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Chapter 7: Q49SE (page 307)

For those of you who don鈥檛 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 鈥淧hysiological and Physical Characteristics of Elite Dragon Boat Paddlers鈥� (J. of Strength and Conditioning, 2013: 137鈥�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鈥檚 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鈥檚 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

Silicone implant augmentation rhinoplasty is used to correct congenital nose deformities. The success of the procedure depends on various biomechanical properties of the human nasal periosteum and fascia. The article 鈥淏iomechanics in Augmentation Rhinoplasty鈥� reported that for a sample of 15 (newly deceased) adults, the mean failure strain (%) was 25.0, and the standard deviation was 3.5.

a. Assuming a normal distribution for failure strain, estimate true average strain in a way that conveys information about precision and reliability.

b. Predict the strain for a single adult in a way that conveys information about precision and reliability. How does the prediction compare to the estimate calculated in part (a)?

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

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\({\rm{v = 10}}\)

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

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.

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?)
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