Question

7. A life testing plan is to be terminated after the seventh failure. Failed items are not replaced during the test. A sample of 14 items is taken with the failures Occ the following times (in hours): 200, 220, 221, 245, 300, 305, and 310. Assume the time to failure is exponentially distributed. 7.1. Estimate the mean life of the items. (7 points) Answer: 7.2. Estimate the failure rate. (3 points) Answer: 7.3. Find a 98% confidence interval for the mean life. (6 points) Answer

Answer 1

1)

n=14

failure no of items = 7

mean = (200+220+221+245+300+305+310 ) / 14

=1801/14

= 128.6429

..............

2) MTBF = 1801 /7

= 257.2857

failure rate = 1 / MTBF

= 0.00389

=0.39%

.............

C)

mean = 128.6429

std dev =31.6464

Level of Significance ,    α =    0.02          
'   '   '          
z value=   z α/2=   2.3263   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   31.6464   / √   14   =   8.457856
margin of error, E=Z*SE =   2.3263   *   8.45786   =   19.675916
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    128.64   -   19.675916   =   108.966984
Interval Upper Limit = x̅ + E =    128.64   -   19.675916   =   148.318816
98%   confidence interval is (   108.97   < µ <   148.32   )

....................

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