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)

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jtellezd jtellezd · Answer 1 · 2021-06-22T06:11:49+02:00

Answer:

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 )