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 = 32 specimens, the sample average toughness was x = 64.3 for the high-purity steel, whereas for n = 37 specimens of commercial steel y = 58.8. 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 01 = 1.4 and 02 1.1, test the relevant hypotheses using a = 0.001. (Use 41 - 42, where 41 is the average toughness for high-purity steel and u2 is the average toughness for commercial steel.) State the relevant hypotheses. O Ho: M1 - H2 = 5 Haill1 - 42 < 5 O Ho: M1 - H2 = 5 Ha: 41 - My + 5 O Ho: 41 - M2 = 5 Ha: 41 - 42 > 5 O HO: 41-42 = 5 Ha: 41 - 4255 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 Ho. The data suggests 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 suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5. O 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. 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) Compute B for the test conducted in part (a) when 41 - H2 = 6. (Round your answer to four decimal places.) B =

Answer 1

оноги, – и, = 5 Наи, – и, > 5

		pop 1	pop 2
sample mean x =		64.30	58.80
std deviation σ=		1.400	1.100
sample size n=		32	37
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.307
test stat z =(x1-x2-Δo)/σx1-x2     =			1.63
p value : =			0.0516	(from excel:1*normsdist(-1.63)

since p value >0.001

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 =			32.000
second sample size =n2 =			37.000
standard error=(√(σ12/n1+σ12/n1))=			0.3065
for 0.001 level and right tail critival Zα=			3.090
rejecf Ho if x>= Δo +Zα*σx or x>=			5.9471
P(Type II error) =P(Xbar<5.947| Δ=6)=P(Z<(5.9471-6)/0.307)=P(Z<-0.17)=0.4325