# ISE361_HW7answer_F03 - Homework Seven 6 Chapter 8 Dec/2001...

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Homework Seven Chapter 8 Dec/2001 6) H 0 : u = 40 vs H a : u = 40, where u is the true average burn-out amperage for this type of fuse. The alternative reflects the fact that a departure from u = 40 in either direction is of concern. Notice that in this formulation, it is initially believed that the value of u is the design value of 40. 20) With H 0 : u = 750, and H a : u < 750 and a significance level of .05, we reject H 0 if z < - 1.645; z = -2.14 < -1.645, so we reject the null hypothesis and do not continue with the purchase. At a significance level of .01, we reject H 0 if z < -2.33; z = -2.14 > -2.33, so we don’t reject the null hypothesis and thus continue with the purchase. 24) H 0 : u = 3,000 vs H a : u = 3,000 ; t = (x – 3,000)/ (s/ (n) 1/2 ) ; reject H 0 if | t | > t .025, 4 = 2.776 ; t = (2,887.6 – 3,000) / 84/(5) 1/2 = -2.99 < -2.776, so we reject H 0 . This requirement is not satisfied. 25)

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## This homework help was uploaded on 04/16/2008 for the course ISE 361 taught by Professor Koon during the Spring '08 term at Binghamton.

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ISE361_HW7answer_F03 - Homework Seven 6 Chapter 8 Dec/2001...

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