/*! 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} Q43 SE The following numbers are observ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Q43 SE (page 714)

The following numbers are observations on tensile strength of synthetic fabric specimens selected from a production process at equally spaced time intervals. Construct appropriate control charts, and comment (assume an assignable cause is identifiable for any out of-control observations).

\(\begin{array}{*{20}{r}}{ 1. }&{51.3}&{51.7}&{49.5}&{12.}&{49.6}&{48.4}&{50.0}\\{ 2. }&{51.0}&{50.0}&{49.3}&{13.}&{49.8}&{51.2}&{49.7}\\{ 3. }&{50.8}&{51.1}&{49.0}&{14.}&{50.4}&{49.9}&{50.7}\\{ 4. }&{50.6}&{51.1}&{49.0}&{15.}&{49.4}&{49.5}&{49.0}\\{ 5. }&{49.6}&{50.5}&{50.9}&{16.}&{50.7}&{49.0}&{50.0}\\{ 6. }&{51.3}&{52.0}&{50.3}&{17.}&{50.8}&{49.5}&{50.9}\\{ 7. }&{49.7}&{50.5}&{50.3}&{18.}&{48.5}&{50.3}&{49.3}\\{ 8. }&{51.8}&{50.3}&{50.0}&{19.}&{49.6}&{50.6}&{49.4}\\{9.}&{48.6}&{50.5}&{50.7}&{20.}&{50.9}&{49.4}&{49.7}\\{ 10. }&{49.6}&{49.8}&{50.5}&{21.}&{54.1}&{49.8}&{48.5}\\{ 11. }&{49.9}&{50.7}&{49.8}&{22.}&{50.2}&{49.6}&{51.5}\end{array}\)

Short Answer

Expert verified

Construct \(s\) chart and \(\bar X\) chart based on \(\bar s\).

Step by step solution

01

Step 1:To Find the Defectives ina unit has lower and upper control

The following table contains value required to compute\(S\)chart and\(\bar X\)chart based on\(\bar s\).

\(\begin{array}{*{20}{c}}i&{{{\bar x}_i}}&{{s_i}}&{{r_i}}\\1&{50.83}&{1.172}&{2.2}\\2&{50.1}&{0.854}&{1.7}\\3&{50.3}&{1.136}&{2.1}\\4&{50.23}&{1.097}&{2.1}\\5&{50.33}&{0.666}&{1.3}\\6&{51.2}&{0.854}&{1.7}\\7&{50.17}&{0.416}&{0.8}\\8&{50.7}&{0.964}&{1.8}\\9&{49.93}&{1.159}&{2.1}\\{10}&{49.97}&{0.473}&{0.9}\\{11}&{50.13}&{0.698}&{0.9}\\{12}&{49.33}&{0.833}&{1.6}\\{13}&{50.23}&{0.839}&{1.5}\\{14}&{50.33}&{0.404}&{0.8}\\{15}&{49.3}&{0.265}&{0.5}\\{16}&{49.9}&{0.854}&{1.7}\\{17}&{50.4}&{0.781}&{1.4}\\{18}&{49.37}&{0.902}&{1.8}\\{19}&{49.87}&{0.643}&{1.2}\\{20}&{50}&{0.794}&{1.5}\\{21}&{50.8}&{2.931}&{5.6}\\{22}&{50.43}&{0.971}&{1.9}\\{\sum {{s_i}} = 19.706}&{\bar s = 0.8957}&{}&{}\\{\sum {\bar x} i = 1103.85}&{\bar x = 50.175}&{}&{}\\{}&{}&{}&{}\end{array}\)

\(S\)chart:

\(LCL = 0.8957 - \frac{{3 \times 0.8957\sqrt {1 - 0.88{6^2}} }}{{0.886}}\mathop = \limits^{(1)} 0;\)

\(UCL = 0.8957 + \frac{{3 \times 0.8957\sqrt {1 - 0.88{6^2}} }}{{0.886}} = 2.3020.\)

(1) : it is lower than zero, so set it to zero.

