Question

An experiment was performed to compare the fracture toughness of high-purity 18 Ni maraging steel with commercial-purity steel of the same type. For m = 33 specimens, the sample average toughness was x = 64.1 for the high-purity steel, whereas for n = 39 specimens of commercial steel y = 58.5. Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercial-purity steel by more than 5. Suppose that both toughness distributions are normal.

(a) Assuming that σ₁ = 1.4 and σ₂ = 1.1, test the relevant hypotheses using α = 0.001. (Use μ₁ − μ₂, where μ₁ is the average toughness for high-purity steel and μ₂is the average toughness for commercial steel.)
State the relevant hypotheses.

H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ ≤ 5

H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ ≠ 5    

H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ > 5

H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ < 5

Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z	=
P-value	=

State the conclusion in the problem context.

Reject H₀. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5.

Fail to reject H₀. The data does not suggest that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5.    

Fail to reject H₀. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5.

Reject H₀. The data does not suggest that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5.

(b) Compute β for the test conducted in part (a) when μ₁ − μ₂ = 6. (Round your answer to four decimal places.)
β =

Answer 1

H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ > 5

Fail to reject H₀. The data does not suggest that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5.    

b)