If a one at a time sequential test yields 10 failures in the fixed time of T 500 hours, then the maximum likelihood estimate of is 500 10 50 hours A 95 percent confidence interval estimate of is 0 (1,000 X025 20, 1,000 X975,20) Running Program 5 8 1b yields that X 025 20 34 17, X975 20 9 66 and so, with 95 percent confidence, 0 (29 27, 103 52) If we wanted to test the hypothesis Ho 000 versus the alternative H 0 00

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Question: If a one-at-a-time sequential test yields 10 failures in the fixed time of T = 500 hours, then the maximum likelihood estimate of is

If a one-at-a-time sequential test yields 10 failures in the fixed time of T = 500 hours, then the maximum likelihood estimate of θ is 500/10 = 50 hours. A 95 percent confidence interval estimate of θ is

0 (1,000/X025.20, 1,000/X975,20) Running Program 5.8.1b yields that X.025.20 = 34.17, X975.20

0 (1,000/X025.20, 1,000/X975,20) Running Program 5.8.1b yields that X.025.20 = 34.17, X975.20 = 9.66 and so, with 95 percent confidence, 0 (29.27, 103.52) If we wanted to test the hypothesis Ho: 000 versus the alternative H:0 00

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