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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

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

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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1. (Amy) 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 cal- culated in part (a)?
Transcribed Image Text:1. (Amy) 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 cal- culated in part (a)?
When X is the mean of a random sample of size n from a normal distribution with mean u, the rv T S/n has a probability distribution called a t distribution with n – 1 degrees of freedom (df). Let ī and s be the sample mean and sample standard deviation computed from the results of a random sample from a normal population with mean u. Then a 100(1 – a)% confidence interval for µ is It ta/2,n-1-8/Vn An upper confidence bound for µ is T+ ta/2,n-1-8//n and replacing + by – in this latter expression gives a lower confidence bound for u, both with confidence level 100(1 – a)%. A prediction interval (PI) for a single observation to be selected from a normal pop- ulation distribution is T± ta/2,n-1.8//1+ The prediction level is 100(1 – a)%. A lower prediction bound results from replacing ta/2 by ta and discarding the + part of the above equation; a similar modification gives an upper prediction bound.
Transcribed Image Text:When X is the mean of a random sample of size n from a normal distribution with mean u, the rv T S/n has a probability distribution called a t distribution with n – 1 degrees of freedom (df). Let ī and s be the sample mean and sample standard deviation computed from the results of a random sample from a normal population with mean u. Then a 100(1 – a)% confidence interval for µ is It ta/2,n-1-8/Vn An upper confidence bound for µ is T+ ta/2,n-1-8//n and replacing + by – in this latter expression gives a lower confidence bound for u, both with confidence level 100(1 – a)%. A prediction interval (PI) for a single observation to be selected from a normal pop- ulation distribution is T± ta/2,n-1.8//1+ The prediction level is 100(1 – a)%. A lower prediction bound results from replacing ta/2 by ta and discarding the + part of the above equation; a similar modification gives an upper prediction bound.
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