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Chapter 7: Q35E (page 303)

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)?

Short Answer

Expert verified

a) The \({\rm{95 confidence interval (23}}{\rm{.0616,26}}{\rm{.9384)}}\)

b) The \({\rm{95\% prediction interval }}\left( {{\rm{17}}{\rm{.2463,32}}{\rm{.7537}}} \right)\)

We note that the prediction interval is much wider than the confidence interval.

Step by step solution

01

To estimate true average strain

Given:

\(\begin{array}{l}{\rm{n = 15}}\\{\rm{\bar x = 25}}{\rm{.0}}\\{\rm{s = 3}}{\rm{.5}}\end{array}\)

(a) Note: I will assume that we need to calculate a $95 \%$ confidence interval, other confidence intervals can be found similarly.

\({\rm{c = 95\% = 0}}{\rm{.95}}\)

Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 15 - 1 = 14}}\) and in the column with \({\rm{\alpha = (1 - c)/2 = 0}}{\rm{.025}}\)in the table of the critical values for t distributions in the appendix:

\(\begin{array}{l}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ = 2}}{\rm{.145}}\\{\rm{The margin of error is then:}}\end{array}\)

\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ = 2}}{\rm{.145 \times }}\frac{{{\rm{3}}{\rm{.5}}}}{{\sqrt {{\rm{15}}} }}{\rm{\gg 1}}{\rm{.9384}}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E = 25}}{\rm{.0 - 1}}{\rm{.9384 = 23}}{\rm{.0616}}\\{\rm{\bar x + E = 25}}{\rm{.0 + 1}}{\rm{.9384 = 26}}{\rm{.9384}}\end{array}\)

Hence the \({\rm{95 confidence interval (23}}{\rm{.0616,26}}{\rm{.9384)}}\)

02

To Predict the strain for a single adult

(b) Note: I will assume that we need to calculate a \(95\% \)prediction interval, other prediction intervals can be found similarly.

\({\rm{c = 95\% = 0}}{\rm{.95}}\)

Determine the t-value by looking in the row starting with degrees of freedom $\({\rm{d f = n - 1 = 15 - 1 = 14}}\)and in the column with \({\rm{\alpha = (1 - c)/2 = 0}}{\rm{.025}}\)in the table of the critical values for t distributions in the appendix:

\({{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ = 2}}{\rm{.145}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} {\rm{ = 2}}{\rm{.145 \times 3}}{\rm{.5}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{15}}}}} {\rm{\gg 7}}{\rm{.7537}}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E = 25}}{\rm{.0 - 7}}{\rm{.7537 = 17}}{\rm{.2463}}\\{\rm{\bar x + E = 25}}{\rm{.0 + 7}}{\rm{.7537 = 32}}{\rm{.7537}}\end{array}\)

Hence the \({\rm{95\% prediction interval }}\left( {{\rm{17}}{\rm{.2463,32}}{\rm{.7537}}} \right)\)

We note that the prediction interval is much wider than the confidence interval.

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Most popular questions from this chapter

a.Use the results of Example 7.5 to obtain a 95% lower confidence bound for the parameter 位 of an exponential distribution, and calculate the bound based on the data given in the example.

b.If lifetime X has an exponential distribution, the probability that lifetime exceeds t is P(X>t) = e-位迟. Use the result of part (a) to obtain a 95% lower confidence bound for the probability that breakdown time exceeds 100 min.

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?)

