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 maraging steel with commercial-purity steel of the same type. For m = 34 specimens, the sample average toughness was x = 64.1 for the high-purity steel, whereas for n = 38 specimens of commercial steel y = 58.7. 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 = 1.3 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.)

H0: 1 2 = 5, Ha: 1 2 < 5

H0: 1 2 = 5, Ha: 1 2 5

H0: 1 2 = 5, Ha: 1 2 5

H0: 1 2 = 5, Ha: 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.)
z=

p-value=

State the conclusion in the problem context.

Fail to reject H0. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5.

Reject H0. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5.   

Reject H0. 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 H0. 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 for the test conducted in part (a) when 1 2 = 6. (Round your answer to four decimal places.)
=

I really need the bold ones

Explanation / Answer

at alpha=0.01 and m1-m2=5, we need to first find the critical value of the difference in the means

stddev = sqrt(s1^2/n1 + s2^2/n2) where s is the standard deviation and n is the number of observations

= sqrt(1.3^2/34 + 1.1^2 / 38) = 0.286

For beta error we must fail to reject null so wqe calculate the statistic at which we fail to reject null

So,P(z>Z) = 1-0.001 = 0.999

Zcritical = 3.09

So, ((x-5)/0.286) = 3.09

x = 5.8824

now we need to check for m1-m2=6

As we fail to reject null when the difference in means is more than 5.8824

So, z = (x-6)/stddev = (5.8824-6)/0.286 = -0.411

Hence beta value for this corresponding probability is 0.6595 , P(z>-0.411)

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