Question

A local company is concerned about the number of days missed by its employees due to illness. A random sample of 10 employees is selected. The number of days absent in one year is listed below. An incentive program is offered in an attempt to decrease the number of days absent. The number of days absent in one year after the incentive program is listed below. Test the claim that the incentive program cuts down on the number of days missed by employees. Use α = 0.025. (Assume that the distribution is normally distributed but be sure to verify the remaining requirements for the test.) Use the P-value Approach for all hypothesis tests. Assume that all samples are randomly obtained.

Employee	A	B	C	D	E	F	G	H	I	J
Days absent before incentive	3	8	7	2	9	4	2	0	7	5
Days absent after incentive	1	7	7	0	8	2	0	1	5	5

Answer 1

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	3	1	2
	8	7	1
	7	7	0
	2	0	2
	9	8	1
	4	2	2
	2	0	2
	0	1	-1
	7	5	2
	5	5	0
Average	4.7	3.6	1.1
St. Dev.	2.983	3.134	1.101
n	10	10	10

For the score differences we have

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ > 0

This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.025, and the degrees of freedom are df=9.

Hence, it is found that the critical value for this right-tailed test is tc​=2.262, for α=0.025 and df=9.

The rejection region for this right-tailed test is R=t:t>2.262.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

The p value = 0.0058

(4) Decision about the null hypothesis

Since it is observed that t=3.161>tc​=2.262, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0058, and since p=0.0058<0.025, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is greater than μ2​, at the 0.025 significance level. So looks like the incentive program worked.

Graphically

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