/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Silicone implant augmentation rh... [FREE SOLUTION] | 91Ó°ÊÓ

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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 "Biomechanics in Augmentation Rhinoplasty" (J. of Med. Engr: and Tech., 2005: 14-17) 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
The mean strain is estimated as (23.06, 26.94)%. Predicted single strain value is (17.25, 32.75)%.

Step by step solution

01

Identify the Problem Type

The task is to find a confidence interval for the population mean and a prediction interval for a single observation, assuming a normal distribution for failure strain.
02

Determine Confidence Interval for the Mean

The confidence interval for the population mean is calculated using the formula: \[ \bar{x} \pm t\left(\frac{s}{\sqrt{n}}\right) \] where \( \bar{x} = 25.0\), \( s = 3.5 \), \( n = 15 \), and \( t \) is the t-score for the selected confidence level with \( n-1 \) degrees of freedom. Assuming a 95% confidence level, find \( t \) from a t-distribution table. For \( n-1 = 14 \), \( t \approx 2.145 \).
03

Calculate Confidence Interval for the Mean

Plug the values into the formula: \[ 25.0 \pm 2.145\left(\frac{3.5}{\sqrt{15}}\right) \]. This calculates the interval: \[ 25.0 \pm 1.94 \approx (23.06, 26.94) \]. The true average strain is estimated to be between 23.06% and 26.94% with 95% confidence.
04

Determine Prediction Interval for a Single Strain Value

The prediction interval for a single future observation uses the formula: \[ \bar{x} \pm t \cdot s \cdot \sqrt{1 + \frac{1}{n}} \]. Using the same values for \( t \), \( s \), and \( n \).
05

Calculate Prediction Interval

Plug the values: \[ 25.0 \pm 2.145 \cdot 3.5 \cdot \sqrt{1 + \frac{1}{15}} \]. Calculate: \[ 25.0 \pm 2.145 \cdot 3.5 \cdot \sqrt{1.0667} = 25.0 \pm 2.145 \cdot 3.5 \cdot 1.033 \approx 25.0 \pm 7.75 \approx (17.25, 32.75) \]. The predicted strain for a single adult is between 17.25% and 32.75% with 95% confidence.
06

Compare Prediction and Confidence Intervals

The confidence interval for the mean is narrower, indicating more precision about the average strain, while the prediction interval is wider, reflecting more variability and less precision about a single observation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval provides a range of values that likely contain the true average of a population parameter, such as the mean strain in our rhinoplasty example. This range is calculated based on sample data and provides a way to estimate the true unknown value with certain reliability. For instance, in the exercise, the confidence interval for the mean failure strain was calculated using the formula:- \[ \bar{x} \pm t\left(\frac{s}{\sqrt{n}}\right) \]Here, \( \bar{x} \) is the sample mean, \( t \) is the t-score from the t-distribution, \( s \) is the standard deviation, and \( n \) is the sample size. For a 95% confidence level, this formula provided an interval of (23.06%, 26.94%), indicating that we are 95% confident the true average strain falls within this range.Confidence intervals are crucial because they offer both precision, in the form of range, and reliability, represented by the confidence level. This helps communicate how certain we are about the estimation of the population mean.
Prediction Interval
A prediction interval is similar to a confidence interval, but instead of estimating a population parameter, it predicts the range where a single new observation is likely to fall. This interval takes into account both the variability within the sample and the uncertainty of predicting a single observation.In the exercise, the prediction interval for a single adult's strain was calculated using:- \[ \bar{x} \pm t \cdot s \cdot \sqrt{1 + \frac{1}{n}} \]Here, the additional factor \( \sqrt{1 + \frac{1}{n}} \) is added to account for extra variation when predicting an individual value. This resulted in an interval of (17.25%, 32.75%) at a 95% confidence level, wider than the confidence interval for the mean.This wider range demonstrates the additional uncertainty of predicting single observations compared to estimating a mean; thus, the prediction interval depicts less precision but still offers a reliable estimate.
Normal Distribution
Normal distribution is a fundamental concept in statistics, often used to model real-world data. It's characterized by a symmetric bell-shaped curve that describes how data values are distributed around a mean.The normal distribution is important because many natural phenomena closely follow this pattern. In statistical analysis, assuming a normal distribution allows for the application of certain statistical methods, including confidence and prediction intervals.The key properties of a normal distribution include:- Symmetry about the mean- Mean, median, and mode are equal- Defined by its mean \( \mu \) and standard deviation \( \sigma \)In our exercise, assuming failure strain follows a normal distribution enabled the use of the t-distribution for constructing these intervals. This simple symmetry and defined shape make predictions and estimations easier and more reliable in various contexts.
T-Distribution
The t-distribution is a statistical distribution that looks similar to the normal distribution but has heavier tails. This means it provides wider confidence intervals and is more suitable for smaller sample sizes when the population variance is unknown. The t-distribution is vital in providing more accurate estimates in cases with small sample sizes. It's commonly used in confidence interval calculations for the mean, especially when data are assumed to be normally distributed but with unknown variance. Characteristics of the t-distribution include: - Flatter and more spread out than the normal distribution - Dependent on degrees of freedom (number of data points minus one) - Approaches normal distribution as the sample size increases In the exercise, given the sample size of 15, the t-distribution provided the t-score needed for the confidence interval calculations. This heavier-tailed distribution helped ensure that the range contains the true mean with the selected confidence level, acknowledging the extra uncertainty from the smaller sample size.

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