Question

An experiment was performed to compare the fracture toughness of high-purity 16 Ni maraging steel with commercial purity steel of the same type Form 30 specimens, the sample average toughness was x = 65.4 for the high-purity steel, whereas for n - 39 specimens of commercial steel y = 59.9. Because the high-punity to 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 a, - 1.4 and a, - 1.0, test the relevant hypotheses using a 0.001. (Use - My, where is the average toughness for high-purity steel and the average toughness for commercial steel.) State the relevant hypotheses. Hoil-H₂-5 HH1 - Hy> 5 O Ho: My - H₂5 H:Hi-H2<5 O Holy - M₂=5 H₂: My - H₂ 55 Hoi Hi-M₂-5 Hoi My - H₂5 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) p-value = State the conclusion in the problem context. 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 Reject Ho. The data suggests 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. 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 #1 - 2 - 6. (Round your answer to four decimal places.)

Answer 1

Ho :   µ1 - µ2 =   5          
Ha :   µ1-µ2 > 5          
                  
Level of Significance ,    α =    0.001   
                  
sample #1   ------->              
mean of sample 1,    x̅1=   65.4000          
population std dev of sample 1,   σ1 =    1.4          
size of sample 1,    n1=   30          
                  
sample #2   --------->              
mean of sample 2,    x̅2=   59.9000          
population std dev of sample 2,   σ2 =    1          
size of sample 2,    n2=   39          
                  
difference in sample means = x̅1 - x̅2 =    65.4   -   59.9   =   5.5
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.3016          
                  
Z-statistic = ((x̅1 - x̅2)-µd)/SE =    5.5   /   0.3016   =   1.66
                  

                  
p-value =        0.0487   [excel function =NORMSDIST(z)]      
Desison:   p-value > α , do not Reject null hypothesis              

Fail to reject Ho. The data does not suggest that the fracture toughness of high-purity steel exceeds that o decimal places)

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