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Cover cases for a personal computer are manufactured by injection molding. Samples of five cases are taken from the process periodically, and the number of defects is noted. Twenty-five samples follow: $$\begin{array}{cccc}\hline \text { Sample } & \text { No. of Defects } & \text { Sample } & \text { No. of Defects } \\\\\hline 1 & 3 & 14 & 8 \\\2 & 2 & 15 & 0 \\\3 & 0 & 16 & 2 \\\4 & 1 & 17 & 4 \\\5 & 4 & 18 & 3 \\\6 & 3 & 19 & 5 \\\7 & 2 & 20 & 0 \\\8 & 4 & 21 & 2 \\\9 & 1 & 22 & 1 \\\10 & 0 & 23 & 9 \\\11 & 2 & 24 & 3 \\\12 & 3 & 25 & 2 \\\13 & 2 & &\end{array}$$ (a) Using all the data, find trial control limits for a \(U\) chart for the process. (b) Use the trial control limits from part (a) to identify out-ofcontrol points. If necessary, revise your control limits. (c) Suppose that instead of samples of five cases, the sample size was 10. Repeat parts (a) and (b). Explain how this change alters your responses to parts (a) and (b).

Short Answer

Expert verified
Calculate \(\bar{u} = 2.56\), \(UCL = 4.805\), \(LCL = 0\). Sample 23 is out-of-control. Revise control limits if needed. Sample size change would adjust the control limits.

Step by step solution

01

Calculate Total Number of Defects and Samples

First, count the total number of defects across all samples and the number of samples. The data shows a total of 25 samples. Sum up the number of defects: \[ 3 + 2 + 0 + 1 + 4 + 3 + 2 + 4 + 1 + 0 + 2 + 3 + 2 + 8 + 0 + 2 + 4 + 3 + 5 + 0 + 2 + 1 + 9 + 3 + 2 = 64 \] So, there are 64 defects in total.
02

Calculate Average Defect Rate

Calculate the average number of defects per sample (\( \bar{u} \)) using the total number of defects and samples. \[ \bar{u} = \frac{64}{25} = 2.56 \]
03

Calculate Control Limits for U Chart

The control limits for a \( U \) chart are given by \[ UCL = \bar{u} + 3 \sqrt{\frac{\bar{u}}{n}} \] \[ LCL = \bar{u} - 3 \sqrt{\frac{\bar{u}}{n}} \] where \( n = 5 \). Substitute the values to find:\[ UCL = 2.56 + 3 \sqrt{\frac{2.56}{5}} = 2.56 + 3(0.715) = 4.805 \] \[ LCL = 2.56 - 3 \sqrt{\frac{2.56}{5}} = 2.56 - 3(0.715) = 0.315 \]Note that LCL is often set to zero if it is negative, as defects cannot be negative.
04

Identify Out-of-Control Points

Check each sample's number of defects against the control limits. Any sample with defects greater than 4.805 or less than 0 is considered out-of-control. Sample 23 having 9 defects exceeds the UCL and is thus out-of-control.
05

Revise Control Limits (if necessary)

Given that sample 23 was identified as out-of-control, remove this sample from the dataset and recalculate \(\bar{u}\) and control limits from Step 2 and Step 3 with the revised dataset.
06

Repeat Steps for Detailed Sample Size Change

If the sample size changes to 10, recalculate the average number of defects and determine new control limits:\[ \bar{u}' = \frac{64}{25} = 2.56 \] Adjust \(UCL\) and \(LCL\) with a sample size of \(n = 10\):\[ UCL' = 2.56 + 3 \sqrt{\frac{2.56}{10}} = 3.52 \] \[ LCL' = 2.56 - 3 \sqrt{\frac{2.56}{10}} = 1.60 \]

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Key Concepts

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

Control Limits
In quality control, particularly when dealing with processes like injection molding, understanding control limits on a U Chart is crucial. Control limits are statistical boundaries that help monitor whether a process is in control or needs attention. These limits help ensure that the quality of the product is maintained by identifying deviations from the expected pattern.

