Question 1175786: Scores (out of 1.00)
C V Q
A 0.62 0.67 0.47
B 0.64 0.58 0.62
C 0.68 0.45 0.61
D 0.64 0.48 0.58
E 0.55 0.23 0.23
F 0.27 0.27 0.00
From the 6 exams, Average scores by C, V and Q are as follows. Using statistical model, can we give a detailed reply which candidate scores best? Also, we also need to give a critical evaluation on all 3 candidates
Hint: One option to solve this problem is using t distribution, but other options can also be used by giving a rationale for the same.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's analyze the candidates' scores using statistical methods to determine who performed best and provide a critical evaluation.
**1. Calculate Average Scores**
First, let's confirm the average scores for each category (C, V, Q):
* **C:** (0.62 + 0.64 + 0.68 + 0.64 + 0.55 + 0.27) / 6 = 3.4 / 6 = 0.5667
* **V:** (0.67 + 0.58 + 0.45 + 0.48 + 0.23 + 0.27) / 6 = 2.68 / 6 = 0.4467
* **Q:** (0.47 + 0.62 + 0.61 + 0.58 + 0.23 + 0.00) / 6 = 2.51 / 6 = 0.4183
**2. Standard Deviation**
Next, let's calculate the standard deviation for each category to understand the variability of scores.
* **C:**
* Variance = Σ(x - μ)² / (n - 1) ≈ 0.021067
* Standard deviation (σ) ≈ √0.021067 ≈ 0.1451
* **V:**
* Variance ≈ 0.025347
* Standard deviation (σ) ≈ √0.025347 ≈ 0.1592
* **Q:**
* Variance ≈ 0.041697
* Standard deviation (σ) ≈ √0.041697 ≈ 0.2042
**3. Z-Scores**
To compare the candidates' performance relative to the average, we can calculate z-scores for each candidate in each category.
* **Z = (x - μ) / σ**
* **Candidate A:**
* C: (0.62 - 0.5667) / 0.1451 ≈ 0.367
* V: (0.67 - 0.4467) / 0.1592 ≈ 1.403
* Q: (0.47 - 0.4183) / 0.2042 ≈ 0.253
* **Candidate B:**
* C: (0.64 - 0.5667) / 0.1451 ≈ 0.471
* V: (0.58 - 0.4467) / 0.1592 ≈ 0.837
* Q: (0.62 - 0.4183) / 0.2042 ≈ 0.998
* **Candidate C:**
* C: (0.68 - 0.5667) / 0.1451 ≈ 0.781
* V: (0.45 - 0.4467) / 0.1592 ≈ 0.021
* Q: (0.61 - 0.4183) / 0.2042 ≈ 0.949
* **Candidate D:**
* C: (0.64 - 0.5667) / 0.1451 ≈ 0.471
* V: (0.48 - 0.4467) / 0.1592 ≈ 0.209
* Q: (0.58 - 0.4183) / 0.2042 ≈ 0.899
* **Candidate E:**
* C: (0.55 - 0.5667) / 0.1451 ≈ -0.115
* V: (0.23 - 0.4467) / 0.1592 ≈ -1.361
* Q: (0.23 - 0.4183) / 0.2042 ≈ -0.922
* **Candidate F:**
* C: (0.27 - 0.5667) / 0.1451 ≈ -2.045
* V: (0.27 - 0.4467) / 0.1592 ≈ -1.110
* Q: (0.00 - 0.4183) / 0.2042 ≈ -2.048
**4. Analysis and Critical Evaluation**
* **Candidate B:**
* Consistently high z-scores across all categories, indicating strong performance relative to the group.
* Demonstrates well rounded abilities.
* **Candidate C:**
* Strong performance in C and Q, but average in V.
* Indicates strength in certain areas, but potential weakness in others.
* **Candidate D:**
* Fairly consistent scores, very similar to candidate B, but slightly lower in each category.
* **Candidate A:**
* Very strong in V, but average in C and Q.
* Indicates a specialization in a specific area.
* **Candidate E and F:**
* Weak performance across all categories.
**5. Conclusion**
* Based on the z-scores, **Candidate B** is the strongest performer overall. They have consistently high scores relative to the group.
* Candidate C is strong in C and Q.
* Candidate A is strong in V.
* Candidate D is fairly consistent.
* Candidates E and F have weak performance.
**Rationale for Using Z-Scores**
* Z-scores allow us to standardize the scores and compare performance across different categories with varying means and standard deviations.
* They provide a measure of how far each candidate's score is from the average in terms of standard deviations.
**Critical Evaluation**
* It's important to note that this analysis is based on a limited dataset (6 exams).
* Other factors, such as the difficulty of the exams, could influence the results.
* Consider the specific requirements of the situation when making a final decision. If a specific category is more important, then adjust accordingly.
* If we were to use the T distribution, we would need to create confidence intervals for each candidate and category, which would be a more robust statistical method, but the Z score method is a good quick comparison.
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