HW6-Sol - 10-20 a) 1) The parameter of interest is the...

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Unformatted text preview: 10-20 a) 1) The parameter of interest is the difference in mean impact strength, , with ∆ = 0 2) H : or 3) H 1 : or 4) The test statistic is 5) Reject the null hypothesis if t < where = 1.714 for α = 0.05 since (truncated) 6) 290 321 12 22 n 1 = 10 n 2 = 16 7) Conclusion: Because - 4.64 < - 1.714 reject the null hypothesis and conclude that supplier 2 provides gears with higher mean impact strength at the 0.05 level of significance. P-value = P( t < - 4.64): P-value < 0.0005 b) 1) The parameter of interest is the difference in mean impact strength, 2) H : 3) H 1 : or 4) The test statistic is 5) Reject the null hypothesis if t > = 1.714 for α = 0.05 where 6)290 321 =25 12 22 n 1 = 10 n 2 = 16 7) Conclusion: Because 0.898 < 1.714, fail to reject the null hypothesis. There is insufficient evidence to conclude that t the mean impact strength from supplier 2 is at least 25 ft-lb higher that supplier 1 using α = 0.05....
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This note was uploaded on 07/08/2011 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .

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HW6-Sol - 10-20 a) 1) The parameter of interest is the...

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