, green, blue
(1 refers to Introverts, 2 refers to Extroverts)
Note: Interpret 0.000 as p < 0.001
Chisquared test in SPSS
Data should be entered in 2 columns, then select
Analyze >Descriptive Statistics>Crosstabs
SPSS can only be used for raw data
Some choices need to be made from the Statistics and Cells buttons in the dialogue box, to get the chisquared test results, and to get the expected frequencies, as shown in the output below. Initially, only the 'Pearson ChiSquare' line needs to be investigated.
Note: The pvalue is printed as .000
This should be interpreted as p< 0.001, and not be taken as exactly 0
Results
The chisquared test statistic is 24.84 with an associated p < 0.001.
Note: .000 should not be interpreted as exactly zero, as in the computer printout.
The null hypothesis is rejected, since p < 0.001, and a conclusion is made that there is a difference in satisfaction of the mothers between the two schemes. Examining the pattern of numbers it is noted that more mothers were satisfied with the scheme at hospital A than with the scheme at hospital B.
A chart illustrates the pattern of responses well.
Bar chart to compare satisfaction responses from mothers in hospitals A and B
Note: If more than one of the expected frequencies is less than 5 (in small tables), or if more than 20% are less than 5 in large tables, cells should be pooled to reduced the number of expected frequencies that are less than 5.
Note: Yates correction and Fisher's exact tests for 2x2 contingency tables are also used.