A company claims that the mean number of sick days taken by its employees is less than 3 days. A random sample of 36 employees is selected, and it is found that they have taken an average of 2.8 sick days with a standard deviation of 0.5. Is there sufficient evidence to support the company's claim at a 0.05 level of significance?

Step 1 ：Step 1: State the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). Here, \(H_0\): \(\mu = 3\) (the mean is 3) and \(H_a\): \(\mu < 3\) (the mean is less than 3).

Step 2 ：Step 2: Compute the test statistic. The formula for the test statistic in a one sample mean test is \(Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}\). Substituting the given values, we get \(Z = \frac{2.8 - 3}{\frac{0.5}{\sqrt{36}}} = -2.4\).

Step 3 ：Step 3: Determine the critical value for a 0.05 level of significance for a one-tailed test. From the Z-table, the critical value is -1.645.

Step 4 ：Step 4: Compare the test statistic with the critical value. Since -2.4 is less than -1.645, we reject the null hypothesis.