Question 1193978: The human resource director is concerned about absenteeism among hourly workers. She decides to sample the records to
determine whether absenteeism is distributed evenly throughout the six-day workweek. Is absenteeism distributed evenly
throughout the week? Use 0.01 level of significance. What is the decision regarding the null hypothesis? Specifically, what does
this indicate to the human resources director?
Number of absent
Monday. 12
Tuesday. 9
Wednesday.11
Thursday. 10
Friday. 9
Saturday. 9
Answer by parmen(42)   (Show Source):
You can put this solution on YOUR website! **1. Set up Hypotheses**
* **Null Hypothesis (H0):** Absenteeism is evenly distributed throughout the workweek.
* **Alternative Hypothesis (H1):** Absenteeism is not evenly distributed throughout the workweek.
**2. Calculate Expected Frequencies**
* **Total Absences:** 12 + 9 + 11 + 10 + 9 + 9 = 60
* **Expected Absences per Day (under the null hypothesis):** 60 absences / 6 days = 10 absences/day
* **Create a table:**
| Day | Observed (O) | Expected (E) | (O - E)² | (O - E)² / E |
|-----------|-------------|-------------|---------|-------------|
| Monday | 12 | 10 | 4 | 0.4 |
| Tuesday | 9 | 10 | 1 | 0.1 |
| Wednesday | 11 | 10 | 1 | 0.1 |
| Thursday | 10 | 10 | 0 | 0 |
| Friday | 9 | 10 | 1 | 0.1 |
| Saturday | 9 | 10 | 1 | 0.1 |
| **Total** | 60 | 60 | | **0.8** |
**3. Calculate the Chi-Square Test Statistic**
* χ² = Σ [(O - E)² / E] = 0.8
**4. Determine Degrees of Freedom**
* Degrees of Freedom (df) = Number of categories - 1 = 6 - 1 = 5
**5. Find the Critical Value**
* Using a chi-square distribution table, find the critical value for α = 0.01 and df = 5.
* The critical value is approximately 15.086.
**6. Make a Decision**
* **Compare the calculated chi-square statistic to the critical value:**
* 0.8 < 15.086
* **Decision:** Since the calculated chi-square statistic (0.8) is less than the critical value (15.086), we **fail to reject the null hypothesis**.
**7. Conclusion**
* There is **not enough evidence** at the 0.01 level of significance to conclude that absenteeism is not evenly distributed throughout the workweek.
**Interpretation for the Human Resources Director:**
* This analysis suggests that there is no significant evidence to indicate that absenteeism is more prevalent on any particular day of the week.
* The observed variations in absenteeism across the days could be due to random chance.
* The HR director may need to investigate other factors that might be contributing to absenteeism, such as employee health, work-life balance, or job satisfaction.
**Note:**
* This analysis assumes that the data meets the assumptions of the chi-square test, such as expected frequencies being sufficiently large (generally, expected frequencies should be greater than 5).
This analysis provides a basic framework for conducting a chi-square goodness-of-fit test.
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