Measuring Things — Question 8

Section 1-1 · Physics pages 26–27

Question 8 Do three laboratories agree? Three laboratories measure the same unchanged plate thickness, using the same operational defini- tion. Their reports are: The reported lengths are 12.40 ± 0.06 mm (A), 12.49 ± 0.05 mm (B), and 12.56 ± 0.04 mm (C). Here each ± value is claimed to be a hard bound: the true thickness must lie in its closed interval. These are not standard deviations or confidence intervals. 12.30 12.40 12.50 12.60 Thickness (mm) A B C Closed bars represent allowed intervals, not probability distributions. (a) Determine which pairs of laboratories have compatible intervals. Give each nonempty pairwise overlap. (b) Is there a thickness compatible with all three reports? Explain why A agreeing with B and B agreeing with C does not settle this question. (c) A manager proposes averaging all three central values to resolve the disagreement. Com- pute that average and assess the proposal. What must be investigated before reporting a trustworthy combined value?
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Question 8 - Solution Do three laboratories agree? 12.30 12.40 12.50 12.60 Thickness (mm) A B C (a) Find the intersections. The allowed intervals are A = [12.34, 12.46], B = [12.44, 12.54], C = [12.52, 12.60] mm. Thus A ∩B = [12.44, 12.46] mm and B ∩C = [12.52, 12.54] mm. A and C are incompatible: A ends at 12.46 mm while C begins at 12.52 mm, leaving a 0.06 mm gap. (b) Require one value to satisfy every report. max(12.34, 12.44, 12.52) = 12.52 > 12.46 = min(12.46, 12.54, 12.60). max(12.34, 12.44, 12.52) = 12.52 > 12.46 = min(12.46, 12.54, 12.60). The common intersection is empty. Compatibility with a broad middle interval is not transitive: it does not make the two outer intervals overlap. (c) An average cannot repair contradictory bounds. x = 12.40 + 12.49 + 12.56 3 = 12.4833 . . . mm. This value lies outside both A’s and C’s claimed intervals. Averaging creates a number, not evidence that it is the true thickness. Given the stated common object and definition, at least one claimed bound is invalid. Investigate calibration, environmental or contact conditions, transcription, and omitted error sources. In practice, also recheck that the object and measurand were genuinely the same. Revised defensible bounds or a justified statistical model are needed before combining the reports.

Question 8

Do three laboratories agree?

Three laboratories measure the same unchanged plate thickness, using the same operational definition. Their reports are: The reported lengths are 12.40±0.06mm12.40\pm0.06\ \mathrm{mm} (A), 12.49±0.05mm12.49\pm0.05\ \mathrm{mm} (B), and 12.56±0.04mm12.56\pm0.04\ \mathrm{mm} (C). Here each ±\pm value is claimed to be a hard bound: the true thickness must lie in its closed interval. These are not standard deviations or confidence intervals.

Diagram shown in the original SVG worksheet.

  1. Determine which pairs of laboratories have compatible intervals. Give each nonempty pairwise overlap.

  2. Is there a thickness compatible with all three reports? Explain why A agreeing with B and B agreeing with C does not settle this question.

  3. A manager proposes averaging all three central values to resolve the disagreement. Compute that average and assess the proposal. What must be investigated before reporting a trustworthy combined value?

Question 8 - Solution

Do three laboratories agree?

Diagram shown in the original SVG worksheet.

(a) Find the intersections.

The allowed intervals are A=[12.34,12.46],B=[12.44,12.54],C=[12.52,12.60]mm.A=[12.34,12.46],\quad B=[12.44,12.54],\quad C=[12.52,12.60]\ \mathrm{mm}. Thus A∩B=[12.44,12.46]\boxed{A\cap B=[12.44,12.46]} mm and B∩C=[12.52,12.54]\boxed{B\cap C=[12.52,12.54]} mm. A and C are incompatible: A ends at 12.4612.46 mm while C begins at 12.5212.52 mm, leaving a 0.060.06 mm gap.

(b) Require one value to satisfy every report.

max⁡(12.34,12.44,12.52)=12.52>12.46=min⁡(12.46,12.54,12.60).\max(12.34,12.44,12.52)=12.52>12.46=\min(12.46,12.54,12.60).

max⁡(12.34,12.44,12.52)=12.52>12.46=min⁡(12.46,12.54,12.60).\max(12.34,12.44,12.52)=12.52>12.46=\min(12.46,12.54,12.60). The common intersection is empty. Compatibility with a broad middle interval is not transitive: it does not make the two outer intervals overlap.

(c) An average cannot repair contradictory bounds.

x¯=12.40+12.49+12.563=12.4833…mm.\overline x=\frac{12.40+12.49+12.56}{3}=12.4833\ldots\ \mathrm{mm}. This value lies outside both A’s and C’s claimed intervals. Averaging creates a number, not evidence that it is the true thickness. Given the stated common object and definition, at least one claimed bound is invalid. Investigate calibration, environmental or contact conditions, transcription, and omitted error sources. In practice, also recheck that the object and measurand were genuinely the same. Revised defensible bounds or a justified statistical model are needed before combining the reports.