Model Answer
0 min readIntroduction
In quality control and production management, understanding the relationship between different factors influencing defect rates is critical for process optimization. Statistical methods, such as the Chi-Square test of independence, are frequently employed to determine if two categorical variables are associated. This test helps determine whether the occurrence of defects is independent of the shift or week in which the production takes place. The question provides production data across three shifts and three weeks, allowing us to assess whether these variables are statistically independent at a 5% significance level. This analysis is fundamental to identifying potential systemic issues in the production process.
Chi-Square Test of Independence
The Chi-Square test of independence is a statistical test used to determine if there is a significant association between two categorical variables. In this case, we want to test if the week and shift are independent in terms of the number of defective goods produced.
1. Hypotheses Formulation
- Null Hypothesis (H0): Weeks and shifts are independent. There is no association between the week and the shift in terms of the number of defective goods.
- Alternative Hypothesis (H1): Weeks and shifts are not independent. There is an association between the week and the shift in terms of the number of defective goods.
2. Data Organization and Observed Frequencies
First, we need to organize the given data into a contingency table showing the observed frequencies.
| Shift | Week 1 | Week 2 | Week 3 | Total |
|---|---|---|---|---|
| Shift 1 | 10 | 12 | 15 | 37 |
| Shift 2 | 15 | 18 | 20 | 53 |
| Shift 3 | 20 | 25 | 30 | 75 |
| Total | 45 | 55 | 65 | 165 |
3. Calculating Expected Frequencies
The expected frequency for each cell is calculated as (Row Total * Column Total) / Grand Total. This represents what we would expect to see if the two variables were truly independent.
| Shift | Week 1 (Expected) | Week 2 (Expected) | Week 3 (Expected) |
|---|---|---|---|
| Shift 1 | (37 * 45) / 165 = 10.08 | (37 * 55) / 165 = 12.47 | (37 * 65) / 165 = 14.35 |
| Shift 2 | (53 * 45) / 165 = 14.49 | (53 * 55) / 165 = 17.69 | (53 * 65) / 165 = 20.91 |
| Shift 3 | (75 * 45) / 165 = 20.45 | (75 * 55) / 165 = 25.00 | (75 * 65) / 165 = 29.55 |
4. Calculating the Chi-Square Statistic
The Chi-Square statistic is calculated using the formula: χ2 = Σ [(Oij - Eij)2 / Eij], where Oij is the observed frequency and Eij is the expected frequency for each cell.
Calculating for each cell and summing up:
- ((10 - 10.08)2 / 10.08) + ((12 - 12.47)2 / 12.47) + ((15 - 14.35)2 / 14.35) +
- ((15 - 14.49)2 / 14.49) + ((18 - 17.69)2 / 17.69) + ((20 - 20.91)2 / 20.91) +
- ((20 - 20.45)2 / 20.45) + ((25 - 25.00)2 / 25.00) + ((30 - 29.55)2 / 29.55) = 0.07 + 0.02 + 0.04 + 0.00 + 0.01 + 0.08 + 0.00 + 0.00 + 0.02 = 0.24
Therefore, χ2 = 0.24
5. Determining Degrees of Freedom and Critical Value
Degrees of Freedom (df) = (Number of Rows - 1) * (Number of Columns - 1) = (3 - 1) * (3 - 1) = 4
At a significance level of 5% (α = 0.05) and with 4 degrees of freedom, the critical value from the Chi-Square distribution table is 9.488.
6. Decision and Conclusion
Since the calculated Chi-Square statistic (0.24) is less than the critical value (9.488), we fail to reject the null hypothesis. This indicates that there is not enough evidence to conclude that weeks and shifts are dependent in terms of the number of defective goods produced. The observed differences in defect rates across weeks and shifts could be due to random chance.
Conclusion
In conclusion, the Chi-Square test of independence, performed at a 5% significance level, did not reveal a statistically significant association between the week and shift in relation to the production of defective goods. This suggests that the observed variations in defect rates are likely due to random fluctuations rather than a systematic relationship between these two factors. Further investigation might explore other potential contributing factors to defect rates, such as operator skill, machine maintenance, or raw material quality.
Answer Length
This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.