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Chapter 7: Q45E (page 306)

Wire electrical-discharge machining (WEDM) is a process used to manufacture conductive hard metal components. It uses a continuously moving wire that serves as an electrode. Coating on the wire electrode allows for

cooling of the wire electrode core and provides an improved cutting performance. The article 鈥淗ighPerformance Wire Electrodes for Wire ElectricalDischarge Machining鈥擜 Review鈥� gave the following sample observations on total coating layer thickness (in mm) of eight wire electrodes used for

WEDM: \({\rm{21 16 29 35 42 24 24 25}}\)

Calculate a \({\rm{99\% }}\)CI for the standard deviation of the coating layer thickness distribution. Is this interval valid whatever the nature of the distribution? Explain.

Short Answer

Expert verified

The boundaries of the confidence interval for the standard deviation is \((4.8248,21.8461)\)

No, the population distribution is a normal distribution

Step by step solution

01

To Calculate a \({\rm{99\% }}\)CI for the standard deviation

Given:

\({\rm{c = 99\% = 0}}{\rm{.99}}\)

\(2116{\rm{ }}29{\rm{ }}35{\rm{ }}42{\rm{ }}24{\rm{ }}24{\rm{ }}25\)

The mean is the sum of all values divided by the number of values:

\({\rm{\bar x = }}\frac{{{\rm{21 + 16 + 29 + 35 + 42 + 24 + 24 + 25}}}}{{\rm{8}}}{\rm{ = }}\frac{{{\rm{216}}}}{{\rm{8}}}{\rm{ = 27}}\)

The variance is the sum of squared deviations from the mean divided by \({\rm{n - 1}}\).The standard deviation is the square root of the variance:

\({\rm{s = }}\sqrt {\frac{{{{{\rm{(21 - 27)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{. + (25 - 27}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{8 - 1}}}}} {\rm{\gg 8}}{\rm{.2115}}\)

02

To Determine the critical values

Determine the critical values using the chi-square table in the appendix, which are given in the row \({\rm{df = n - 1 = 8 - 1 = 7}}\)and in the columns of \(\frac{{{\rm{1 - c}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.005 and 1 - }}\frac{{{\rm{1 - c}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.995}}\) :

\(\begin{array}{l}{\rm{\chi }}_{{\rm{1 - 0}}{\rm{.005}}}^{\rm{2}}{\rm{ = \chi }}_{{\rm{0}}{\rm{.995}}}^{\rm{2}}{\rm{ = 0}}{\rm{.989 }}\\{\rm{\chi }}_{{\rm{0}}{\rm{.005}}}^{\rm{2}}{\rm{ = 20}}{\rm{.276}}\end{array}\)

The boundaries of the confidence interval for the standard deviation are then:

\(\begin{array}{l}\sqrt {\frac{{{\rm{n - 1}}}}{{{\rm{\chi }}_{{\rm{\alpha /2}}}^{\rm{2}}}}} {\rm{ \times s = }}\sqrt {\frac{{{\rm{8 - 1}}}}{{{\rm{20}}{\rm{.276}}}}} {\rm{ \times 8}}{\rm{.2115\gg 4}}{\rm{.8248}}\\\sqrt {\frac{{{\rm{n - 1}}}}{{{\rm{\chi }}_{{\rm{1 - \alpha /2}}}^{\rm{2}}}}} {\rm{ \times s = }}\sqrt {\frac{{{\rm{8 - 1}}}}{{{\rm{0}}{\rm{.989}}}}} {\rm{ \times 8}}{\rm{.2115\gg 21}}{\rm{.8461}}\end{array}\)

This interval is NOT valid whatever the nature of the distribution, because we require that the population distribution is a normal distribution.

Hence the boundaries of the confidence interval for the standard deviation is \((4.8248,21.8461)\)

No, the population distribution is a normal distribution

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Most popular questions from this chapter

The article 鈥淢easuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers鈥� contained the following observations on degree of polymerization for paper specimens for which viscosity tim\({\rm{418 421 421 422 425 427 431 434 437 439 446 447 448 453 454 463 465}}\)es concentration fell in a certain middle range:

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c. Predict the resulting work of adhesion value resulting from a single future replication of the experiment by calculating a \({\rm{95\% }}\)prediction interval, and compare the width of this interval to the width of the CI from (b).

d. Calculate an interval for which you can have a high degree of confidence that at least \({\rm{95\% }}\)of all UHPC specimens adhered to steel will have work of adhesion values between the limits of the interval.

The negative effects of ambient air pollution on children鈥檚 lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of 鈥淚ndoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities鈥� investigated the relationship between indoor air-pollution metrics and lung function growth among children ages \({\rm{6--13}}\)years living in four Chinese cities. For each subject in the study, the authors measured an important lung-capacity index known as FEV1, the forced volume (in ml) of air that is exhaled in \({\rm{1}}\) second. Higher FEV1 values are associated with greater lung capacity. Among the children in the study, \({\rm{514}}\) came from households that used coal for cooking or heating or both. Their FEV1 mean was \({\rm{1427}}\)with a standard deviation of \({\rm{325}}\). (A complex statistical procedure was used to show that burning coal had a clear negative effect on mean FEV1 levels.)

a. Calculate and interpret a \({\rm{95\% }}\) (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. Does it appear that the parameter of interest has been accurately estimated?

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