Answer:
The new incentive program cuts down on the number of days missed by employees.
Step-by-step explanation:
In this case we need to test whether the new incentive program cuts down on the number of days missed by employees.
The hypothesis can e defined as follows:
<em>H₀</em>: The new incentive program does not cuts down on the number of days missed by employees, i.e. <em>d</em> ≥ 0.
<em>Hₐ</em>: The new incentive program cuts down on the number of days missed by employees, i.e. <em>d</em> < 0.
The paired <em>t</em>-test would be used in this case as the data provided is a matched paired data.
The mean and standard deviation of the differences are computed in the Excel sheet.
Compute the test statistic as follows:
![t=\frac{\bar d}{SD/\sqrt{n}}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20d%7D%7BSD%2F%5Csqrt%7Bn%7D%7D)
![=\frac{1.10}{1.1005/\sqrt{10}}\\\\=3.161](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1.10%7D%7B1.1005%2F%5Csqrt%7B10%7D%7D%5C%5C%5C%5C%3D3.161)
The test statistic value is 3.161.
The degrees of freedom is:
df = n - 1 = 10 - 1 = 9
Compute the <em>p</em>-value of the test as follows:
![p-value=P(t_{9}](https://tex.z-dn.net/?f=p-value%3DP%28t_%7B9%7D%3C3.161%29%3D0.0058)
*Use a <em>t</em>-table.
The <em>p</em>-value of the test is 0.0058.
Decision rule:
If the <em>p</em>-value of the test is less than the significance level, 0.05, the null hypothesis will be rejected.
<em>p</em>-value = 0.0058 < 0.05
The null hypothesis will be rejected.
Hence, it can be concluded that the new incentive program cuts down on the number of days missed by employees.