Control limits consist of the Upper Control Limit (UCL) and Lower Control Limit (LCL). In the case of a U Chart, which monitors attribute data like defects per sample unit, these are calculated based on the average rate of defects and sample size. The general formula for the control limits of a U Chart is:
  • Upper Control Limit (UCL): \[ UCL = \bar{u} + 3 \sqrt{\frac{\bar{u}}{n}} \]
  • Lower Control Limit (LCL): \[ LCL = \bar{u} - 3 \sqrt{\frac{\bar{u}}{n}} \]
It's important to note that if the LCL is a negative number, it is typically set to zero because it is not possible to have a negative number of defects.

By comparing the number of defects in each sample against these control limits, you can quickly identify whether a sample falls outside the expected range, which signals an out-of-control condition. This helps in taking timely corrective measures to improve process stability and control.
Defect Rate
The defect rate is a fundamental concept in quality control that measures the frequency of defects in a process. Calculating the defect rate allows businesses to assess the effectiveness of their production processes, identify areas for improvement, and maintain high-quality standards.

The defect rate can be expressed as the average number of defects per unit. For a series of samples, this is represented as \( \bar{u} \), calculated by dividing the total number of defects by the number of samples collected:
\[ \bar{u} = \frac{\text{Total Defects}}{\text{Total Samples}} \]

In the example of injection molding, where samples are taken periodically, each sample's number of defects is counted and the average is computed across all samples. This average is crucial as it acts as the benchmark for setting the control limits on the U Chart. When you have a grip on the defect rate, it becomes easier to monitor changes and unexpected trends in the production line.

Understanding the defect rate is not only vital for measurement purposes but also plays a role in continuous improvement practices. By striving to reduce defect rates, companies can enhance product quality, reduce waste, and ensure more satisfied customers.
Injection Molding Quality Control
Injection molding is a popular manufacturing process for producing parts with complex shapes from plastic materials. Like any other manufacturing process, it is subject to variations and potential defects. Quality control in injection molding focuses on maintaining consistent production quality, reducing defects, and ensuring that the final product meets design specifications.

To achieve effective quality control in injection molding, it is important to regularly monitor the number of defects. This typically involves collecting samples at different stages of the process and utilizing statistical tools such as U Charts to track performance. By using these tools, manufacturers can quickly detect any variation that may need correcting.

The role of U Charts in injection molding quality control is significant as they provide a visual representation of the process stability over time. By plotting defects per sample on the chart and comparing against control limits, managers can determine whether the process is in control or if adjustments are needed.

Regularly updating control limits based on accurate and current data helps better accommodate process variations and can lead to continual improvements in quality. Through systematic quality control practices like these, manufacturers are able to ensure consistent delivery of high-quality products, reduce waste, and enhance customer satisfaction.

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Most popular questions from this chapter

Consider a \(P\) -chart with subgroup size \(n=50\) and center line at 0.12 . (a) Calculate the \(L C L\) and \(U C L\). (b) Suppose that the true proportion defective changes from 0.12 to \(0.18 .\) What is the ARL after the shift? Assume that the sample proportions are approximately normally distributed. (c) Rework part (a) and (b) with \(n=100\) and comment on the difference in ARL. Does the increased sample size change the ARL substantially?

