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Chapter 15: Problem 2

Twenty-five samples of size 5 are drawn from a process at one-hour intervals, and the following data are obtained: $$ \sum_{i=1}^{25} \bar{x}_{i}=362.75 \quad \sum_{i=1}^{25} r_{i}=8.60 \quad \sum_{i=1}^{25} s_{i}=3.64 $$ a. Calculate trial control limits for \(\bar{X}\) and \(R\) charts. b. Repeat part (a) for \(\bar{X}\) and \(S\) charts.

Short Answer

Expert verified
Control limits for \(\bar{X}\) and \(R\)-charts: UCL = 14.709, LCL = 14.311; \(R\)-chart UCL = 0.727, LCL = 0. Control limits for \(\bar{X}\) and \(S\)-charts: UCL = 14.717, LCL = 14.303; \(S\)-chart UCL = 0.304, LCL = 0.

Step by step solution

01

Calculate Overall Process Mean

The overall process mean \( \bar{\bar{x}} \) for \( \bar{X} \)-chart is calculated as the average of the sample means. Use the formula:\[\bar{\bar{x}} = \frac{\sum \bar{x}_i}{n}\]where \( \sum \bar{x}_i = 362.75 \) and \( n = 25 \). Hence,\[ \bar{\bar{x}} = \frac{362.75}{25} = 14.51 \]
02

Calculate Average Range

Calculate the average range \( \bar{R} \) for the \( R \)-chart using the formula:\[\bar{R} = \frac{\sum r_i}{n}\]where \( \sum r_i = 8.60 \). Hence,\[ \bar{R} = \frac{8.60}{25} = 0.344 \]
03

Determine Control Limits for \(\bar{X}\)-chart

Use the formula for \( \bar{X} \)-chart control limits:\[UCL_{\bar{X}} = \bar{\bar{x}} + A_2 \bar{R} \quad \text{and} \quad LCL_{\bar{X}} = \bar{\bar{x}} - A_2 \bar{R}\]Given that the sample size is 5, from control chart constants, \( A_2 = 0.577 \). Therefore,\[UCL_{\bar{X}} = 14.51 + 0.577 \times 0.344 = 14.709\]\[LCL_{\bar{X}} = 14.51 - 0.577 \times 0.344 = 14.311\]
04

Determine Control Limits for \(R\)-chart

For the \( R \)-chart, use the formulas:\[UCL_R = D_4 \bar{R} \quad \text{and} \quad LCL_R = D_3 \bar{R}\]Given that \( D_3 = 0 \) and \( D_4 = 2.114 \) for sample size 5, \[UCL_R = 2.114 \times 0.344 = 0.727\]\[LCL_R = 0 \times 0.344 = 0\]
05

Calculate Average Standard Deviation

Calculate the average standard deviation \( \bar{s} \) for the \( S \)-chart using:\[\bar{s} = \frac{\sum s_i}{n}\]where \( \sum s_i = 3.64 \). Thus,\[ \bar{s} = \frac{3.64}{25} = 0.1456 \]
06

Determine Control Limits for \(\bar{X}\)-chart Using \(S\)

The \( \bar{X} \)-chart control limits using standard deviation are:\[UCL_{\bar{X}} = \bar{\bar{x}} + A_3 \bar{s} \quad \text{and} \quad LCL_{\bar{X}} = \bar{\bar{x}} - A_3 \bar{s}\]With \( A_3 = 1.427 \) for sample size 5,\[UCL_{\bar{X}} = 14.51 + 1.427 \times 0.1456 = 14.717\]\[LCL_{\bar{X}} = 14.51 - 1.427 \times 0.1456 = 14.303\]
07

Determine Control Limits for \(S\)-chart

For the \( S \)-chart, use:\[UCL_S = B_4 \bar{s} \quad \text{and} \quad LCL_S = B_3 \bar{s} \]With constants \( B_3 = 0 \) and \( B_4 = 2.089 \) for sample size 5,\[UCL_S = 2.089 \times 0.1456 = 0.304 \]\[LCL_S = 0 \times 0.1456 = 0 \]

