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 = 38 specimens of commercial steel y = 58.6. 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 on = 1.4 and 0% = 1.1, test the relevant hypotheses using a = 0.001. (Use My - Hly, where y, is the average toughness for high-purity steel and My is the average toughness for commercial steel.) State the relevant hypotheses. HO: H1 - H2 = 5 H:H1-425 O Ho: 42 - 42 = 5 HO: H1 - M2 ,<5 O HO:41 - M2- 5 H:Hi-M25 O HO: 41 - H2 = 5 H:H1 - My 55 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. O Reject H. The data does not suggest that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5. O Fail to reject Ho. 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 Ho. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5. O Reject Ho. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5. (b) Compute B for the test conducted in part (a) when wy - My = 6. (Round your answer to four decimal places.) B = You may need to use the appropriate table in the Appendix of Tables to answer this question.

Will give instant thumbs up, please show some work and equations if done in excel.

Answer 1

a)

но: 4. - 42 = 5 на: 11 - 12 > 5

		pop 1	pop 2
sample mean x =		64.10	58.60
std deviation σ=		1.400	1.100
sample size n=		33	38
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)

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

b)

Hypothesized mean difference Δo       =			5
true mean difference Δ                                                          =			6
first sample standard deviation =σ1    =			1.400
second sample std deviation =σ2=			1.100
first sample size =n1 =			33.000
second sample size =n2 =			38.000
standard error=(√(σ12/n1+σ12/n1))=			0.3021
for 0.001 level and right tail critival Zα=			3.090
rejecf Ho if x>= Δo +Zα*σx or x>=			5.9333
P(Type II error) =P(Xbar<5.933| Δ=6)=P(Z<(5.9333-6)/0.302)=P(Z<-0.22)=0.4129