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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.)

Short Answer

Expert verified

The interval is \((228.533,234.733)\).

Step by step solution

01

Step 1:Confidence bound.

Upper confidence bound for \(\mu \) and lower confidence bound for \(\mu \) are given by,

\(\overline x + {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-upper bound,

\(\overline x - {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-lower bound.

With confidence level of \(100(1 - \alpha )\% \). Then distribution the random sample is taken from is normal.

02

Step 2:Sample standard deviation.

In order to obtain confidence interval with \(\mu \) confidence level there are a couple of values that are missing , \(\overline x ,s,{t_{\alpha /2,n - 1}}\). From the resulting interval \((229.764,233.504)\). The sample mean can be obtained as the centre of the interval,

\(\begin{array}{l}\overline x = \frac{{229.764 + 233.504}}{2}\\\overline x = 231.633\end{array}\)

The sample standard deviation can be obtained using the width of the interval,

\(w = 2 \cdot {t_{\alpha /2}} \cdot \frac{s}{{\sqrt n }}\)

Where in case of \(95\% \) confidence level, \(\alpha \) is \(0.05\), which means that \(\alpha /2 = 0.025\) and the sample size is \(5\). From the appendix table, the \(t\) value can be obtained and it is ,

\({t_{0.025,4}} = 2.776\)

The width \(w\)is the difference between upper and lower limit

\(\begin{array}{l}w = 233.504 - 229.764\\w = 3.738\end{array}\)

Now the sample standard deviation can be obtained from

\(w = 2 \cdot {t_{\alpha /2}} \cdot \frac{s}{{\sqrt n }}\)

As follows,

\(\begin{array}{l}s = \frac{{w\sqrt n }}{{2{t_{\alpha /2}}}}\\s = \frac{{3.738\sqrt 5 }}{{2(2.776)}}\\s = 1.5055\end{array}\)

03

Step 3:Confidence interval.

To obtain the confidence interval for \(\mu \) with \(\mu \) confidence level, the only missing value is \({t_{\alpha /2,n - 1}}\) which for

\(\begin{array}{l}\alpha = 0.01\\(\alpha /2) = 0.005\end{array}\)

And

\(\begin{array}{l}n - 1 = 5 - 1\\ = 4\end{array}\)

From the table in the appendix of the book. Therefore the interval becomes,

\(\begin{array}{l}\left( {231.633 - 4.604\frac{{1.5055}}{{\sqrt 5 }},231.633 + 4.604\frac{{1.5055}}{{\sqrt 5 }}} \right)\\ = (228.533,234.733)\end{array}\)

Hence, The interval is \((228.533,234.733)\).

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Most popular questions from this chapter

Suppose that a random sample of \({\rm{50}}\) bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \({\rm{\mu }}\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \({\rm{95\% }}\)confidence interval is \({\rm{(7}}{\rm{.8,9}}{\rm{.4)}}\). a. Would a \({\rm{90\% }}\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \({\rm{95\% }}\) chance that \({\rm{\mu }}\) is between \({\rm{7}}{\rm{.8}}\) and \({\rm{9}}{\rm{.4}}\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \({\rm{95\% }}\) of all bottles of this type of cough syrup have an alcohol content that is between \({\rm{7}}{\rm{.8}}\) and \({\rm{9}}{\rm{.4}}\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size \({\rm{50}}\) and then computing the corresponding \({\rm{95\% }}\) interval is repeated \({\rm{100}}\)times, \({\rm{95}}\) of the resulting intervals will include \({\rm{\mu }}\). Is this statement correct? Why or why not?

Determine the following:

a. The 95th percentile of the chi-squared distribution with \({\rm{v = 10}}\)

b. The 5th percentile of the chi-squared distribution with\({\rm{v = 10}}\)

\(\begin{array}{l}{\rm{c}}{\rm{.P}}\left( {{\rm{10}}{\rm{.98£ }}{{\rm{\chi }}^{\rm{2}}}{\rm{£ 36}}{\rm{.78}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi - squared rv with \nu = 22}}\\{\rm{d}}{\rm{.P}}\left( {{{\rm{\chi }}^{\rm{2}}}{\rm{ < 14}}{\rm{.611}}} \right.{\rm{ or }}\left. {{{\rm{\chi }}^{\rm{2}}}{\rm{ > 37}}{\rm{.652}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi squared rv with \nu = 25}}\end{array}\)

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?)

Determine the values of the following quantities

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

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?

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