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Key Concepts

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

Process Mean
Understanding the process mean is essential in control charts as it represents the central tendency of a process. The process mean, denoted as \( \bar{\bar{x}} \), is calculated by averaging the individual sample means collected from the process. In our example, we have 25 sample means, and their sum is given as 362.75. To find \( \bar{\bar{x}} \), you divide the sum of the sample means by the number of samples:
  • Process Mean Formula: \( \bar{\bar{x}} = \frac{\sum \bar{x}_i}{n} \)
  • From our data: \( \bar{\bar{x}} = \frac{362.75}{25} = 14.51 \)
This mean serves as the target value for control chart calculations and helps identify any variations that may indicate potential issues.
Average Range
The average range, \( \bar{R} \), is a measure used in the \( R \)-chart to understand the dispersion of the process. It provides insights into the variability between different samples. It is calculated by dividing the total sum of the ranges \( \sum r_i \) by the number of samples \( n \):
  • Average Range Formula: \( \bar{R} = \frac{\sum r_i}{n} \)
  • Example Calculation: \( \bar{R} = \frac{8.60}{25} = 0.344 \)
Knowing the average range helps in determining the components of the control limits for the \( R \)-chart and supports understanding how much the process varies within a sample group.
Standard Deviation
Standard deviation is a crucial statistical tool that measures the amount of variation or dispersion in a set of values. In the context of control charts, especially \( S \)-charts, the average standard deviation, \( \bar{s} \), allows us to quantify variability more accurately than the range. It is calculated by dividing the sum of the standard deviations of each sample \( \sum s_i \) by the number of samples \( n \):
  • Average Standard Deviation Formula: \( \bar{s} = \frac{\sum s_i}{n} \)
  • Our Example: \( \bar{s} = \frac{3.64}{25} = 0.1456 \)
A lower standard deviation reflects less variability, indicating a more consistent process, whereas a higher standard deviation suggests greater variability and possible quality issues.
UCL and LCL
UCL (Upper Control Limit) and LCL (Lower Control Limit) are critical components in control charts to determine process stability. They define the acceptable range of variation in the process. For the \( \bar{X} \)-chart using range, limits are given by:
  • \( UCL_{\bar{X}} = \bar{\bar{x}} + A_2 \bar{R} \)
  • \( LCL_{\bar{X}} = \bar{\bar{x}} - A_2 \bar{R} \)
Using standard deviation, the formulas change to:
  • \( UCL_{\bar{X}} = \bar{\bar{x}} + A_3 \bar{s} \)
  • \( LCL_{\bar{X}} = \bar{\bar{x}} - A_3 \bar{s} \)
These limits help identify when a process is out of control. Points outside these limits or exhibiting non-random patterns indicate that corrective actions might be needed.
Sample Size
Sample size is a fundamental element in control charts, impacting the calculation of limits and their sensitivity. In our exercise, each sample consists of 5 observations. Sample size affects key constants like \( A_2, A_3, D_3, D_4, B_3, \) and \( B_4 \), which vary depending on how large each sample is. These constants modify how we calculate UCL and LCL:
  • For a sample size of 5, we use certain constants to compute accurate limits.
  • Larger sample sizes generally lead to narrower control limits, making the chart more sensitive to small shifts.
Understanding the role of sample size ensures that the data analyzed accurately reflects process behavior, influencing decision-making and quality control adjustments.

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Most popular questions from this chapter

Consider a \(P\) -chart with subgroup size \(n=50\) and center line at 0.12 a. Calculate the \(L C L\) and \(U C L\). b. Suppose that the true proportion defective changes from 0.12 to 0.18 . What is the ARL after the shift? Assume that the sample proportions are approximately normally distributed. c. Rework part (a) and (b) with \(n=100\) and comment on the difference in ARL. Does the increased sample size change the ARL substantially?

Suppose that a quality characteristic is normally distributed with specifications at \(100 \pm 20 .\) The process standard deviation is 6 a. Suppose that the process mean is \(100 .\) 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 is \(106 .\) What are the natural tolerance limits? What is the fraction defective? Calculate \(P C R\) and \(P C R_{k}\) and interpret these ratios.

The pull strength of a wire-bonded lead for an integrated circuit is monitored. The following table provides data for 20 samples each of size 3. $$ \begin{array}{cccc} \text { Sample Number } & x_{1} & x_{2} & x_{3} \\ 1 & 15.4 & 15.6 & 15.3 \\ 2 & 15.4 & 17.1 & 15.2 \\ 3 & 16.1 & 16.1 & 13.5 \\ 4 & 13.5 & 12.5 & 10.2 \\ 5 & 18.3 & 16.1 & 17.0 \\ 6 & 19.2 & 17.2 & 19.4 \\\ 7 & 14.1 & 12.4 & 11.7 \\ 8 & 15.6 & 13.3 & 13.6 \\ 9 & 13.9 & 14.9 & 15.5 \\\ 10 & 18.7 & 21.2 & 20.1 \\ 11 & 15.3 & 13.1 & 13.7 \\ 12 & 16.6 & 18.0 & 18.0 \\ 13 & 17.0 & 15.2 & 18.1 \\ 14 & 16.3 & 16.5 & 17.7 \\ 15 & 8.4 & 7.7 & 8.4 \\ 16 & 11.1 & 13.8 & 11.9 \\ 17 & 16.5 & 17.1 & 18.5 \\ 18 & 18.0 & 14.1 & 15.9 \\ 19 & 17.8 & 17.3 & 12.0 \\\20 & 11.5 & 10.8 & 11.2 \end{array} $$a. Use all the data to determine trial control limits for \(\bar{X}\) and \(R\) charts, construct the control limits, and plot the data. b. Use the control limits from part (a) to identify outof-control points. If necessary, revise your control limits assuming that any samples that plot outside of the control limits can be eliminated. c. Repeat parts (a) and (b) for \(\bar{X}\) and \(S\) charts.

Suppose that a quality characteristic is normally distributed with specifications from 20 to 32 units. a. What value is needed for \(\sigma\) to achieve a \(P C R\) of \(1.5 ?\) b. What value for the process mean minimizes the fraction defective? Does this choice for the mean depend on the value of \(\sigma ?\)

An article in Quality \& Safety in Health Care \(\left[{ }^{\mu} \mathrm{Sta}-\right.\) tistical 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 turnaround 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 on the control of the process. If necessary, assume that assignable causes can be found, eliminate suspect points, and revise the control limits. $$ \begin{array}{ccccccccccccccc} t & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \text { TAT } & 51 & 73 & 28 & 52 & 65 & 49 & 51 & 50 & 25 & 39 & 40 & 30 & 49 & 31 \end{array} $$
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