Comparing two packing lines: coefficient of variation and a tracing a data-entry error
Sunrise Foods runs two packing lines. Over 40 sampled shifts on Line A the mean output was 250 packets per shift with a standard deviation of 20. Over 60 sampled shifts on Line B the mean was 220 packets with a standard deviation of 25. These standard deviations are population values for the shifts recorded, i.e. the divisor used was n. The plant manager must recommend one line for a large export order where consistency of daily despatch matters more than peak volume, and he has been told that 'Line A is better because its standard deviation is smaller'. A second issue has surfaced: one Line A shift was entered in the register as 300 packets, but the supervisor's log shows 200. The manager wants to know how far that single error travels - through the means, the standard deviations and the consistency comparison - before he writes his recommendation to the export cell.
1. Compare the consistency of the two lines and evaluate the statement that 'Line A is better because its standard deviation is smaller'.
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The two lines have different means, so raw standard deviations are not directly comparable. The coefficient of variation (CV = SD/mean × 100) expresses variability relative to the level of output:
- CV for Line A = 20/250 × 100 = 8.00%
- CV for Line B = 25/220 × 100 = 11.36%
Line A is more consistent in relative terms - its shift-to-shift output deviates by about 8% of its own mean against 11.4% for Line B.
The manager's adviser is right in conclusion but wrong in reasoning. Here the absolute SDs also favour A (20 < 25), so the conclusion survives; but the SD comparison alone is misleading because the means differ by 30 packets. Had Line B's mean been 100 with SD 25 (CV 25%), the SD ranking and the CV ranking would disagree, and only the CV would answer the question asked. The general rule: use CV whenever the means differ materially or the units differ; use SD alone only when comparing like with like.
2. Compute the mean and standard deviation of output for the two lines taken together, and explain why the combined SD exceeds both individual SDs.
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Combined mean: (40 × 250 + 60 × 220)/100 = (10,000 + 13,200)/100 = 232 packets.
Use the combined-variance formula: σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂), where dᵢ is each group mean minus the combined mean.
d₁ = 250 − 232 = 18; d₂ = 220 − 232 = −12.
σ² = [40(400 + 324) + 60(625 + 144)]/100 = [40 × 724 + 60 × 769]/100 = [28,960 + 46,140]/100 = 75,100/100 = 751
σ = √751 ≈ 27.40 packets.
Why this exceeds 20 and 25: the combined variance has two sources - variation within each line (400 and 625) and variation between the line means (the d² terms, 324 and 144). The second source is created by the fact that the two lines run at different average levels, so the pooled data spread out more than either line does on its own. A candidate who simply averages 20 and 25 gets 22.5 and misses both the between-group term and the unequal sample sizes.
3. Correct the 300/200 data-entry error and state whether it changes the recommendation.
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Line A totals can be recovered from the summary because the SD is a population SD given with n = 40:
- Σx = 40 × 250 = 10,000; Σx² = 40(20² + 250²) = 40(400 + 62,500) = 25,16,000.
Correcting 300 → 200: Σx becomes 9,900 and Σx² becomes 25,16,000 − 90,000 + 40,000 = 24,66,000.
- Corrected Line A mean = 9,900/40 = 247.5 packets.
- Corrected Line A variance = 24,66,000/40 − (247.5)² = 61,650 − 61,256.25 = 393.75 → SD ≈ 19.84.
- Corrected CV_A = 19.84/247.5 × 100 ≈ 8.02%.
- Corrected combined mean = (9,900 + 13,200)/100 = 231 packets.
The recommendation does not change: Line A stays at roughly 8% against Line B's 11.36%. The error moved the mean by 2.5 packets and the SD by 0.16, but because it was a single observation in a sample of 40 near the centre of the data, it had little leverage. Two judgement points: an error near the mean is far less damaging than one in the tail; and here the correction was computable only because the SD was supplied with its divisor - with only a mean and no Σx², the corrected SD could not be recovered, and the honest answer would be to say so.