Interpreting a Quality Deviation on a Small Production Batch
A factory produces 500 brackets per day on a single machining cell. The drawing specifies a bore diameter of 25.00 mm with a tolerance of plus or minus 0.05 mm. A sample of 30 brackets taken at midday gives a mean of 25.06 mm and a standard deviation of 0.015 mm. The process was set up that morning after a tool change. The operator says the parts are 'within drawing tolerance' because the last five measured parts were all between 24.98 and 25.10 mm. The quality engineer notes that the customer requires a process capability index Cpk of at least 1.33 and that the last audit flagged the cell. Production is due to ship 500 brackets this evening.
1. Calculate Cpk for the sample and interpret the result against the customer requirement.
Read the worked answer
Cpk = min[(USL - mean) / (3 sigma), (mean - LSL) / (3 sigma)].
USL = 25.05, LSL = 24.95, mean = 25.06, sigma = 0.015.
Upper side: (25.05 - 25.06) / (3 x 0.015) = (-0.01) / 0.045 = -0.22. Lower side: (25.06 - 24.95) / 0.045 = 0.11 / 0.045 = 2.44.
Cpk = -0.22.
A negative Cpk means the process mean lies outside the upper specification limit. On average the parts are already oversize, and roughly half the output will exceed 25.05 mm. The process fails the 1.33 requirement by a wide margin.
The operator's claim that parts are 'within tolerance' reflects the last five parts only; with a mean above the USL, a short run of acceptable parts is expected by chance and does not demonstrate capability.
2. The operator argues the sample is too small and the parts are fine. How should the engineer respond, and what immediate action is justified before the evening shipment?
Read the worked answer
The engineer should separate the two issues: sample size and process centring.
Sample size affects the confidence in the estimate, not the direction of the error. With n = 30, the standard error of the mean is 0.015 / sqrt(30) = 0.0027 mm. The mean of 25.06 is therefore about 3.7 standard errors above the USL of 25.05 - the centring error is statistically clear, not a sampling artefact. A larger sample would tighten the interval, not move the mean back inside the limit.
The operator's five-part check is a conformance check, not a capability check. Individual parts can be inside tolerance while the process as a whole is not capable.
Immediate action: stop the cell and re-centre the process (the morning tool change is the likely cause), then re-sample before releasing product. The 500 brackets due to ship should be quarantined and either 100% inspected or held pending the re-centred process result. Shipping on the basis of five conforming parts would transfer a known capability failure to the customer.
3. After re-centring, a new sample of 30 gives a mean of 25.00 mm and the same standard deviation of 0.015 mm. Recalculate Cpk and state whether shipment can proceed.
Read the worked answer
With mean = 25.00, sigma = 0.015:
Upper side: (25.05 - 25.00) / 0.045 = 0.05 / 0.045 = 1.11. Lower side: (25.00 - 24.95) / 0.045 = 0.05 / 0.045 = 1.11.
Cpk = 1.11.
The process is now centred, and both sides are equal, which confirms the centring is correct. However, 1.11 is still below the customer's 1.33 requirement. The shortfall is now due to spread, not position: the standard deviation of 0.015 mm is too large for a tolerance band of 0.10 mm. To reach Cpk 1.33 the standard deviation would need to fall to about 0.0125 mm (0.05 / (3 x 1.33)).
Shipment can proceed only if the customer accepts a deviation, or if the product is 100% inspected and sorted. The engineer should not declare the process capable; the honest position is that centring is fixed but variation reduction is still required. A concession request with the measured data is the defensible route.