One value is larger than the upper control limit\(UCL\)

\(UCL = 2.3020 < 2.931 = {s_{21}}\)

thus, because in the exercise it says that an assignable cause is assumed to have been identified remove this value and repeat procedure. It is easy to obtain new required value to compute new\(LCL\)and\(UCL\)

\(\begin{array}{l}\sum {{s_i}} = 16.775;\\\bar s = 0.7998;\\\bar \bar x = 50.145. \\New S chart:\\ LCL = 0.7998 - \frac{{3 \times 0.7998\sqrt {1 - 0.88{6^2}} }}{{0.886}}\mathop = \limits^{(1)} 0;\\UCL = 0.7998 + \frac{{3 \times 0.7998\sqrt {1 - 0.88{6^2}} }}{{0.886}} = 2.0529.\end{array}\)

This process is in-control, every\({s_i}\)value is between the limits.

\(\bar X\)chart based on\(\bar s\):

\(\begin{array}{l}LCL = 50.145 - \frac{{3 \times 0.7988}}{{0.886 \times \sqrt 3 }} = 48.58\\UCL = 50.145 + \frac{{3 \times 0.7988}}{{0.886 \times \sqrt 3 }} = 51.71\end{array}\)

Values\({\bar x_i}\)are between the limit; thus the process is in-control.

One has all required value to compute a \(r\) chart in the table.

02

Step 2:Final proof

Construct \(s\) chart and \(\bar X\) chart based on \(\bar s\).

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

Let \(\alpha \) be a number between 0 and 1 , and define a sequence \({W_1},{W_2},{W_3}, \ldots \) by \({W_0} = \mu \)and \({W_t} = \alpha {\bar X_t} + (1 - \alpha ){W_{t - 1}}\) for \(t = 1,2, \ldots .\)Substituting for \({W_{t - 1}}\) its representation in terms of \({\bar X_{t - 1}}\) and \({W_{t - 2}}\), then substituting for \({W_{t - 2}}\), and so on, results in

\(\begin{array}{l}{W_t} = \alpha {{\bar X}_t} + \alpha (1 - \alpha ){{\bar X}_{t - 1}} + \ldots \\ + \alpha {(1 - \alpha )^{t - 1}}{{\bar X}_1} + {(1 - \alpha )^t}\mu \end{array}\)

The fact that \({W_t}\) depends not only on \({\bar X_t}\) but also on averages for past time points, albeit with (exponentially) decreasing weights, suggests that changes in the process mean will be more quickly reflected in the \({W_t}\) 's than in the individual \({\bar X_t}\) 's.

a. Show that \(E\left( {{W_t}} \right) = \mu \).

b. Let \(\sigma _t^2 = V\left( {{W_t}} \right)\), and show that

\(\sigma _t^2 = \frac{{\alpha \left( {1 - {{(1 - \alpha )}^{2\eta }}} \right)}}{{2 - \alpha }} \times \frac{{{\sigma ^2}}}{n}\)

c. An exponentially weighted moving-average control chart plots the \({W_t}\)'s and uses control limits \({\mu _0} \pm 3{\sigma _t}\) (or \(\bar \bar x\) in place of \({\mu _0}\) ). Construct such a chart for the data of Example 16.9, using \({\mu _0} = 40\).

A cork intended for use in a wine bottle is considered acceptable if its diameter is between 2.9 cm and 3.1 cm (so the lower specification limit is LSL 5 2.9 and the upper specification limit is USL 5 3.1).

a. If cork diameter is a normally distributed variable with mean value 3.04 cm and standard deviation .02 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)?

When the out-of-control ARL corresponds to a shift of 1 standard deviation in the process mean, what are the characteristics of the CUSUM procedure that has ARLs of 250 and \(4.8\), respectively, for the in-control and out of-control conditions?

