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 20 (newly deceased) adults, the mean failure strain (%) was 24.0, 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.)( , )(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.)( , )(c)How does the prediction compare to the estimate calculated in part (a)? (Select the answer from 1~3)1.The prediction interval is much wider than the confidence interval in part (a).2.The prediction interval is the same as the confidence interval in part (a).3.The prediction interval is much narrower than the confidence interval in part (a).

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The solution to this question is given below:

From the information given:

n=15

x bar=25.0

s=3.5

(a) I'll suppose that calculating a 95% confidence interval is necessary; other confidence intervals can be calculated in a similar way.

c=0.95 or 95%

In the table with the critical values for tt distributions in the appendix, locate the row beginning with degrees of freedom.

The maximum number of logically independent values—that is, values with the freedom to change—in the data sample is referred to as the degree of freedom.

df=n-1

=15-1

=14

and the column with alpha=(1-c)/2=0.025 =(1c)/2 =0.025 to find the t-value:

t α/2 =2.145

Therefore, the error margin is:

E=t α/2 × s/√n​

=2.145× 3.5/√15​

≈1.9384

The confidence interval's outer limits are thus:

xbar-E=25.0-1.9384 =23.0616

xbar+E=25.0+1.9384 = 26.9384

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