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:
<u>Step 1</u>:-
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.
<u>Step 2</u><u>:</u>-
By using 't' distribution of confidence intervals
x⁻± tₐ (S/√n)
The 95% of confidence intervals are
(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