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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:

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

(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 Pew Forum on Religion and Public Life reported on \({\rm{Dec}}{\rm{. 9, 2009}}\), that in a survey of \({\rm{2003}}\) American adults, \({\rm{25\% }}\)said they believed in astrology.

a. Calculate and interpret a confidence interval at the \({\rm{99\% }}\)confidence level for the proportion of all adult Americans who believe in astrology.

b. What sample size would be required for the width of a \({\rm{99\% }}\)CI to be at most .05 irrespective of the value of p?

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 \({\rm{450}}\)?

Let\({{\rm{\alpha }}_{\rm{1}}}{\rm{ > 0,}}{{\rm{\alpha }}_{\rm{2}}}{\rm{ > 0,}}\)with\({{\rm{\alpha }}_{\rm{1}}}{\rm{ + }}{{\rm{\alpha }}_{\rm{2}}}{\rm{ = 0}}\)

\({\rm{P}}\left( {{\rm{ - }}{{\rm{z}}_{{\alpha _{\rm{1}}}}}{\rm{ < }}\frac{{{\rm{\bar X - m}}}}{{{\rm{s/}}\sqrt {\rm{n}} }}{\rm{ < }}{{\rm{z}}_{{\alpha _{\rm{2}}}}}} \right){\rm{ = 1 - }}\alpha \)

a. Use this equation to derive a more general expression for a \({\rm{100(1 - \alpha )\% }}\)CI for \({\rm{\mu }}\)of which the interval (7.5) is a special case.

b. Let \({\rm{\alpha = 0}}{\rm{.5}}\)and \({{\rm{\alpha }}_{\rm{1}}}{\rm{ = \alpha /4,}}\)\({{\rm{\alpha }}_{\rm{2}}}{\rm{ = 3\alpha /4}}{\rm{.}}\)Does this result in a narrower or wider interval than the interval (7.5)?

A CI is desired for the true average stray-load loss \({\rm{\mu }}\) (watts) for a certain type of induction motor when the line current is held at \({\rm{10 amps}}\) for a speed of \({\rm{1500 rpm}}\). Assume that stray-load loss is normally distributed with \({\rm{\sigma = 3}}{\rm{.0}}\). a. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 25}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). b. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). c. Compute a \({\rm{99\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). d. Compute an \({\rm{82\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). e. How large must n be if the width of the \({\rm{99\% }}\) interval for \({\rm{\mu }}\) is to be \({\rm{1}}{\rm{.0}}\)?

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%?
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