SOLUTION: The masses of boxes of apples are normally distributed such that 20% of them are greater than 5.62 kg. Estimate the mean and standard deviation of the mases.
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-> SOLUTION: The masses of boxes of apples are normally distributed such that 20% of them are greater than 5.62 kg. Estimate the mean and standard deviation of the mases.
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Question 1193877: The masses of boxes of apples are normally distributed such that 20% of them are greater than 5.62 kg. Estimate the mean and standard deviation of the mases. Found 2 solutions by parmen, ikleyn: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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