An experiment was performed to compare the fracture toughnes

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 = 33 specimens, the sample average toughness was x = 65.3 for the high-purity steel, whereas for n = 36 specimens of commercial steel y = 59.9. 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 = 1.4 and ₂ = 1.1, test the relevant hypotheses using = 0.001. (Use _{1 2}, where ₁ is the average toughness for high-purity steel and ₂ is the average toughness for commercial steel.)
State the relevant hypotheses.

A.) H₀: _{1 2} = 5
B.) H_a: _{1 2} > 5 H₀: _{1 2} = 5
C.) H_a: _{1 2} 5     H₀: _{1 2} = 5
D.) H_a: _{1 2} 5 H₀: _{1 2} = 5
E.) H_a: _{1 2} < 5

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.)

    (b) Compute for the test conducted in part (a) when _{1 2} = 6. (Round your answer to four decimal places.)

=

Solution

Hypotheses would be

B.) Ha: 1 2 > 5 H0: 1 2 = 5

Right tailed test for difference of means

The mean of Group One minus Group Two equals 5.400

df = 33+36-2 = 67

standard error of difference = 0.302  

Test statistic t = mean difference/std error = 5.4/0.302 = 17.8927

p value <0.0001

As p value is less than alpha, reject H0.

Mu1 is greater than mu2 by more than 5, is statistically evident.

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When mu1-mu2 =6 t statistic would be 6/0.302 = 19.867

p value <0.0001

Beta = 1-p

Hence Beta >0.9999