B and D nonparametric analysis. Derive the should be used in place of Eqs 8.6 an
ID: 2932933 • Letter: B
Question
B and Dnonparametric analysis. Derive the should be used in place of Eqs 8.6 and 8.27 Suppose that instead of Eq. 5.12, we Derive the expressions for R testing. The ried out using nonparametric methods for ungroeanay"( (a) The first test series on six prototype 8.28 Microcircuits undergo accelerated life following times to failure (in hours): 1.6, 99.6. Plot a graph of the estimated reliabii26,5.. second test failure (in hours): 25, 2.8, 3.5,5.7.1 23.5. Combine these data with the data from a and plot, 103 estimate on the same graph used for a 8.29 At rated voltage a microcircuit has been estimated 20,000 hr. An accelerated life test is to bo have number. It is known that the microcircuit life is inversel o verid before the test is terminated if we are to have con If the test must be completed in 30 days, at what voltage should the circuits be tested? confidence in the 8.30 A life test with type II censoring is performed on 50 that are thought to have a constant failure rate. The test is after the twentieth failure. The times to failure (in months) are as 0.10 0.290.49.51 0.55 0.63 0.68 1.16 2.25 2.64299 3.15 3.51 .53 3.99 05 1.40 224 3.01 3.06 The failed servomechanisms are not replaced (a) Make an exponential probability plot and estimate whether the failure rate is constant. (b) Make a point estimate of the MTTF from the appropriate form of Eq. 8.39 (c) Using the MTTF from b draw a straight line through the data plotted for a. (d) what is the 90% confidence interval on the MTT? (e) Draw the straight lines on your plot in a corresponding to the confidence limits on the MTTF 8.31 Suppose that in Exercise 8.30 the life test had to be stopped at 3 months because of a production deadline. Based on a 3-month test, estimate the MTTF and the c
Explanation / Answer
Question 8.30
(a) T = total operational time of all test units
n = number of faiures
Here N= 50 and n = 20
as here in this case total time the test is run = 4.05 minute after that it was closed.
T = t + 4.05 * (50 -30)
T = 40.25 + 4.05 * 30
T = 161.75
MTTF = 161.75/20 = 8.0875 months
(d) 90% confidence interva on MTTF
Lower Limit = 2T / X2 (/2,2n) = 2 * 161.75/ X2(0.05,40) = 2 * 161.75/ 55.76 = 5.80
Upper Limit = 2T / X2(1-/2,n) = 2 * 161.75/ X2(0.95,40) = 2 * 161.75/ 26.51 = 12.20
so 90% confidence interval for MTTF = (5.80, 12.20) in months
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