/*! 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 47 A process mean is centered betwe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A process mean is centered between the specification limits and \(P C R=1.33 .\) Assume that the process mean increases by \(1.5 \sigma\). (a) Calculate \(P C R\) and \(P C R_{k}\) for the shifted process. (b) Calculate the estimated fallout from the shifted process and compare your result to those in Table \(15-4 .\) Assume a normal distribution for the measurement.

Short Answer

Expert verified
(a) \(\text{PCR} = 1.33\), \(\text{PCR}_k = -0.17\). (b) Estimated fallout is approximately 3.4 DPMO with a 1.5\(\sigma\) shift.

Step by step solution

01

Understand the Problem

We need to determine the new Process Capability Ratio (\(\text{PCR}\)) and the Process Capability Ratio adjusted for the position of the process mean (\(\text{PCR}_k\)) for a process whose mean has shifted by \(1.5\sigma\). We then need to compare the estimated fallout for this shifted process to an existing data table.
02

Initial Calculations of Shift in Process Mean

Initially, the process mean is centered and \(\text{PCR} = 1.33\). A shift of \(1.5\sigma\) implies that the process mean moves away from the center of the specification limits by \(1.5\sigma\).
03

Calculate New PCR_k

The formula for \(\text{PCR}_k\) is \(\text{PCR}_k = \frac{\text{PCR} - k}{\sigma}\), where \(k = \frac{|\mu - T|}{\sigma/2}\) is the shift in process mean from the target \(T\). Given the mean shifts by \(1.5\sigma\), \(k = 1.5\). Therefore, \(\text{PCR}_k = 1.33 - 1.5 = -0.17\).
04

Calculate New PCR After Shift

Since the process mean's shift does not directly impact the nominal PCR if not adjusted for the mean, the unadjusted new \(\text{PCR}\) stays the same as before, \(\text{PCR} = 1.33\) since it calculates capability irrespective of the mean shift.
05

Estimate Fallout from the Shifted Process

To estimate fallout, calculate the probability of defect outside the new limits after the shift of \(1.5\sigma\). Look up the areas falling outside of \([-3\sigma, 3\sigma]\), compare these calculations to Table 15-4, finding corresponding entries when a \(1.5\sigma\) shift is implemented, typically fallout increases substantially for such shifts.
06

Compare Estimated Fallout with Table 15-4

Table 15-4 provides standard fallout rates for various \(\sigma\) shifts. By calculating or referencing the standard normal distribution tables, find that a \(1.5\sigma\) shift typically results in fallout comparable to 3.4 defects per million opportunities, assuming \(\sigma = 1\) in standard units.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Process Capability Ratio
The Process Capability Ratio, often abbreviated as PCR, is a key metric in process capability analysis. It measures how well a process can produce outputs within specified limits, given that the process mean is perfectly centered between those limits. The formula for PCR is given by:\[PCR = \frac{USL - LSL}{6\sigma}\]Where:
  • \( USL \) = Upper Specification Limit
  • \( LSL \) = Lower Specification Limit
  • \( \sigma \) = Standard Deviation
A PCR of 1 means the process variation just fits within the specification limits, while a PCR greater than 1 indicates a process that consistently produces outputs within limits. In simpler terms, the higher the PCR, the more capable the process is of delivering products that meet quality standards.
Shift in Process Mean
A shift in process mean refers to any movement of the average output of the process from its target value. This shift can greatly affect both the outcome of the production and the process capability analysis. In this specific exercise, the process mean is noted to have shifted by 1.5 standard deviations (\( 1.5\sigma \)) from its original centered position.When such a shift occurs:
  • The process might start producing outputs closer to one of the specification limits.
  • This shift directly impacts the Process Capability Ratio adjusted for mean position, denoted as \( PCR_k \).
The shift penalizes the process capability, as seen with the decrease in the adjusted PCR value from the nominal PCR when not accounting for shifts.
Normal Distribution
The Normal Distribution is a fundamental concept in statistics and process capability analysis. It is a bell-shaped curve that describes the distribution of a continuous set of data points. Characteristics of this distribution include its symmetry around the mean and the spread determined by the standard deviation (\( \sigma \)).For process capability analysis:
  • It is often assumed that the measurement data of a process follows a normal distribution.
  • This assumption allows the use of standard statistical methods and tables to predict process outcomes and fallout rates.
In scenarios where the mean shifts, like in our exercise, knowledge of the normal distribution aids in estimating the defect probabilities and understanding how shifts affect the overall distribution of data points.
Defect Probability
Defect Probability refers to the likelihood of a process producing outputs that fall outside the specified limits. In the context of a shifted process, it becomes crucial as shifts often result in increased defect probabilities.To calculate this:1. Determine the new distribution of the process after the shift.2. Use standard normal distribution tables to find the probability associated with areas beyond the specification limits.For the given exercise, the shift of \(1.5\sigma\) increases the likelihood of defects. Typically, such a shift increases defect probability to around 3.4 defects per million, a standard benchmark for a process operating at Six Sigma quality level. Understanding defect probabilities helps in improving processes and minimizing waste and errors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a control chart for individuals applied to a continuous 24 -hour chemical process with observations taken every hour. (a) If the chart has 3-sigma limits, verify that the in-control ARL is \(370 .\) How many false alarms would occur each 30 -day month, on the average, with this chart? (b) Suppose that the chart has 2 -sigma limits. Does this reduce the ARL for detecting a shift in the mean of magnitude \(\sigma\) ? (Recall that the ARL for detecting this shift with 3 -sigma limits is \(43.9 .)\) (c) Find the in-control ARL if 2 -sigma limits are used on the chart. How many false alarms would occur each month with this chart? Is this in-control ARL performance satisfactory? Explain your answer.

Consider an \(\bar{X}\) control chart with \(\hat{\sigma}=1.40, U C L\) \(=21.71, L C L=18.29,\) and \(n=6 .\) Suppose that the mean shifts to \(17 .\) (a) What is the probability that this shift is detected on the next sample? (b) What is the ARL after the shift?

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.

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. $$\begin{array}{cccc}\hline \text { Wafer } & {x} & \text { Wafer } & {x} \\\\\hline 1 & 16.8 & 16 & 15.4 \\\2 & 14.9 & 17 & 14.3 \\\3 & 18.3 & 18 & 16.1 \\\4 & 16.5 & 19 & 15.8 \\\5 & 17.1 & 20 & 15.9 \\\6 & 17.4 & 21 & 15.2 \\\7 & 15.9 & 22 & 16.7 \\\8 & 14.4 & 23 & 15.2 \\\9 & 15.0 & 24 & 14.7 \\\10 & 15.7 & 25 & 17.9 \\\11 & 17.1 & 26 & 14.8 \\\12 & 15.9 & 27 & 17.0 \\\13 & 16.4 & 28 & 16.2 \\\14 & 15.8 & 29 & 15.6 \\\15 & 15.4 & 30 & 16.3\end{array}$$

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?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.