Question

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 "Biomechnics in
Augmentation Rhinoplasty" reported that for a sample of 15 (newly deceased)
adults, the mean failure strain (%) was 25.0, and the standard deviation was 3.5.
(p. 277 #35)
a. Assuming a normal distribution for failure strain, estimate true average strain in
a way that converys 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)?

Answer 1

Concepts and reason

Confidence intervals: A confidence interval provides a range of values which is likely to contain the population parameter of interest.

The population standard deviation is unknown and the sample size is less than 30 use one sample $t$ -test.

Fundamentals

The formula for confidence interval for single mean is as shown below:

$\bar x - {t_{\alpha /2,n - 1}}\left( {\frac{s}{{\sqrt n }}} \right) < \mu < \bar x + {t_{\alpha /2,n - 1}}\left( {\frac{s}{{\sqrt n }}} \right)$

Here $\bar x$ is the sample mean, s is the standard deviation, $\mu$ is the population mean, ${t_{\alpha /2,n - 1}}$ is the critical value and n is the sample size.

The formula for prediction interval is as shown below:

$\bar x - {t_{\alpha /2,n - 1}}s\sqrt {1 + \frac{1}{n}} < \mu < \bar x + {t_{\alpha /2,n - 1}}s\sqrt {1 + \frac{1}{n}}$

The formula for degrees of freedom is, $df = n - 1$

(a)

The available information is $n = 15,\bar x = 25{\rm{ and }}s = 3.5$

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 15 - 1\\\\ = 14\\\end{array}$

From the t-distribution tables, the critical value at 5% level of the significance for 14 degrees of freedom about two tailed is 2.145.

The confidence interval estimate of true mean is as shown below:

$\begin{array}{c}\\\bar x - {t_{\alpha /2,n - 1}}\left( {\frac{s}{{\sqrt n }}} \right) < \mu < \bar x + {t_{\alpha /2,n - 1}}\left( {\frac{s}{{\sqrt n }}} \right)\\\\25 - 2.145\left( {\frac{{3.5}}{{\sqrt {15} }}} \right) < \mu < 25 + 2.145\left( {\frac{{3.5}}{{\sqrt {15} }}} \right)\\\\23.0616 < \mu < 26.9384\\\end{array}$

(b)

The available information is $n = 15,\bar x = 25{\rm{ and }}s = 3.5$

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 15 - 1\\\\ = 14\\\end{array}$

From the t-distribution tables, the critical value at 5% level of the significance for 14 degrees of freedom about two tailed is 2.145.

The predicted interval for a is as shown below:

$\begin{array}{c}\\\bar x - {t_{\alpha /2,n - 1}}s\sqrt {1 + \frac{1}{n}} < \mu < \bar x + {t_{\alpha /2,n - 1}}s\sqrt {1 + \frac{1}{n}} \\\\25 - 2.145\left( {3.5} \right)\sqrt {1 + \frac{1}{{15}}} < \mu < 25 + 2.145\left( {3.5} \right)\sqrt {1 + \frac{1}{{15}}} \\\\17.2463 < \mu < 32.7537\\\end{array}$

From the result of part (a), the difference between upper and lower limit is approximately 4. In part (b), the difference between the upper and the lower limit is approximately 16.

Therefore, from this it can be said that the prediction interval is roughly four times as wide.

Ans: Part a

The confidence interval for the true mean is (23.0616,26.9384).

Part b

The prediction interval for the true mean is (17.2463, 32.7537).

The prediction interval is roughly four times as wide.