/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 An early example of SPC was desc... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(n\).

Short Answer

Expert verified
For both sample sizes, the process is statistically in control. The control limits are narrower for larger samples (n=1500) compared to smaller ones (n=100), affecting the sensitivity of detecting shifts.

Step by step solution

01

Calculate Average Fraction Defective

The average fraction defective, \( \bar{p} \), is calculated by summing up all the given fractions defective and then dividing by the number of samples. \[ \bar{p} = \frac{0.11 + 0.06 + 0.1 + 0.11 + 0.14 + 0.11 + 0.14 + 0.03 + 0.02 + 0.03 + 0.03 + 0.03 + 0.04 + 0.07 + 0.04 + 0.04 + 0.04 + 0.03 + 0.06 + 0.06}{20} \approx 0.066 \]
02

Calculate Control Limits for n=1500

Control limits for a \(P\) chart are given by:\[ UCL = \bar{p} + 3\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \]\[ LCL = \bar{p} - 3\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \]Substitute \(\bar{p}\) and \(n=1500\) to calculate.\[ UCL \approx 0.066 + 3\sqrt{\frac{0.066 \times 0.934}{1500}} \approx 0.090 \]\[ LCL \approx 0.066 - 3\sqrt{\frac{0.066 \times 0.934}{1500}} \approx 0.042 \]
03

Evaluate Process Control for n=1500

Plot each sample's fraction defective against the control limits. If all points lie within the limits, the process is in control. In this case, all points fall between 0.042 and 0.090, indicating that the process is in statistical control with \(n=1500\).
04

Calculate Control Limits for n=100

Recalculate the control limits using \(n=100\) with the same steps as in Step 2.\[ UCL \approx \bar{p} + 3\sqrt{\frac{\bar{p}(1-\bar{p})}{100}} \approx 0.147 \]\[ LCL \approx \bar{p} - 3\sqrt{\frac{\bar{p}(1-\bar{p})}{100}} \approx -0.015 \]The LCL is less than 0, so we set it to 0.
05

Evaluate Process Control for n=100

Plot each sample's fraction defective against the new (revised) control limits (0 to 0.147). With \(n=100\), none of the points exceed the new UCL, meaning the process is still in control for this sample size.
06

Compare Control Limits for Different n

For larger \(n=1500\), control limits are narrower compared to \(n=100\). A larger sample size reduces variability in \(\bar{p}\), leading to tighter control limits. This means the assessment on whether the process is in control varies due to the differing sensitivity of the control limits to sample size.

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Key Concepts

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

P chart
A P chart, or proportion chart, is a type of control chart used to monitor the proportion of defective items in a process, based on sample data. This is particularly useful in quality control because it helps to identify variations in a process that could indicate a loss of control or performance. The P chart uses the fraction defective, represented as \(p\), calculated by dividing the number of defective items by the total number of items in a sample.

In the exercise example, we're working with fractions defective over several samples, and each sample helps in assessing whether the process is stable. The P chart considers both the variability inherent in the process and the sample size while setting the limits within which the samples should ideally fall. When all points on a P chart lie within the control limits, it suggests the process is operating normally and predictably without any signals of out-of-control situations.
Control limits
Control limits are the boundaries set on a control chart, like a P chart, which help in identifying whether a process is in control or not. These limits are set at three standard deviations (±3σ) from the process average, providing us with the Upper Control Limit (UCL) and the Lower Control Limit (LCL). The formula for the control limits in a P chart involves the average fraction defective \( \bar{p} \) and the sample size \( n \).

The equations used to calculate the control limits are:
  • UCL: \( \bar{p} + 3\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \)
  • LCL: \( \bar{p} - 3\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \)
For practical purposes, if the calculated LCL turns out negative, it is set to zero. These limits are crucial as they guide whether observed process behavior can be attributed to random variation or indicates potential issues needing correction.
Fraction defective
The fraction defective represents the part or proportion of items in a sample that do not meet quality standards or specifications. It is a primary focus in quality control, as monitoring these defects helps in maintaining the desired level of quality in production.

