Question

The U.S. Army commissioned a study to assess how deeply a bullet penetrates ceramic body armor. In the standard test, a cylindrical clay model is layered under the armor vest. A projectile is then fired, causing an indentation in the clay. The deepest impression in the clay is measured (in mm) as an indication of the survivability of someone wearing the armor. Out of 20 different samples of the body armor, the average indentation in the clay was ¯x = 33.370 mm with standard deviation s = 5.268 mm. (a) Calculate and interpret a 95% upper confidence bound for the true average indentation depth. What, if any, assumptions did you make in order to compute this interval, and why did you need to make them? (b) Calculate and interpret a 95% confidence interval for the true average indentation depth, under the same assumptions you used in part (a). (c) Calculate and interpret a 95% prediction interval for the indentation depth of the next piece of body armor produced, under the same assumptions you used in previous parts.

Answer 1

Answer:

a)The 95% of confidence intervals

(30.905 , 35.835)

the average indentation in the clay was ¯x = 33.370 mm between in this

95% of confidence intervals

Step-by-step explanation:

Step 1:-

given 20 different samples of the body armor

n =20

Given sample mean ¯x = 33.370

sample standard deviation (S) = 5.268

degrees of freedom γ =n-1 = 20-1 =19

see from "t' table the tabulated value tₐ = 2.093 at 19 degrees of freedom at 95% level of significance.

Step 2:-

By using 't' distribution of confidence intervals

x⁻± tₐ (S/√n)

The 95% of confidence intervals are

$(x^{-} - t_{\alpha } (\frac{S}{\sqrt{n} } ,(x^{-} +t_{\alpha } (\frac{S}{\sqrt{n} } ))$

$(33.370-2.093(\frac{5.268)}{\sqrt{20} } ,33.370+2.093(\frac{5.268}{\sqrt{20} } )$

(33.370 - 2.465, 33.370 + 2.465)

(30.905 , 35.835)

the average indentation in the clay was ¯x = 33.370 mm between in this

95% of confidence intervals