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Suppose that a \(P\) chart with center line at \(\bar{p}\) and \(k\) -sigma control limits is used to control a process. What is the smallest sample size that can be used on this control chart to ensure that the lower control limit is positive?

Short Answer

Expert verified
The smallest sample size is the smallest integer greater than \( \frac{k^2(1-\bar{p})}{\bar{p}} \).

Step by step solution

01

Understanding the Problem

The problem involves determining the sample size for a P-chart such that the lower control limit (LCL) is positive. The P-chart is typically used for monitoring the proportion of defective items in a process.
02

Formulas Involved

The center line for a P-chart is given by \( \bar{p} \). The control limits are defined by: \( LCL = \bar{p} - k\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \) and \( UCL = \bar{p} + k\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \). We need \( LCL > 0 \).
03

Setting Up the Inequality

To ensure the LCL is positive, set up the inequality: \( \bar{p} - k\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} > 0 \). Simplifying, we get: \( k\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} < \bar{p} \).
04

Solving for Sample Size

Rearrange the inequality to solve for \( n \): \( \sqrt{\frac{\bar{p}(1-\bar{p})}{n}} < \frac{\bar{p}}{k} \). Squaring both sides gives: \( \frac{\bar{p}(1-\bar{p})}{n} < \frac{\bar{p}^2}{k^2} \). Solving for \( n \), we have: \( n > \frac{k^2(1-\bar{p})}{\bar{p}} \).
05

Determine Smallest Sample Size

The smallest integer value satisfying the inequality \( n > \frac{k^2(1-\bar{p})}{\bar{p}} \) ensures that the LCL remains positive.

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Key Concepts

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

Control Limits
In the context of a P-chart, control limits are essential for monitoring the process's stability over time. These are the boundaries within which the process is considered to be in control. In simpler terms, control limits help us understand whether the process is working as expected or if there's something unusual happening.

For a P-chart, which is used specifically to track the proportion of defective items, the control limits are calculated using the formula:
  • Lower Control Limit (LCL): \( \bar{p} - k\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \)
  • Upper Control Limit (UCL): \( \bar{p} + k\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \)
Here, \( \bar{p} \) is the average proportion of defectives, \( k \) is the standard deviation multiplier, and \( n \) is the sample size. The value of \( k \) commonly used is 3, which corresponds to a 99.73% confidence interval under a normal distribution assumption.

To ensure the LCL is positive, which means that no natural variation would lead to negative proportions (since negative proportions don't make sense in reality), choosing an appropriate sample size is crucial. The inequality \( \bar{p} - k\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} > 0 \) guides us to find a suitable \( n \). This ensures the control limits are meaningful and the control chart is functional.
Process Monitoring
Process monitoring is essential to ensure that a product or service is produced according to quality standards. In the context of using a P-chart, it involves tracking the proportion of defective items over time and determining whether the process is consistent with past performance or if corrective actions are necessary.

By setting control limits on a P-chart, businesses and quality control specialists can visually track whether the proportion of defective items stays within the expected range. A common approach for effective monitoring includes:
  • Collecting consistent sample sizes at regular intervals.
  • Calculating the proportion of defectives in each sample.
  • Plotting these proportions against the control limits.
  • Taking corrective action when a data point falls outside the control limits, signalling a potential process issue.
Process monitoring helps avoid surprises by identifying trends or shifts in the process early. Thus, action can be taken before issues become significant, reducing defects and ensuring customer satisfaction.
Proportion of Defectives
In quality control, the proportion of defectives refers to the percentage of items in a sample that do not meet specified quality criteria. It is a critical metric used in P-charts to monitor process performance.

For instance, if a manufacturing process is designed to produce a certain component within specific tolerances, the proportion of defectives is the rate at which components fall outside these tolerances. This proportion is depicted by \( \bar{p} \) in the context of control charts.

To calculate \( \bar{p} \):
  • Count the number of defective items in the sample.
  • Divide by the total number of items to find the proportion.
This measure helps in understanding how well a process is performing. Lower proportions generally indicate better quality control, while higher proportions may signal a quality issue that needs attention.

Proportion of defectives is used to set the centerline on a P-chart, helping in the visual representation of whether a process is stable and in control.

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Most popular questions from this chapter

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 Quality \& Safety in Health Care ["Statistical Process Control as a Tool for Research and Healthcare Improvement," (2003) Vol. \(12,\) pp. \(458-464]\) considered a number of control charts in healthcare. The following approximate data were used to construct \(\bar{X}-S\) charts for the turn around time (TAT) for complete blood counts (in minutes). The subgroup size is \(n=3\) per shift, and the mean standard deviation is \(21 .\) Construct the \(\bar{X}\) chart and comment $$\begin{array}{cc|c|c|c|c|c|c|c|c|c|c|c|c|c|}t & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\\\\hline \text { TAT } & 51 & 73 & 28 & 52 & 65 & 49 & 51 & 50 & 25 & 39 & 40 & 30 & 49 & 31 \\\\\hline\end{array}$$ on the control of the process. If necessary, assume that assignable causes can be found, eliminate suspect points, and revise the control limits.

\(.\operatorname{An} \bar{X}\) chart uses samples of size \(4 .\) The center line is at 100 , and the upper and lower 3 -sigma control limits are at 106 and \(94,\) respectively. (a) What is the process \(\sigma\) ? (b) Suppose that the process mean shifts to \(96 .\) Find the probability that this shift is detected on the next sample. (c) Find the ARL to detect the shift in part (b).

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 \(n\).

In a semiconductor manufacturing process, CVD metal thickness was measured on 30 wafers obtained over approximately two weeks. Data are shown in the following table. (a) Using all the data, compute trial control limits for individual observations and moving-range charts. Construct the chart and plot the data. 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. $$\begin{array}{cccc}\hline \text { Wafer } & {x} & \text { Wafer } & {x} \\\\\hline 1 & 16.8 & 16 & 15.4 \\\2 & 14.9 & 17 & 14.3 \\\3 & 18.3 & 18 & 16.1 \\\4 & 16.5 & 19 & 15.8 \\\5 & 17.1 & 20 & 15.9 \\\6 & 17.4 & 21 & 15.2 \\\7 & 15.9 & 22 & 16.7 \\\8 & 14.4 & 23 & 15.2 \\\9 & 15.0 & 24 & 14.7 \\\10 & 15.7 & 25 & 17.9 \\\11 & 17.1 & 26 & 14.8 \\\12 & 15.9 & 27 & 17.0 \\\13 & 16.4 & 28 & 16.2 \\\14 & 15.8 & 29 & 15.6 \\\15 & 15.4 & 30 & 16.3\end{array}$$

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