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 = 31 specimens, the sample average toughness was x = 65.3 for the high-purity steel, whereas for n = 38 specimens of commercial steel y = 59.5. 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 = 1.0, test the relevant hypotheses using a = 0.001. (Use 41 - U2, where uy is the average toughness for high-purity steel and uy is the average toughness for commercial steel.) State the relevant hypotheses. O Ho: M1 - U2 = 5 Ha: M1 - U2 > 5 O Ho: 41 - U2 = 5 Hai M1 - 4275 O Ho: U1 - U2 = 5 Ha: M1 - U2 <5 O Ho: M1 - U2 = 5 Ha: M1 - M2 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. 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. 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 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 41 - M2 = 6. (Round your answer to four decimal places.) B =

Answer 1

Solutions From the Given data, ,- 65.3, 0 - 14 m-31 on = 101 - 38 er 59.5 -0.001 Aim of the hypothesi's test is to obtain the true average fracture, toughness is higher than that of commercial Steel Ę it is appropriate to use right tailed test, Mi" Represents mean fracture 14, Represents the mean commercial steel. Null hypothesis: to: M, - M₂ = 5 alternative hypothesis: Hai M, M₂ >5 option CAD correct Answer Test Statistics z=(2, 2) - (M-M2) 0,25 (65.3 – 59.5) = 5 (liyjar Crom 31 38 ni 2 + - (5.8-5) 0.063 225 +0.0263157 Z 2.67349 P-value po P(Z >121) PE P(Z > 2.67349) "P(Z <2.67349) 1- 0.9962 0.00 38

p-value00038 p-value: (0.0038) > level of significance (&-0.001) Reject Ho So, we cannot failed to reject 'Hé, data does not suggest that fracture toughness of high purity steel exceeds that of commercial. purity Steel by more than 5 option () correct Answer; b) 'p" for the test conducted in part ca] when Mi-M₂ = 6 Z- cã, - X) -(M-M2) + ne 5.8 - 6 ( 65.3 – 59.5) – (6) (l.48 (10) + 10:063225+0.0263157 31 38 -0.66837 P- Value Plz7-0.66837) sil-p(ZC -0.66837) 1 - 0 2519 0.7481