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Chapter 15: Problem 46

A control chart for individual observations has 3 -sigma control limits \(U C L=1.80\) and \(L C L=1.62 .\) The process specification limits are (1.64,1.84) . (a) Estimate the process standard deviation. (b) Calculate \(P C R\) and \(P C R_{k}\) for the process.

Short Answer

Expert verified
(a) Standard deviation is 0.06. (b) PCR is 0.5556 and PCRk is 0.3889.

Step by step solution

01

Gather Required Formulas

To estimate the process standard deviation and calculate the Process Capability Ratio (PCR) and the Process Capability Ratio adjusted for centering (PCRk), we need the following formulas:- Process Standard Deviation: \[ \sigma = \frac{UCL - LCL}{3} \]- PCR: \[ PCR = \frac{USL - LSL}{6\sigma} \]- PCRk: \[ PCR_k = \min\left(\frac{USL - \bar{x}}{3\sigma}, \frac{\bar{x} - LSL}{3\sigma}\right) \]
02

Estimate the Process Standard Deviation

Utilize the control limits to estimate the standard deviation:\[ \sigma = \frac{1.80 - 1.62}{3} = \frac{0.18}{3} = 0.06 \]
03

Calculate PCR

Use the process specification limits and the calculated standard deviation to find the PCR:- USL (Upper Specification Limit) = 1.84- LSL (Lower Specification Limit) = 1.64\[ PCR = \frac{1.84 - 1.64}{6 \times 0.06} = \frac{0.20}{0.36} = 0.5556 \]
04

Calculate PCRk

First, calculate the centered process mean \(\bar{x}\) which is the midpoint of the control limits:\[ \bar{x} = \frac{1.80 + 1.62}{2} = 1.71 \]Next, compute the PCRk using the standard deviation and the mean:\[ PCR_k = \min\left(\frac{1.84 - 1.71}{3 \times 0.06}, \frac{1.71 - 1.64}{3 \times 0.06}\right) = \min(0.7222, 0.3889) = 0.3889 \]

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Key Concepts

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

Control Charts
Control charts are an essential tool in statistical quality control. They help organizations monitor the stability and consistency of their processes over time. The chart consists of data points plotted in time order, which show insights into process variations. By examining these charts, any anomalies or deviations from the norm can be quickly detected, allowing for corrective actions.
  • Upper Control Limit (UCL) and Lower Control Limit (LCL) mark the boundaries of expected variation.
  • Data points falling outside these limits indicate potential process issues.
Control charts not only track data but also guide improvement efforts by pinpointing variation causes. They play a pivotal role in decision-making processes concerning quality improvements.
Process Standard Deviation
The concept of process standard deviation is crucial for understanding the spread or variability in a process. Standard deviation quantifies the variation present in a set of data. In process capability analysis, it helps determine how much variability exists in a manufacturing process.
Given the control limits in the exercise, the formula for estimating the process standard deviation is \[ \sigma = \frac{UCL - LCL}{3} \] This calculation provides an estimate of the variability inherent in the process. The result can then be used to evaluate how consistently a process can meet specification requirements based on this level of variation.
Process Capability Ratio (PCR)
The Process Capability Ratio (PCR) is a common metric used in quality control to assess how well a process can produce outputs within specified limits. It provides insight into the process's ability to produce products to meet specifications and is a key indicator of process potential.
The formula for PCR is \[ PCR = \frac{USL - LSL}{6\sigma} \] where USL is the Upper Specification Limit and LSL is the Lower Specification Limit. A PCR value greater than 1 indicates that the process is capable of producing outputs within specification limits consistently. However, if the PCR is less than 1, it suggests the process will frequently produce defects.
Specification Limits
Specification limits are the defined range of values that a process output should meet to satisfy customer requirements. These limits are essential in determining whether a product is acceptable or if it needs adjustments.
  • Upper Specification Limit (USL) is the maximum acceptable value for the process output.
  • Lower Specification Limit (LSL) is the minimum acceptable value.
These limits differ from control limits as they are based on customer needs and not statistical control. In the context of capability analysis, comparing process performance against these specification limits shows how well the process meets customer expectations. Ensuring outputs fall within these limits underpins product quality and customer satisfaction.

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Most popular questions from this chapter

\(.\operatorname{An} \bar{X}\) 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 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.

Suppose that the average number of defects in a unit is known to be 8 . If the mean number of defects in a unit shifts to \(16,\) what is the probability that it is detected by a \(U\) chart on the first sample following the shift (a) if the sample size is \(n=4 ?\) (b) if the sample size is \(n=10 ?\) Use a normal approximation for \(U\).

Consider an \(\bar{X}\) control chart with \(\hat{\sigma}=2.466, U C L\) \(=37.404, L C L=30.780,\) and \(n=5 .\) Suppose that the mean shifts to 36 (a) What is the probability that this shift is detected on the next sample? (b) What is the ARL after the shift?

Suppose that a quality characteristic is normally distributed with specifications from 10 to 30 units. The process standard deviation is 2 units. (a) Calculate the natural tolerance limits, fraction defective, \(P C R,\) and \(P C R_{k}\) when the process mean is \(20 .\) (b) Suppose that the process mean shifts higher by 1.5 standard deviations. Recalculate the quantities in part (b). (c) Compare the results in parts (a) and (b) and comment on any differences.
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