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 = 32 specimens, the sample average toughness was x = 63.7 for the high-purity steel, whereas for n = 36 specimens of commercial steel y = 57.9. 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 = 1.2 and 2 = 1.0, 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. Ho: Aa1 H2 Ha: 1 Ho: Au 1 Ha: 1- A12 5 Ho: Aa1 H2 Ha: H2. Ho: Au 1 Ha: 1 Round your test statistic to two decimal places and your P-value to four decimal places.) Calculate the test statistic and determine the P-value. P-value State the conclusion in the problem context 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 of high-purity steel exceeds that of commercial-purity steel by more than 5 O Reject Ho. The data suggests that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5 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. (b) Compute B for the test conducted in part (a) when u1 u2 6. Round your answer to four decimal places

Explanation / Answer

Part-a

Hypothesis is option-2

From following output

z= 2.97

p-value=0.0015

State the conclusion-option-3

Part-b

Beta=P(Z<2.97/mu1-mu2=6)

= 0.2292

Two sample Z-tes

n1

32

n2

36

xbar1

63.7

xbar2

57.9

Mean Difference

5.8

sigma1

1.2

sigma2

1

SE

0.269773568

Hypothesise difference

5

Z

2.97

p-value(two tailed)

0.0030

p-value(one-taield)

0.0015

Two sample Z-tes

n1

32

n2

36

xbar1

63.7

xbar2

57.9

Mean Difference

5.8

sigma1

1.2

sigma2

1

SE

0.269773568

Hypothesise difference

5

Z

2.97

p-value(two tailed)

0.0030

p-value(one-taield)

0.0015

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