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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

presented a sample of \({\rm{n = 153}}\)observations on ultimate tensile strength, the previous section gave summary quantities and requested a large-sample confidence interval. Because the sample size is large, no assumptions about the population distribution are required for the validity of the CI.

a. Is any assumption about the tensile-strength distribution required prior to calculating a lower prediction bound for the tensile strength of the next specimen selected using the method described in this section? Explain.

b. Use a statistical software package to investigate the plausibility of a normal population distribution.

c. Calculate a lower prediction bound with a prediction level of \({\rm{95\% }}\)for the ultimate tensile strength of the next specimen selected.

In Example 6.8, we introduced the concept of a censored experiment in which n components are put on test and the experiment terminates as soon as r of the components have failed. Suppose component lifetimes are independent, each having an exponential distribution with parameter 位. Let Y1 denote the time at which the first failure occurs, Y2 the time at which the second failure occurs, and so on, so that Tr= Y1 + 鈥� + Yr+ (n- r)Yr is the total accumulated lifetime at termination. Then it can be shown that 2位罢r has a chi-squared distribution with 2r df. Use this fact to develop a 100(1 - 伪)% CI formula for true average lifetime \(\frac{{\bf{1}}}{{\bf{\lambda }}}\) . Compute a 95% CI from the data in Example 6.8.

Exercise 72 of Chapter 1 gave the following observations on a receptor binding measure (adjusted distribution volume) for a sample of 13 healthy individuals: 23, 39, 40, 41, 43, 47, 51, 58, 63, 66, 67, 69, 72.

a. Is it plausible that the population distribution from which this sample was selected is normal?

b. Calculate an interval for which you can be 95% confident that at least 95% of all healthy individuals in the population have adjusted distribution volumes lying between the limits of the interval.

c. Predict the adjusted distribution volume of a single healthy individual by calculating a 95% prediction interval. How does this interval鈥檚 width compare to the width of the interval calculated in part (b)?

A more extensive tabulation of t critical values than what appears in this book shows that for the t distribution with

\({\rm{20}}\)df, the areas to the right of the values \({\rm{.687, }}{\rm{.860, and 1}}{\rm{.064 are }}{\rm{.25, }}{\rm{.20, and }}{\rm{.15,}}\)respectively. What is the confidence level for each of the following three confidence intervals for the mean m of a normal population distribution? Which of the three intervals would you recommend be used, and why?

\(\begin{array}{l}{\rm{a}}{\rm{.(\bar x - }}{\rm{.687s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.725s/}}\sqrt {{\rm{21}}} {\rm{)}}\\{\rm{b}}{\rm{.(\bar x - }}{\rm{.860\;s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.325s/}}\sqrt {{\rm{21}}} {\rm{)}}\\{\rm{c}}{\rm{.(\bar x - 1}}{\rm{.064s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.064s/}}\sqrt {{\rm{21}}} {\rm{)}}\end{array}\)

A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n =\({\rm{20}}\) randomly selected healthy men.

\({\rm{.95 }}{\rm{.85 }}{\rm{.92 }}{\rm{.95 }}{\rm{.93 }}{\rm{.86 1}}{\rm{.00 }}{\rm{.92 }}{\rm{.85 }}{\rm{.81 }}{\rm{.78 }}{\rm{.93 }}{\rm{.93 1}}{\rm{.05 }}{\rm{.93 1}}{\rm{.06 1}}{\rm{.06 }}{\rm{.96 }}{\rm{.81 }}{\rm{.96}}\)

A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows:

Variable N Mean Median TrMean StDev SEMean cadence

\({\rm{20 0}}{\rm{.9255 0}}{\rm{.9300 0}}{\rm{.9261 0}}{\rm{.0809 0}}{\rm{.0181}}\)

Variable Min Max Q1 Q3 cadence

\({\rm{0}}{\rm{.7800 1}}{\rm{.0600 0}}{\rm{.8525 0}}{\rm{.9600}}\)

a. Calculate and interpret a \({\rm{95\% }}\) confidence interval for population mean cadence.

b.Calculate and interpret a \({\rm{95\% }}\)prediction interval for the cadence of a single individual randomly selected from this population.

c. Calculate an interval that includes at least \({\rm{99\% }}\)of the cadences in the population distribution using a confidence level of \({\rm{95\% }}\)
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