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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

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

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.)

The negative effects of ambient air pollution on children鈥檚 lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of 鈥淚ndoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities鈥� investigated the relationship between indoor air-pollution metrics and lung function growth among children ages \({\rm{6--13}}\)years living in four Chinese cities. For each subject in the study, the authors measured an important lung-capacity index known as FEV1, the forced volume (in ml) of air that is exhaled in \({\rm{1}}\) second. Higher FEV1 values are associated with greater lung capacity. Among the children in the study, \({\rm{514}}\) came from households that used coal for cooking or heating or both. Their FEV1 mean was \({\rm{1427}}\)with a standard deviation of \({\rm{325}}\). (A complex statistical procedure was used to show that burning coal had a clear negative effect on mean FEV1 levels.)

a. Calculate and interpret a \({\rm{95\% }}\) (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. Does it appear that the parameter of interest has been accurately estimated?

b. Suppose the investigators had made a rough guess of \({\rm{320}}\) for the value of s before collecting data. What sample size would be necessary to obtain an interval width of \({\rm{50}}\)ml for a confidence level of \({\rm{95\% }}\)?

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?

It is important that face masks used by fire fighters be able to withstand high temperatures because fire fighters commonly work in temperatures of 200鈥�500掳F. In a test of one type of mask, 11 of 55 masks had lenses pop out at 250掳. Construct a 90% upper confidence bound for the true proportion of masks of this type whose lenses would pop out at 250掳.
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