Question

An article gave the accompanying data on ultimate load (kN) for two different types of beams.

Type Sample Size Sample Mean Sample SD

Fiberglass grid 27    33.4 2.1

Commercial carbon grid 27 42.8 4.4

Assuming that the underlying distributions are normal, calculate a 99% CI for the difference between true average load for the fiberglass beams and carbon beams, μfiberglass − μcarbon. (Round your answer to two decimal places.)

( , )

Note: I've tried other answers from the same problem (with different numbers) and keep getting a rounded answer of (-12,-6.8) and that is not correct

Answer 1

let sample 1 = Fiberglass grid

let sample 2 = Commercial carbon grid

we have given

Sample 1 Mean (M1):	33.4
Sample 1 Size (n1):	27
Standard Deviation 1 (s1):	2.1
Sample 2 Mean (M2):	42.8
Sample 2 Size (n2):	47
Standard Deviation 2 (s2):	4.4

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 618.02 / 52 = 11.89

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((11.89/27) + (11.89/27)) = 0.94

Confidence Interval
μ_Fiberglass - μ_carbon = (M₁ - M₂) ± ts_{(M1 - M2)} = 9.4 ± (2.67 * 0.94) = 9.4 ± 2.509

μ_Fiberglass - μ_{carbon =} [6.89, 11.91].