The diameter of fuse pins used in an aircraft engine application is an important quality characteristic. Twenty-five samples of three pins each are shown as follows: $$\begin{array}{cccc}\hline \begin{array}{c}\text { Sample } \\\\\text { Number }\end{array} & & \text { Diameter } & \\\\\hline 1 & 64.030 & 64.002 & 64.019 \\\2 & 63.995 & 63.992 & 64.001 \\\3 & 63.988 & 64.024 & 64.021 \\\4 & 64.002 & 63.996 & 63.993 \\\5 & 63.992 & 64.007 & 64.015 \\\6 & 64.009 &63.994 & 63.997 \\\7 & 63.995 & 64.006 & 63.994 \\\8 & 63.985 & 64.003 & 63.993 \\ 9 & 64.008 & 63.995 & 64.009 \\\10 & 63.998 & 74.000 & 63.990 \\\11 & 63.994 & 63.998 & 63.994 \\\12 & 64.004 & 64.000 & 64.007 \\\13 & 63.983 & 64.002 & 63.998 \\\14 & 64.006 & 63.967 & 63.994 \\\15 & 64.012 & 64.014 & 63.998 \\\16 & 64.000 & 63.984 & 64.005 \\\17 & 63.994 & 64.012 & 63.986 \\\18 & 64.006 & 64.010 & 64.018 \\\19 & 63.984 & 64.002 & 64.003 \\\20 & 64.000 & 64.010 & 64.013 \\\21 & 63.988 & 64.001 & 64.009 \\\22 & 64.004 & 63.999 & 63.990 \\\23 & 64.010 & 63.989 &63.990\\\24 & 64.015 & 64.008 & 63.993 \\\25 & 63.982 & 63.984 & 63.995\end{array}$$ (a) Set up \(\bar{X}\) and \(R\) charts for this process. If necessary, revise limits so that no observations are out of control. (b) Estimate the process mean and standard deviation. (c) Suppose that the process specifications are at \(64 \pm 0.02\). Calculate an estimate of \(P C R\). Does the process meet a minimum capability level of \(P C R \geq 1.33 ?\) (d) Calculate an estimate of \(P C R_{k}\). Use this ratio to draw conclusions about process capability. (e) To make this process a 6-sigma process, the variance \(\sigma^{2}\) would have to be decreased such that \(P C R_{k}=2.0 .\) What should this new variance value be? (f) Suppose that the mean shifts to \(64.01 .\) What is the probability that this shift is detected on the next sample? What is the ARL after the shift?

Suppose that a quality characteristic is normally distributed with specifications at \(100 \pm 20 .\) The process standard deviation is 6 . (a) Suppose that the process mean is \(100 .\) 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 is \(106 .\) What are the natural tolerance limits? What is the fraction defective? Calculate \(P C R\) and \(P C R_{c}\) and interpret these ratios.

An early example of SPC was described in Industrial Quality Control ["The Introduction of Quality Control at Colonial Radio Corporation" (1944, Vol. 1(1), pp. \(4-9\) ) . The following are the fractions defective of shaft and washer assemblies during the month of April in samples of \(n=1500\) each $$\begin{array}{cccc}\hline & \text { Fraction } & & \text { Fraction } \\\\\text { Sample } & \text { Defective } & \text { Sample } & \text { Defective } \\\\\hline 1 & 0.11 & 11 & 0.03 \\\2 & 0.06 & 12 & 0.03 \\\3 & 0.1 & 13 & 0.04 \\\4 & 0.11 & 14 & 0.07 \\\5 & 0.14 & 15 & 0.04 \\\6 & 0.11 & 16 & 0.04 \\\7 & 0.14 & 17 & 0.04 \\\8 & 0.03 & 18 & 0.03 \\\9 & 0.02 & 19 & 0.06 \\\10 & 0.03 & 20 & 0.06\end{array}$$ (a) Set up a \(P\) chart for this process. Is this process in statistical control? (b) Suppose that instead of \(n=1500, n=100\). Use the data given to set up a \(P\) chart for this process. Revise the control limits if necessary. (c) Compare your control limits for the \(P\) charts in parts (a) and (b). Explain why they differ. Also, explain why your assessment about statistical control differs for the two sizes of \(n\).

The following data are the number of spelling errors detected for every 1000 words on a news Web site over 20 weeks. $$\begin{array}{cccc}\hline \text { Week } & \begin{array}{l}\text { No. of Spelling } \\ \text { Errors }\end{array} & \text { Week } & \begin{array}{l}\text { No. of Spelling } \\\\\text { Errors }\end{array} \\\\\hline 1 & 3 & 11 & 1 \\\2 & 6 & 12 & 6 \\\3 & 0 & 13 & 9 \\\4 & 5 & 14 & 8 \\\5 & 9 & 15 & 6 \\\6 & 5 & 16 & 4 \\\7 & 2 & 17 & 13 \\\8 & 2 & 18 & 3 \\\9 & 3 & 19 & 0 \\\10 & 2 & 20 & 7 \\\\\hline\end{array}$$ (a) What control chart is most appropriate for these data? (b) Using all the data, compute trial control limits for the chart in part (a), construct the chart, and plot the data. (c) Determine whether the process is in statistical control. If not, assume that assignable causes can be found and outof-control points eliminated. Revise the control limits.

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