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Chapter 7: Q37E (page 303)

A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n =\({\rm{20}}\) randomly selected healthy men.

\({\rm{.95 }}{\rm{.85 }}{\rm{.92 }}{\rm{.95 }}{\rm{.93 }}{\rm{.86 1}}{\rm{.00 }}{\rm{.92 }}{\rm{.85 }}{\rm{.81 }}{\rm{.78 }}{\rm{.93 }}{\rm{.93 1}}{\rm{.05 }}{\rm{.93 1}}{\rm{.06 1}}{\rm{.06 }}{\rm{.96 }}{\rm{.81 }}{\rm{.96}}\)

A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows:

Variable N Mean Median TrMean StDev SEMean cadence

\({\rm{20 0}}{\rm{.9255 0}}{\rm{.9300 0}}{\rm{.9261 0}}{\rm{.0809 0}}{\rm{.0181}}\)

Variable Min Max Q1 Q3 cadence

\({\rm{0}}{\rm{.7800 1}}{\rm{.0600 0}}{\rm{.8525 0}}{\rm{.9600}}\)

a. Calculate and interpret a \({\rm{95\% }}\) confidence interval for population mean cadence.

b.Calculate and interpret a \({\rm{95\% }}\)prediction interval for the cadence of a single individual randomly selected from this population.

c. Calculate an interval that includes at least \({\rm{99\% }}\)of the cadences in the population distribution using a confidence level of \({\rm{95\% }}\)

Short Answer

Expert verified

a) The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.0379 = 0}}{\rm{.8876}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.0379 = 0}}{\rm{.9634}}\end{aligned}\)

b) The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.2925 = 0}}{\rm{.6330}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.2925 = 1}}{\rm{.2180}}\end{aligned}\)

c) The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.2925 = 0}}{\rm{.6330}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.2925 = 1}}{\rm{.2180}}\end{aligned}\)

Step by step solution

01

To Calculate and interpret a \({\rm{95\% }}\)confidence interval

Given:

\(\begin{align}\text x虅 &=0 \text{.9255} \\ \text s&=0 \text{.0809} \\ \text n&=20 c=95 \end{align}\)

(a)

Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 20 - 1 = 19}}\) and in the column with \({\rm{\alpha = (1 - c)/2 = (1 - 0}}{\rm{.95)/2 = 0}}{\rm{.025}}\) in table A. 5 :

\({{\rm{t}}_{{\rm{0}}{\rm{.025,19}}}}{\rm{ = 2}}{\rm{.093}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ = 2}}{\rm{.093 \times }}\frac{{{\rm{0}}{\rm{.0809}}}}{{\sqrt {{\rm{20}}} }}{\rm{\gg 0}}{\rm{.0379}}\)

Hence The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.0379 = 0}}{\rm{.8876}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.0379 = 0}}{\rm{.9634}}\end{aligned}\)

02

Step 2:To Calculate and interpret a \({\rm{95\% }}\)prediction interval

(b)

Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 20 - 1 = 19}}\)nd in the column with \({\rm{\alpha = (1 - c)/2 = (1 - 0}}{\rm{.95)/2 = 0}}{\rm{.025}}\)in tableA. 5 :

\({{\rm{t}}_{{\rm{0}}{\rm{.02x,19}}}}{\rm{ = 2}}{\rm{.093}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} {\rm{ = 2}}{\rm{.093 \times 0}}{\rm{.0809}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{20}}}}} {\rm{\gg 0}}{\rm{.1735}}\)

Hence The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.1735 = 0}}{\rm{.7520}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.1735 = 1}}{\rm{.0990}}\end{aligned}\)

03

To calculate an interval that includes at least \({\rm{99\% }}\)

(c)

Given:

\({\rm{k = 99\% = 0}}{\rm{.99}}\)

The tolerance critical value is given in the row with $n=20$ and in the column with confidence level \(95\% \)and\(\% \)of population captured \( \ge 99\% \)

\({\rm{tol = 3}}{\rm{.615}}\)

The margin of error is then the product of the tolerance critical value and the standard deviation:

\({\rm{E = tol \times s = 3}}{\rm{.615 \times 0}}{\rm{.0809\gg 0}}{\rm{.2925}}\)

Hence The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.2925 = 0}}{\rm{.6330}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.2925 = 1}}{\rm{.2180}}\end{aligned}\)

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Most popular questions from this chapter

The article 鈥淒istributions of Compressive Strength Obtained from Various Diameter Cores鈥� (ACI Materials J., 2012: 597鈥�606) described a study in which compressive strengths were determined for concrete specimens of various types, core diameters, and length -to-diameter ratios. For one particular type, diameter, and l/d ratio, the 18 tested specimens resulted in a sample mean compressive strength of 64.41 MPa and a sample standard deviation of 10.32 MPa. Normality of the compressive strength distribution was judged to be quite plausible.

a.Calculate a confidence interval with confidence level 98% for the true average compressive strength under these circumstances.

b.Calculate a 98% lower prediction bound for the compressive strength of a single future specimen tested under the given circumstances. (Hint: t.02,17 = 2.224.)

A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of \({\rm{8}}{\rm{.48}}\)MPa and a sample standard deviation of . \({\rm{79}}\)MPa a. Calculate and interpret a \({\rm{95\% }}\)lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress?

b. Calculate and interpret a \({\rm{95\% }}\)lower prediction bound for the proportional limit stress of a single joint of this type.

A normal probability plot of the n=\({\rm{26}}\) observations on escape time shows a substantial linear pattern; the sample mean and sample standard deviation are \({\rm{370}}{\rm{.69 and 24}}{\rm{.36}}\), respectively.

a. Calculate an upper confidence bound for population mean escape time using a confidence level of \({\rm{95\% }}\)

b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of \({\rm{95\% }}\). How does this bound compare with the confidence bound of part (a)?

c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by \({\rm{X27 and X28}}\), and let X new denote the average of these two values. Modify the formula for a PI for a single x value to obtain a PI for X new, and calculate a 95% two-sided interval based on the given escape data.

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.)

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?
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