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Chapter 16: Problem 23

When \(n=150\), what is the smallest value of \(\bar{p}\) for which the LCL in a \(p\) chart is positive?

Short Answer

Expert verified
The smallest value of \(\bar{p}\) is approximately 0.0567.

Step by step solution

01

Define the LCL formula for a p-chart

The Lower Control Limit (LCL) for a \(p\) chart is given by the formula: \( LCL = \bar{p} - z\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \), where \(\bar{p}\) is the average proportion of defects, \(z\) is the standard normal deviant for a given confidence level (typically \(z = 3\) for control chart applications), and \(n\) is the sample size.
02

Set up the equation for a positive LCL

To ensure that the LCL is positive, we need the equation: \( \bar{p} - 3\sqrt{\frac{\bar{p}(1-\bar{p})}{150}} > 0 \). This can be rewritten to find the smallest value of \(\bar{p}\) that satisfies this condition.
03

Simplify the inequality

Solve the inequality for \(\bar{p}\): \( \bar{p} > 3 \sqrt{\frac{\bar{p}(1-\bar{p})}{150}} \). This involves algebraic manipulation to isolate \(\bar{p}\) on one side of the inequality.
04

Square both sides to eliminate the square root

Squares both sides to eliminate the square root: \( \bar{p}^2 > 9 \times \frac{\bar{p}(1-\bar{p})}{150} \), which simplifies further to \(150\bar{p}^2 > 9\bar{p} - 9\bar{p}^2 \).
05

Solve for \(\bar{p}\)

Rearrange the terms: \(159\bar{p}^2 > 9\bar{p} \), which simplifies to \(159\bar{p}^2 - 9\bar{p} > 0 \). Divide by \(\bar{p}\) (since \(\bar{p} > 0\)) to get \(159\bar{p} > 9\), leading to \(\bar{p} > \frac{9}{159} = \frac{1}{17.6667} \approx 0.0567\).
06

Verify the solution

Plug \(\bar{p} = 0.0567\) back into the LCL formula: Calculate the LCL and ensure it's positive. For \(\bar{p} = 0.0567\), the LCL should indeed be positive, confirming our solution.

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Key Concepts

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

Control Limits
Control limits are the boundaries set within a control chart to determine whether a process is within control. In a p-chart, which is used to monitor the proportion of defects in samples over time, there are usually three control limits:
  • The Centerline (\( \bar{p} \)), which represents the average proportion of defects.
  • The Upper Control Limit (UCL), set by adding a factor multiplied by the standard deviation to the centerline.
  • The Lower Control Limit (LCL), calculated by subtracting this factor from the centerline.
These control limits help in identifying any out-of-control process signals, so corrective actions can be taken promptly. By maintaining the process within these limits, businesses ensure quality stays consistent and predictable.
Proportion of Defects
The proportion of defects (\( \bar{p} \)) is a critical measure in quality control, representing the average defect rate in a production process. This metric is vital for constructing a p-chart. Here's how it works:
  • \( \bar{p} \) is calculated by dividing the number of defects found by the total items inspected in a sample.
  • This value helps identify trends and variations in the defect rate over time.
  • Achieving an accurate \( \bar{p} \) ensures that control limit calculations on the p-chart are valid, aiding reliable decision-making.
A meaningful interpretation of \( \bar{p} \) helps quality managers assess perfection levels and pinpoint improvement areas within operations.
Sample Size
Sample size (\( n \)) in the context of p-charts affects both the accuracy of the \( \bar{p} \) calculation and the precision of control limits. The steps involved include:
  • Larger samples result in more reliable statistical estimates of the proportion of defects.
  • With a given sample size, you can calculate necessary adjustments for control limits to reflect real process performance accurately.
  • Sample size directly influences the control chart's ability to detect small shifts in the process.
Deciding on the appropriate sample size involves balancing the cost of gathering data against the need for accuracy, ensuring a stable and effective monitoring system.
Standard Normal Deviant
The standard normal deviant (\( z \)) is a vital component in setting control limits on a p-chart. It reflects the confidence level stakeholders have in the limits. Here's how it is used:
  • The value of \( z \) is typically chosen based on desired confidence levels, commonly set at 3 for quality control charts, representing roughly 99.7% confidence in a normal distribution.
  • This integer multiplies the standard deviation calculated from the sample proportion, to define precisely how far from the centerline the control limits are placed.
  • A higher \( z \) value typically results in wider control limits, suggesting more tolerance for variation within the process.
Using the correct \( z \) value helps prevent false alarms of out-of-control conditions, ensuring precise tracking of process stability.

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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 $$ \begin{aligned} u_{i} &=\frac{\text { the number of defects observed at time } i}{\text { size of entity inspected at time } i} \\ &=\frac{x_{i}}{g_{i}} \end{aligned} $$ where "size" might refer to area, length, volume, or simply the number of items inspected. Then a \(u\) chart plots \(u_{1}, u_{2}, \ldots\), has center line \(\bar{u}\), and the control limits for the \(i\) th observations are \(\bar{u} \pm 3 \sqrt{\bar{u} / g_{i}}\). 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. $$ \begin{array}{rcr} \text { Panel } & \begin{array}{c} \text { Area } \\ \text { Examined } \end{array} & \begin{array}{c} \text { No. of } \\ \text { Blemishes } \end{array} \\ \hline 1 & .8 & 3 \\ 2 & .6 & 2 \\ 3 & .8 & 3 \\ 4 & .8 & 2 \\ 5 & 1.0 & 5 \\ 6 & 1.0 & 5 \\ 7 & .8 & 10 \\ 8 & 1.0 & 12 \\ 9 & .6 & 4 \\ 10 & .6 & 2 \\ 11 & .6 & 1 \\ 12 & .8 & 3 \\ 13 & .8 & 5 \\ 14 & 1.0 & 4 \\ 15 & 1.0 & 6 \\ 16 & 1.0 & 12 \\ 17 & .8 & 3 \\ 18 & .6 & 3 \\ 19 & .6 & 5 \\ 20 & .6 & 1 \\ \hline \end{array} $$

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 \(.005\) ?

A cork intended for use in a wine bottle is considered acceptable if its diameter is between \(2.9 \mathrm{~cm}\) and \(3.1 \mathrm{~cm}\) (so the lower specification limit is \(\mathrm{LSL}=2.9\) and the upper specification limit is \(\mathrm{USL}=3.1)\). a. If cork diameter is a normally distributed variable with mean value \(3.04 \mathrm{~cm}\) and standard deviation \(.02 \mathrm{~cm}\), what is the probability that a randomly selected cork will conform to specification? b. If instead the mean value is \(3.00\) and the standard deviation is \(.05\), is the probability of conforming to specification smaller or larger than it was in (a)?

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.

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?
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