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?

Question

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&#x27;s claim at a 0.05 level of significance?

Accepted Answer

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.

Answer

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.

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