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 = 34 specimens, the sample average toughness was x = 63.4 for the high-purity steel, whereas for n = 37 specimens of commercial steel y = 57.8. 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 σ₁ = 1.2 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.)
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	=

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

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

a)

Ho :   µ1 - µ2 =   5          
Ha :   µ1-µ2 > 5          
                  
Level of Significance ,    α =    0.001          
                  
sample #1   ------->              
mean of sample 1,    x̅1=   63.4          
population std dev of sample 1,   σ1 =    1.2          
size of sample 1,    n1=   34          
                  
sample #2   --------->              
mean of sample 2,    x̅2=   57.8          
population std dev of sample 2,   σ2 =    1.1          
size of sample 2,    n2=   37          
                  
difference in sample means = x̅1 - x̅2 =    63.4   -   57.8   =   5.6
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.2740          
                  
Z-statistic = ((x̅1 - x̅2)-µd)/SE = (5.6-5) /   0.2740   =   2.19

p-value = 0.0143 [Excel function =NORMSDIST(-z)  

b)

ß = P(Z< Zα/2 - true difference in mean/SE)  - P(Z < -Zα/2 - true difference in mean/SE)
Zα =       3.0902   (right tailed test)

ß = P(Z< 3.0902 - (6-5)/0.2740) - P(Z< -3.0902 - (6-5)/0.2740) = P(Z<-0.5594 ) - P(Z< -6.7398) = 0.2879 - 0.000 = 0.2879(answer)