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 15 (newly deceased) adults, the mean failure strain (%) was 24.0, 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. (Use a 95% confidence interval. Round your answers to two decimal places.) 1.96 X %, 2.57 (b) Predict the strain for a single adult in a way that conveys information about precision and reliability. (Use a 95% prediction interval. Round your answers to two decimal places.) %, 21.86

Answer 1

95% CI for μ using t-dist Sample Mean 24 Sample Standard deviations 3.5 Sample Size = n-15 Significance level α-1-0.95-0.05 Degrees of freedom for t-distribution, d.f.-n -1 14 Critical Value - ta/2,df to.025,df-14 2.145 (from t-table, two-tails, d.f. - 14 3.5 Margin of Error = E-ta/2,df × E 2.145 x 0.903696 -= 2.145 × Vn Margin of Error, E 1.938428 Limits of 95% confidence interval are given by: Lower limit E-24-1.938428 22.061572 Upper limit = + E = 24 + 1.938428 25.938428 22.06 25.94 95% confidence interval is: x - (22.061572, 25.938428) E-24士1.938428 95% CI using t-dist: 22.06 <μ< 25.94 Note that exact answer using technology is: (24 3.5/SQRT (15) *T. INV (1-(1-0.95)/2,15-1 , 24 3.5/SQRT (15) T.INV(1- (1-0.95)/2,15-1) (22.0617646045, 25.9382353955)

95% PI for xo using t-dist Sample Mean 24 Sample Standard deviations 3.5 Sample Size -n-15 Significance level-α-1-0.95-0.05 Degrees of freedom for t-distribution, d.f.-n -1 14 Critical Value-to/2df-t0.025df-14-2.145 (from t-table, two-tails. d.f.-14 MIargin of Prediction Error = E-ta/2.df × 8x × ( /1 +-) = 2.145 × 3.5 × 15 PE-2.145 × 3.614784 Margin of Prediction Error, PE 7.753713 Limits of 95% prediction interval are given by Lower limit E-24-7.753713 ะ 16.246287 Upper limitE 24 7.753713 31.753713 16.25 31.75 95% prediction interval is:- E-24 7.753713 (16.246287, 31.753713) 95%PI using t-dist: 16.25 <Zo < 31.75 Note that exact answer using technology is: (24 3.5*SQRT (1+1/15) T. INV (1-(1-0.95)/2,15-1) , 24 3.5 SQRT (1+1/15) *T. INV(1-(1-0.95)/2,15-1 ) (16.2470584181, 31.7529415819) @seoThank ,youJ0户