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 = 30 specimens, the sample average toughness was x = 65.5 for the high-purity steel, whereas for n = 39 specimens of commercial steel y = 59.9. 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 = 1.3 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 μ₂ is the average toughness for commercial steel.)
State the relevant hypotheses.

H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ ≤ 5H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ > 5    H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ < 5H₀: 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.

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

Answer 1

a)

H₀: μ₁ − μ₂ = 5
H_a: μ₁ − μ₂ > 5

b)

		pop 1	pop 2
sample mean x =		65.50	59.90
std deviation σ=		1.300	1.100
sample size n=		30	39
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.296
test stat z =(x1-x2-Δo)/σx1-x2     =			2.03
p value : =			0.0212	(from excel:1*normsdist(-2.03)

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_)

standard error=(√(σ12/n1+σ12/n1))=			0.2956
for 0.001 level and right tail critival Zα=			3.090
rejecf Ho if x>= Δo +Zα*σx or x>=			5.9133
P(Type II error) =P(Xbar<5.913| Δ=6)=P(Z<(5.9133-6)/0.296)=P(Z<-0.29)=0.3859