Model Answer
0 min readIntroduction
Performance evaluation is a crucial aspect of organizational development, particularly within law enforcement agencies. Training programs are implemented to enhance skills and improve performance, but their effectiveness must be rigorously assessed. Statistical methods provide a systematic way to determine whether observed improvements are due to the training or simply due to chance. This question presents a scenario where the shooting scores of police personnel were recorded before and after a training program, requiring us to analyze the data to ascertain if the training had a statistically significant positive impact on their performance. The core principle here is to move beyond subjective impressions and rely on objective data analysis.
Understanding the Problem & Hypothesis
The problem at hand is to determine if a month-long shooting training program improved the shooting skills of eleven police personnel. We can formulate the following hypotheses:
- Null Hypothesis (H0): The training program had no effect on the shooting scores. (μ1 = μ2, where μ1 is the mean score before training and μ2 is the mean score after training).
- Alternative Hypothesis (H1): The training program improved the shooting scores. (μ1 < μ2). This is a one-tailed test.
Data Analysis
Let's assume the following marks were awarded (this data is needed to proceed, as it wasn't provided in the question. We will create sample data for demonstration):
| Personnel | Test 1 (Before Training) | Test 2 (After Training) | Difference (Test 2 - Test 1) |
|---|---|---|---|
| 1 | 65 | 75 | 10 |
| 2 | 70 | 80 | 10 |
| 3 | 55 | 60 | 5 |
| 4 | 80 | 85 | 5 |
| 5 | 60 | 70 | 10 |
| 6 | 75 | 80 | 5 |
| 7 | 50 | 65 | 15 |
| 8 | 85 | 90 | 5 |
| 9 | 65 | 70 | 5 |
| 10 | 70 | 75 | 5 |
| 11 | 55 | 60 | 5 |
To determine if the training benefitted the personnel, a paired t-test is the most appropriate statistical test. This test is used when comparing the means of two related samples (in this case, the same individuals tested twice). The paired t-test focuses on the differences between the paired observations.
Steps in Performing the Paired T-Test:
- Calculate the difference (d) for each individual (as shown in the table above).
- Calculate the mean difference (d̄): (10+10+5+5+10+5+15+5+5+5+5)/11 = 7.73
- Calculate the standard deviation of the differences (sd): Approximately 3.82 (calculated using standard statistical formulas).
- Calculate the t-statistic: t = d̄ / (sd / √n) = 7.73 / (3.82 / √11) = 6.74
- Determine the degrees of freedom (df): df = n - 1 = 11 - 1 = 10
- Find the p-value: Using a t-distribution table or statistical software, with df = 10 and t = 6.74, the p-value is less than 0.001.
Interpretation of Results
The p-value (less than 0.001) is less than the significance level (typically α = 0.05). Therefore, we reject the null hypothesis (H0) and accept the alternative hypothesis (H1). This indicates that there is statistically significant evidence to conclude that the training program improved the shooting scores of the police personnel.
It's important to note that statistical significance doesn't necessarily imply practical significance. While the scores improved significantly, the magnitude of the improvement needs to be considered in the context of operational requirements.
Conclusion
Based on the statistical analysis using a paired t-test, the data provides strong evidence that the month-long shooting training program benefitted the police personnel. The observed improvement in scores was statistically significant, suggesting the training was effective. However, further analysis considering the practical implications of the score improvements is recommended. Future evaluations should also incorporate control groups and consider other factors that might influence shooting performance, such as individual aptitude and equipment 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.