Question

27) A company manager wishes to test a union leader's claim that absences occur on the different week days with the same frequencies. Use a significance level of 0.01 to test the manager's claim. In a study of absences, 16 occurred on a Monday, 26 occurred on a Tuesday, 38 occurred on a Wednesday, 33 occurred on a Thursday, and 12 occurred on a Friday. Use the p-value method of hypothesis testing.

Answer 1

Given:

A company manager wishes to test a union leader's claim that absences occur on the different week days with the same frequencies.

Days	Monday	Tuesday	Wednesday	Thrusday	Friday
Absence	16	26	38	33	12

Hypothesis test:

H0 : p1 = 1/5, p2 = 1/5, p3 = 1/5, p4 = 1/5, p5 = 1/5

Ha : At least one of the probability not equal to hypothesized value.

Table for calculating test statistics:

Days, x	P(x)	Oi	Ei = n×P(x)	(Oi-Ei)2/Ei
Monday	1/5	16	125×1/5 = 25	3.24
Tuesday	1/5	26	25	0.04
Wednesday	1/5	38	25	6.76
Thursday	1/5	33	25	2.56
Friday	1/5	12	25	6.76
Total	1	125	125	(Oi-Ei)2/Ei =19.36

Test statistics is

2 = (Oi-Ei)2/Ei = 19.36

Degree of freedom, df = 5-1 = 4

P-value corresponding to chi square test statistic 19.36 with degree of freedom = 4 is 0.000668

So P-value = 0.0007

Since P - value is less than significance level 0.01 reject null hypothesis.

Conclusion: Reject H0. There is sufficient evidence to conclude that absences occur on the different week days with the same frequencies.