Comparing two training batches with summary statistics
A skill-training centre runs two batches for a 100-mark aptitude test. Batch A has 40 trainees with mean 62 and standard deviation 8. Batch B has 60 trainees with mean 68 and standard deviation 12. The centre plans to merge the batches and wants a single summary of average performance and spread. A merit prize is to be given for the stronger relative performance: one trainee from A scored 75 and one from B scored 72. The coordinator also claims that Batch B is clearly the better batch because its mean is higher, and that a score of 70 in either batch represents the same relative standing. The centre has only the summary statistics, not the individual score lists, and needs to know what can and cannot be concluded.
1. Compute the combined mean and combined standard deviation of the merged group, and interpret the result.
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Combined mean: [ \bar{x} = \frac{40(62) + 60(68)}{100} = \frac{2,480 + 4,080}{100} = 65.6. ]
For the combined variance, use deviations from the combined mean: d_A = 62 - 65.6 = -3.6 and d_B = 68 - 65.6 = 2.4. [ s^2 = \frac{40(8^2 + 3.6^2) + 60(12^2 + 2.4^2)}{100} = \frac{40(64 + 12.96) + 60(144 + 5.76)}{100} = \frac{3,078.4 + 8,985.6}{100} = 120.64. ] So s = √120.64 ≈ 10.98.
The merged group has mean about 65.6 and standard deviation about 11.0. Its spread is larger than Batch A's but smaller than Batch B's, reflecting the mixing of a tighter low-mean group with a wider high-mean group.
2. Compare the consistency of the two batches using coefficients of variation, and decide which trainee has the stronger relative performance.
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Coefficient of variation:
- Batch A: CV = (8/62) × 100 ≈ 12.90%.
- Batch B: CV = (12/68) × 100 ≈ 17.65%.
Batch A is more consistent relative to its mean because its CV is smaller, even though its absolute standard deviation is also smaller.
Relative performance via z-scores:
- A trainee scoring 75: z = (75 - 62)/8 = 13/8 = 1.625.
- B trainee scoring 72: z = (72 - 68)/12 = 4/12 ≈ 0.333.
The A trainee is 1.625 standard deviations above the A mean, while the B trainee is only about 0.33 standard deviations above the B mean. The A trainee has the stronger relative performance; the absolute score is also higher, so the merit prize should go to the A trainee.
3. The coordinator claims a score of 70 represents the same relative standing in both batches, and that Batch B is clearly better because its mean is higher. Assess these claims.
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A score of 70 is not the same relative standing:
- In A: z = (70 - 62)/8 = 1.00.
- In B: z = (70 - 68)/12 ≈ 0.167.
A score of 70 is one full standard deviation above the A mean but only about one-sixth of a standard deviation above the B mean. It is therefore a much stronger relative performance in A.
A higher mean for B does not by itself make B 'better' on every criterion. B also has a much larger standard deviation and CV, so it is less consistent. Moreover, summary statistics alone cannot tell us how many trainees scored above 70 in each batch. Two distributions with the same mean and standard deviation can have very different proportions above a cutoff. Individual scores or the full distribution would be needed for that claim.