To calculate the fraction defective for a given sample, divide the number of defective items by the total number of items in that sample. This simple but essential calculation allows us to quantify how often defects occur, which is crucial for setting up control charts like the P chart. Understanding the fraction defective helps in evaluating whether a process meets required quality norms and identifying any need for improvements or interventions to enhance process reliability.
Sample size effects
Sample size plays a critical role in statistical process control, especially when using a P chart. The size of a sample, denoted by \( n \), directly influences the variability we observe and thus affects the control limits. A larger sample size tends to provide a more accurate estimate of the process average and results in narrower control limits because it reduces the effect of variability.

In contrast, smaller sample sizes may lead to wider control limits, as seen in the original exercise with sample sizes of \( n = 1500 \) and \( n = 100 \). For \( n = 1500 \), the control limits are tighter, reducing the chance of misinterpreting random variability as out-of-control conditions. With \( n = 100 \), the limits are broader, making the process appear more variable and potentially showing different control status due to the increased sensitivity to variability. Understanding these effects helps in selecting the appropriate sample size for accurate monitoring and control of processes.

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Most popular questions from this chapter

A process is controlled by a \(P\) chart using samples of size 100 . The center line on the chart is \(0.05 .\) What is the probability that the control chart detects a shift to 0.08 on the first sample following the shift? What is the probability that the shift is detected by at least the third sample following the shift?

The following dataset was considered in Quality Engineering ["Analytic Examination of Variance Components" \((1994-1995,\) Vol. 7(2)\(] .\) A quality characteristic for cement mortar briquettes was monitored. Samples of size \(n=6\) were taken from the process, and 25 samples from the process are shown in the following table: (a) Using all the data, calculate trial control limits for \(\bar{X}\) and \(S\) charts. Is the process in control? $$\begin{array}{ccc}\hline \text { Batch } & \bar{X} & s \\\\\hline 1 & 572.00 & 73.25 \\\2 & 583.83 & 79.30 \\\3 & 720.50 & 86.44 \\\4 & 368.67 & 98.62 \\\5 & 374.00 & 92.36 \\\6 & 580.33 & 93.50 \\\7 & 388.33 & 110.23 \\\8 & 559.33 & 74.79 \\\9 & 562.00 & 76.53 \\\10 & 729.00 & 49.80 \\\11 & 469.00 & 40.52\end{array}$$ $$\begin{array}{lll}12 & 566.67 & 113.82 \\\13 & 578.33 & 58.03 \\\14 & 485.67 & 103.33 \\\15 & 746.33 & 107.88 \\\16 & 436.33 & 98.69 \\\17 & 556.83 & 99.25 \\\18 & 390.33 & 117.35 \\\19 & 562.33 & 75.69 \\\20 & 675.00 & 90.10 \\\21 & 416.50 & 89.27 \\\22 & 568.33 & 61.36 \\\23 & 762.67 & 105.94 \\\24 & 786.17 & 65.05 \\\25 & 530.67 & 99.42\end{array}$$ (b) Suppose that the specifications are at \(580 \pm 250 .\) What statements can you make about process capability? Compute estimates of the appropriate process capability ratios. (c) 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? (d) Suppose the mean shifts to \(600 .\) What is the probability that this shift is detected on the next sample? What is the ARL after the shift?

Suppose that a \(P\) chart with center line at \(\bar{p}\) with \(k\) -sigma control limits is used to control a process. There is a critical fraction defective \(p_{c}\) that must be detected with probability 0.50 on the first sample following the shift to this state. Derive a general formula for the sample size that should be used on this chart.

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.

The production manager of a soap manufacturing company wants to monitor the weights of the bars produced on the line. Twenty bars are taken during a stable period of the process. The weights of the bars are shown in the following table. $$\begin{array}{cccc}\hline \text { Bar } & \text { Weight }(\mathrm{g}) & \text { Bar } & \text { Weight }(\mathrm{g}) \\\\\hline 1 & 74 & 11 & 99 \\\2 & 82 & 12 & 75 \\\3 & 97 & 13 & 77 \\\4 & 86 & 14 & 82 \\\5 & 71 & 15 & 93 \\\6 & 68 & 16 & 70 \\\7 & 83 & 17 & 87 \\\8 & 90 & 18 & 76 \\\9 & 88 & 19 & 84 \\\10 & 64 & 20 & 94\end{array}$$ (a) Using all the data, compute trial control limits for individual observations and moving-range charts. Determine whether the process is in statistical control. If not, assume that assignable causes can be found to eliminate these samples, and revise the control limits. (b) Estimate the process mean and standard deviation for the in-control process.

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