An experiment was performed to compare the fracture toughness of high-purity 18
ID: 3316015 • Letter: A
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- 63.4 for the high-purit, steel, whereas for n 39 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 excceds 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 a = 0 0 1. Use 1 State the relevant hypotheses 2 where 1 s the average toughness for high purity steel and 2 s the average toughness for commercial steel. 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) P-value = State the conclusion in the problem context. O 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 O Reject Ho. The dete 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 for the test conducted in part (a) when 1-12-6. (Round your answer to rour decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question.Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2< 5
Alternative hypothesis: 1 - 2 > 5
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.001. Using sample data, we will conduct a two-sample z - test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.2895
z = [ (x1 - x2) - d ] / SE
z = 1.73
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a z statistic of 1.73. We use the z Distribution Calculator to find P(z > 1.73).
Therefore, the P-value in this analysis is 0.0418
Interpret results. Since the P-value (0.0418) is greater than the significance level (0.001), we have to accept the null hypothesis.
b) Beta for the test if u1-u2 = 6 is 0.49.
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