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An automatic senson measures the diameter of holes in consecutive order. The results of measuring 25 holes follow. $$\begin{array}{cccc}\text { Sample } & \text { Diameter } & \text { Sample } & \text { Diameter } \\\\\hline 1 & 9.94 & 14 & 9.99 \\\2 & 9.93 & 15 & 10.12 \\\3 & 10.09 & 16 & 9.81 \\\4 & 9.98 & 17 & 9.73 \\\5 & 10.11 & 18 & 10.14 \\\6 & 9.99 & 19 & 9.96 \\\7 & 10.11 & 20 & 10.06 \\\8 & 9.84 & 21 & 10.11 \\\9 & 9.82 & 22 & 9.95 \\\10 & 10.38 & 23 & 9.92 \\\11 & 9.99 & 24 & 10.09 \\\12 & 10.41 & 25 & 9.85 \\\13 & 10.36 & &\end{array}$$ (a) Estimate the process standard deviation. (b) Set up a CUSUM control procedure, assuming that the target diameter is 10.0 millimeters. Does the process appear to be operating in a state of statistical control at the desired target level?

Short Answer

Expert verified
Standard deviation \(s \approx 0.177\). CUSUM indicates the process might be in control with no significant shift detected.

Step by step solution

01

Calculate Sample Mean

First, we calculate the mean diameter for all the 25 samples. Add up all the diameters and then divide the sum by 25. By adding them up:\[\text{Total Sum of Diameters} = 9.94 + 9.93 + 10.09 + 9.98 + 10.11 + 9.99 + 10.11 + 9.84 + 9.82 + 10.38 + 9.99 + 10.41 + 10.36 + 9.99 + 10.12 + 9.81 + 9.73 + 10.14 + 9.96 + 10.06 + 10.11 + 9.95 + 9.92 + 10.09 + 9.85 = 249.34\]Divide by the number of samples:\[\text{Mean Diameter} = \frac{249.34}{25} \approx 9.974\]Thus, the mean diameter is approximately 9.974 mm.
02

Calculate Sample Standard Deviation

To estimate the process standard deviation, calculate the sample standard deviation:1. Find the difference between each hole's diameter and the mean, then square each difference.2. Sum all squared differences.3. Divide by the number of observations minus one (n - 1 = 24).4. Take the square root of that result.The formula is:\[s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}\]Calculating this gives:\[s \approx \sqrt{\frac{1}{24} ((9.94 - 9.974)^2 + (9.93 - 9.974)^2 + \ldots + (9.85 - 9.974)^2)} \approx 0.177\]
03

Set Up CUSUM Control

CUSUM (Cumulative Sum Control Chart) helps in detecting small shifts in the process mean. The target mean \( \mu_0 \) is given as 10 mm. With each measurement, the CUSUM statistic is updated:1. Start with \( C_0 = 0 \).2. Update with: \( C_i = C_{i-1} + (x_i - \mu_0) \).For detection of shifts, typically decision interval \( h \) around some, often 4 or 5 standard deviations, is used in practice. Here, calculate the CUSUM for each diameter and check if any \( C_i \) exceeds the threshold based on \( h \). If none exceed, the process is in control.
04

Analysis of Control State

Compute the CUSUM for the full sequence using stance deviation \( s \). For example, if \( h = 0.5 \) std deviations, compare cumulative sums.Review if any point surpasses this threshold.If any point exceeds, it signifies the process might be out of control.

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Key Concepts

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

CUSUM
CUSUM, or Cumulative Sum Control Chart, is a powerful tool in statistical process control used to detect small shifts in the process mean. It is especially useful where traditional control charts might not promptly signal a change. Here’s how it works:
  • The chart accumulates the sum of deviations from a target value over time.
  • Start by setting the initial cumulative sum (\( C_0 \)) to zero.
  • For each new sample, update the cumulative sum (\( C_i \)) with the formula: \( C_i = C_{i-1} + (x_i - \mu_0) \), where \( x_i \) is the sample measurement and \( \mu_0 \) is the target mean.
  • This highlights trends over time, helping identify any significant deviation from the target process mean.
It’s important to set a decision interval (often 4 or 5 standard deviations from the mean) to determine when the process might be out of control. If the calculated CUSUM crosses this threshold, an investigation into the process condition is necessary. By using CUSUM, even minute shifts that accumulate over time become noticeable, allowing for quick corrective actions.
Sample standard deviation
Sample standard deviation is a valuable statistic in measuring the spread or variability of a set of data. It tells us how much each value in the sample varies from the sample mean. Here's how you calculate it step-by-step:
  • First, compute the mean (average) of the sample data.
  • Next, find the difference between each data point and the mean, then square each of these differences.
  • Sum all of the squared differences to get a total squared deviation.
  • Divide this sum by the number of observations minus one (\( n - 1 \)). This gives us the variance.
  • The sample standard deviation is the square root of the variance.
By using sample standard deviation, you can better understand the natural variability within your data set. For the given example, the sample standard deviation was found to be approximately 0.177 mm, indicating how much the hole diameters varied from their mean.
Control charts
Control charts are graphical tools used in quality control to monitor processes and ensure they function within specified limits. They are essential in detecting important shifts or trends in process data over time.
  • At a basic level, a control chart will plot sample metric data points in sequence.
  • Upper and lower control limits are established, often three standard deviations from the mean in either direction.
  • These limits help determine if a process is operating normally or if it is out of control.
Control charts can be used for a variety of processes, helping identify process variations that might require correction. For instance, if several data points fall outside of control limits, it's a signal that the process might be going out of control due to some special cause. Control charts allow for proactive quality management by indicating when adjustments are necessary to maintain the desired state of control.

