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Chapter 7: Q49SE (page 307)

For those of you who don’t already know, dragon boat racing is a competitive water sport that involves 20 paddlers propelling a boat across various race distances. It has become increasingly popular over the last few years. The article “Physiological and Physical Characteristics of Elite Dragon Boat Paddlers” (J. of Strength and Conditioning, 2013: 137–145) summarized an extensive statistical analysis of data obtained from a sample of 11 paddlers. It reported that a 95% confidence interval for true average force (N) during a simulated 200-m race was (60.2, 70.6). Obtain a 95% prediction interval for the force of a single randomly selected dragon boat paddler undergoing the simulated race.

Short Answer

Expert verified

The boundaries of the confidence interval is \((47.4,83.4)\).

Step by step solution

01

Given information

\(\begin{array}{l}{\rm{Sample size }}n = 11\\{\rm{Confidence interval }}c = 95\% \end{array}\)

The boundaries of the confidence interval are\((60.2,70.6)\).

02

Margin error.

The margin of error is for the true average comprehensive strength:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }}\)

The margin of error is for the comprehensive strength of a single future specimen tested:

\(E = {t_{\alpha /2}} \times s\sqrt {1 + \frac{1}{n}} \)

03

To find the sample mean.

Determine the t-value by looking in the row starting with degrees of freedom \(\begin{array}{c}df = n - 1\\ = 11 - 1\\ = 10\end{array}\)

And in the column with \((1 - c)/2 = 0.025\) in the table of the student’s T distribution:

\({t_{\alpha /2}} = 2.228\)

The sample mean is the centre of the confidence interval and thus is the average of the boundaries of the confidence interval:

\(\begin{array}{c}\overline x = \frac{{60.2 + 70.6}}{2}\\ = 65.4\end{array}\)

04

To find the margin error.

The margin of error is half the width of the confidence interval:

\(\begin{array}{c}E = \frac{{70.6 - 60.2}}{2}\\ = 5.2\end{array}\)

The margin of error is also given by the formula,

\(\begin{array}{c}E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }}\\ = 5.2\end{array}\)

Solve the equation to the standard deviation s:

\(s = 5.2\frac{{\sqrt n }}{{{t_{\alpha /2}}}}\)

Substitute the respective values and evaluate:

\(\begin{array}{l}s = 5.2\frac{{\sqrt {11} }}{{2.228}}\\s \approx 7.7408\end{array}\)

05

To find the Prediction interval.

Determine the t-value by looking in the row starting with degrees of freedom \(\begin{array}{c}df = n - 1\\ = 11 - 1\\ = 10\end{array}\)

And in the column with \((1 - c)/2 = 0.025\) in the table of the student’s T distribution:

\({t_{\alpha /2}} = 2.228\)

The margin of error is then,

\(\begin{array}{c}E = {t_{\alpha /2}} \times s\sqrt {1 + \frac{1}{n}} \\ = 2.228 \times 7.7408\sqrt {1 + \frac{1}{{11}}} \\ \approx 18.0134\end{array}\)

The boundaries of the confidence interval, then become:

\(\begin{array}{c}\overline x - E = 65.4 - 18.0134\\ = 47.3866\\\overline x + E = 65.4 + 18.0134\\ = 83.4134\end{array}\)

Hence, the boundaries of the confidence interval is \((47.4,83.4)\).

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Most popular questions from this chapter

Consider a normal population distribution with the value of \({\rm{\sigma }}\) known. a. What is the confidence level for the interval \({\rm{\bar x \pm 2}}{\rm{.81\sigma /}}\sqrt {\rm{n}} \)? b. What is the confidence level for the interval \({\rm{\bar x \pm 1}}{\rm{.44\sigma /}}\sqrt {\rm{n}} \)? c. What value of \({{\rm{z}}_{{\rm{\alpha /2}}}}\) in the CI formula (\({\rm{7}}{\rm{.5}}\)) results in a confidence level of \({\rm{99}}{\rm{.7\% }}\)? d. Answer the question posed in part (c) for a confidence level of \({\rm{75\% }}\).

Let X1, X2,…, Xn be a random sample from a continuous probability distribution having median \(\widetilde {\bf{\mu }}\) (so that \({\bf{P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = }}{\bf{.5}}\)).

a. Show that

\({\bf{P(min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < }}\widetilde {\bf{\mu }}{\bf{ < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)) = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\)

So that \({\bf{(min(}}{{\bf{x}}_{\bf{i}}}{\bf{),max(}}{{\bf{x}}_{\bf{i}}}{\bf{))}}\)is a \({\bf{100(1 - \alpha )\% }}\) confidence interval for \(\widetilde {\bf{\mu }}\) with (\({\bf{\alpha = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\).Hint :The complement of the event \(\left\{ {{\bf{min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < \mu < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}}} \right\}\)is \({\bf{\{ max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}{\bf{\} }} \cup {\bf{\{ min(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \ge \widetilde {\bf{\mu }}{\bf{\} }}\). But \({\bf{max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\) iff \({\bf{(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\)for all i.

b.For each of six normal male infants, the amount of the amino acid alanine (mg/100 mL) was determined while the infants were on an isoleucine-free diet,

resulting in the following data:

2.84 3.54 2.80 1.44 2.94 2.70

Compute a 97% CI for the true median amount of alanine for infants on such a diet(“The Essential Amino Acid Requirements of Infants,” Amer. J. Of Nutrition, 1964: 322–330).

c.Let x(2)denote the second smallest of the xi’s and x(n-1) denote the second largest of the xi’s. What is the confidence level of the interval(x(2), x(n-1)) for \(\widetilde \mu \)?

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with\({\rm{\sigma = 100,}}\)The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of\({\rm{\sigma }}\).

a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8439 lb, compute a 90% CI for the true average yield point of the modified bar.

b. How would you modify the interval in part (a) to obtain a confidence level of 92%?

The article “Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers” contained the following observations on degree of polymerization for paper specimens for which viscosity tim\({\rm{418 421 421 422 425 427 431 434 437 439 446 447 448 453 454 463 465}}\)es concentration fell in a certain middle range:

a. Construct a boxplot of the data and comment on any interesting features.

b. Is it plausible that the given sample observations were selected from a normal distribution?

c. Calculate a two-sided \({\rm{95\% }}\)confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that \({\rm{440}}\) is a plausible value for true average degree of polymerization? What about \({\rm{450}}\)?

Determine the values of the following quantities

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