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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

The article 鈥淐oncrete Pressure on Formwork鈥� (Mag. of Concrete Res., 2009: 407鈥�417) gave the following observations on maximum concrete pressure (kN/m2 ):

33.2 41.8 37.3 40.2 36.7 39.1 36.2 41.8

36.0 35.2 36.7 38.9 35.8 35.2 40.1

a.Is it plausible that this sample was selected from a normal population distribution?

b. Calculate an upper confidence bound with confidence level 95% for the population standard deviation of maximum pressure.

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation .75.

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average porosity of 4.56.

c. How large a sample size is necessary if the width of the 95% interval is to be .40?

d. What sample size is necessary to estimate true average porosity to within .2 with 99% confidence?

Consider a normal population distribution with the value of \({\rm{\sigma }}\) known. a. What is the confidence level for the interval \({\rm{\bar x \pm 2}}{\rm{.81\sigma /}}\sqrt {\rm{n}} \)? b. What is the confidence level for the interval \({\rm{\bar x \pm 1}}{\rm{.44\sigma /}}\sqrt {\rm{n}} \)? c. What value of \({{\rm{z}}_{{\rm{\alpha /2}}}}\) in the CI formula (\({\rm{7}}{\rm{.5}}\)) results in a confidence level of \({\rm{99}}{\rm{.7\% }}\)? d. Answer the question posed in part (c) for a confidence level of \({\rm{75\% }}\).

Determine the confidence level for each of the following large-sample one-sided confidence bounds:

\(\begin{array}{l}{\rm{a}}{\rm{.Upperbound:\bar x + }}{\rm{.84s/}}\sqrt {\rm{n}} \\{\rm{b}}{\rm{.Lowerbound:\bar x - 2}}{\rm{.05s/}}\sqrt {\rm{n}} \\{\rm{c}}{\rm{.Upperbound:\bar x + }}{\rm{.67\;s/}}\sqrt {\rm{n}} \end{array}\)

Determine the values of the following quantities

\(\begin{array}{l}{\rm{a}}{\rm{.}}{{\rm{x}}^{\rm{2}}}{\rm{,1,15}}\\{\rm{b}}{\rm{.}}{{\rm{X}}^{\rm{3}}}{\rm{,125}}\\{\rm{c}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{01,25}}\\{\rm{d}}{\rm{.}}{{\rm{X}}^{\rm{2}}}{\rm{00525}}\\{\rm{e}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{9925}}\\{\rm{f}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{995,25}}\end{array}\)
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