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 12 (newly deceased) adults, the mean failure strain (in percentage) was 24.0%, and the standard deviation was 3.4%.

(a) Assuming a normal distribution for failure strain, estimate the true average strain in a way that conveys information about precision and reliability. (Use a [tex]$95\%$[/tex] confidence interval. Round your answers to two decimal places.)

[tex]\[ \text{Confidence Interval: } 22.46\% \text{, } 25.54\% \][/tex]

(b) Predict the strain for a single adult in a way that conveys information about precision and reliability. (Use a [tex]$95\%$[/tex] prediction interval. Round your answers to two decimal places.)

[tex]\[ \text{Prediction Interval: } 16.93\% \text{, } 31.07\% \][/tex]

How does the prediction compare to the estimate calculated in part (a)?

A. The prediction interval is much narrower than the confidence interval in part (a).
B. The prediction interval is the same as the confidence interval in part (a).
C. The prediction interval is much wider than the confidence interval in part (a).



Answer :

Let's start solving the given problems step by step.

### Part (a): Calculating the 95% Confidence Interval for the Mean Failure Strain

1. Given Data:
- Sample size (\( n \)): 12
- Sample mean (\( \bar{x} \)): 24.0%
- Sample standard deviation (\( s \)): 3.4
- Confidence level: 95%

2. Step-by-Step Solution:

To estimate the true average strain in a way that conveys precision and reliability, we need to calculate a 95% confidence interval for the mean failure strain. This interval is calculated using the following steps:

a. Determine the t-critical value:
The degrees of freedom (\( df \)) is \( n - 1 = 11 \). Using a t-distribution table or calculator for a 95% confidence level and 11 degrees of freedom, we find the t-critical value.

b. Calculate the standard error of the mean (SE):
[tex]\[ SE = \frac{s}{\sqrt{n}} = \frac{3.4}{\sqrt{12}} \][/tex]

c. Calculate the margin of error (ME):
[tex]\[ ME = t_{\text{critical}} \times SE \][/tex]

d. Construct the confidence interval:
[tex]\[ \left( \bar{x} - ME, \bar{x} + ME \right) \][/tex]

Using the above steps, we find the 95% confidence interval for the mean failure strain to be approximately:
[tex]\[ (21.84\%, 26.16\%) \][/tex]

### Part (b): Calculating the 95% Prediction Interval for the Failure Strain of a Single Adult

1. Given Data:
- The same data as part (a).

2. Step-by-Step Solution:

To predict the strain for a single adult with a 95% prediction interval, we do the following:

a. Determine the t-critical value:
Using the same t-critical value as in part (a).

b. Calculate the prediction interval margin of error (PE):
[tex]\[ PE = t_{\text{critical}} \times s \times \sqrt{1 + \frac{1}{n}} \][/tex]

c. Construct the prediction interval:
[tex]\[ \left( \bar{x} - PE, \bar{x} + PE \right) \][/tex]

Using the above steps, we find the 95% prediction interval for the failure strain of a single adult to be approximately:
[tex]\[ (16.21\%, 31.79\%) \][/tex]

### Comparison and Explanation

- Confidence Interval vs. Prediction Interval:
The confidence interval from part (a) conveys the precision of the estimate of the population mean failure strain. It gives a range within which we are 95% confident that the true mean lies.

- Prediction Interval:
The prediction interval from part (b) provides a range within which we expect the strain for a single new observation to fall 95% of the time.

- The Comparison:
The prediction interval is much wider than the confidence interval because it accounts for both the variability in the sample mean (as the confidence interval does) and the additional variability of individual observations.

Therefore, the correct comparison statement is:
- The prediction interval is much wider than the confidence interval in part (a).