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An \(\bar{X}\) control chart with three-sigma control limits has \(U C L=48.75\) and \(L C L=42.71 .\) Suppose that the process standard deviation is \(\sigma=2.25 .\) What subgroup size was used for the chart?

Short Answer

Expert verified
The subgroup size used for the chart is 5.

Step by step solution

01

Understand Control Limits Formula

For an \(\bar{X}\) control chart, the control limits can be calculated using the formula: \[ UCL = \mu + z \cdot \frac{\sigma}{\sqrt{n}}, \quad LCL = \mu - z \cdot \frac{\sigma}{\sqrt{n}} \] where \(\mu\) is the process mean, \(z\) is the number of standard deviations from the mean (in this case, 3 for three-sigma limits), \(\sigma\) is the process standard deviation, and \(n\) is the subgroup size.
02

Calculate Range for Mean

The difference between the \(UCL\) and \(LCL\) is related to \(3 \cdot \frac{\sigma}{\sqrt{n}}\).\[ UCL - LCL = 2z \cdot \frac{\sigma}{\sqrt{n}} = 48.75 - 42.71 = 6.04 \] Thus, \(2 \cdot 3 \cdot \frac{2.25}{\sqrt{n}} = 6.04\).
03

Solve for Subgroup Size n

Rearrange the equation from the previous step to solve for \(n\):\[ 6 \cdot \frac{2.25}{\sqrt{n}} = 6.04 \] Divide both sides by 6 to isolate \(\frac{2.25}{\sqrt{n}}\) and solve for \(n\):\[ \frac{2.25}{\sqrt{n}} = 1.0067 \] \[ \sqrt{n} = \frac{2.25}{1.0067} \approx 2.235 \] \[ n = (2.235)^2 \approx 4.994 \] Rounding up to the nearest whole number gives us \(n = 5\).

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Key Concepts

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

Three-Sigma Control Limits
Control charts are essential tools in statistical process control (SPC) to monitor process stability. When we talk about "three-sigma control limits," we're referencing how these limits are established by using three standard deviations from the process mean. This is significant because it provides a wide range encompassing 99.73% of the data if the process is statistically stable.

Key points about three-sigma control limits:
  • The **Upper Control Limit (UCL)** is calculated by adding three times the process standard deviation (σ) divided by the square root of subgroup size (n) to the process mean (µ).
  • The **Lower Control Limit (LCL)**, by contrast, is determined by subtracting the same value from the mean.
  • Three sigma limits help identify when a process may be out of control due to irregular variations that deviate significantly from the mean.
In the given exercise, the UCL is 48.75 and the LCL is 42.71. These values are crucial thresholds that define the boundaries of acceptable process variation based on statistical theory.
Subgroup Size
The subgroup size, noted as "n" in control chart formulas, is a critical component. It determines how many observations are grouped together to calculate the control limits.

Why is subgroup size important?
  • The larger the subgroup size, the smaller the variability between the subgroups, leading to narrower control limits. This can help in detecting smaller shifts in the process mean.
  • Smaller subgroups might have more variability, resulting in wider control limits, which could miss subtle variations that might indicate a process drifting from its intended target.
In solving the provided exercise, we calculated the subgroup size as 5. This means five consecutive data points were used to compute the \( \bar{X} \) values and establish control limits.
This setup helps balance sensitivity to changes and practicality in data collection.
Process Standard Deviation
The process standard deviation, denoted by σ, reflects the amount of variation within a process. Understanding this concept is pivotal when setting control charts, as it influences the positioning of the control limits. A higher process standard deviation indicates greater variability, potentially requiring broader control limits for effective monitoring.

Considerations for process standard deviation:
  • Standard deviation is used in computing control limits and essentially defines how variable each subgroup's mean might be from the process mean as shown by \( \frac{\sigma}{\sqrt{n}} \).
  • A precise estimation of σ is critical, as inaccuracies can lead to either too many false alarms (if σ is underestimated) or missed signals of process issues (if overestimated).
In the example problem, σ was given as 2.25. This value served as an anchor for calculating the control limits and explained how subgroup means could fluctuate in a controlled process.
An accurate standard deviation ensures we set realistic and meaningful thresholds for process evaluation.

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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.

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 \(U\).

The following are the numbers of defective solder joints found during successive samples of 500 solder joints: $$\begin{array}{cccc}\hline \text { Day } & \text { No. of Defectives } & \text { Day } & \text { No. ofDefectives } \\\\\hline 1 & 106 & 12 & 37 \\\2 & 116 & 13 & 25 \\\3 & 164 & 14 & 88 \\\4 & 89 & 15 & 101 \\\5 & 99 & 16 & 64 \\\6 & 40 & 17 & 51 \\\7 & 112 & 18 & 74 \\\8 & 36 & 19 & 71 \\\9 & 69 & 20 & 43 \\\10 & 74 & 21 & 80 \\\11 & 42 & &\end{array}$$ (a) Using all the data, compute trial control limits for a fraction-defective control chart, construct the chart, and plot the data. (b) Determine whether the process is in statistical control. If not, assume that assignable causes can be found and outof-control points eliminated. Revise the control limits.

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 following are the number of defects observed on 15 samples of transmission units in an automotive manufacturing company. Each lot contains five transmission units. (a) Using all the data, compute trial control limits for a \(U\) control chart, construct the chart, and plot the data. (b) Determine whether the process is in statistical control. If not, assume assignable causes can be found and out-ofcontrol points eliminated. Revise the control limits. $$\begin{array}{cccc}\hline \text { Sample } & \text { No. of Defects } & \text { Sample } & \text { No. of Defects } \\\\\hline 1 & 8 & 11 & 6 \\\2 & 10 & 12 & 10 \\\3 & 24 & 13 & 11 \\\4 & 6 & 14 & 17 \\\5 & 5 & 15 & 9 \\\6 & 21 & & \\\7 & 10 & & \\\8 & 7 & & \\\9 & 9 & & \\\10 & 15 & & \\\& &\end{array}$$

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