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

Short Answer

Expert verified
(a) \(\sigma = 4\). (b) Probability is approximately 0.1587. (c) ARL is approximately 6.30.

Step by step solution

01

Understand the Control Chart

The \( \bar{X} \) chart is used to monitor the mean of a process. When using samples of size \( n = 4 \), the central line (mean of the process) is \(\mu = 100\). The control limits are \( UCL = 106 \) and \( LCL = 94 \). These limits are determined by the equation \( \mu \pm 3\frac{\sigma}{\sqrt{n}}\).
02

Calculate the Process Standard Deviation (σ)

To find the process standard deviation \( \sigma \), set up the equation for the control limits: \[UCL = \mu + 3\frac{\sigma}{\sqrt{n}} = 106\] Substitute \( \mu = 100 \) and \( n = 4 \) into the equation:\[106 = 100 + 3\frac{\sigma}{\sqrt{4}} \106 = 100 + 3\frac{\sigma}{2} \6 = 1.5\sigma \\sigma = 4 \]
03

Determine the Detection of Mean Shift to 96

When the mean shifts to 96, calculate the Z-score to determine the probability of detecting this shift. Use:\[Z = \frac{\text{Observed Mean} - \text{New Process Mean}}{\sigma/\sqrt{n}} \]For detection, we use the LCL since the mean decreased:\[Z = \frac{94 - 96}{4/2} = \frac{-2}{2} = -1\]Find the probability \(P(Z < -1)\) using the standard normal distribution table, which gives approximately 0.1587.
04

Calculate ARL (Average Run Length)

The ARL can be calculated as the reciprocal of the probability of detecting the shift:\[ARL = \frac{1}{P(Z < -1)} = \frac{1}{0.1587} \approx 6.30\]This means that on average, it will take about 6.30 samples to detect the shift.

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Key Concepts

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

Process Standard Deviation Calculation
In a control chart analysis, understanding the calculation of process standard deviation is crucial.
It helps in setting accurate control limits which tell us when a process may be going out of control.
For an \( \bar{X} \) chart, these control limits are calculated using the equation \( \mu \pm 3\frac{\sigma}{\sqrt{n}} \), where \( \mu \) is the mean and \( n \) is the sample size.
To find \( \sigma \), we use the known values of the upper (UCL) and lower (LCL) control limits.Let's say UCL is 106 and LCL is 94, with \(\mu = 100\) and \( n = 4 \).
We plug these into the formula for the UCL:
  • \( 106 = 100 + 3\frac{\sigma}{2} \)
  • This simplifies to \( 6 = 1.5\sigma \)
  • Solving gives \( \sigma = 4 \)
This standard deviation reflects the inherent variability of the process under control conditions.
It's important because it tells us how much variation is expected when the process is in control.
Mean Shift Detection
Detecting shifts in the process mean is a key aspect of control chart analysis.
When the process mean shifts, it could indicate a problem or change in the process that needs to be addressed.
This detection usually involves calculating a Z-score, which shows how far from the mean a given point is.In this scenario, after a mean shift to 96 from 100, we use the lower control limit (94) to detect the shift:
  • The formula for Z-score is \( Z = \frac{\text{Observed Mean} - \text{New Process Mean}}{\sigma/\sqrt{n}} \)
  • Using \( Z = \frac{94 - 96}{4/2} \)
  • This simplifies to \( Z = -1 \)
The Z-score tells us how many standard deviations the observed mean is from the new mean.
In a standard normal distribution table, \( P(Z < -1) \approx 0.1587 \), which represents the probability that the mean shift will be detected.
This means there is about a 15.87% chance of detecting this shift at the next sample.
Average Run Length Calculation
The Average Run Length (ARL) is an important metric in control chart analysis.
It predicts how many samples, on average, will be taken before an out-of-control signal is detected.
The ARL helps practitioners estimate responsiveness of the control chart.To calculate ARL, use the reciprocal of the probability of detecting a mean shift:
  • The formula is \( ARL = \frac{1}{P} \), where \( P \) is the probability of detection.
  • In the mean shift scenario with \( P(Z < -1) \approx 0.1587 \), we compute \( ARL = \frac{1}{0.1587} \)
  • This results in approximately \( ARL = 6.30 \)
This ARL indicates that it will take about 6.3 samples, on average, to detect the mean shift.
Understanding ARL allows quality engineers to balance the speed of detection against false alarms.
A low ARL offers quick detection but might increase false alarms, whereas a high ARL reduces false alarms but delays detection.

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Most popular questions from this chapter

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 \(P C R\) for a measurement is 1.5 and the control limits for an \(\bar{X}\) chart with \(n=4\) are 24.6 and 32.6 . (a) Estimate the process standard deviation \(\sigma\). (b) Assume that the specification limits are centered around the process mean. Calculate the specification limits.

Control charts are to be constructed for samples of size \(n=4,\) and \(\bar{x}\) and \(s\) are computed for each of 20 preliminary samples as follows: \(\sum_{i=1}^{20} \bar{x}_{i}=4460 \quad \sum_{i=1}^{20} s_{i}=271.6\) (a) Calculate trial control limits for \(\bar{X}\) and \(S\) charts. (b) Assuming the process is in control, estimate the process mean and standard deviation.

Suppose that a process is in control and an \(\bar{X}\) chart is used with a sample size of 4 to monitor the process. Suddenly there is a mean shift of \(1.5 \sigma .\) (a) If 3 -sigma control limits are used on the \(\bar{X}\) chart, what is the probability that this shift remains undetected for three consecutive samples? (b) If 2 -sigma control limits are in use on the \(\bar{X}\) chart, what is the probability that this shift remains undetected for three consecutive samples? (c) Compare your answers to parts (a) and (b) and explain why they differ. Also, which limits you would recommend using and why?

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?

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