/*! 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 49 The sample means were calculated... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 49

The sample means were calculated for 40 samples of size \(n=5\) for a process that was judged to be in control. The means of the 40 values and the standard deviation of the combined 200 measurements were \(\overline{\bar{x}}=155.9\) and \(s=4.3,\) respectively. a. Use the data to determine the upper and lower control limits for an \(\bar{x}\) chart. b. Construct an \(\bar{x}\) chart for the process and explain how it can be used.

Short Answer

Expert verified
Answer: The calculated upper control limit (UCL) is approximately 157.01, and the lower control limit (LCL) is approximately 154.79. By plotting the sample means on the \(\bar{x}\) chart and comparing them to the UCL and LCL, we can monitor the process performance over time. If most sample means fall within the control limits, the process is considered to be in control. Points falling outside these limits or any patterns and trends in the plotted points may indicate potential problems or assignable causes that should be investigated.

Step by step solution

01

Determine the value of A2 and calculate standard error

The first step is to determine the value of A2 for n = 5. You can find this in a statistical control chart constants table or use online resources. Generally, for n = 5, A2 鈮� 0.58. Next, we will calculate the standard error (SE) of the sample mean using the given standard deviation (s = 4.3). The SE is given by the formula: \(SE = \frac{s}{\sqrt{n}}\) Plugging in our values, \(SE = \frac{4.3}{\sqrt{5}} \approx 1.92\)
02

Calculate the upper control limit (UCL) and lower control limit (LCL)

Now that we have the value of A2 and the standard error, we can calculate the upper and lower control limits using the formulas: \(UCL = \overline{\bar{x}} + A_2 \frac{s}{\sqrt{n}}\) \(LCL = \overline{\bar{x}} - A_2 \frac{s}{\sqrt{n}}\) Plugging in the given values, \(UCL = 155.9 + 0.58(1.92) \approx 157.01\) \(LCL = 155.9 - 0.58(1.92) \approx 154.79\) The upper control limit (UCL) is approximately 157.01, and the lower control limit (LCL) is approximately 154.79.
03

Construct the \(\bar{x}\) chart

Now that we have calculated the UCL and LCL, we can construct the \(\bar{x}\) chart. For this, plot the points of the control chart as follows: 1. On the y-axis, indicate the mean of the sample means (\(\overline{\bar{x}} = 155.9\)). 2. Draw a horizontal line at the upper control limit (UCL = 157.01). 3. Draw a horizontal line at the lower control limit (LCL = 154.79). 4. Plot the 40 sample means on the control chart.
04

Interpret the \(\bar{x}\) chart

Once the control chart has been constructed, you can use it to monitor the process performance over time. If most of the plotted sample means fall within the control limits (between UCL and LCL), the process is considered to be in control. If any points fall outside these control limits, they indicate potential problems or assignable causes that should be investigated. You can also look for patterns or trends in the plotted points, such as runs above or below the centerline of the chart or continuous trends in one direction, which may indicate a systematic problem in the process.

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 91影视!

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

Suppose that college faculty with the rank of professor at two-year institutions earn an average of \(\$ 64,571\) per year \(^{7}\) with a standard deviation of \(\$ 4,000 .\) In an attempt to verify this salary level, a random sample of 60 professors was selected from a personnel database for all two- year institutions in the United States. a. Describe the sampling distribution of the sample \(\operatorname{mean} \bar{x}\) b. Within what limits would you expect the sample average to lie, with probability \(.95 ?\) c. Calculate the probability that the sample mean \(\bar{x}\) is greater than \(\$ 66,000 ?\) d. If your random sample actually produced a sample mean of \(\$ 66,000,\) would you consider this unusual? What conclusion might you draw?

Suppose a random sample of \(n=25\) observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 12 a. Give the mean and the standard deviation of the sampling distribution of the sample mean \(\bar{x}\). b. Find the probability that \(\bar{x}\) exceeds \(110 .\) c. Find the probability that the sample mean deviates from the population mean \(\mu=106\) by no more than 4

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 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?

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, \(x\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation
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.