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Determine the values of the following quantities

\(\begin{array}{l}{\rm{a}}{\rm{.}}{{\rm{x}}^{\rm{2}}}{\rm{,1,15}}\\{\rm{b}}{\rm{.}}{{\rm{X}}^{\rm{3}}}{\rm{,125}}\\{\rm{c}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{01,25}}\\{\rm{d}}{\rm{.}}{{\rm{X}}^{\rm{2}}}{\rm{00525}}\\{\rm{e}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{9925}}\\{\rm{f}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{995,25}}\end{array}\)

Short Answer

Expert verified

The values are

\(\begin{aligned}{{\rm{a}}{\rm{.\;22}}{\rm{.307}}}\\{{\rm{b}}{\rm{.\;34}}{\rm{.381}}}\\{{\rm{c}}{\rm{.\;44}}{\rm{.313}}}\\{{\rm{d}}{\rm{. 46}}{\rm{.925}}}\\{{\rm{e}}{\rm{. 11}}{\rm{.524}}}\\{{\rm{f}}{\rm{. 10}}{\rm{.519}}}\\{\rm{\;}}\end{aligned}\)

Step by step solution

01

To Determine the values

The values of the quantities can be obtained from the appendix of the book. The \({\rm{\alpha or 1 - \alpha }}\)can be found in a column part and the degrees of freedom in a row part. Next to every value there is a picture with the area of a tail.

(a):

For \({{\rm{\chi }}_{{\rm{0}}{\rm{.1,15,}}}}{\rm{\alpha = 0}}{\rm{.1 and \nu = 15}}\)degrees of freedom, therefore, from the appendix of the book, Table A.7, the value is

\({{\rm{\chi }}_{{\rm{0}}{\rm{.1,15}}}}{\rm{ = 22}}{\rm{.307}}{\rm{.}}\)

(b):

For \({{\rm{\chi }}_{{\rm{0}}{\rm{.1,25}}}}{\rm{,\alpha = 0}}{\rm{.1 and \nu = 25}}\)degrees of freedom, therefore, from the appendix of the book, Table A.7, the value is \({{\rm{\chi }}_{{\rm{0}}{\rm{.1,25}}}}{\rm{ = 34}}{\rm{.381}}{\rm{.}}\)

(c):

For \({{\rm{\chi }}_{{\rm{0}}{\rm{.01,25}}}}{\rm{,\alpha = 0}}{\rm{.01 and \nu = 25}}\) degrees of freedom, therefore, from the appendix of, the value is \({{\rm{\chi }}_{{\rm{0}}{\rm{.01,25}}}}{\rm{ = 44}}{\rm{.313}}\)

02

To Determine the values

\({\rm{(d):}}\)

For \({{\rm{\chi }}_{{\rm{0}}{\rm{.005,25,}}}}{\rm{\alpha = 0}}{\rm{.01 and \nu = 25}}\)degrees of freedom, therefore, from the appendix of the book, Table A.7, the value is\({{\rm{\chi }}_{{\rm{0}}{\rm{.005,25}}}}{\rm{ = 46}}{\rm{.925}}{\rm{.}}\)

(e):

For \({{\rm{\chi }}_{{\rm{0}}{\rm{.99,25,}}}}{\rm{1 - \alpha = 0}}{\rm{.99 and \nu = 25}}\) degrees of freedom, therefore, from the appendix of the book,the value is \({{\rm{\chi }}_{{\rm{0}}{\rm{.99,25}}}}{\rm{ = 11}}{\rm{.523}}{\rm{.}}\)

(f):

For \({{\rm{\chi }}_{{\rm{0}}{\rm{.995,25r}}}}{\rm{,\alpha = 0}}{\rm{.995 and \nu = 25}}\) degrees of freedom, therefore, from the appendix of the book, the value is \({{\rm{\chi }}_{{\rm{0}}{\rm{.995,25}}}}{\rm{ = 10}}{\rm{.519}}{\rm{.}}\)

Hence

\(\begin{aligned}{{\rm{a}}{\rm{.\;22}}{\rm{.307}}}\\{{\rm{b}}{\rm{.\;34}}{\rm{.381}}}\\{{\rm{c}}{\rm{.\;44}}{\rm{.313}}}\\{{\rm{d}}{\rm{. 46}}{\rm{.925}}}\\{{\rm{e}}{\rm{. 11}}{\rm{.524}}}\\{{\rm{f}}{\rm{. 10}}{\rm{.519}}}\\{\rm{\;}}\end{aligned}\)

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Most popular questions from this chapter

A normal probability plot of the n=\({\rm{26}}\) observations on escape time shows a substantial linear pattern; the sample mean and sample standard deviation are \({\rm{370}}{\rm{.69 and 24}}{\rm{.36}}\), respectively.

a. Calculate an upper confidence bound for population mean escape time using a confidence level of \({\rm{95\% }}\)

b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of \({\rm{95\% }}\). How does this bound compare with the confidence bound of part (a)?

c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by \({\rm{X27 and X28}}\), and let X new denote the average of these two values. Modify the formula for a PI for a single x value to obtain a PI for X new, and calculate a 95% two-sided interval based on the given escape data.

A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.764, 233.504). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine \(\overline x \) and s.)

Each of the following is a confidence interval for \({\rm{\mu = }}\) true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: \({\rm{(114}}{\rm{.4,115}}{\rm{.6)(114}}{\rm{.1,115}}{\rm{.9)}}\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \({\rm{90\% }}\) and for the other is \({\rm{99\% }}\). Which of the intervals has the \({\rm{90\% }}\) confidence level, and why?

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’s width compare to the width of the interval calculated in part (b)?

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

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