An engineering consulting firm wanted to evaluate the diameter of rivet heads. The following data represent the diameters (in hundredths of an inch) for a random sample of 25 rivet heads: (20 pts) 6.81 6.79 6.69 6.59 6.65 6.60 6.74 6.70 6.76 6.84 6.81 6.71 6.66 6.76 6.76 6.77 6.72 6.68 6.71 6.79 6.72 6.72 6.72 6.79 6.83
a. Set up a 95% confidence interval estimate of the average diameter of rivet heads (in hundredths of an inch)
b. Set up a 95% CI estimate of the standard deviation of the diameter rivet heads (in hundredths of an inch)

a)CI 95 %  =  ( 6.7063  ;  6.7593)  ( in hundredths of an inch)

b) CI 95 %  =  ( 0.05 <  σ  <  0.089 ) ( in hundredths of an inch)

Step-by-step explanation:

From the problem statement, we have a manufacturing  process and we we assume a normal distribution, from sample data:

x =  6.7328        and  s  =  0.0644

a) CI 95 %  =  (  x ±  t(c) * s/√n )

t(c)    df =  n  -1    df  =  25 - 1     df = 24

CI  =  95 %   α = 5 %     α  = 0.05     α /2  =  0.025

Then  from t-student table   t(c) = 2.060

s/√n  =  0.0644/ √ 25      s/√n = 0.01288

CI 95 %  =  (  x ±  t(c) * s/√n )  =  ( 6.7328  ± 2.060*0.01288)

CI 95 %  = ( 6.7328 ±  0.02653 )

CI 95 %  =  ( 6.7063  ;  6.7593)  ( in hundredths of an inch)

b) CI 95 % of the variance  is:

CI 95 %  =  (  ( n - 1 ) * s² / χ²₁ - α/₂ ≤  σ²  ≤   ( n - 1 )*s² / χ²α/₂ )

( n - 1 ) * s²   = ( 25 - 1 ) * (0.0644)² =  24* 0.00414

( n - 1 ) * s²   = 0.09936

And from  χ²  table we look for values of

χ² α/₂ ,df    df = 24   and  α/2 =  0.025

χ² (0.025,24)  = 12.40      and    χ²₁ - α/₂  =   χ² (0.975, 24)

χ² (0.975, 24) = 39.36

Then

CI 95 %  = ( 0.09936 / 39.36  ≤  σ²  ≤   0.09936 / 12.40)

CI 95 %  = ( 0.0025  ≤   σ²  ≤  0.0080)

Then for the standard deviation,  we take the square root of that interval

CI 95 %  =  ( 0.05 <  σ  <  0.089 )

over the argument its not good to fight.