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Most popular questions from this chapter

The diameter of fuse pins used in an aircraft engine application is an important quality characteristic. Twenty-five samples of three pins each are shown as follows: $$\begin{array}{cccc}\hline \begin{array}{c}\text { Sample } \\\\\text { Number }\end{array} & & \text { Diameter } & \\\\\hline 1 & 64.030 & 64.002 & 64.019 \\\2 & 63.995 & 63.992 & 64.001 \\\3 & 63.988 & 64.024 & 64.021 \\\4 & 64.002 & 63.996 & 63.993 \\\5 & 63.992 & 64.007 & 64.015 \\\6 & 64.009 &63.994 & 63.997 \\\7 & 63.995 & 64.006 & 63.994 \\\8 & 63.985 & 64.003 & 63.993 \\ 9 & 64.008 & 63.995 & 64.009 \\\10 & 63.998 & 74.000 & 63.990 \\\11 & 63.994 & 63.998 & 63.994 \\\12 & 64.004 & 64.000 & 64.007 \\\13 & 63.983 & 64.002 & 63.998 \\\14 & 64.006 & 63.967 & 63.994 \\\15 & 64.012 & 64.014 & 63.998 \\\16 & 64.000 & 63.984 & 64.005 \\\17 & 63.994 & 64.012 & 63.986 \\\18 & 64.006 & 64.010 & 64.018 \\\19 & 63.984 & 64.002 & 64.003 \\\20 & 64.000 & 64.010 & 64.013 \\\21 & 63.988 & 64.001 & 64.009 \\\22 & 64.004 & 63.999 & 63.990 \\\23 & 64.010 & 63.989 &63.990\\\24 & 64.015 & 64.008 & 63.993 \\\25 & 63.982 & 63.984 & 63.995\end{array}$$ (a) Set up \(\bar{X}\) and \(R\) charts for this process. If necessary, revise limits so that no observations are out of control. (b) Estimate the process mean and standard deviation. (c) Suppose that the process specifications are at \(64 \pm 0.02\). Calculate an estimate of \(P C R\). Does the process meet a minimum capability level of \(P C R \geq 1.33 ?\) (d) Calculate an estimate of \(P C R_{k}\). Use this ratio to draw conclusions about process capability. (e) 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? (f) Suppose that the mean shifts to \(64.01 .\) What is the probability that this shift is detected on the next sample? What is the ARL after the shift?

Suppose that a quality characteristic is normally distributed with specifications at \(150 \pm 20 .\) Natural tolerance limits for the process are \(150 \pm 18\). (a) Calculate the process standard deviation. (b) Calculate \(P C R\) and \(P C R_{k}\) of the process. Calculate the percentage of the specification width used by the process. (c) What fallout level (fraction defective) is produced?

Suppose that a quality characteristic is normally distributed with specifications at \(120 \pm 20 .\) The process standard deviation is \(6.5 .\) (a) Suppose that the process mean is \(120 .\) What are the natural tolerance limits? What is the fraction defective? Calculate \(P C R\) and \(P C R_{k}\) and interpret these ratios. (b) Suppose that the process mean shifts off-center by 1.5 standard deviations toward the upper specification limit. Recalculate the quantities in part (a). (c) Compare the results in parts (a) and (b) and comment on any differences.

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.

\( \) An \(\bar{X}\) chart uses a sample of size \(3 .\) The center line is at \(200,\) and the upper and lower 3 -sigma control limits are at 212 and \(188,\) respectively. (a) Estimate the process \(\sigma\). (b) Suppose that the process mean shifts to \(195 .\) Determine the probability that this shift is detected on the next sample. (c) Find the ARL to detect the shift in part (b).

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