Example of Cross Tabulation and Chi-Square

Interpret the results

The engineer uses the Pearson test and the Likelihood-Ratio test to determine whether an association between machine and shift exists. Because the p-values for the Pearson test and the Likelihood-Ratio test are less than 0.05, the engineer rejects the null hypothesis and concludes that there is an association between the variables of machine and shift.

In these results, there were a total of 408 defective umbrella handles. 143 defective handles were made by Machine 1, 155 defective handles were made by Machine 2, and 110 defective handles were made by Machine 3. Also, more of the defective handles were made during the first shift than during the other shifts. A total of 160 defective handles were made on the first shift, 134 defective handles were made on the second shift, and 114 defective handles were made on the third shift.

In each cell, Minitab displays the actual count, the expected count, and the standardized residual, which indicates the magnitude and direction of the difference between the actual and expected counts. For instance, from Machine 3, during the 3rd shift, 34 defective handles were made, and 30.74 were expected. The small positive residual indicates that the actual and expected counts are fairly close. But from Machine 2, during the 3rd shift, 32 defective handles were made, and 43.31 were expected. The larger negative residual indicates that less defective handles were produced than expected.