MATH SOLVE

3 months ago

Q:
# 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 (%) was 25.0, and the standard deviation was 3.3. (a) Assuming a normal distribution for failure strain, estimate true average strain in a way that conveys information about precision and reliability. (Use a 95% confidence interval. Round your answers to two decimal places.)

Accepted Solution

A:

Answer with explanation:Let [tex]\mu[/tex] be the population mean strain in a way that conveys information about precision and reliability.The sample mean is the best point estimate of the true population mean .As per given , we haveSample size : n= 12degree of freedom : [tex]df=12-1=11[/tex]Sample mean : [tex]\overline{x}=25.0[/tex]The true average strain in a way that conveys information about precision and reliability= 25.0sample standard deviation : s= 3.3Significance level : [tex]\alpha= 0.05[/tex]Since sample population standard deviation is unknown , so we use t-test.Critical t-value for t : [tex]t_{\alpha/2, df}=t_{0.025, 11}=2.201[/tex]95% Confidence interval for true average strain in a way that conveys information about precision and reliability:[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex][tex]25.0\pm (2.201)\dfrac{3.3}{\sqrt{12}}\\\\=\approx 25.0\pm2.10\\\\=(25.0-2.10, 25.0+2.10)=(22.9,\ 27.1)[/tex]The 95% Confidence interval for true average strain in a way that conveys information about precision and reliability: [tex](22.9,\ 27.1)[/tex]We 95% confident that the true population average strain in a way that conveys information about precision and reliability lies between 22.9 and 27.1 .