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Resistance observations (ohms) for subgroups of a certain type of register gave the following summary quantities: $$ \begin{array}{cccc|cccc} \boldsymbol{i} & \boldsymbol{n}_{i} & \overline{\boldsymbol{x}}_{\boldsymbol{i}} & \boldsymbol{s}_{\boldsymbol{i}} & \boldsymbol{i} & \boldsymbol{n}_{i} & \overline{\boldsymbol{x}}_{\boldsymbol{i}} & \boldsymbol{s}_{\boldsymbol{i}} \\\ \hline 1 & 4 & 430.0 & 22.5 & 7 & 4 & 420.8 & 25.4 \\ 2 & 4 & 418.2 & 20.6 & 8 & 4 & 431.4 & 24.0 \\ 3 & 3 & 435.5 & 25.1 & 9 & 4 & 428.7 & 21.2 \\ 4 & 4 & 427.6 & 22.3 & 10 & 4 & 440.1 & 25.8 \\ 5 & 4 & 444.0 & 21.5 & 11 & 4 & 445.2 & 27.3 \\ 6 & 3 & 431.4 & 28.9 & 12 & 4 & 430.1 & 22.2 \end{array} $$ $$ \begin{array}{cccc|cccc} i & \boldsymbol{n}_{i} & \overline{\boldsymbol{x}}_{\boldsymbol{t}} & \boldsymbol{s}_{\boldsymbol{t}} & \boldsymbol{i} & \boldsymbol{n}_{\boldsymbol{t}} & \overline{\boldsymbol{x}}_{\boldsymbol{i}} & \boldsymbol{s}_{i} \\ \hline 13 & 4 & 427.2 & 24.0 & 17 & 3 & 447.0 & 19.8 \\ 14 & 4 & 439.6 & 23.3 & 18 & 4 & 434.4 & 23.7 \\ 15 & 3 & 415.9 & 31.2 & 19 & 4 & 422.2 & 25.1 \\ 16 & 4 & 419.8 & 27.5 & 20 & 4 & 425.7 & 24.4 \\ \hline \end{array} $$ Construct appropriate control limits. [Hint: Use \(\bar{x}=\sum n_{i} \bar{x}_{i} / \sum n_{i}\) and \(\left.s^{2}=\sum\left(n_{i}-1\right) s_{i}^{2} / \sum\left(n_{i}-1\right) .\right]\)

Short Answer

Expert verified
Control limits: UCL ≈ 470.03 ohms, LCL ≈ 392.45 ohms.

Step by step solution

01

Calculate the Grand Mean

To calculate the grand mean \( \bar{x} \), sum the products of each subgroup size \( n_i \) and its corresponding mean \( \bar{x}_i \), and then divide by the total number of observations. Calculate the total number of observations by \( \sum n_i = (4+4+3+4+4+3+4+4+4+4+4+4+4+4+3+4+3+4+4+4) = 75 \).Now calculate \( \bar{x} = \frac{(4 \times 430.0 + 4 \times 418.2 + 3 \times 435.5 + \ldots + 4 \times 425.7)}{75} = 431.24 \mathrm{\ ohms} \).
02

Calculate the Pooled Standard Deviation

The pooled standard deviation \( s \) is derived from the formula for \( s^2 \). First, calculate \( s^2 \) using the formula \( s^2 = \frac{\sum (n_i - 1) s_i^2}{\sum(n_i - 1)} \). For the sum of \((n_i - 1) s_i^2 \), calculate and sum up: \( (3 \times 22.5^2 + 3 \times 20.6^2 + 2 \times 25.1^2 + \dots + 3 \times 24.4^2) = 38014.15 \). The degrees of freedom total \( \sum(n_i - 1) = 54 \). Thus, \( s^2 = \frac{38014.15}{54} = 703.96 \), leading to \( s = \sqrt{703.96} = 26.53 \mathrm{\ ohms} \).
03

Determine the Control Limits

Control limits are usually established using statistical principles that include a central line (process mean) and upper/lower control limits. The limits are determined using the mean and standard deviation.For example, the control limits for subgroup means can be established as:- Upper Control Limit (UCL) = \( \bar{x} + 3\frac{s}{\sqrt{\bar{n}}} \)- Lower Control Limit (LCL) = \( \bar{x} - 3\frac{s}{\sqrt{\bar{n}}} \)Here, the average subgroup size \( \bar{n} \approx 4 \) (noting the mix slightly above and below 4 per subgroup). So, UCL \( \approx 431.24 + 3 \times \frac{26.53}{\sqrt{4}} = 470.03 \mathrm{\ ohms} \) and LCL \( \approx 431.24 - 3 \times \frac{26.53}{\sqrt{4}} = 392.45 \mathrm{\ ohms} \).

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Key Concepts

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

Control Limits
To maintain high-quality in manufacturing processes, control limits are essential tools. These limits help in identifying variations within a process, pointing out if any unusual changes occur that need attention. Statistical Process Control (SPC) utilizes these limits to ensure product consistency.

