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 20 (newly deceased) adults, the mean failure strain (%) was 26.0, and the standard deviation was 3.4. (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.)

Answer 1

The true average strain in a way that conveys information about precision and reliability is 27.314.

Given that sample size is 20, mean is 26%, standard deviation is 3.4, confidence level is 95%.

We have to calculate the true average strain in a way that conveys information about precision and reliability.

We have to use t test in our problem because sample size is less than 30.

Weknow that,

t=x bar-μ/s$\sqrt{n}$

where μ is sample mean,

s is sample standard deviation.

Degree of freedom=n-1

=20-1

=19

T value at 95% confidence interval=1.7281

put the values in the formula of t given above.

1.7291=x bar-26/3.4/$\sqrt{20}$

1.7291=xbar-26/3.4/4.47

1.7291= x bar-26/0.76

1.7291*0.76=x bar-26

1.314=x bar-26

x bar=26+1.314

x bar=27.314

After rounding it will be 27.31

Hence the true average strain in a way that conveys information about precision and reliability is 27.31%.

Learn more about t test at brainly.com/question/6589776

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