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. An article reported that for a sample of 10 (newly deceased) adults, the mean failure strain (%) was 24.0, and the standard deviation was 3.2.

Required:
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

Given information :

A sample of n = 10 adults

The mean failure was 24 and the standard deviation was 3.2

a). The formula to calculate the 95% confidence interval is given by :

`$\overline x \pm t_(\alpha/2,-1) * (s)/(\sqrt n)$`

Here,
`$t_(\alpha/2,n-1) = t_(0.05/2,10-1)$`

= 2.145

Substitute the values

`$24 \pm 2.145 * \frac{3.2}{\sqrt {10}}$`

(26.17, 21.83)

When the
`\text{sampling of the same size}` is repeated from the
`\text{population}`
`n` infinite number of
`\text{times}`, and the
`\text{confidence intervals}` are constructed, then
`95\%` of them contains the
`\text{true value of the population mean}`, μ in between
`(26.17, 21.83)`

b). The formula to calculate 95% prediction interval is given by :

`$\overline x \pm t_(\alpha/2,-1) * s \sqrt{1+(1)/(n)}$`

`$24 \pm 2.145 * 3.2 \sqrt{1+(1)/(10)}$`

(31.13, 16.87)