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Samples of \(n=100\) items were selected hourly over a 100 -hour period, and the sample proportion of defectives was calculated each hour. The mean of the 100 sample proportions was .035 a. Use the data to find the upper and lower control limits for a \(p\) chart. b. Construct a \(p\) chart for the process and explain how it can be used.

Short Answer

Expert verified
Answer: The control limits for the p-chart with a mean sample proportion of 0.035 from 100 samples are an Upper Control Limit (UCL) of 0.0896 and a Lower Control Limit (LCL) of 0. The p-chart is used to monitor the proportion of defectives in the process and helps identify trends, patterns, or changes in the process over time. It also assists in maintaining the process within acceptable control limits and indicates when further investigation and corrective actions may be necessary.

Step by step solution

01

Calculate the centerline (average proportion of defectives)

The centerline of the p-chart is the average proportion of defectives, which is given in the problem as \(0.035\).
02

Calculate the standard deviation of the sample proportions

The standard deviation of sample proportions can be calculated using the formula: \(\sigma_p = \sqrt{ \frac{\bar{p} (1- \bar{p}) } {n}}\). In this case, \(\bar{p}=0.035\) and \(n=100\). Plug these values into the formula and calculate the standard deviation: \(\sigma_p = \sqrt{ \frac{0.035 (1- 0.035) } {100}} \approx 0.0182\).
03

Calculate the upper and lower control limits

To calculate the control limits, we use the following formulas: - Upper Control Limit (UCL) = Centerline + 3 * Standard Deviation (\(\bar{p} + 3\sigma_p\)) - Lower Control Limit (LCL) = Centerline - 3 * Standard Deviation (\(\bar{p} - 3\sigma_p\)) Plug the values from step 1 and step 2 into the formulas: - UCL = \(0.035 + 3 * 0.0182 \approx 0.0896\) - LCL = \(0.035 - 3 * 0.0182 \approx -0.0196\) Since the lower control limit cannot be negative, we set the LCL to 0: - LCL = \(0\) Thus, the control limits for the p-chart are: - Upper Control Limit (UCL) = \(0.0896\) - Lower Control Limit (LCL) = \(0\)
04

Construct the p-chart and explain its usage

A p-chart is a control chart that displays the proportion of defectives for each sample, with a centerline at the average proportion of defectives (\(\bar{p} = 0.035\)) and control limits at \(0\) (LCL) and \(0.0896\) (UCL). To construct the p-chart: 1. Plot the sample proportions on the vertical axis and the sample number (hour) on the horizontal axis. 2. Draw a horizontal line at the centerline (\(0.035\)). 3. Draw horizontal lines at the upper control limit (\(0.0896\)) and the lower control limit (\(0\)). The p-chart can be used to monitor the proportion of defectives in the process. If a sample proportion falls above the UCL or below the LCL, it indicates that the process may be out of control, and further investigation and corrective actions may be necessary. The p-chart helps to identify trends, patterns, or changes in the process over time, and to maintain the process within acceptable control limits.

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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 an essential part of a p-chart. They help us determine whether a process is under control or not. By setting bounds within which the process should operate, control limits serve as a guide to identify unusual variations that might require adjustments.

For a p-chart, which is used to monitor the proportion of defectives, we use the upper control limit (UCL) and the lower control limit (LCL). These limits are calculated using the average proportion of defectives and its standard deviation. The formulas for control limits ensure that most of the data (about 99.7% if we're using \(3\sigma\) limits) should fall between them:
  • Upper Control Limit (UCL) = Centerline + 3 * Standard Deviation
  • Lower Control Limit (LCL) = Centerline - 3 * Standard Deviation
If any point falls outside these limits, it's a signal that the process might be "out of control." Keep in mind that for proportions, LCL cannot be negative, and if calculated as such, it is set to zero instead. This adjustment prevents illogical conclusions, such as having a negative proportion of defectives.
Standard Deviation
Standard deviation is a statistic that measures the dispersion of a dataset relative to its mean. In simpler terms, it tells us how much the data points differ from the average.

In the context of a p-chart, the standard deviation helps calculate the control limits. It gives us an idea of how variable the sample proportions are. To find the standard deviation of sample proportions, we use the formula:\[\sigma_p = \sqrt{ \frac{\bar{p} (1- \bar{p}) } {n}}\]where \(\bar{p}\) is the average proportion of defectives, and \(n\) is the sample size. The standard deviation helps in setting accurate bounds for the UCL and LCL, ensuring that they reflect the variability in the process.

Understanding standard deviation is critical because it provides insight into the stability and reliability of the process. A smaller standard deviation indicates consistent quality, while a larger one could signal possible problems with the process stability.
Proportion of Defectives
The proportion of defectives is a core metric for quality control, especially when using a p-chart. It represents the fraction of the items in a sample that do not meet quality standards.

To find the average proportion of defectives, you take the mean of the sample proportions, sometimes denoted as \( \bar{p} \). In our original problem, this was provided as \(0.035\). This average serves as the centerline on a p-chart.

Monitoring the proportion of defectives is crucial for maintaining product quality. It allows you to see if the process consistently produces within acceptable standards. In control charting, we analyze how the proportion of defectives in each sample compares to the centerline and control limits.
  • If the points stay within the limits and around the centerline, the process is considered "in control."
  • Significant deviations or patterns might indicate an out-of-control process.
This can prompt a deeper look into process variations and help to improve overall quality control by identifying specific problem areas.

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Most popular questions from this chapter

Telephone Service Suppose a telephone company executive wishes to select a random sample of \(n=20\) (a small number is used to simplify the exercise) out of 7000 customers for a survey of customer attitudes concerning service. If the customers are numbered for identification purposes, indicate the customers whom you will include in your sample. Use the random number table and explain how you selected your sample.

Explain the difference between an \(\bar{x}\) chart and a \(p\) chart.

A producer of brass rivets randomly samples 400 rivets each hour and calculates the proportion of defectives in the sample. The mean sample proportion calculated from 200 samples was equal to .021. Construct a control chart for the proportion of defectives in samples of 400 rivets. Explain how the control chart can be of value to a manager.

You take a random sample of size \(n=40\) from a distribution with mean \(\mu=100\) and \(\sigma=20 .\) The sampling distribution of \(\bar{x}\) will be approximately _____ with a mean of ____ and a standard deviation (or standard error) of _____

A manufacturer of paper used for packaging requires a minimum strength of 20 pounds per square inch. To check on the quality of the paper, a random sample of 10 pieces of paper is selected each hour from the previous hour's production and a strength measurement is recorded for each. The standard deviation \(\sigma\) of the strength measurements, computed by pooling the sum of squares of deviations of many samples, is known to equal 2 pounds per square inch, and the strength measurements are normally distributed. a. What is the approximate sampling distribution of the sample mean of \(n=10\) test pieces of paper? b. If the mean of the population of strength measurements is 21 pounds per square inch, what is the approximate probability that, for a random sample of \(n=10\) test pieces of paper, \(\bar{x}<20 ?\) c. What value would you select for the mean paper strength \(\mu\) in order that \(P(\bar{x}<20)\) be equal to \(.001 ?\)

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