Control limits are established based on the process mean (also known as the central line) and the variation within the process, measured using the standard deviation. Specifically, we use the upper control limit (UCL) and lower control limit (LCL) to form a range indicating acceptable performance levels.
  • Upper Control Limit (UCL): This boundary is set above the process mean and usually defined as the mean plus three times the process standard deviation divided by the square root of the average subgroup size. This helps in detecting any unusually high process values.
  • Lower Control Limit (LCL): Found below the process mean, this is calculated as the mean minus three times the standard deviation over the square root of the average subgroup size. It catches any unusual low values.
If measurement data falls within these limits, it indicates the process is in control and functioning as expected. When points fall outside these limits, it signals potential issues that might require investigating.
Pooled Standard Deviation
The pooled standard deviation is a measure that assesses the combined spread or variability within data from different subgroups of a sample.

In scenarios where your sample consists of multiple subgroups, each with its own variability (or standard deviation), pooled standard deviation becomes useful. It gives a singular, overall measure of variability, making it easier to understand the process's behavioral consistency across all units.To calculate this, you derive a pooled variance first using the formula:\[ s^2 = \frac{\sum (n_i - 1) s_i^2}{\sum (n_i - 1)} \]Here:
  • \( n_i \) represents the size of each subgroup
  • \( s_i \) is the standard deviation of that subgroup
  • The sums are taken over all subgroups, adjusting weights according to the subgroup sizes
Finally, the pooled standard deviation \( s \) is the square root of the pooled variance, providing an overall spread measure vital for accurate control limit determinations.
Grand Mean
The grand mean, also known as the overall mean, helps in understanding the central tendency of an entire dataset composed of several subgroups. It provides a single metric that summarizes the data's average value across all observations.

In statistical process control, calculating the grand mean involves considering each subgroup's mean, accounting for their respective sizes. By doing this, you ensure each subgroup's mean contributes proportionally to the overall mean. This is done using:\[ \bar{x} = \frac{\sum n_i \bar{x}_i}{\sum n_i} \]Where:
  • \( \bar{x}_i \) is the mean of each subgroup
  • \( n_i \) is the number of observations in each subgroup
The grand mean acts as the central line or reference point around which control limits are determined. It informs whether the process is centered according to expectations. Calculating this value accurately is fundamental for drawing reliable control charts and ensuring precise monitoring in quality management contexts.

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Most popular questions from this chapter

In some situations, the sizes of sampled specimens vary, and larger specimens are expected to have more defects than smaller ones. For example, sizes of fabric samples inspected for flaws might vary over time. Alternatively, the number of items inspected might change with time. Let $$ u_{i}=\frac{\text { the number of defects observed at time } i}{\text { size of entity inspected at time } i} $$ where "size" might refer to area, length, volume, or simply the number of items inspected. Then a \(\boldsymbol{u}\) chart plots \(u_{1}, u_{2}, \ldots\), has center line \(\bar{u}\), and the control limits for the \(i\) th observations are \(\bar{u} \pm 3 \sqrt{u} / g_{i}\) Painted panels were examined in time sequence, and for each one, the number of blemishes in a specified sampling region was determined. The surface area \(\left(\mathrm{ft}^{2}\right)\) of the region examined varied from panel to panel. Results are given below. Construct a \(u\) chart. $$ \begin{array}{rrr} \text { Panel } & \begin{array}{c} \text { Area } \\ \text { Examined } \end{array} & \begin{array}{c} \text { No. of } \\ \text { Blemishes } \end{array} \\ \hline 1 & .8 & 3 \\ 2 & .6 & 2 \\ 3 & .8 & 3 \\ 4 & .8 & 2 \\ 5 & 1.0 & 5 \\ 6 & 1.0 & 5 \\ 7 & .8 & 10 \\ 8 & 1.0 & 12 \\ 9 & .6 & 4 \\ 10 & .6 & 2 \\ 11 & .6 & 1 \\ 12 & .8 & 3 \\ 13 & .8 & 5 \\ 14 & 1.0 & 4 \\ 15 & 1.0 & 6 \\ 16 & 1.0 & 12 \\ 17 & .8 & 3 \\ 18 & .6 & 3 \\ 19 & .6 & 5 \\ 20 & .6 & 1 \\ \hline \end{array} $$

The accompanying observations are numbers of defects in 25 1-square-yard specimens of woven fabric of a certain type: \(3,7,5,3,4,2,8,4,3,3,6,7,2,3,2,4,7,3,2,4,4\), \(1,5,4,6\). Construct a \(c\) chart for the number of defects.

Consider the single-sample plan with \(c=2\) and \(n=50\), as discussed in Example 16.11, but now suppose that the lot size is \(N=500\). Calculate \(P(A)\), the probability of accepting the lot, for \(p=.01, .02, \ldots, .10\) using the hypergeometric distribution. Does the binomial approximation give satisfactory results in this case?

Consider the double-sampling plan for which both sample sizes are 50. The lot is accepted after the first sample if the number of defectives is at most 1 , rejected if the number of defectives is at least 4 , and rejected after the second sample if the total number of defectives is 6 or more. Calculate the probability of accepting the lot when \(p=.01, .05\), and .10.

In the case of known \(\mu\) and \(\sigma\), what control limits are necessary for the probability of a single point being outside the limits for an in-control process to be \(.005\) ?

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