johnny2585
20.04.2020 •
Mathematics
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 32 specimens, the sample average toughness was 65.6 for the high-purity steel, whereas for 38 specimens of commercial steel the sample average was 59.8. 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. Assume that both toughness distributions are normal, σ = 1.2 and σ = 1.1.
(a) Test the relevant hypothesis using α = .001.
(b) Compute β for the test conducted in part (a) when μ1 - μ2 = 6.
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Ответ:
(a) Fail to reject the null hypothesis.
(b) The value of β is 0.00015.
Step-by-step explanation:
(a)
Here we need to test whether the difference between the two population means exceeds 5 or not.
The hypotheses are:
H₀: µ₁ - µ₂ ≤ 5 vs. Hₐ: µ₁ - µ₂ > 5.
Since the population standard deviations are provided we will use a z-test.
The information provided is:
Compute the value of the test statistic as follows:
The test statistic value is 2.89.
Decision rule:
Reject the null hypothesis if p-value is less than the significance level, α = 0.001.
Compute the p-value as follows:
The p-value of the test is 0.00193.
p-value = 0.00193 > α = 0.001.
Thus, we fail to reject the null hypothesis at α = 0.001.
Conclusion:
The fracture toughness does not exceeds that of commercial-purity steel by more than 5.
(b)
A type II error is a statistical word used within the circumstance of hypothesis testing that defines the error that take place when one is unsuccessful to discard a null hypothesis that is truly false. It is symbolized by β i.e.
β = Probability of accepting H₀ when H₀ is false.
β = P (µ₁ - µ₂ ≤ 5 | µ₁ - µ₂ = 6)
*Use a z-table for the probability.
Thus the value of β is 0.00015.
Ответ:
Step-by-step explanation:
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