Question

An experiment was performed to compare the fracture toughness of high-purity 18 Ni maraging weel with commercial purity of the for average toughness was 65.4 for the high-purity steel, whereas for - 39 specimens of commercially-59.9. Because the certain application can be justified only if its fracture toughness exceeds that of commercial purity Weel by more than 5. Suppose that the (a) Assuming that - 1.4 anda, - 1.0, test the relevant hypotheses using a 0.001.( UM), where the average for hurt average toughness for commercial steel) State the relevant hypotheses HoiHI-H₂-5 My > 5 Ho HI-H₂-5 Hoil-H₂-5 Ho: My-H₂ 55 Hoito-P₂-5 Hoi Hy - H₂5 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your value to four decimal places) p-value State the conclusion in the problem context. Reject No. The data does not suggest that the fracture toughness of high-purity steel exceeds that of commercial purity stes by more than Reject Ho. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more the Fail to reject Ho. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than Fail to reject No. 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 for the best conducted in part (a) when - - 6. (Round your answer to four decimal places.) 0
question b

Answer 1

Ho 2 s не - > s

		pop 1	pop 2
sample mean x =		65.40	59.90
std deviation σ=		1.400	1.000
sample size n=		30	39
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.302
test stat z =(x1-x2-Δo)/σx1-x2     =			1.66
p value : =			0.0485	(from excel:1*normsdist(-1.66)

fail to reject Ho ; the data does not suggest........

b)

Hypothesized mean difference Δo       =			5
true mean difference Δ                                                          =			6
first sample standard deviation =σ1    =			1.400
second sample std deviation =σ2=			1.000
first sample size =n1 =			30.000
second sample size =n2 =			39.000
standard error=(√(σ12/n1+σ12/n1))=			0.3016
for 0.001 level and right tail critival Zα=			3.090
rejecf Ho if x>= Δo +Zα*σx or x>=			5.932
P(Type II error) =P(Xbar<5.932| Δ=6)=P(Z<(5.932-6)/0.302)=P(Z<-0.23)=0.4090