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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

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time.

b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p ˆ?

Consider the next \({\rm{1000}}\) 95% CIs for m that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these\({\rm{1000}}\)intervals do you expect to capture the corresponding value of m? What is the probability that between \({\rm{940 and 960}}\)Do these intervals contain the corresponding value of m? (Hint: Let Y = the number among the \({\rm{1000}}\) intervals that contain m. What kind of random variable is Y?)

a.Use the results of Example 7.5 to obtain a 95% lower confidence bound for the parameter λ of an exponential distribution, and calculate the bound based on the data given in the example.

b.If lifetime X has an exponential distribution, the probability that lifetime exceeds t is P(X>t) = e-λ³Ù. Use the result of part (a) to obtain a 95% lower confidence bound for the probability that breakdown time exceeds 100 min.

A random sample of n=\({\rm{15}}\)heat pumps of a certain type yielded the following observations on lifetime (in years):

\(\begin{array}{*{20}{l}}{{\rm{2}}{\rm{.0 1}}{\rm{.3 6}}{\rm{.0 1}}{\rm{.9 5}}{\rm{.1 }}{\rm{.4 1}}{\rm{.0 5}}{\rm{.3}}}\\{{\rm{15}}{\rm{.7 }}{\rm{.7 4}}{\rm{.8 }}{\rm{.9 12}}{\rm{.2 5}}{\rm{.3 }}{\rm{.6}}}\end{array}\)

a. Assume that the lifetime distribution is exponential and use an argument parallel to that to obtain a \({\rm{95\% }}\) CI for expected (true average) lifetime.

b. How should the interval of part (a) be altered to achieve a confidence level of \({\rm{99\% }}\)?

c. What is a \({\rm{95\% }}\)CI for the standard deviation of the lifetime distribution? (Hint: What is the standard deviation of an exponential random variable?)

By how much must the sample size n be increased if the width of the CI (7.5) is to be halved? If the sample size is increased by a factor of 25, what effect will this have on the width of the interval? Justify your assertions.

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