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 13 (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

Answer:

The 95% confidence interval for the true average strain is between 23.94% and 28.06%.

Step-by-step explanation:

We have the standard deviation of the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 13 - 1 = 12

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 12 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.18

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.18\frac{3.4}{\sqrt{13}} = 2.06$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 26 - 2.06 = 23.94

The upper end of the interval is the sample mean added to M. So it is 26 + 2.06 = 28.06

The 95% confidence interval for the true average strain is between 23.94% and 28.06%.