/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 A producer of brass rivets rando... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
Answer: A control chart for monitoring the proportion of defectives consists of a horizontal axis showing the sample number, a vertical axis showing the proportion of defectives, a centerline representing the mean sample proportion, and upper and lower control limits calculated based on the standard deviation of the sampling distribution. The control chart is valuable to a manager because it provides a visual representation of the process variability and indicates whether the process is stable and in control. If the proportion of defectives consistently falls within the control limits, the process is stable and product quality is consistent. If the proportion falls outside the control limits, it may indicate a quality issue requiring investigation, helping the manager make informed decisions about product quality and process improvements.

Step by step solution

01

Calculate the centerline of the control chart

To calculate the centerline, we will use the mean sample proportion calculated from the 200 samples, which is given as 0.021. The centerline of the control chart is given by: Centerline = p̄ = .021
02

Determine the standard deviation of the sampling distribution

To find the standard deviation of the sampling distribution of the sample proportion, we use the following formula: Standard Deviation = σ_p̂ = √(p̄ * (1-p̄) / n) In our case, we have p̄ = .021 and n = 400: σ_p̂ = √(.021 * (1 - .021) / 400) ≈ 0.00714
03

Calculate the control limits

To calculate the control limits, we use the formula: Upper Control Limit (UCL) = p̄ + 3 * σ_p̂ Lower Control Limit (LCL) = p̄ - 3 * σ_p̂ Substitute the value of p̄ = .021 and σ_p̂ ≈ 0.00714 into the formula: UCL = .021 + 3 * 0.00714 ≈ 0.04142 LCL = .021 - 3 * 0.00714 ≈ 0.00058
04

Construct the control chart

The control chart will consist of a horizontal axis showing the sample number and a vertical axis showing the proportion of defectives. The centerline and the control limits will then be plotted as horizontal lines: - Centerline at p̄ = 0.021 - Upper Control Limit at UCL ≈ 0.04142 - Lower Control Limit at LCL ≈ 0.00058
05

Explain how the control chart can be of value to a manager

A control chart provides a visual representation of the variability in the proportion of defectives for a process. It helps the manager to monitor the process over time and understand whether the process is stable and in control. If the proportion of defectives consistently falls within the control limits, it means the process is stable, and the manager can be confident that the product quality is consistent. However, if the proportion of defectives falls outside the control limits, it indicates that the process is not in control, and there may be a quality issue requiring investigation. Thus, the control chart can help the manager detect problems early on and make informed decisions about product quality and process improvements.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Statistical Process Control
Statistical process control (SPC) is a method that uses statistical techniques to monitor and control production processes. The objective of SPC is to detect any significant deviations from the production process, which could lead to substandard products. By collecting data about a process and using control charts, a producer can visualize process performance over time and pinpoint any variations that fall outside predetermined control limits.

One of the main tools in SPC is the control chart, which is a graphical representation displaying process data in a time-ordered sequence. Control charts have a central line, usually the process mean, and one or more control limits that define the bounds of acceptable variation. If the process data stays within these limits, the process is considered to be in control; if not, it may indicate a problem that requires further investigation.
Sample Proportion
When analyzing data like the proportion of defective items in a production batch, the sample proportion is a key metric. It represents the fraction of items in a sample that have a particular attribute – in this case, rivets that are defective. To calculate it, you divide the number of defectives by the total sample size. For instance, if you sample 400 rivets and find 10 defectives, the sample proportion would be 10/400, or 0.025.

The sample proportion is an estimate of the population proportion and is used to make inferences about the overall quality of the production process. By sampling different batches of the product and calculating the sample proportion of each batch, manufacturers can assess the process's stability and identify trends or shifts in quality.
Standard Deviation
The standard deviation is an important concept in statistics, measuring the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation shows that the values are spread out over a wider range.

In our exercise, the standard deviation of the sample proportion (denoted as \(\sigma_\hat{p}\)) provides an estimate of the variability in the proportion of defective rivets in the samples. This information is crucial when creating a control chart, as it is used to determine the control limits. These control limits are typically set at three standard deviations (\(3\sigma\)) from the mean. The control limits are boundaries which, when breached, suggest that the process may be out of control, prompting further investigation.
Quality Management
Quality management is an umbrella term that encompasses all the activities and tasks required to maintain a certain level of excellence. It includes the determination of a quality policy, creating and implementing quality planning and assurance, and quality control and improvement. One of the tools at the heart of quality management is the control chart, as discussed before, which tracks whether a process is in a state of control.

Understanding and applying quality management practices, such as utilizing control charts like in our exercise, enable managers to assure the process consistency and product quality. Moreover, it contributes to timely detection and prevention of quality issues, continuous process improvement, and customer satisfaction, all of which are key factors for a successful business operation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A bottler of soft drinks packages cans in six-packs. Suppose that the fill per can has an approximate normal distribution with a mean of 12 fluid ounces and a standard deviation of 0.2 fluid ounces. a. What is the distribution of the total fill for a case of 24 cans? b. What is the probability that the total fill for a case is less than 286 fluid ounces? c. If a six-pack of soda can be considered a random sample of size \(n=6\) from the population, what is the probability that the average fill per can for a six-pack of soda is less than 11.8 fluid ounces?

Random samples of size \(n=75\) were selected from a binomial population with \(p=.4 .\) Use the normal distribution to approximate the following probabilities: a. \(P(\hat{p} \leq .43)\) b. \(P(.35 \leq \hat{p} \leq .43)\)

News reports tell us that the average American is overweight. Many of us have tried to trim down to our weight when we finished high school or college. And, in fact, only \(19 \%\) of adults say they do not suffer from weight-loss woes. Suppose that the \(19 \%\) figure is correct, and that a random sample of \(n=100\) adults is selected. a. Does the distribution of \(\hat{p},\) the sample proportion of adults who do not suffer from excess weight, have an approximate normal distribution? If so, what is its mean and standard deviation? b. What is the probability that the sample proportion \(\hat{p}\) exceeds .25? c. What is the probability that \(\hat{p}\) lies within the interval .25 to \(.30 ?\) d. What might you conclude about \(p\) if the sample proportion exceeded .30?

The total daily sales, \(x\), in the deli section of a local market is the sum of the sales generated by a fixed number of customers who make purchases on a given day. a. What kind of probability distribution do you expect the total daily sales to have? Explain. b. For this particular market, the average sale per customer in the deli section is \(\$ 8.50\) with \(\sigma=\$ 2.50\). If 30 customers make deli purchases on a given day, give the mean and standard deviation of the probability distribution of the total daily sales, \(x\).

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 _____

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.