The U.S. Army commissioned a study to assess how deeply a bullet penetrates ceramic body armor. In the standard test, a cylindrical clay model is layered under the armor vest. A projectile is then fired, causing an indentation in the clay. The deepest impression in the clay is measured as an indication of survivability of someone wearing the armor. Here is data from one testing organization under particular experimental conditions; measurements (in mm) were made using a manually controlled digital caliper:

\(\begin{array}{l}{\rm{22}}{\rm{.4 23}}{\rm{.6 24}}{\rm{.0 24}}{\rm{.9 25}}{\rm{.5 25}}{\rm{.6 25}}{\rm{.8 26}}{\rm{.1 26}}{\rm{.4 26}}{\rm{.7 27}}{\rm{.4 27}}{\rm{.6 28}}{\rm{.3 29}}{\rm{.0}}\\{\rm{ 29}}{\rm{.1 29}}{\rm{.6 29}}{\rm{.7 29}}{\rm{.8 29}}{\rm{.9 30}}{\rm{.0 30}}{\rm{.4 30}}{\rm{.5 30}}{\rm{.7 30}}{\rm{.7 31}}{\rm{.0 31}}{\rm{.0 31}}{\rm{.4 31}}{\rm{.6 31}}{\rm{.7 31}}{\rm{.9 31}}{\rm{.9 }}\\{\rm{32}}{\rm{.0 32}}{\rm{.1 32}}{\rm{.4 32}}{\rm{.5 32}}{\rm{.5 32}}{\rm{.6 32}}{\rm{.9 33}}{\rm{.1 33}}{\rm{.3 33}}{\rm{.5 33}}{\rm{.5 33}}{\rm{.5 33}}{\rm{.5 33}}{\rm{.6 33}}{\rm{.6 33}}{\rm{.8 33}}{\rm{.9 }}\\{\rm{34}}{\rm{.1 34}}{\rm{.2 34}}{\rm{.6 34}}{\rm{.6 35}}{\rm{.0 35}}{\rm{.2 35}}{\rm{.2 35}}{\rm{.4 35}}{\rm{.4 35}}{\rm{.4 35}}{\rm{.5 35}}{\rm{.7 35}}{\rm{.8 36}}{\rm{.0 36}}{\rm{.0 36}}{\rm{.0 36}}{\rm{.1 36}}{\rm{.1 }}\\{\rm{36}}{\rm{.2 36}}{\rm{.4 36}}{\rm{.6 37}}{\rm{.0 37}}{\rm{.4 37}}{\rm{.5 37}}{\rm{.5 38}}{\rm{.0 38}}{\rm{.7 38}}{\rm{.8 39}}{\rm{.8 41}}{\rm{.0 42}}{\rm{.0 42}}{\rm{.1 44}}{\rm{.6 48}}{\rm{.3 55}}{\rm{.0}}\end{array}\)

a. Construct a box plot of the data and comment on interesting features.

b. Construct a normal probability plot. Is it plausible that impression depth is normally distributed? Is a normal distribution assumption needed in order to calculate a confidence interval or bound for the true average depth m using the foregoing data? Explain.

c. Use the accompanying Minitab output as a basis for calculating and interpreting an upper confidence bound for m with a confidence level of\({\rm{99\% }}\)

Variable Count Mean SE Mean StDev Depth\({\rm{83 33}}{\rm{.370 0}}{\rm{.578 5}}{\rm{.268}}\)

Q1 Median Q3 IQR

\({\rm{30}}{\rm{.400 33}}{\rm{.500 36}}{\rm{.000 5}}{\rm{.600}}\)

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?

a. Under the same conditions as those leading to the interval\({\rm{(7}}{\rm{.5),p((}}\overline {\rm{X}} {\rm{ - \mu )/(\sigma /}}\sqrt {\rm{n}} {\rm{) < 1}}{\rm{.645 = }}{\rm{.95}}{\rm{.}}\)Use this to derive a one-sided interval for\({\rm{\mu }}\)that has infinite width and provides a lower confidence bound on m. What is this interval for the data in Exercise 5(a)?

b. Generalize the result of part (a) to obtain a lower bound with confidence level\({\rm{100(1 - \alpha )\% }}\)

c. What is an analogous interval to that of part (b) that provides an upper bound on\({\rm{\mu }}\)? Compute this 99% interval for the data of Exercise 4(a).
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