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Chapter 15: Problem 43

Suppose that a quality characteristic is normally distributed with specifications at \(120 \pm 20 .\) The process standard deviation is \(6.5 .\) (a) Suppose that the process mean is \(120 .\) What are the natural tolerance limits? What is the fraction defective? Calculate \(P C R\) and \(P C R_{k}\) and interpret these ratios. (b) Suppose that the process mean shifts off-center by 1.5 standard deviations toward the upper specification limit. Recalculate the quantities in part (a). (c) Compare the results in parts (a) and (b) and comment on any differences.

Short Answer

Expert verified
After the mean shift, process capability (PCR) remains, but centering (PCRk) worsens, increasing defect rate.

Step by step solution

01

Calculate Natural Tolerance Limits for (a)

The natural tolerance limits can be calculated using the formula: \[ LTL = \mu - 3\sigma \] and \[ UTL = \mu + 3\sigma \]. Given \( \mu = 120 \) and \( \sigma = 6.5 \), the lower natural tolerance limit (LTL) is: \[ LTL = 120 - (3 \times 6.5) = 120 - 19.5 = 100.5 \] and the upper natural tolerance limit (UTL) is: \[ UTL = 120 + (3 \times 6.5) = 120 + 19.5 = 139.5 \]. Therefore, the natural tolerance limits are \([100.5, 139.5]\).
02

Calculate Fraction Defective for (a)

The fraction defective occurs when the process produces output outside the specification limits. The specification limits are given as \([100, 140]\). We calculate \(P(X < 100)\) and \(P(X > 140)\) using the z-score formula. \[ Z = \frac{X - \mu}{\sigma} \]For \(X = 100\): \( Z = \frac{100 - 120}{6.5} \approx -3.08 \).For \(X = 140\): \( Z = \frac{140 - 120}{6.5} \approx 3.08 \).Using the standard normal distribution table, the probabilities are: \( P(X < 100) \approx 0.001 \) and \( P(X > 140) \approx 0.001 \). Therefore, the total fraction defective is \(2 \times 0.001 = 0.002\).
03

Calculate PCR and PCRk for (a)

The Process Capability Ratio (PCR) is: \[ PCR = \frac{USL - LSL}{6\sigma} \]\(USL = 140\), \(LSL = 100\) and \(\sigma = 6.5\).\[ PCR = \frac{140 - 100}{6 \times 6.5} = \frac{40}{39} \approx 1.03 \]The Process Capability Ratio adjusted for centering (PCRk) is: \[ PCR_{k} = \min\left( \frac{USL - \mu}{3\sigma}, \frac{\mu - LSL}{3\sigma} \right) \]\[ PCR_{k} = \min\left(\frac{140 - 120}{19.5}, \frac{120 - 100}{19.5}\right) = \min(1.03, 1.03) = 1.03 \].Both PCR and PCRk indicate that the process is capable since they are greater than 1.
04

Recalculate Natural Tolerance Limits for (b)

If the process mean shifts by 1.5 standard deviations toward the upper specification limit, the new mean is \(\mu = 120 + (1.5 \times 6.5) = 129.75\).Recalculate the natural tolerance limits:\[LTL = \mu - 3\sigma = 129.75 - 19.5 = 110.25 \]\[UTL = \mu + 3\sigma = 129.75 + 19.5 = 149.25 \]The new natural tolerance limits are \([110.25, 149.25]\).
05

Calculate Fraction Defective for (b)

With the shifted mean, calculate \(P(X < 100)\) and \(P(X > 140)\).For \(X = 100\): \(Z = \frac{100 - 129.75}{6.5}\approx -4.58\).\(P(X < 100)\approx 0\) (very close to zero due to low z-value).For \(X = 140\): \(Z = \frac{140 - 129.75}{6.5}\approx 1.58\).\(P(X > 140)\approx 0.057\).The total fraction defective is approximately \(0.057\).
06

Recalculate PCR and PCRk for (b)

Recalculate PCR:\[PCR = \frac{USL - LSL}{6\sigma} = \frac{140 - 100}{39} \approx 1.03\]Recalculate PCRk:\[PCR_{k} = \min\left(\frac{140 - 129.75}{19.5}, \frac{129.75 - 100}{19.5}\right) = \min(0.52, 1.53) = 0.52\].PCR remains above 1 indicating the ability to meet specification, but PCRk has reduced, indicating deterioration in process centering.
07

