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Chi square test for goodness of fit for Absenteeism

The personal manager of a firm is concerned about absenteeism. She decides to sample the records to determine if absenteeism is distributed evenly through the six-day work week. The null hypothesis to be tested is: Absenteeism is distributed evenly throughout the week. The 0.01 level is to be used. The sample results are: (Chi-Square Test)

Days Number absent
Monday 12
Tuesday 9
Wednesday 11
Thursday 10
Friday 9
Saturday 9

The manager wanted to know the answers for the following questions:

1) What is the chi-square critical value at the (1) percent level ?
2) Using the chi-square test of significance. Compute X^.
3) Is the null hypothesis rejected or not, and why.

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Solution Summary

The solution provides step by step method for the calculation of Chi square test for goodness of fit . Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.