# CHI-SQUARE TEST FOR CROSS TABS

## CHI-SQUARE TEST FOR CROSS TABS

Chi-square test for Cross tabs is used to test the association of two variables under study.  The test forms a bivariate cross tabulation of the variables with the frequency for each cell.

CASE ANALYSIS-1

We are attempting to find out if there is any significant relationship between the income and the number of times of using a credit card. The following data set has been prepared from 40 selected people of the different income groups (< 3 lakhs, 3-5 lakhs, 5-7 lakhs, and > 7 lakhs) who are using the credit card in daily life.

The hypotheses for the analysis are:

Null hypothesis-H0: Credit card usage is independent of income.

Alternative Hypothesis- H1: Credit card usage pattern is associated with income.

Coded variable

The codes used for income are:  (Codes: < 3 lakhs =1, 3-5 lakhs= 2, 5-7 lakhs= 3, > 7 lakhs = 4) and the codes for credit card usage pattern are: (Codes: < 2 times = 1, 2-5 times = 3, > 7 times = 3). The following table is treated as the input data.

Table-1: Input Out

Performing the Analysis with SPSS

For SPSS Version 11, click on   Analyze ⇒ Descriptive Statistics ⇒ Crosstabs.  This will bring up the SPSS screen dialogue box as shown below.

After clicking Crosstabs, this will bring up the following SPSS screen dialogue box

Select the row and column variables and move them to Row(s) and Column (s) box.

Select Statistics, click Chi-square and then click Continue to return to the Cross tabs dialogue box.

Click OK of Cross tabs dialogue box to get the output.

SPSS Output

The SPSS outputs of the analysis are depicted in table-2 to table-4

Table-1: Case Processing Summary

Table-2: Count Income * Credit Cross tabulation

Table-3: Chi-Square Tests

a  9 cells (75.0%) have expected count less than 5. The minimum expected count is .90.

From the output Chi-Square = 12.979

DECISION

Reject the null hypothesis if p-value ≤ 0.05

INTERPRETATION

The p-value is 0.043 and it is less than 0.05 (5% level of significance). Therefore we reject the null hypothesis and conclude that the credit card usage pattern is associated with the income.

CASE ANALYSIS-2

A management educational institute is interested to test whether there exists any relationship between the educational back ground and the chance of getting jobs with good salary package after the degree.

The hypotheses for the analysis are:

Null hypothesis-H0: The chance of getting jobs with good package is independent of the educational back ground.

Alternative Hypothesis- H1: The chance of getting jobs with good package is associated with the educational back ground.

Coded variable

Codes for educational background are: Commerce = 1, Engineering = 2, Arts = 3, Science =4

Salary Package codes are: Less than 4 lakhs per annum =1, 4-8 lakhs =2, More than 8 lakhs per annum = 3. The following table is used as the input data for the analysis.

Table-1: Input Out

SPSS Output

Table -2: Case Processing Summary

Table -3: Edn*Salary Cross tabulation

Count

Table -3: Chi-Square Tests

a  12 cells (100.0%) have expected count less than 5. The minimum expected count is 1.20.

From the output Chi-Square = 5.807

DECISION

Reject the null hypothesis if p-value ≤ 0.05

INTERPRETATION

The p-value is 0.445 and it is more than 0.05 (5% level of significance). Therefore we accept the null hypothesis and conclude that the chance of getting jobs with good salary package is independent of the educational back ground.

SPSS Commands

1. Click on ANALYZE at the SPSS menu bar (in older versions of SPSS, click on STATISTICS instead of ANALYZE).
2. Click on DESCRIPTIVE followed by CROSS TABS.
3. Select the row and column variables and move them to ROW(S) and COLUMS(S) box
4. Select SATISTICS, click CHI-SQUARE and then CONTINUE to return the main dialogue box.
5. Now click on the button OK.

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