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 )
