Bimodal Capability Data, a Recalculated p-Chart, and a 95/95 Reliability Demonstration
A line fills 500 mL bottles. The label fill volume is 505 mL, with upper specification limit 515 mL and lower specification limit 495 mL. A capability study on 125 consecutive bottles from a single shift gives a mean of 505.8 mL and a sample standard deviation of 2.4 mL. A histogram of the data is clearly bimodal, and the two modes correspond to the two filling heads on the line. Separately, the p-chart for label defects shows nine consecutive points above the centre line starting shortly after a new label applicator was installed. The line supervisor has recalculated the control limits excluding those nine points so that the chart 'looks in control'. A customer requirement states 95% reliability at 95% confidence with zero failures permitted in the demonstration test; the test laboratory has proposed a sample of 30 units.
1. Calculate Cp and Cpk from the study and assess how much confidence you can place in them.
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Calculation.
- Cp = (USL - LSL) / (6s) = (515 - 495) / (6 × 2.4) = 20 / 14.4 = 1.39
- Upper side: (515 - 505.8) / (3 × 2.4) = 9.2 / 7.2 = 1.28
- Lower side: (505.8 - 495) / 7.2 = 10.8 / 7.2 = 1.50
- Cpk = min(1.28, 1.50) = 1.28
So the line is off-centre toward the upper limit, and Cpk is below the commonly required 1.33 for an established process.
Why the numbers should not be trusted yet.
- Bimodality. Capability indices assume a single, approximately normal distribution. Two filling heads typically produce two offsets; pooling them inflates the standard deviation and can disguise a head sitting close to or over a limit while the other is well-centred.
- Stability. Indices are only meaningful for a process in statistical control. A pooled study of two heads with no stratification cannot establish that.
- Sampling. 125 consecutive bottles from one shift captures no shift-to-shift, head-to-head (if heads alternate by run), or lot-to-lot variation.
Correct method. Chart each head separately (individuals/moving range or Xbar-R), confirm control, then compute per-head capability. Check the measurement system with a gauge R&R before acting on s. If both heads are individually capable but offset, the fix is a head calibration/offset adjustment, not a specification change.
2. Assess the supervisor's recalculation of the p-chart. What is the correct analysis and action?
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Nine consecutive points above the centre line is a signal, not noise. Run rules exist precisely to detect sustained shifts that individual points would miss. The pattern here is the classic signature of a process level change, and it began immediately after a known event: the new label applicator.
The recalculation is data manipulation. Control limits are computed from the process as it behaves, to detect change. Removing the points that carry the signal and recomputing limits from the quieter periods builds a chart that can only confirm what the supervisor already believes. It also creates a documentation problem: the revised chart misrepresents the process history if anyone later uses it to demonstrate control.
Correct sequence.
- Verify the data first: same defect definition, same inspection method, same subgroup size, no change in who inspects. A definition change can fabricate a shift.
- Treat the applicator as the likely assignable cause and investigate it: registration, alignment, adhesive/tension settings, speed, and commissioning records.
- If the applicator is fixed and defects return to baseline, recompute limits from post-fix subgroups and note the intervention on the chart.
- If the new level persists, decide whether it is acceptable. If it is, recompute limits from the post-change data with a documented rationale and an annotated chart, and update the control plan. Do not delete the transition points.
- Where the elevated defect rate has cost or customer implications, open a corrective action rather than closing the chart discussion.
3. Determine the sample size needed for the 95/95 zero-failure demonstration and evaluate the laboratory's proposal of 30 units.
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Model. With zero failures permitted, the binomial relationship is R^n = 1 - C, where R is the reliability to be demonstrated and C the confidence.
n = ln(1 - C) / ln(R) = ln(0.05) / ln(0.95) = (-2.9957) / (-0.05129) = 58.4, so 59 units
With 59 units and zero failures, the demonstration supports 95% reliability at 95% confidence.
Evaluating 30 units. Two equivalent ways to state the shortfall:
- Confidence achieved at 95% reliability: 1 - 0.95^30 = 1 - 0.2146 = 78.5%, not 95%.
- Reliability demonstrated at 95% confidence with 30 units: 0.05^(1/30), giving approximately 90.5%, not 95%.
Either way, 30 units does not meet the requirement, and the gap is large: roughly half the required sample.
Assumptions to state explicitly before the test. The binomial model requires independent units, a constant failure probability, and a representative sample from production. The requirement must also define what 'reliability' means here - cycles, time, or actuations at the stated life - and whether the test is time-terminated or failure-terminated. If the analysis is a Weibull demonstration rather than zero-failure binomial, the sample size and acceptance number change, and the customer's requirement wording should be confirmed before testing begins. Testing 30 units and reporting a pass would be a defensible-looking result that does not meet the stated criterion.