When installing a bath faucet, it is important to properly fasten the threaded end of the faucet stem to the water supply line. The threaded stem dimensions must meet product specifications, otherwise malfunction and leakage may occur. Authors of "Improving the Process Capability of a Boring Operation by the Application of Statistical Techniques" (Intl. J. Sci. Engr. Research, Vol. 3, Issue\(5\), May\(2012\)) investigated the production process of a particular bath faucet manufactured in India. The article reported the threaded stem diameter (target value being\(13\;mm\)) of each faucet in\(25\)samples of size\(4\)as shown here:

\(\begin{aligned}{*{20}{c}}{Subgroup}&{x\_(1)}&{x\_(2)}&{x\_(3)}&{x\_(4)}\\1&{13.02}&{12.95}&{12.92}&{12.99}\\2&{13.02}&{13.10}&{12.96}&{12.96}\\3&{13.04}&{13.08}&{13.05}&{13.10}\\4&{13.04}&{12.96}&{12.96}&{12.97}\\5&{12.96}&{12.97}&{12.90}&{13.05}\\6&{12.90}&{12.88}&{13.00}&{13.05}\\7&{12.97}&{12.96}&{12.96}&{12.99}\\8&{13.04}&{13.02}&{13.05}&{12.97}\\9&{13.05}&{13.10}&{12.98}&{12.96}\\{10}&{12.96}&{13.00}&{12.99}&{12.99}\\{11}&{12.90}&{13.05}&{12.98}&{12.88}\\{12}&{12.96}&{12.98}&{12.97}&{13.02}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{ }&{}&{}&{}&{}\\{13}&{13.00}&{12.96}&{12.99}&{12.90}\\{14}&{12.88}&{12.94}&{13.05}&{13.00}\\{15}&{12.96}&{12.96}&{13.04}&{12.98}\\{16}&{12.99}&{12.94}&{13.00}&{13.05}\\{17}&{13.05}&{13.02}&{12.88}&{12.96}\\{18}&{13.08}&{13.06}&{13.10}&{13.05}\\{19}&{13.02}&{13.05}&{13.04}&{12.97}\\{20}&{12.96}&{12.90}&{12.97}&{13.05}\\{21}&{12.98}&{12.99}&{12.96}&{13.00}\\{22}&{12.97}&{13.02}&{12.96}&{12.99}\\{23}&{13.04}&{13.00}&{12.98}&{13.10}\\{24}&{13.02}&{12.90}&{13.05}&{12.97}\\{25}&{12.93}&{12.88}&{12.91}&{12.90}\end{aligned}\)

Three-dimensional (3D) printing is a manufacturing technology that allows the production of three-dimensional solid objects through a meticulous layering process performed by a\(3D\)printer.\(3D\)printing has rapidly become a time-saving and economical way to create a wide variety of products such as medical implants, furniture, tools, and even jewelry. The article "Process Capability Analysis of Cost Effective Rapid Casting Solution Based on Three Dimensional Printing" (MIT Intl. J. Mech. Engr., \(\;2012: 31 - 38\)) considered the production process of metal castings by using a\(3D\)printer. Data was collected on\(16\)batches (each having two castings), where the outer diameter of each casting (in\(mm\)) was recorded. The target diameter of each casting was\(60\;mm\). The resulting data is given here:

\(\begin{aligned}{*{20}{c}}{ Batch }&{{x_1}}&{{x_2}}\\1&{59.664}&{59.675}\\2&{59.661}&{59.648}\\3&{59.679}&{59.652}\\4&{59.665}&{59.654}\\5&{59.667}&{59.678}\\6&{59.673}&{59.657}\\7&{59.676}&{59.661}\\8&{59.648}&{59.651}\\9&{59.681}&{59.675}\\{10}&{59.655}&{59.672}\\{11}&{59.691}&{59.676}\\{12}&{59.682}&{59.651}\\{13}&{59.651}&{59.682}\\{14}&{59.668}&{59.685}\\{15}&{59.691}&{59.682}\\{16}&{59.661}&{59.673}\\{}&{}&{}\end{aligned}\)

Apply the supplemental rules suggested in the text to the data. Are there any out-of-control signals?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

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.