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 oughness was x = 62.1 for the high-purity steel, whereas for n = 37 specimens of commercial steel y = 56.5. Because the high-purity steel is more expensive, its use for a certain application can ustified 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 a₁ = 1.2 and ₂ = 1.0, test the relevant hypotheses using a = 0.001. (Use ₁-₂, where #₁ is the average toughness for high-purity steel and #₂ is the average tough for commercial steel.) State the relevant hypotheses. Hỏi thịty 5 H₂H₁ H₂5 Ho: ₁₂=5 H₂i M₁ - H₂5 0 0 1 - 2 = 5 H₂H₁-H₂> 5 но: M1 -M2 =5 H₂H₁ H₂ S5 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 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 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 by more than 5.

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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 = 62.1 for the high-purity steel, whereas for n = 37 specimens of commercial steel y = 56.5. 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 ₁ = 1.2 and ₂ = 1.0, test the relevant hypotheses using a = 0.001. (Use ₁-₂, where μ₁ is the average toughness for high-purity steel and
for commercial steel.)
State the relevant hypotheses.
Но из-H2=5
H₂: H₂H₂ 5
Hoi 11 - 12 = 5
Hy My H₂5
Ho: M₁ M₂ = 5
Hai H₁
H₂> 5
Но М1 -H2=5
H₂H₁ H₂ 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 does not suggest that the fracture toughness 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.
Reject Ho. The data does not suggest that the fracture toughness f high-purity steel exceeds that of commercial-purity steel by more than 5.
Reject Ho. The data suggests that the fracture toughness high-purity steel exceeds that of commercial-purity steel by more than 5.
(b) Compute for the test conducted in part (a) when #₁ - #₂ = 6. (Round your answer to four decimal places.)
B =
is the average toughness
Transcribed Image Text: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 = 62.1 for the high-purity steel, whereas for n = 37 specimens of commercial steel y = 56.5. 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 ₁ = 1.2 and ₂ = 1.0, test the relevant hypotheses using a = 0.001. (Use ₁-₂, where μ₁ is the average toughness for high-purity steel and for commercial steel.) State the relevant hypotheses. Но из-H2=5 H₂: H₂H₂ 5 Hoi 11 - 12 = 5 Hy My H₂5 Ho: M₁ M₂ = 5 Hai H₁ H₂> 5 Но М1 -H2=5 H₂H₁ H₂ 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 does not suggest that the fracture toughness 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. Reject Ho. The data does not suggest that the fracture toughness f high-purity steel exceeds that of commercial-purity steel by more than 5. Reject Ho. The data suggests that the fracture toughness high-purity steel exceeds that of commercial-purity steel by more than 5. (b) Compute for the test conducted in part (a) when #₁ - #₂ = 6. (Round your answer to four decimal places.) B = is the average toughness
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