Question 2 of 6: Combining Product, Gage and Inspector Variability

Question 2: Combining Product, Gage and Inspector Variability (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Each source of variation is quoted as a six-standard-deviation spread in units of ten-thousandths of an inch, on the same scale as the total tolerance of 7.0 units.

Find. The observed (as-measured) variability of the ring diameter for the best and for the worst combination of product, gage and inspector, and in each case the long-run percentage of rings that will be rejected against the printed tolerance.

Approach. Product variation, gage variation and inspector variation are independent sources acting on the same reported reading, so their variances add; the best and worst combinations are obtained by pairing the fixed product spread with the smallest and the largest of the gage and inspector spreads, and the reject fraction then follows from the normal distribution with the tolerance at ±3.5 units about the nominal.

1. State the rule that combines the sources. The reported reading is the true diameter plus a gage error plus an inspector error, and the three are independent, so $$\sigma_{obs}^{2} = \sigma_{prod}^{2} + \sigma_{gage}^{2} + \sigma_{insp}^{2}$$ Because every figure in the table is a six-standard-deviation spread, the same relation holds directly between the tabulated numbers, and the working can stay in units of ten-thousandths of an inch throughout.
2. Part (a) — identify and combine the best combination. The product is not a matter of choice, so the best attainable case pairs it with the air gage (1.1, against 4.4 for the mechanical gage) and with inspector B (3.7, the smallest of the four): $$6\sigma_{obs} = \sqrt{4.6^{2}+1.1^{2}+3.7^{2}} = \sqrt{21.16+1.21+13.69} = \sqrt{36.06}$$ $$\boxed{6\sigma_{obs} = 6.00\ \text{units} = 0.00060\ \text{in}}$$
3. Convert to a reject percentage. One standard deviation is $6.00/6 = 1.0008$ units, and a ring is rejected when its reported diameter falls more than half the tolerance from the nominal, that is beyond $\pm 3.5$ units: $$z = \frac{3.5}{1.0008} = 3.497 \qquad p = 2\left[1-\Phi(3.497)\right] = 0.00047$$ $$\boxed{p_{best} = 0.047\ \text{per cent, about 1 ring in 2,100}}$$
4. Part (b) — combine the worst combination. The worst case pairs the same product with the mechanical gage (4.4) and with inspector D (9.3), whose spread alone is larger than the whole tolerance: $$6\sigma_{obs} = \sqrt{4.6^{2}+4.4^{2}+9.3^{2}} = \sqrt{21.16+19.36+86.49} = \sqrt{127.01}$$ $$\boxed{6\sigma_{obs} = 11.27\ \text{units} = 0.00113\ \text{in}}$$
5. Convert the worst case to a reject percentage. Now $\sigma_{obs} = 11.27/6 = 1.8783$ units, so $$z = \frac{3.5}{1.8783} = 1.863 \qquad p = 2\left[1-\Phi(1.863)\right] = 0.0624$$ $$\boxed{p_{worst} = 6.24\ \text{per cent, about 1 ring in 16}}$$ The two answers differ by a factor of more than a hundred although the product being measured is identical.
6. Interpret the result as a gage capability statement. Comparing each spread with the tolerance gives $6\sigma_{obs}/T = 85.8$ per cent for the best combination and 161 per cent for the worst, or equivalently $C_{p} = T/6\sigma_{obs} = 1.17$ and 0.62. The product on its own would give $C_{p} = 7.0/4.6 = 1.52$ and a reject rate of 0.0005 per cent, so essentially every rejection in the worst case is manufactured by the measuring system rather than by the process: the measurement spread alone is $\sqrt{4.4^{2}+9.3^{2}} = 10.29$ units against 3.86 units for the best combination.