An experiment was performed to compare the fracture toughness of high-purity 18

Question

An experiment was performed to compare the fracture toughness of high-purity 18 Ni managing steel with commercial purity steel of the same type. For m = 30 specimens, the sample average toughness was X = 65.1 for the high-purity h-purity steel, whereas for n = 39 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 sigma_1 = 1.4 and sigma_2 = 1.0, test the relevant hypotheses using alpha = 0.001 (Use mu_1 - mu_2, where u is the average toughness for high-purity steel and mu_2 is the average toughness for commercial steel.) State the relevant hypotheses. H_0: mu_1 - mu_2 = 5 H_a:mu_1 - mu_2 lessthanorequalto 5 H_0: mu_1 - mu_2 = 5 H_a: mu_1 - mu_2 > 5 H_0: mu_1 - mu_2 - 5 H_a: mu_1 - mu_2 - 5 H_0: mu_1 - mu_2 = 5 H_a: mu_1 - mu_2 notequalto 5 H_0: mu_1 - mu_2 - 5 H_a: mu_1 - mu_2

Explanation / Answer

Power and Sample Size

2-Sample t Test

Testing mean 1 = mean 2 (versus >)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.001 Assumed standard deviation = 1.1898

Sample
Difference Size Power
6 30 1
6 39 1

The sample size is for each group.

Beta = 1 -power

therefore Beta = 0

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