Compare and Comment on Results

In part (a), both PCR and PCRk are 1.03, indicating a capable and centered process. In part (b), after the mean shift, PCR remains the same indicating capability, but PCRk drops to 0.52, reflecting a less centered process and an increased fraction defective. This demonstrates that centering is important along with capability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Tolerance Limits
Natural tolerance limits help us see the range where most of the process data should fall if everything is running smoothly. They are calculated based on the standard deviation (\( \sigma \)) and the process mean (\( \mu \)). To find these limits, we use the formula:
  • Lower Tolerance Limit (LTL) = \( \mu - 3\sigma \)
  • Upper Tolerance Limit (UTL) = \( \mu + 3\sigma \)
For example, imagine a process with a mean of 120 and a standard deviation of 6.5. You'd calculate:
\[ LTL = 120 - (3 \times 6.5) = 100.5 \]
\[ UTL = 120 + (3 \times 6.5) = 139.5 \]
Thus, most data points should be between 100.5 and 139.5, ensuring the process stays within control.
Fraction Defective
The fraction defective refers to the portion of outputs that don't meet the pre-set specification limits, often referred to as defects. It's critical to calculate this to understand how much of what you're producing doesn't fit the desired quality standards. To determine this, you look at the probability of values lying outside the specification limits.
  • Use the Z-score to calculate the probability for lower and upper limits.
  • Z = \( \frac{X - \mu}{\sigma} \)
If the specification limits are between 100 and 140, calculate the Z-score for both limits and use a standard normal distribution table to find probabilities. For a mean of 120 and standard deviation of 6.5, this might result in a tiny fraction defective like 0.002, meaning only 0.2% of products are faulty鈥攓uite efficient!
Process Capability Ratio (PCR)
The Process Capability Ratio (PCR) is a simple yet powerful metric to understand how well a process performs relative to defined limits. It provides a quick snapshot of the process's ability to produce within specifications.To calculate PCR:
  • PCR = \( \frac{USL - LSL}{6\sigma} \)
  • Where USL = Upper Specification Limit, and LSL = Lower Specification Limit.
A PCR greater than 1 generally indicates a capable process that meets specifications most of the time. For example, with USL = 140, LSL = 100, and \( \sigma = 6.5 \), the PCR would be around 1.03. This shows that the process is adequately equipped to meet quality expectations.
Process Capability Index (PCRk)
The Process Capability Index (PCRk) is an extension of the PCR. It accounts for whether a process is centered within the specification limits. While PCR measures the overall capability, PCRk evaluates capability with an eye on centering. Calculate PCRk using:
  • PCRk = \( \min\left( \frac{USL - \mu}{3\sigma}, \frac{\mu - LSL}{3\sigma} \right) \)
If the process mean drifts, PCRk reflects this shift. For instance, if the process mean shifts to 129.75, this changes the centering, and PCRk drops to around 0.52. This shows that centering impacts quality, as a shift off-center can lead to more defects even if the overall capability (PCR) stays the same.

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Most popular questions from this chapter

Heart rate (in counts/minute) is measured every 30 minutes. The results of 20 consecutive measurements are as follows: $$\begin{array}{cccc}\hline \text { Sample } & \text { Heart Rate } & \text { Sample } & \text { Heart Rate } \\\\\hline 1 & 68 & 11 & 79 \\\2 & 71 & 12 & 79 \\\3 & 67 & 13 & 78 \\\4 & 69 & 14 & 78 \\\5 & 71 & 15 & 78 \\\6 & 70 & 16 & 79 \\\7 & 69 & 17 & 79 \\\8 & 67 & 18 & 82 \\\9 & 70 & 19 & 82 \\\10 & 70 & 20 & 81 \\\\\hline\end{array}$$ Suppose that the standard deviation of the heart rate is \(\sigma=3\) and the target value is 70 . (a) Design a CUSUM scheme for the heart rate process. Does the process appear to be in control at the target? (b) How many samples on average would be required to detect a shift of the mean heart rate to \(80 ?\)

