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Acceptance Sampling of Conductor Resistance

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Acceptance Sampling of Conductor Resistance

A QA engineer measures the dc resistance of eight 100 m samples cut from a production run of 250 mm2 copper conductor. The contract limits the mean resistance to 5.00 milliohm per 100 m at 20 deg C. The results are 4.82, 4.91, 4.78, 4.88, 4.95, 4.85, 4.90 and 4.80 milliohm per 100 m. The supplier claims the run is compliant. The engineer must decide whether to accept the run using a one-sided 95% upper confidence bound on the mean, and must also judge whether a single spot reading of 5.02 milliohm per 100 m would indicate a nonconforming batch. Assume the measurements are independent and approximately normal.

1. Compute the sample mean and the sample standard deviation.

Read the worked answer

Sum = 4.82 + 4.91 + 4.78 + 4.88 + 4.95 + 4.85 + 4.90 + 4.80 = 38.89

Mean = 38.89/8 = 4.861 milliohm/100 m.

Squared deviations from the mean: 0.00170, 0.00238, 0.00660, 0.00035, 0.00788, 0.00013, 0.00150, 0.00375; sum = 0.02429.

s = sqrt(0.02429/7) = sqrt(0.003470) = 0.0589 milliohm/100 m.

2. Compute the one-sided 95% upper confidence bound on the mean using t(0.05, 7) = 1.895 and decide whether to accept the run.

Read the worked answer

Standard error = s/sqrt(n) = 0.0589/2.828 = 0.0208.

Upper bound = 4.861 + 1.895(0.0208) = 4.861 + 0.0394 = 4.900 milliohm/100 m.

Since 4.900 < 5.00, there is 95% confidence that the true mean is below the contract limit, so the run can be accepted on the basis of the mean.

3. A single spot reading of 5.02 milliohm/100 m is obtained. Use a 95% prediction interval to judge whether this is unusual.

Read the worked answer

Prediction interval half-width = tssqrt(1 + 1/n) = 1.895(0.0589)(sqrt(1.125)) = 1.895(0.0589)(1.0607) = 0.118.

Interval = 4.861 +/- 0.118 = 4.743 to 4.980 milliohm/100 m.

The reading 5.02 lies above 4.980, so it is outside the 95% prediction interval for a single future measurement. It is unusual and warrants investigation (calibration, temperature correction, or local defect), but one reading does not by itself prove the batch mean exceeds 5.00; the statistical decision rests on the confidence bound, not on a single spot value.