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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

A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions.

a. What sample size is necessary if the \({\rm{95\% }}\)CI for p is to have a width of at most . \({\rm{10}}\)irrespective of p?

b. If the legislator has strong reason to believe that at least \({\rm{2/3}}\)of the electorate know of her position, how large a sample size would you recommend?

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\(\begin{array}{*{20}{l}}{{\rm{115 181 255 418 441 461 516 739 743 789 807}}}\\{{\rm{865 924 983 1025 1062 1063 1165 1191 1222 1222 1251}}}\\{{\rm{1277 1290 1357 1369 1408 1455 1478 1519 1578 1578 1599}}}\\{{\rm{1603 1605 1696 1735 1799 1815 1852 1899 1925 1965}}}\end{array}\)

a. Can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. (Note: A normal probability plot of the data exhibits a reasonably linear pattern.)

b. Calculate and interpret a confidence interval with a \({\rm{99\% }}\)confidence level for true average lifetime.

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?

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

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