A company manager wishes to test a union leader's claim that
absences occur on the different week days with the same
frequencies.
To test the claim, the following data is collected:
Day |
Mon Tue
Wed Thur Fri
Absences | 37 15 12 23 43
Use this data to test the manager's claim at the 0.05 level of significance;
include all necessary steps for hypothesis testing.
We will do a Chi-square test for this.
H0: Absences occur on the different week days with the same frequencies.
H1: Absences do not occur on the different week days with the same frequencies.
Contingency table | ||||||
Day | Mon | Tue | Wed | Thur | Fri | Total |
Original (Oi) | 37 | 15 | 12 | 23 | 43 | 130 |
Expected (Ei) | 26 | 26 | 26 | 26 | 26 | 130 |
(Oi-Ei)2 | 121 | 121 | 196 | 9 | 289 | 736 |
Now, the degree of freedom is
The test statistics is given by
Then, the test statistics is
As the chi-square test is always a right tailed test, the
critical value of
for 0.05 level of significance is 9.49.
As the test statistics is greater than the critical value at 0.05 level of significance, we have sufficient evidence to reject H0.
We do not have sufficient evidence to conclude that absences occur on the different week days with the same frequencies.
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