The thickness of a metal part is an important quality parameter. Data on thickness (in inches) are given in the following table, for 25 samples of five parts each. $$\begin{array}{cccccc}\hline \begin{array}{l}\text { Sample } \\\\\text { Number }\end{array} & x_{1} & x_{2} & x_{3} & x_{4} & x_{5} \\\\\hline 1 & 0.0629 & 0.0636 & 0.0640 & 0.0635 & 0.0640 \\\2 & 0.0630 & 0.0631 & 0.0622 & 0.0625 & 0.0627 \\\3 & 0.0628 & 0.0631 & 0.0633 & 0.0633 & 0.0630 \\\4 & 0.0634 & 0.0630 & 0.0631 & 0.0632 & 0.0633 \\\5 & 0.0619 & 0.0628 & 0.0630 & 0.0619 & 0.0625 \\\6& 0.0613 & 0.0629 & 0.0634 & 0.0625 & 0.0628 \\\7 & 0.0630 & 0.0639 & 0.0625 & 0.0629 & 0.0627 \\\8 & 0.0628 & 0.0627 & 0.0622 & 0.0625 & 0.0627 \\\9 & 0.0623 & 0.0626 & 0.0633 & 0.0630 & 0.0624 \\\10 & 0.0631 & 0.0631 & 0.0633 & 0.0631 & 0.0630 \\\11 & 0.0635 & 0.0630 & 0.0638 & 0.0635 & 0.0633 \\\12 & 0.0623 & 0.0630 & 0.0630 & 0.0627 & 0.0629 \\\13 & 0.0635 & 0.0631 & 0.0630 & 0.0630 & 0.0630 \\\14 & 0.0645 & 0.0640 & 0.0631 & 0.0640 & 0.0642 \\\15 & 0.0619 & 0.0644 & 0.0632 & 0.0622 & 0.0635 \\\16 & 0.0631 & 0.0627 & 0.0630 & 0.0628 & 0.0629 \\\17 & 0.0616 & 0.0623 & 0.0631 & 0.0620 & 0.0625 \\\18 & 0.0630 & 0.0630 & 0.0626 & 0.0629 & 0.0628 \\\19 & 0.0636 & 0.0631 & 0.0629 & 0.0635 & 0.0634 \\\20 & 0.0640 & 0.0635 & 0.0629 & 0.0635 & 0.0634 \\\21 & 0.0628 & 0.0625 & 0.0616 & 0.0620 & 0.0623 \\\22 & 0.0615 & 0.0625 & 0.0619 & 0.0619 & 0.0622 \\\23 & 0.0630 & 0.0632 & 0.0630 & 0.0631 & 0.0630 \\\24 & 0.0635 & 0.0629 & 0.0635 & 0.0631 & 0.0633 \\\25 & 0.0623 & 0.0629 & 0.0630 & 0.0626 & 0.0628 \\\\\hline\end{array}$$ (a) Using all the data, find trial control limits for \(\bar{X}\) and \(R\) charts, construct the chart, and plot the data. Is the process in statistical control? (b) Use the trial control limits from part (a) to identify outof-control points. If necessary, revise your control limits assuming that any samples that plot outside the control limits can be eliminated. (c) Repeat parts (a) and (b) for \(\bar{X}\) and \(S\) charts.

An article in Journal of the Operational Research Society ["A Quality Control Approach for Monitoring Inventory Stock Levels" (1993, pp. \(1115-1127\) ) ] reported on a control chart to monitor the accuracy of an inventory management system. Inventory accuracy at time \(t, A C(t),\) is defined as the difference between the recorded and actual inventory (in absolute value) divided by the recorded inventory. Consequently, \(A C(t)\) ranges between 0 and 1 with lower values better. Extracted data are shown in the following table. (a) Calculate individuals and moving-range charts for these data. (b) Comment on the control of the process. If necessary, assume that assignable causes can be found, eliminate suspect points, and revise the control limits. $$\begin{array}{lcccccccccc}\hline t & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\\A C(t) & 0.190 & 0.050 & 0.095 & 0.055 & 0.090 & 0.200 & 0.030 & 0.105 & 0.115 & 0.103 & 0.121 \\\t & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\\A C(t) & 0.089 & 0.180 & 0.122 & 0.098 & 0.173 & 0.298 &0.075 & 0.083 & 0.115 & 0.147 & 0.079\end{array}$$

A article in Graefe's Archive for Clinical and Experimental Ophthalmology ["Statistical Process Control Charts for Ophthalmology," (2011, Vol. 249, pp. \(1103-1105\) ) ] considered the number of cataract surgery cases by month. The data are shown in the following table. (a) What type of control chart is appropriate for these data? Construct this chart. (b) Comment on the control of the process. (c) If necessary, assume that assignable causes can be found, eliminate suspect points, and revise the control limits. (d) In the publication, the data were approximated as normally distributed and an individuals chart was constructed. Construct this chart and compare it to the attribute chart you built in part (a). Why might an individuals chart be reasonable? $$\begin{array}{ccccccc}\hline \text { January } & \text { February } & \text { March } & \text { April } & \text {May } & \text { June } & \text { July } \\\61 & 88 & 80 & 68 & 80 & 70 & 60 \\\\\text { August } & \text { September } & \text { October } & \text { November } & \text { December } & & \\\56 & 72 & 118 & 106 & 60 & & \\\\\hline\end{array}$$

The \(P C R\) for a measurement is 1.5 and the control limits for an \(\bar{X}\) chart with \(n=4\) are 24.6 and 32.6 . (a) Estimate the process standard deviation \(\sigma\). (b) Assume that the specification limits are centered around the process mean. Calculate the specification limits.
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